Chapter 1 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
1 2
mV  ekmolecular
2
3
kT  ekmolecular
2

3kT
V 
m
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m  5.14  10 26 kg .
Substitute and solve:

V  487.6 m/s
The molecules are traveling really, fast (around the length of five football fields every second).
Comment:
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is
sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have:
 m 
f (v)dv  4 

 2kT 
3/ 2
 m 2 2
v v dv
exp
 2kT 
where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed
by integrating the expression above


V 
 f (v)vdv
0

 f (v)dv

8kT
 449 m/s
m
0
2
1.3
Derive the following expressions by combining Equations 1.4 and 1.5:

3kT
Va2 
ma

3kT
Vb2 
mb
Therefore,

Va2 mb
 
Vb2 ma
Since mb is larger than ma, the molecules of species A move faster on average.
3
1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
0 º C, 0 º Reamur  and 100 º C, 80 º Reamur 
Create an equation using the two points:
T º Reamur   0.8 T º Celsius 
At 22 ºC,
T  17.6 º Reamur
4
1.5
(a)
After a short time, the temperature gradient in the copper block is changing (unsteady state), so
the system is not in equilibrium.
(b)
After a long time, the temperature gradient in the copper block will become constant (steady
state), but because the temperature is not uniform everywhere, the system is not in equilibrium.
(c)
After a very long time, the temperature of the reservoirs will equilibrate; The system is then
homogenous in temperature. The system is in thermal equilibrium.
5
1.6
We assume the temperature is constant at 0 ºC. The molecular weight of air is
MW  29 g/mol  0.029 kg/mol
Find the pressure at the top of Mount Everest:




 0.029 kg/mol9.81 m/s8848 m  
P  1 atm exp 



 J 


2
73.15
K

 8.314 



 mol  K  



P  0.330 atm  33.4 kPa
Interpolate steam table data:
T sat  71.4 º C
for P sat  33.4 kPa
Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the
temperature remains constant. In reality the temperature decreases with height as we go up the
mountain. However, a solution in which T and P vary with height is not as straight-forward.
6
1.7
To solve these problems, the steam tables were used. The values given for each part constrain
the water to a certain state. In most cases we can look at the saturated table, to determine the
state.
(a)
(b)
(c)
(d)
Subcooled liquid
Explanation: the saturation pressure at T = 170 [oC] is 0.79 [MPa] (see page 508);
Since the pressure of this state, 10 [bar],
is greater than the saturation
pressure, water is a liquid.
Saturated vapor-liquid mixture
Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04
[m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume
of this state, 3 [m3/kg], is in between these values we have a saturated vaporliquid mixture.
Superheated vapor
Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa],
is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since
the volume of this state, 0.05 [m3/kg], is greater than this value, it is a vapor.
Superheated vapor
Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5
[MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state,
7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the
superheated water vapor tables for P = 500 [kPa], we see the state is constrained
to T = 200 [oC].
7
1.8
From the steam tables in Appendix B.1:
 m3 
vˆ critical  0.003155 

 kg 
T  374.15 º C, P  22.089 MPa 
At 10 bar, we find in the steam tables
vˆlsat
 m3 
 0.001127 

 kg 
 m3 
vˆvsat  0.19444 

 kg 
Because the total mass and volume of the closed, rigid system remain constant as the water
condenses, we can develop the following expression:
vˆ critical  1  x vˆlsat  xvˆvsat
where x is the quality of the water. Substituting values and solving for the quality, we obtain
x  0.0105 or 1.05 %
A very small percentage of mass in the final state is vapor.
8
1.9
The calculation methods will be shown for part (a), but not parts (b) and (c)
(a)
Use the following equation to estimate the specific volume:
vˆ1.9 MPa, 250 º C   vˆ1.8 MPa, 250 º C   0.5vˆ2.0 MPa, 250 º C   vˆ1.8 MPa, 250 º C 
Substituting data from the steam tables,
 m3 
vˆ1.9 MPa, 250 º C   0.11821 

 kg 
From the NIST website:
 m3 
vˆ NIST 1.9 MPa, 250 º C   0.11791 

 kg 
Therefore, assuming the result from NIST is more accurate
 vˆ  vˆ NIST

% Difference  
 100 %  0.254 %
 vˆ NIST

(b)
Linear interpolation:
 m3 
vˆ1.9 MPa, 300 º C   0.13284 

 kg 
NIST website:
 m3 
vˆ NIST 1.9 MPa, 300 º C   0.13249 

 kg 
Therefore,
% Difference  0.264 %
(c)
Linear interpolation:
 m3 
vˆ1.9 MPa, 270 º C   0.12406 

 kg 
9
Note: Double interpolation is required to determine this value. First, find the molar volumes at
270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from
the previous step to find the molar volume at 270 ºC and 1.9 MPa.
NIST website:
 m3 
vˆ NIST 1.9 MPa, 270 º C   0.12389 

 kg 
Therefore,
% Difference  0.137 %
With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website
agree very well. The discrepancies will not significantly affect the accuracy of any subsequent
calculations.
10
1.10
For saturated temperature data at 25 ºC in the steam tables,
 m3 
L
vˆlsat  0.001003    1.003  
 kg 
 kg 
Determine the mass:
m
V

vˆlsat
1 L
 0.997 kg
L
1.003  
 kg 
Because molar volumes of liquids do not depend strongly on pressure, the mass of water at 25 ºC
and atmospheric pressure in a one liter should approximately be equal to the mass calculated
above unless the pressure is very, very large.
11
1.11
First, find the overall specific volume of the water in the container:
vˆ 
 m3 
1 m3
 0.2 

5 kg
 kg 
Examining the data in the saturated steam tables, we find
at P sat  2 bar
vˆlsat  vˆ  vˆvsat
Therefore, the system contains saturated water and the temperature is
T  T sat  120.23 º C
Let ml represent the mass of the water in the container that is liquid, and mv represent the mass
of the water in the container that is gas. These two masses are constrained as follows:
ml  mv  5 kg
We also have the extensive volume of the system equal the extensive volume of each phase
ml vˆlsat  mv vˆvsat  V

 

ml 0.001061 m 3 / kg  mv 0.8857 m 3 /kg  1 m 3
Solving these two equations simultaneously, we obtain
ml  3.88 kg
mv  1.12 kg
Thus, the quality is
x
mv 1.12 kg

 0.224
m
5 kg
The internal energy relative to the reference state in the saturated steam table is
U  ml uˆlsat  mv uˆvsat
From the steam tables:
12
uˆlsat  504.47 kJ/kg 
uˆvsat  2529.5 kJ/kg 
Therefore,
U  3.88 kg 504.47 kJ/kg   1.12 kg 2529.5 kJ/kg 
U  4790.4 kJ
13
1.12
First, determine the total mass of water in the container. Since we know that 10 % of the mass is
vapor, we can write the following expression

V  m 0.9 vˆlsat  0.1vˆvsat

From the saturated steam tables for P sat  1 MPa
 m3 
sat
vˆl  0.001127 

kg


 m3 
vˆvsat  0.1944 

 kg 
Therefore,
m
V
0.9vˆlsat  0.1vˆvsat
m  48.9 kg

1 m3


 m3  
 m3  
0.9 0.001127 

   0.1 0.1944 



kg
kg
 
 




and

Vv  0.1mvˆvsat  0.148.9 kg  0.1944 m 3 / kg
Vv  0.950 m 3
14

1.13
In a spreadsheet, make two columns: one for the specific volume of water and another for the
pressure. First, copy the specific volumes of liquid water from the steam tables and the
corresponding saturation pressures. After you have finished tabulating pressure/volume data for
liquid water, list the specific volumes and saturation pressures of water vapor. Every data point
is not required, but be sure to include extra points near the critical values. The data when plotted
on a logarithmic scale should look like the following plot.
The Vapor-Liquid Dome for H2O
100.0000
Critical Point
Pressure, P (MPa)
10.0000
Liquid
1.0000
0.0001
Vapor-Liquid
0.001
0.01
Vapor
0.1
1
10
0.1000
0.0100
0.0010
Specific Volume, v (m 3/kg)
15
100
1000
1.14
The ideal gas law can be rewritten as
v
RT
P
For each part in the problem, the appropriate values are substituted into this equation where
 L  bar 
R  0.08314 
 mol  K 
is used. The values are then converted to the units used in the steam table.
(a)
 L 
v  30.7 
 mol 
 L 
30.7 
3
3
v
 mol  10 3  m   1.70  m 
vˆ 

 kg 
 L 
MW
 kg 
 
 
0.018 

 mol 
For part (a), the calculation of the percent error will be demonstrated. The percent error will be
based on percent error from steam table data, which should be more accurate than using the ideal
gas law. The following equation is used:
 vˆ  vˆST

 100 %
%Error   IG
vˆST


where v IG is the value calculated using the ideal gas law and vST is the value from the steam
table.
3
3

 1.70  m   1.6729  m 

 kg 
 kg 
%Error  

 m3 

1.6729 

kg 







%  1.62%




This result suggests that you would introduce an error of around 1.6% if you characterize boiling
water at atmospheric pressure as an ideal gas.
16
(b)
 L 
v  64.3 
 mol 
 m3 
vˆ  3.57 

 kg 
% Error  0.126%
For a given pressure, when the temperature is raised the gas behaves more like an ideal gas.
(c)
 L 
v  0.643 
 mol 
 m3 
vˆ  0.0357 

 kg 
%Error = 8.87%
(d)
 L 
v  1.06 
 mol 
 m3 
vˆ  0.0588 

 kg 
%Error = 0.823%
The largest error occurs at high P and low T.
17
1.15
The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall –
your answer should vary. The volume of the room is
V  12 ft 16 ft 10 ft   1920 ft 3  54.4 m 3
The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles
using the ideal gas law:



1.01325  10 5 Pa 54.4 m 3
PV

RT 
 J 
295.15 K 
 8.314 
 mol  K  

n  2246 mol
n
Use the molecular weight of air to obtain the mass:
m  nMW   2246.27 mol0.029 kg/mol  65.2 kg
That’s pretty heavy.
18
1.16
First, find the total volume that one mole of gas occupies. Use the ideal gas law:
RT

v
P

 J 
 8.314 
293.15 K 
 mol  K  

1105 Pa 
 m3 
 0.0244 

 mol 
Now calculate the volume occupied by the molecules:
 Number of molecules  Volume of 
4

  N A  πr 3 
  
v occupied  
per mole
3

  one molcule

 6.022  10 23 molecules   4
3
 π 1.5  10 10 m 
v occupied  

  3

1 mol



 m3 
v occupied  8.5110  6 

 mol 


Determine the percentage of the total volume occupied by the molecules:
 v occupied

Percentage  
100 %  0.0349 %
v


A very small amount of space, indeed.
19
1.17
Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the
maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF
which satisfies equilibrium conditions. In other words, we must find the saturation density of
water vapor at 40 ºF. From the steam tables,
 m3 
vˆvsat  153.74 

 kg 
(sat. water vapor at 40 ºF = 4.44 ºC)
We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the
ideal gas law.
vˆvsat
,277.6 K
T277.6
vˆ
 294.3 K
T294.3


 m3 
 m3  
T
294.3 K 


vˆ294.3 K  294.3 vˆvsat

153
.
74
163




, 277.6 K
T277.6
277.6 K 
 kg 
 kg  

Therefore,
ˆ 
1
vˆvsat
 kg 
 6.13  10  3 

 m3 


If the density at 70 ºF is greater than or equal to 6.13  10 3 kg/m 3 , the density at night when the
temperature is 40 ºF will be greater than the saturation density, so water will condense onto the


wall. However, if the density is less than 6.13  10 3 kg/m 3 , then saturation conditions will not
be obtained, and water will not condense onto the walls.
20
1.18
We consider the air inside the soccer ball as the system. We can answer this question by looking
at the ideal gas law:
Pv  RT
If we assume it is a closed system, it will have the same number of moles of air in the winter as
the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv
will be lower and the balls will be under inflated.
Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over
time. Thus we have an open system and the number of moles decrease with time – leading to the
under inflation.
21
1.19
First, write an equation for the volume of the system in its initial state:
ml vˆlsat  mv vˆvsat  V
Substitute ml  1  x m and mv  xm :


m 1  x vˆlsat  xvˆvsat  V
At the critical point
mvˆ critical  V
Since the mass and volume don’t change
vˆ critical  1  x vˆlsat  xvˆvsat
From the steam tables in Appendix B.1:
 m3 
vˆ critical  0.003155 

 kg 
T  374.15 º C, P  22.089 MPa 
At 0.1 MPa, we find in the steam tables
 m3 
vˆlsat  0.001043 

 kg 
 m3 
sat
vˆv  1.6940 

 kg 
Therefore, solving for the quality, we get
x  0.00125 or 0.125 %
99.875% of the water is liquid.
22
1.20
As defined in the problem statement, the relative humidity can be calculated as follows
Relative Humidity 
mass of water
mass of water at saturation
We can obtain the saturation pressure at each temperature from the steam tables. At 10 [oC], the
saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the
vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is:
pWater  0.9 1.22  1.10 [kPa]
This value represents the vapor pressure of water in the air. At 30 [oC], the saturation pressure of
water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is:
pWater  0.5  5.63  2.12 [kPa]
Since the total pressure in each case is the same (atmospheric), the partial pressure is
proportional to the mass of water in the vapor. We conclude there is about twice the amount of
water in the air in the latter case.
23
1.21
(a)
When you have extensive variables, you do not need to know how many moles of each substance
are present. The volumes can simply be summed.
V1  Va  Vb
(b)
The molar volume, v1, is equal to the total volume divided by the total number of moles.
v1 
V1 Va  Vb

ntot na  nb
We can rewrite the above equation to include molar volumes for species a and b.
nb
n v  nb vb
na
vb
va 
v1  a a

na  nb
na  nb
na  nb
v1  xa va  xb vb
(c)
The relationship developed in Part (a) holds true for all extensive variables.
K1  K a  K b
(d)
k1  xa k a  xb kb
24
Chapter 2 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
2.1
There are many possible solutions to this problem. Assumptions must be made to solve the
problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the
towel when you dry yourself. In other words, let
m H 2 O  0.5 kg 
Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy
balance and neglecting potential and kinetic energy effects reveals
qˆ  hˆ
Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find
the minimum energy required for drying the towel, assume that the temperature of the towel
remains constant at T  25 º C  298.15 K . In the drying process, the absorbed water is
vaporized into steam. Therefore, the expression for heat is
v
l
qˆ  hˆH
 hˆH
2O
2O
v
is the specific enthalpy of water vapor at P  1.01 bar and T  298.15 K and
where is hˆH
O
2
hˆ l
is the specific enthalpy of liquid water at P  1.01 bar and T  298.15 K . A hypothetical
H 2O
path must be used to calculate the change in enthalpy. Refer to the diagram below
P
hˆ
liquid
1 atm
vapor
P = 1 [atm]
hˆ3
hˆ1
P
3.17 kPa
sat
hˆ2
liquid
vapor
By adding up each step of the hypothetical path, the expression for heat is
qˆ  h1  h2  h3
 hˆ l ,sat 25 º C   hˆ l

 25 º C, 1.01 bar  hˆHv,satO 25 º C  hˆHl ,satO 25 º C
 hˆHv O 25 º C, 1.01 bar   hˆHv ,sat
O 25 º C 
H 2O
H 2O
2
2
2
2
2
However, the calculation of heat can be simplified by treating the water vapor as an ideal gas,
which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on
temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is
zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the
hypothetical path containing the pressure change of the liquid can be neglected. This leaves
v
25 º C, 3.17 kPa   hˆHl 2 O 25 º C, 3.17 kPa 
qˆ  hˆH
O
2
From the steam tables:
 kJ 
v, sat

hˆH
2547
.
2
 kg 
O
2
 
 kJ 
l , sat
hˆH

104
.
87
 kg 
O
2
 
(sat. H2O vapor at 25 ºC)
(sat. H2O liquid at 25 ºC)
which upon substitution gives
 kJ 
q̂  2442.3  
 kg 
Therefore,

 kJ  
Q  0.5 kg  2442.3     1221.2 kJ 
 kg  

To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20
minutes (1200 s) to dry the towel. From the definition of electrical work,
W  IVt  30 A 208 V 1200 s  7488 kJ 
Therefore, the efficiency is
 1221.2 kJ 

Q

 100 %  
 100 %  16.3%
W

 7488 kJ 

 
There are a number of ways to improve the drying process. A few are listed below.
 Dry the towel outside in the sun.
 Use a smaller volume dryer so that less air needs to be heated.
 Dry more than one towel at a time since one towel can’t absorb all of the available
heat. With more towels, more of the heat will be utilized.
3
2.3
In answering this question, we must distinguish between potential energy and internal energy.
The potential energy of a system is the energy the macroscopic system, as a whole, contains
relative to position. The internal energy represents the energy of the individual atoms and
molecules in the system, which can have contributions from both molecular kinetic energy and
molecular potential energy. Consider the compression of a spring from an initial uncompressed
state as shown below.
uncompressed
compressed
x
State 1
M
State 2
Since it requires energy to compress the spring, we know that some kind of energy must be
stored within the spring. Since this change in energy can be attributed to a change of the
macroscopic position of the system and is not related to changes on the molecular scale, we
determine the form of energy to be potential energy. In this case, the spring’s tendency to restore
its original shape is the driving force that is analogous to the gravity for gravitational potential
energy.
This argument can be enhanced by the form of the expression that the increased energy takes. If
we consider the spring as the system, the energy it acquires in a reversible, compression from its
initial uncompressed state may be obtained from an energy balance. Assuming the process is
adiabatic, we obtain:
E  Q  W  W
We have left the energy in terms of the total energy, E. The work can be obtained by integrating
the force over the distance of the compression:
W    F  dx   kxdx 
1 2
kx
2
Hence:
E 
1 2
kx
2
We see that the increase in energy depends on macroscopic position through the term x.
It should be noted that there is a school of thought that assigns this increased energy to internal
energy. This approach is all right as long as it is consistently done throughout the energy
balances on systems containing springs.
4
2.4
For the first situation, let the rubber band represent the system. In the second situation, the gas is
the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the
energy balance becomes
U  W
Since work must be done on the rubber band to stretch it, the value of the work is positive. From
the energy balance, the change in internal energy is positive, which means that the temperature
of the system rises.
When a gas expands in a piston-cylinder assembly, the system must do work to expand against
the piston and atmosphere. Therefore, the value of work is negative, so the change in internal
energy is negative. Hence, the temperature decreases.
In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the
rubber band goes to increase its potential energy; however, a part of it goes into stretching the
polymer molecules which make up the rubber band, and the qualitative argument given above
still is valid.
5
2.5
To explain this phenomenon, you must realize that the water droplet is heated from the bottom.
At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The
water vapor forms an insulation layer between the skillet and the water droplet. At low
temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower
through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures.
6
2.6
Apartment
Surr.
HOT
System
Fridge
Q
+
W -
If the entire apartment is treated as the system, then only the energy flowing across the apartment
boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the
refrigerator is not explicitly accounted for in the energy balance because it is within the system.
By neglecting kinetic and potential energy effects, the energy balance becomes
U  Q  W
The Q term represents the heat from outside passing through the apartment’s walls. The W term
represents the electrical energy that must be supplied to operate the refrigerator.
To determine whether opening the refrigerator door is a good idea, the energy balance with the
door open should be compared to the energy balance with the door closed. In both situations, Q
is approximately the same. However, the values of W will be different. With the door open,
more electrical energy must be supplied to the refrigerator to compensate for heat loss to the
apartment interior. Therefore,
Wajar  Wshut
where the subscript “ajar” refers the situation where the door is open and the subscript “shut”
refers to the situation where the door is closed. Since,
Qajar  Qshut
U ajar  Qajar  Wajar  U shut  Qshut  Wshut
 Tajar  Tshut
The refrigerator door should remain closed.
7
2.7
The two cases are depicted below.
Heater off
Heater on
Surr.
Surr.
System
System
Heater
Q
Heater
Q
W
Let’s consider the property changes in your house between the following states. State 1, when
you leave in the morning, and state, the state of your home after you have returned home and
heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2,
the state of the system is the same and U must be zero, so
U  Q  W  0
or
Q W
where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that
must be delivered to the heater. The case where more heat escapes will require more work and
result in higher energy bills. When the heater is on during the day, the temperature in the system
is greater than when it is left off. Since heat transfer is driven by difference in temperature, the
heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off
when you are gone.
8
2.8
The amount of work done at constant pressure can be calculated by applying Equation 2.57
H  Q
Hence,
H  Q  mhˆ
where the specific internal energy is used in anticipation of obtaining data from the steam tables.
The mass can be found from the known volume, as follows:
 3 

L
 


 1.0 kg 

 m3  
 0.0010 



kg
 



m
V

vˆ
1L 0.001 m
As in Example 2.2, we use values from the saturated steam tables at the same temperature for
subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1:
 kJ 
 kJ 
 kJ 
uˆ  uˆl ,2 at 100 o C  uˆl ,1 at 25 o C  419.02    104.87    314.15  
 kg 
 kg 
 kg 

 
  
Solve for heat:

 kJ  
Q  muˆ  1.0 kg  314.05     314.15 kJ 
 kg  

and heat rate:
Q
Q  
t
314.15 kJ 
 0.52 kW 
 60 s 
10 min.

 min 
This value is the equivalent of five strong light bulbs.
9
2.9
(a)
From Steam Tables:
 kJ 
uˆ1  2967.8  
 kg 
 kJ 
uˆ 2  2659.8  
 kg 
(100 kPa, 400 ºC)
(50 kPa, 200 ºC)
 kJ 
uˆ  uˆ 2  uˆ1  308.0  
 kg 
(b)
From Equations 2.53 and 2.63
T2
T2
T1
T1
u  u 2  u1   cv dT 
 c P  R dT
From Appendix A.2
c P  R( A  BT  CT 2  DT 2  ET 3 )
T2


u  R  A  BT  CT 2  DT  2  ET 3  1 dT
T1
Integrating


E
1 1
B
C
u  R ( A  1)(T2  T1 )  (T2 2  T12 )  (T23  T13 )  D(  )  (T2 4  T14 )
2
3
4
T2 T1


The following values were found in Table A.2.1
A  3.470
B  1.45  10  3
C 0
D  1.21 10 4
E 0
Substituting these values and using
10
 J 
R  8.314 
 mol  K 
T1  (400  273.15 K)  673.15 K
T2  (200  273.15 K)  473.15 K
provides
 J 
u  5551 
 mol 
 kJ 

 J   1 [mol H 2 O]  1000 g  1 kJ 




uˆ    5551 
308
.
1





 kg 




 mol   18.0148 [g H 2 O]  1 kg  1000 J 

 
The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis
for calculating the percent difference since steam table data should be more accurate.
% Difference 
 308   308.1
 100 %  0.03%
 308.0
11
2.10
(a)
Referring to the energy balance for closed systems where kinetic and potential energy are
neglected, Equation 2.30 states
U  Q  W
(b)
Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the
process is isothermal
U  0
According to Equation 2.77
P
W  nRT ln  2
 P1

P
  n1 RT1 ln  2

 P1



From the ideal gas law:
n1RT1  P1V1
P 
W  P1V1 ln 2 
 P1 
Substitution of the values from the problem statement yields



 5 bar 
W  8  105 Pa 2.5  10  3 m 3 ln

 8 bar 
W  940 J 
The energy balance is
0 J  Q W
 Q  940 J 
(c)
Since the process is adiabatic
Q0
The energy balance reduces to
U  W
12
The system must do work on the surroundings to expand. Therefore, the work will be negative
and
U  0
T2
U  n  cv T  0
T1
 T2  T1
T2 will be less than 30 ºC
13
2.11
(a)
(i).
1
Path B
Pa
A
th
P [bar]
3
2
2
1
0.01
0.02
0.03
3
v [m /mol]
(ii).
Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the
process is isothermal
u  0
Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore,
h  0
Performing an energy balance and neglecting potential and kinetic energy produces
u  q  w  0
For an isothermal, adiabatic process, Equation 2.77 states
P 
W  nRT ln 2 
 P1 
or
w
P 
W
 RT ln 2 
n
 P1 
Substituting the values from the problem statement gives
14

 1 bar 
 J 
w   8.314 
(88  273.15) K ln


 3 bar 
 mol  K  

 J 
w  3299 
 mol 
Using the energy balance above
 J 
q   w  3299 
 mol 
(b)
(i).
See path on diagram in part (a)
(ii).
Since the overall process is isothermal and u and h are state functions
u  0
h  0
The definition of work is
w    PE dv
During the constant volume part of the process, no work is done. The work must be solved for
the constant pressure step. Since it is constant pressure, the above equation simplifies to
w   PE  dv   PE (v2  v1 )
The ideal gas law can be used to solve for v 2 and v1

 J  mol  
 8.314 
(88  273.15) K 
 m3 
RT2 
 K  


v2 
0
.
030


P2
1 10 5 Pa
 mol 

 J  mol  
 8.314 
(88  273.15) K 
 m3 
RT1 
 K  

0
.
010
v1 



P1
3  10 5 Pa
 mol 
Substituting in these values and realizing that PE  P1 since the process is isobaric produces
15

 m3 
 m3  
w  (3  10 5 Pa) 0.030 
  0.010 



mol
mol



 

 J 
w  6000 
 mol 
Performing an energy balance and neglecting potential and kinetic energy results in
u  q  w  0
 J 
 q   w  6000 
 mol 
16
2.12
First, perform an energy balance. No work is done, and the kinetic and potential energies can be
neglected. The energy balance reduces to
U  Q
We can use Equation 2.53 to get
T2
Q  n  cv dT
T1
which can be rewritten as
T2
Q  n  c P dT
T1
since the aluminum is a solid. Using the atomic mass of aluminum we find
n
5 kg
 185.3 mol
 kg 
0.02698 
 mol 
Upon substitution of known values and heat capacity data from Table A.2.3, we get

 J 
Q  185.3 mol 8.314 
  2.486  1.49  10 3 T dT


mol
K

 294.15 K

Q  131.61 kJ 


323.15 K
17
2.13
First, start with the energy balance. Potential and kinetic energy effects can be neglected.
Therefore, the energy balance becomes
U  Q  W
The value of the work will be used to obtain the final temperature. The definition of work
(Equation 2.7) is
V2
W    PE dV
V1
Since the piston expands at constant pressure, the above relationship becomes
W   PE V2  V1 
From the steam tables
 m3 
vˆ1  0.02641 

 kg 
(10 MPa, 400 ºC)
 

 m3  
3

ˆ
V1  m1v1  (3 kg) 0.02641 
   0.07923 m


 kg  

Now V2 and v2 are found as follows
V2  V1 
V
vˆ2  2
m2
 
 748740 J
W
 0.4536 m3
 0.07923 m3 
6
PE
2.0  10 Pa
 m3 
0.4536 m 3

 0.1512 

3 kg 
 kg 
 
Since v̂2 and P2 are known, state 2 is constrained. From the steam tables:
T2  400 º C

 3 
 20 bar, 0.1512  m  


 kg  

Now U will be evaluated, which is necessary for calculating Q . From the steam tables:
18
 kJ 
uˆ 2  2945.2  
 kg 

 3 
 20 bar, 0.1512  m  


 kg  

 kJ 
uˆ1  2832.4  
 kg 
100 bar, 400 º C

 kJ 
 kJ  
U  m1 uˆ 2  uˆ1   3 kg  2945.2    2832.4     338.4 kJ 
 kg 
 kg  

Substituting the values of U and W into the energy equation allows calculation of Q
Q  U  W
Q  338400 [J]   748740 J   1.09  10 6 J 
19
2.14
In a reversible process, the system is never out of equilibrium by more than an infinitesimal
amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of
1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible.
To solve for the final temperature of the system, the energy balance will be written. The pistoncylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore,
potential and kinetic energy effects can be neglected. The energy balance simplifies to
U  W
Conservation of mass requires
n1  n2
Let n  n1  n2
The above energy balance can be rewritten as
T2
V2
T1
V1
n  cv dT    PE dV
Since cv and PE are constant:
ncv T2  T1    PE V2  V1 
V2 and T1 can be rewritten using the ideal gas law
nRT2
P2
PV
T1  1 1
nR
V2 
Substituting these expressions into the energy balance, realizing that PE  P2 , and simplifying
the equation gives
5 

 P2  P1 V1
2 
T2  
7
nR
2
Using the following values
20
P1  2 bar 
P2  1 bar 
V1  10 L
n  1.0 mol
 L  bar 
R  0.08314 
 mol  K 
results in
T2  206 K 
To find the value for work, the energy balance can be used
W  U  ncv T2  T1 
Before the work can be calculated, T1 must be calculated
PV
T1  1 1 
nR
2 bar 10 L
1 mol 0.08314  L  bar  
 mol  K  

 241 K 
Using the values shown above
W  727 J 
21
2.15
The maximum work can be obtained through a reversible expansion of the gas in the piston.
Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston
assembly is well-insulated, so the heat transfer contribution to the energy balance can be
neglected, in addition to potential and kinetic energy effects. The energy balance reduces to
U  W
In this problem, the process is a reversible, adiabatic expansion. For this type of process,
Equation 2.90 states
W
1
P2V2  P1V1 
k 1
From the problem statement (refer to problem 2.13),
P1  2 bar 
V1  10 L
P2  1 bar 
To calculate W, V2 must be found. For adiabatic, reversible processes, the following
relationship (Equation 2.89) holds:
PV k  const
where k is defined in the text. Therefore,
1
P
k
V2   1 V1k 
 P2

c
7
Noting that k  P  and substituting the proper values provides
cv 5
V2  16.4 L
Now all of the needed values are available for calculating the work.
W  9 L  bar   900 J 
From the above energy balance,
U  900 J 
22
The change in internal energy can also be written according to Equation 2.53:
T2
U  n  cv dT
T1
Since cv is constant, the integrated form of the above expression is
5 
U  n R T2  T1 
2 
Using the ideal gas law and knowledge of P1 and V1 ,
T1  240.6 K 
and
T2  197.3 K 
The temperature is lower because more work is performed during the reversible expansion.
Review the energy balance. As more work is performed, the cooler the gas will become.
23
2.16
Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible.
Furthermore, no work is done on the system and potential and kinetic energy effects can be
neglected. Therefore, the energy balance becomes
uˆ  0
or
uˆ 2  uˆ1
From the steam tables
 kJ 
uˆ1  2619.2  
 kg 
 kJ 
 uˆ 2  2619.2  
 kg 
(200 bar, 400 ºC)
The values of û 2 and P2 constrain the system. The temperature can be found from the steam
tables using linear interpolation:
T2  327.5 º C

 kJ  
100 bar , uˆ 2  2619.2   


 kg  

Also at this state,
 m3 
vˆ2  0.02012 

 kg 
Therefore,
 

 m3  
3
Vvessel  mv2  1.0 kg  0.02012 
   0.020 m


kg

 

24
2.17
Let the entire tank represent the system. Since no heat or work crosses the system boundaries,
and potential and kinetic energies effects are neglected, the energy balance is
u  0
Since the tank contains an ideal gas
T2  T1  0
T2  T1  300 K
The final pressure can be found using a combination of the ideal gas law and conservation of
mass.
T1
T
 2
P1V1 P2V2
We also know
V2  2V1
Therefore,
P
P2  1  5 bar
2
25
2.18
(a)
First, as always, simplify the energy balance. Potential and kinetic energy effects can be
neglected. Therefore, the energy balance is
U  Q  W
Since, this system contains water, we can the use the steam tables. Enough thermodynamic
properties are known to constrain the initial state, but only one thermodynamic property is
known for the final state: the pressure. Therefore, the pressure-volume relationship will be used
to find the specific volume of the final state. Since the specific volume is equal to the molar
volume multiplied by the molecular weight and the molecular weight is constant, the given
expression can be written
Pvˆ1.5  const
This equation can be used to solve for v̂2 .
1
 P 
 1.5
vˆ2    1 vˆ11.5 
  P2 

Using
P1  20 bar 
vˆ1 
 
 m3 
1.0 m 3
 0.1  
10 kg 
 kg 
P2  100 bar 
gives
 m3 
vˆ2  0.0342 

 kg 
Now that the final state is constrained, the steam tables can be used to find the specific internal
energy and temperature.
T2  524.7 K 
 kJ 
uˆ 2  3094.6  
 kg 
To solve for the work, refer to the definition (Equation 2.7).
26
V2
W    PE dV
V1
or
vˆ 2
wˆ    PE dvˆ
vˆ1
Since the process is reversible, the external pressure must never differ from the internal pressure
by more than an infinitesimal amount. Therefore, an expression for the pressure must be
developed. From the relationship in the problem statement,
Pvˆ1.5  P1vˆ11.5  const
Therefore, the expression for work becomes
vˆ 2
wˆ   
vˆ1
P1vˆ11.5
vˆ1.5
dvˆ   P1vˆ11.5
vˆ 2
1
 vˆ1.5 dvˆ
vˆ1
Integration and substitution of proper values provides
 bar  m 3 
 kJ 
wˆ  2.840 
  284  
 kg 
 kg 

 kJ  
W  10 kg  284     2840 kJ 
 kg  

A graphical solution is given below:
P [bar]
2
100
60
1
20
Work
0.02
0.06
27
0.10
v [m 3/kg]
To solve for Q , U must first be found, then the energy balance can be used.

 kJ 
 kJ  
U  muˆ 2  uˆ1   10 kg  3094.6    2602.8     4918 kJ 
 kg 
 kg  

Now Q can be found,
Q  U  W  4918 kJ   2840 kJ   2078 kJ 
(b)
Since the final state is the same as in Part (a), U remains the same because it is a state function.
The energy balance is also the same, but the calculation of work changes. The pressure from the
weight of the large block and the piston must equal the final pressure of the system since
mechanical equilibrium is reached. The calculation of work becomes:
vˆ 2
W  mPE  dvˆ
vˆ1
All of the values are known since they are the same as in Part (a), but the following relationship
should be noted
PE  P2
Substituting the appropriate values results in
W  6580 kJ 
Again we can represent this process graphically:
P [bar]
2
100
Work
60
1
20
0.02
0.06
28
0.10
v [m 3/kg]
Now Q can be solved.
Q  U  W  4918 kJ   6580 kJ   1662 kJ 
(c)
This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is
characteristic of design problems there are many possible alternative solutions. We first refer to
the energy balance. The value of heat transfer will be zero when
U  W
For the same initial and final states as in Parts (a) and (b),
W  U  4918 kJ 
There are many processed we can construct that give this value of work. We show two
alternatives which we could use:
Design 1:
If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a
reversible compression followed by an irreversible compression. Let the subscript “i" represent
the intermediate state where the process switches from a reversible process to an irreversible
process. The equation for the work then becomes
vˆi


1
1
.
5

W  m  P2 (vˆ2  vˆi )  P1vˆ1 
dvˆ   4918000 J 
1
.
5


vˆ1 vˆ


Substituting in known values (be sure to use consistent units) allows calculation of v̂i :
 m3 
vˆi  0.0781 

 kg 
The pressure can be calculated for this state using the expression from part (a) and substituting
the necessary values.
1.5
v 
Pi  P1  1 
 vi 
Pi  29.0 bar 
29
Now that both Pi and v̂i are known, the process can be plotted on a P-v graph, as follows:
P [bar]
2
100
Work
60
i
1
20
0.02
0.06
0.10
v [m 3/kg]
Design 2:
In an alternative design, we can use two irreversible processes. First we drop an intermediate
weight on the piston to compress it to an intermediate state. This step is followed by a step
similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this
case, we again must find the intermediate state. Writing the equation for work:
W  m Pi (vˆi  vˆ1 )  P2 (vˆ2  vˆi )  4918000 J 
However, we again have the relationship:
v 
Pi  P1  1 
 vi 
1.5
Substitution gives one equation with one unknown vi:
1.5


 v1 
W  m  P1   (vˆi  vˆ1 )  P2 (vˆ2  vˆi )  4918000 J 


 vi 
There are two possible values vi to the above equation.
Solution A:
 m3 
vˆi  0.043  
 kg 
30
which gives
Pi  70.8 bar 
This solution is graphically shown below:
P [bar]
2
100
i
60
20
Work
1
0.06
0.10
v [m 3/kg]
0.02
Solution B
 m3 
vˆi  0.0762  
 kg 
which gives
Pi  30.0 bar 
This solution is graphically shown below:
P [bar]
2
100
60
i
20
Work
1
0.06
0.10
v [m 3/kg]
0.02
31
2.19
(a)
Force balance to find k:
Fspring=kx
Fatm=PatmA
Piston
Fmass=mg
Fgas=PgasA
mg
Pgas  Patm  A  kx
A
since V=Ax
Pgas  Patm  A  k2V
A
mg
since V is negative. Now solve:
2  10 5 Pa  1 10 5 Pa 
2040 kg 9.81 m/s 2   k 0.02 m 3 
0.1 m 2 2
0 .1 m 2
N
 k  5.01 10 4  
m
Work can be found graphically (see P-V plot) or analytically as follows: Substituting the
expression in the force balance above:
P
k V 
W A   
atm  A  A 2 dV
mg
V f
Vf
mg 
kV

W A    Patm 
d V 
dV  
2
A 

A
Vi
Vi


2


5.01 10 4 N/m  0.02 m 3  0 2 
W A  3.00  10 Pa 0.03  0.05 m 
2


2
0 .1 m 2



5

2


32

W A  5  103 J 
-W
0.5kJ
= 10squares square
= 5 kJ
3
P [bar]
k V
2
A
2
mg
A
2
Work
1
1
Patm
0.01
0.03
0.05
V [m3 ]
(b)
You need to find how far the spring extends in the intermediate (int) position. Assume
PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer
slightly). Since you know P and V for each state in Part (a), you can calculate n.
Pi
Pf
 V n
 V i 
or
f 
10 5 Pa
2 10 5 Pa
n
0.03 m 3 
 
3 

0.05 m
 n  1.35
Now using the force balance
mg
Pgas,int  Patm  A  kV
A2
with the above equation yields:
 P
V
Pint  Pi V i
int
n
mg k Vint  V i 
atm  A 
A2
This last equality represents 1 equation. and 1 unknown (we know k), which gives
Vint  0.0385 m 3
Work can be found graphically (see P-V plot) or analytically using:
33
V int
Vf
Vi
Vint
W B   Pgas dV   PgasdV
expanding as in Part (a)
V
int
5 N 
W B   
2  10 m2 dV 
Vi
Vint
V
V if
V i
Vint
Vint
f
6 N

5
V



5

10
d
V

 3  10 mN2 
5
dV 
m
6
 5  10 N5 V d V 
m
Therefore,
WB  3.85 kJ 
just like we got graphically.
P [bar]
3
2
2
mg
A
mg
A
1
1
Work
Patm
0.01
0.03
0.05
V [m3 ]
(c)
The least amount of work is required by adding differential amounts of mass to the piston. This
is a reversible compression. For our assumption that PVn = const, we have the following
expression:
WC , rev 
Vf
Vf
Vi
Vi
const
 Pgas dV   V 1.35 dV
Calculate the constant from the initial state
34


const.  1 105 Pa 0.05 m 3
1.35  1.75 103
Therefore,
.03
WC,rev   1.75  103 dV
1.35
V
.05
10 V
  1.750.35
3
0 .35 .03
.05
 2800 J 
P [bar]
3
2
2
1
1
Workrev
Patm
0.01
0.03
0.05
V [m3 ]
35
2.20
Before this problem is solved, a few words must be said about the notation used. The system
was initially broken up into two parts: the constant volume container and the constant pressure
piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers
the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f”
denotes the final state.
To begin the solution, the mass of water present in each part of the system will be calculated.
The mass will be conserved during the expansion process. Since the water in the rigid tank is
saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is
constrained to a specific state. From the steam tables,
 kJ 
vˆ1l, i  0.001156  
 kg 
 kJ 
vˆ1v, i  0.12736  
 kg 
 kJ 
uˆ1l, i  850.64  
 kg 
 kJ 
uˆ1v, i  2595.3  
 kg 
(Sat. water at 200 ºC)
P sat  1553.8 kPa 
Knowledge of the quality of the water and the overall volume of the rigid container can be used
to calculate the mass present in the container.
 
 
V1  0.05m1 vˆ1l, i  0.95m1 vˆ1v,i
 
Using the values from the steam table and V1  0.5 m 3 provides
m1  4.13 kg 
Using the water quality specification,
m1v  0.95m1  3.92 kg 
m1l  0.05m1  0.207 kg 
For the piston-cylinder assembly, both P and T are known. From the steam tables
36
 m3 
vˆ2, i  0.35202 

 kg 
 kJ 
uˆ 2, i  2638.9  
 kg 
(600 kPa, 200 ºC)
Enough information is available to calculate the mass of water in the piston assembly.
V
m2  2  0.284 kg 
vˆ2
Now that the initial state has been characterized, the final state of the system must be determined.
It helps to consider what physically happens when the valve is opened. The initial pressure of
the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank
and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is
constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final
temperature and pressure of the entire system will match the surroundings’. In other words,
 kJ 
uˆ f  uˆ 2, i  2638.9  
 kg 
(600 kPa, 200 ºC)
Thus, the change in internal energy is given by
U  (m2  m1v, i  m1l, i )uˆ f  m2 uˆ 2, i  m1v, i uˆ1v, i  m1l, i uˆ1l, i
Substituting the appropriate values reveals
U  541.0 kJ 
To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa
(weight of the piston was assumed negligible). From Equation 2.7,
Vf
W   PE  dV   PE (V f  Vi )
Vi
where
PE  600000 Pa 
 
V f  (m2  m1v, i  m1l,i )vˆ2, i  1.55 m 3
   
 
Vi  0.1 m 3  0.5 m 3  0.6 m 3
37
Note: vˆ2, i was used to calculate V f because the temperature and pressure are the same
for the final state of the entire system and the initial state of the piston-cylinder assembly.
The value of W can now be evaluated.
W  570 kJ 
The energy balance is used to obtain Q.
Q  U  W  541.0 kJ    570 kJ   1111 kJ 
38
2.21
A sketch of the process follows:
system
boundary
system
Process
B
p 1,B = 20 bar
T1,B = 250 o C
A
p 1,A = 10 bar
T1,A = 700 o C
H2 O
boundary
H2 O
50 cm
A
p2
T2
H2 O
B
p2
T2
H2 O
10 cm
The initial states are constrained. Using the steam tables, we get the following:
p
T
v
u
V
m
V
v
State 1,A
10 [bar]
700 [oC]
0.44779 [m3/kg]
3475.35 [kJ/kg]
0.01 m3
State 1,B
20 [bar]
250 [oC]
0.11144 [m3/kg]
2679.58 [kJ/kg]
0.05 m3
0.11 [kg]
0.090 [kg]
All the properties in the final state are equal. We need two properties to constrain the system:
We can find the specific volume since we know the total volume and the mass:
v2 
V1, A  V1, B
m1, A  m1, B

 
 m3 
0.06 m 3
 0.30 

0.20 kg 
 kg 
We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is
rigid, W=0. An energy balance gives:
U  Q  W  0
Thus,
U 2  U1  m1,A u1, A  m1,Bu1,B
or
u2 
 kJ 
U 2 m1, Au1, A  m1, B u1, B

 3121  
m2
m1, A  m1, B
 kg 
We have constrained the system with u2 and v2, and can find the other properties from the steam
Tables. Very close to
T2 = 500 [oC] and
P2 = 1200 [kPa]
Thus,
39

 m3  
V2, A  v2, A m2, A   0.30 
 0.09 kg   0.267 m and


kg

 

x  0.267  0.1  0.167 m
40
2.22
We start by defining the system as a bubble of vapor rising through the can. We assume the
initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable
estimate for T1? A schematic of the process gives:
State 2:
T 2= ?
P2= 1.01 bar
State 1:
T 1 = 278 K
P1 = 3 bar
where the initial state is labeled state 1, and the final state is labeled state 2. To find the final
temperature, we perform an energy balance on the system, where the mass of the system (CO2 in
the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy
effects are negligible, the energy balance is
u  w
Expressions for work and internal energy can be substituted to provide
cv T2  T1     PE dv   PE v 2  v1 
where cv = cP – R. Since CO2 is assumed an ideal gas, the expression can be rewritten as
T

PT 
T 
cv T2  T1    PE R  2  1    R T2  2 1 
P1 
 P2 P1 

where the equation was simplified since the final pressure, P2, is equal to the external pressure,
PE. Simplifying, we get:


RP 
R
T2 1    T1 1  2 
 c v P1 
 cv 
or

RP  c
T2  T1 1  2  v  237 K
 cv P1  c P
41
2.23
The required amount of work is calculated as follows:
W   P V
The initial volume is zero, and the final volume is calculated as follows:
4
4
V  πr 3  π 0.5 ft 3  1.54 ft 3  0.0436 m 3
3
3
Assuming that the pressure is 1 atm, we calculate that



W  1.01325  105 Pa 0.0436 m 3  0 m 3  4417 J
This doesn’t account for all of the work because work is required to stretch the rubber that the
balloon is made of.
42
2.24
(a)
Since the water is at its critical point, the system is constrained to a specific temperature,
pressure, and molar volume. From Appendix B.1
 m3 
vˆc  0.003155 

 kg 
Therefore,
m
V

vˆc
 
0.01 m 3
 3.17 kg 
 m3 
0.003155 

 kg 
(b)
The quality of the water is defined as the percentage of the water that is vapor. The total volume
of the vessel can be found using specific volumes as follows
V  m l vˆ l  m v vˆ v  1  x mvˆ l   xm vˆ v
where x is the quality of the water. To solve for the quality, realize that starting with saturated
water at a pressure of 1 bar constrains the water. From the steam tables,
 m3 
vˆ v  1.6940 

 kg 
 m3 
ˆv l  0.001043 

 kg 
(sat. H2O at P = 1 bar)
Now the quality can be found
x  0.00125
Thus, the quality of the water is 0.125%.
(c)
To determine the required heat input, perform an energy balance. Potential and kinetic energy
effects can be neglected, and no work is done. Therefore,
U  Q
43
where

U  muˆ 2  1  x muˆ1l   xm uˆ1v

From the steam tables
 kJ 
uˆ 2  2029.58  
 kg 
 m3 
l
ˆ
u1  417.33 

 kg 
 m3 
uˆ1v  2506.1 

 kg 
(H2O at its critical point)
(sat. H2O at P=1 bar)
Evaluation of the expression reveals
U  5102.6 kJ   5.10  10 6 J 
44
2.25
(a)
Consider the air in ChE Hall to be the system. The system is constant volume, and potential and
kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is
du
 q
dt
since the temperature of the system changes over time. Using the given expression for heat
transfer and the definition of dU , the expression becomes
cv dT
 hT  Tsurr 
dt
We used a negative sign since heat transfer occurs from the system to the surroundings. If cv is
assumed constant, integration provides
cv ln T  Tsurr   ht  C
where C is the integration constant. Therefore,
T  Tsurr  C1e
 h 
  t 
 cv 
where C1 is a constant. Examining this equation reveals that the temperature is an exponential
function of time. Since the temperature is decreasing, we know that the plot of temperature vs.
time shows exponential decay.
Temperature
T0
Tsurr
time
45
(b)
Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is
22 ºC. Therefore,
T  Tsurr  C1e
 h 
 t
 c 
 v 
22 º C  2 º C  C1e (0)
C1  20 º C
After 10 PM, ( t  4 hr ), the temperature is 12 ºC.
12 º C  2 º C  20 º C e
h
  0.173 hr -1
cv
 h

  ( 4 hr) 
 c

 v

At 6 AM, t  12 hr . Substitution of this value into the expression for temperature results in
T  4 .5 º C
46
2.26
The gas leaving the tank does flow work as it exits the valve. This work decreases the internal
energy of the gas – lowering the temperature. During this process, water from the atmosphere
will become supersaturated and condense. When the temperature drops below the freezing point
of water, the water forms a solid.
Attractive interactions between the compressed gas molecules can also contribute to this
phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of
these interactions in Chapter 4.
47
2.27
Mass balance
dm
 m in  m out  m in
dt
Separating variables and integrating:
m2
t
m1
0
 dm   m in dt
or
t
m2  m1   m in dt
0
Energy balance
Since the potential and kinetic energy effects can be neglected, the open system, unsteady state
energy balance is
 dU 

   m out hout   m in hin  Q  W s
 dt  sys out
in
The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream
and not outlet stream. Therefore, the energy balance simplifies to
 dU

 dt

  m in hin
 sys
The following math is performed
U2
t
t
U1
0
0
 dU   m in hin dt  hin  m in dt
U 2  U 1  m2 uˆ 2  m1uˆ1  m2  m1 hˆin
where the results of the mass balance were used. Both m2 and m1 can be calculated by dividing
the tank volume by the specific volume
48
m2 
V
vˆ2
m1 
V
vˆ1
Substitution of these relationships and simplification results in
uˆ 2  hin  uˆ1  hin 
vˆ2

vˆ1
0
From the steam tables:
 kJ 
uˆ1  2583.6  
 kg 
 m3 
vˆ1  0.19444 

 kg 
 kJ 
hˆin  3177.2  
 kg 
(sat. H2O vapor at 1 MPa)
(6 MPa, 400 ºC)
There are still two unknowns for this one equation, but the specific volume and internal energy
are coupled to each other. To solve this problem, guess a temperature and then find the
corresponding volume and internal energy values in the steam tables at 6 MPa. The correct
temperature is the one where the above relationship holds.
T  600 º C : Expression = 4427.6
T  500 º C : Expression = 1375.9
T  450 º C : Expression = -558.6
Interpolation between 500 ºC and 450 ºC reveals that the final temperature is
T2  464.4 º C
49
2.28
We can pick room temperature to be 295 K
Tin  T1  295 K
Mass balance
dn
 n in  n out  n in
dt
Separating variables and integrating:
n2
t
n1
0
 dn   nin dt
or
t
n2  n1   nin dt
0
Energy balance
Neglecting ke and pe, he unsteady energy balance, written in molar units is written as:
 dU 
  nin hin  nout hout  Q  W

dt
 sys

The terms associated with flow out, heat and work are zero.
 dU 
  nin hin

 dt  sys
Integrating both sides with respect to time from the initial state where the pressure is 10 bar to
the final state when the tank is at a pressure of 50 bar gives:
U2
t
t
U1
0
0
 dU   nin hin dt  hin  nin dt
since the enthalpy of the inlet stream remains constant throughout the process. Integrating and
using the mass balance above:
n2u2  n1u1  n2  n1 hin
50
Now we do some math:
n2u 2  n1u1  n2  n1 hin
n2 u 2  hin   n1 u1  hin 
By the definition of h
hin  uin  Pin vin  uin  RTin  u1  RT1
so
n2 u 2  u1   n2 RT1  n1 u1  u1   n1R T1
n2 cv T2  T1   n2 RT1  n1RT1
Since cv  c P  R 
n2
3
R
2
3
T2  T1  n2T1  n1T1
2
or
3n2T2  5n2T1  2n1T1
dividing by n1:
n
n
3 2 T2  5 2 T1  2T1
n1
n1
Using the ideal gas law:
n2 P2T1

n1 P1T2
so
P T 
P T 
3 2 1 T2  5 2 1 T1  2T1
 P1T2 
 P1T2 
or
51
P T 
5 2 1 
 P1   434 [K]
T2 
  P2  
3   2
  P1  
(b) Closed system
u  q  w  q
uˆ 
cv
T2  T1   5R T2  T1   28.9  kJ 
MW
2MW
 kg 
 kJ 
q  28.9  
 kg 
(c)
P2T2  P3T3
PT
P3  2 2  34 bar 
T3
52
2.29
Mass balance
dn
 n in  n out  n in
dt
Separating variables and integrating:
n2
t
n1  0
0
 dn   nin dt
or
t
n2   nin dt
0
Energy balance
Neglecting ke and pe, the unsteady energy balance, in molar units, is written as:
 dU 

  nin hin  nout hout  Q  W
dt

 sys
The terms associated with flow out and heat are zero.
 dU 

  nin hin  W
 dt  sys
Integrating both sides with respect to time from the empty initial state to the final state gives:
U2
t
t
t
U1
0
0
0
 dU   nin hin dt   W dt  hin  nin dt  W  hin n2  W
since the enthalpy of the inlet stream remains constant throughout the process. The work is
given by:
W   n2 Pext (v2  v1 )   n2 Pext v2
53
n2u2  n2 hin  Pext v2 
Rearranging,
u 2  hin  Pext v2  uin  Pin vin  Pext v2
u 2  uin  Pin vin  Pext v2
P T
cv T2  Tin   RTin  R ext 2
P2
so
T2 
cv  R 


P
 cv  ext R 
P2 

Tin  333 K 
54
2.30
valve maintains
pressure in system
constant
v
T1 = 200 oC
x1 = 0.4
V = 0.01 m3
l
Mass balance
dm
 m in  m out  m out
dt
Separating variables and integrating:
m2
t
m1
0
 dm   m out dt
or
t
m2  m1    m out dt
0
Energy balance
 dU 
   m out hˆout  Q

dt

 sys
Integrating
m 2 uˆ 2
t
m1uˆ1
0


t
t
0
0
 dU    m out hˆout  Q dt  hˆout  m out dt   Q dt
Substituting in the mass balance and solving for Q
Q  m2uˆ2  m1uˆ1  m2  m1 hˆout
55
We can look up property data for state 1 and state 2 from the steam tables:
 m3 
vˆ1  (1  x)vˆ f  xvˆg  0.6  .001  0.4  0.1274  0.051  
 kg 
 m3 
vˆ2  0.1274  
 kg 
So the mass in each state is:
 
m1 
0.01 m 3
V1
 0.196 kg 

vˆ1
 m3 
0.051  
 kg 
m2 
V2

vˆ2
 
0.01 m 3
 0.0785 kg 
 m3 
0.1274  
 kg 
m2  m1  0.1175 kg 
And for energy and enthalpy
 kJ 
uˆ1  (1  x)uˆ f  xuˆ g  0.6  850.64  0.4  2597.5  1549  
 kg 
 kJ 
uˆ 2  2595.3  
 kg 
 kJ 
hˆout  2793.2  
 kg 
Solving for heat, we get
Q  m2uˆ2  m1uˆ1  m2  m1 hˆout  228 kJ 
56
2.31
Consider the tank as the system.
Mass balance
dm
 m in  m out  m in
dt
Separating variables and integrating:
m2
t
m1
0
 dm   m in dt
or
t
m2  m1   m in dt
0
Energy balance
Since the potential and kinetic energy effects can be neglected, the open system, unsteady state
energy balance is
 dU 

   m out hout   m in hin  Q  W s
 dt  sys out
in
The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream
and not outlet stream. Therefore, the energy balance simplifies to
 dU 

  m in hin
 dt  sys
The following math is performed
U2
t
U1  0
0
 dU   m
t
h dt  hin  m in dt
in in
0
U 2  m2 uˆ 2  m2 hˆin
where the results of the mass balance were used. Thus,
uˆ 2  hˆin
From the steam tables
57
 kJ 
uˆ 2  3632.5  
 kg 
(9 MPa, 800 ºC)
so
 kJ 
hˆin  3632.5  
 kg 
We can use the value of hin and the fact that the steam in the pipe is at 9 MPa to find the
temperature.
Tin  600 º C
58
2.32
(a)
First, the energy balance must be developed. Since the problem asks how much energy is stored
in the battery after 10 hours of operation, the process is not steady-state. Let the battery be the
system. Potential and kinetic energy effects can be neglected. Furthermore, heating of the
battery as it is charged can be ignored. The energy balance is
 dU 

  Q  W s
dt

 sys
No shaft work is performed, but electrical is supplied to the battery, which must be accounted for
in W s . The value of Q is given explicitly in the problem statement. Both of these values remain
constant over time, so integration provides


U  Q  W s t
From the problem statement
W s  5 kW 
Q  1 kW 
t  36000 s
Substituting these values allows the calculation of the amount of energy stored:
U  144,000 kJ   144 MJ 
(b)
To calculate the velocity of the falling water, an energy balance must be developed with the
water passing through the electricity generator (probably a turbine) as the system, where the

water enters with a velocity V1 and leaves with a negligible velocity, which will be approximated
as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the
temperature of the water does not change in the process, so the change in internal energy is zero.
Also, view the process as adiabatic. The energy balance reduces to
E K  W river
where W river is the power of the flowing water. The actual power being provided by the stream
can be calculated using the efficiency information. Let  represent the efficiency.
59
W
  s
Wriver
W
5 kW 
W river  s 
 10 kW 
0.5

The value of W river should be negative since the water is supplying work that is stored electrical
energy. Therefore, the energy balance becomes
E K  10000 W 
This expression can be rewritten as


1 2 2
m V2 V1  10000 W 
2
From the problem statement and the assumptions made,
 kg 
m  200  
 s 

m
V2  0  
s
Therefore,

m
V1  10  
s
There are a number of reasons for the low conversion efficiency. A possible potential energy
loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is
lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag)
effects.
60
2.33
Considering the turbine to be the system, rearrangement of the steady-state, open system energy
balance provides
 nout (h  ek  e p )out   nin (h  ek  e p )in  Q  W s
out
in
Performing a mass balance reveals
n in  n1  n out  n 2
Assuming the rate of heat transfer and potential energy effects are negligible and realizing that
there is one inlet and one outlet allows the simplification of the above equation to
W s  n 2 h2  h1   e K , 2  e K ,1 
h2  h1  can be rewritten using Equations 2.58 and Appendix A.2
h2  h1    c p dT  R  A  BT  CT 2  DT  2  ET 3 dT
T2
T1


Since the quantity ek ,2  ek ,1 is multiplied by n , it is rewritten as follows for dimensional
homogeneity
e
k ,2

 2 2
1
 ek ,1   ( MW ) air V2  V1
2

To solve for n , the ideal gas law is used
P2V2  n 2 RT21
P V
n 2  2 2
RT2
To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional
area

 D2 2V2 


V2  

 4 
The energy balance is now
61


 D 2V   T2
 2 2 
P
 1
2
3
2
2
2
2




Ws 
R A  BT  CT  DT  ET dT   ( MW ) air V2  V1 

RT2 
4   
 2
 T1


Substituting values from Table A.2.1 and the problem statement results in


W  4.84  10 6 [W]  -4.84 [MW]
62


2.34
First, a sketch of the process is useful:
30 bar
20 bar
100 oC
150 oC
q
To find the heat in we will apply the 1st law. Assuming steady state, the open system energy
balance with one stream in and one stream out can be written:
0  n h1  h2   Q
which upon rearranging is:
Q
 h2  h1
n
Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from
the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values
for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an
expression for the ideal gas heat capacity:
cp
R
 1.424  14.394  10  3 T  4.392  10  6 T 2
with T in (K). Since this expression is limited to ideal gases any change in temperature must be
under ideal conditions. From the definition of heat capacity:


T2
Q
 h2  h1   1.424  14.394 10  3 T  4.392 10  6 T 2 dT
n
T1
By integrating and substituting the temperatures, we obtain:
Q
 J 
 5590 
n
 mol 
63
2.35
A schematic of the process follows:
WÝs
T1 = 350 oC
T1 = 308 oC
P2= 1 atm
v1 = 600 cm3/mol
To solve for W s / n we need a first law balance. With negligible eK and eP, the 1st law for a
steady state process becomes:
0  n h1  h2   Q  W s
If heat transfer is negligible,
W s
 h
n
We can calculate the change in enthalpy from ideal gas heat capacity data provided in the
Appendix.


T2
T2
W s
 h   c p dT R  1.213  28.785  10  3 T  8.824  10  6 T 2 dT
n
T1
T1
Integrate and evaluate:
W s
 J 
 5358 
n
 mol 
64
2.36
(a)
First start with the energy balance. Nothing is mentioned about shaft work, so the term can be
eliminated from the energy balance. The potential and kinetic energy effects can also be
neglected. Since there is one inlet and one outlet, the energy balance reduces to
Q  n 2 h2  n1h1
A mass balance shows
n 2  n1
so the energy balance reduces to
Q  n1 h2  h1 
Using the expressions from Appendix A.1, the energy balance becomes
T2


Q  n1R  A  BT  CT 2  DT  2  ET 3 dT
T1
Using
A  3.376
B  0.557  10  3
C 0
D  0.031 105
E 0
 J 
R  8.314 
 mol  K 
 mol 
n1  20 
 s 
T1  373.15 K 
T2  773.15 K 
gives
Q  245063 W   245.1 kW 
65
(b)
To answer this question, think about the structure of n-hexane and carbon monoxide. N-hexane
is composed of 20 atoms, but carbon monoxide has two. One would expect the heat capacity to
be greater for n-hexane since there are more modes for molecular kinetic energy (translational,
kinetic, and vibrational). Because the heat capacity is greater and the rate of heat transfer is the
same, the final temperature will be less.
66
2.37
First start with the energy balance around the nozzle. Assume that heat transfer and potential
energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance
reduces to
n 2 (h  e K ) 2  n1 (h  e K )1  0
A mass balance shows
n1  n 2
On a mass basis, the energy balance is


1 
hˆ2  hˆ1  eˆK ,1  eˆK ,2  V12  V2 2
2

Since the steam outlet velocity is much greater than the velocity of the inlet, the above
expression is approximately equal to
 
1 
hˆ2  hˆ1   V2 2
2
The change in enthalpy can be calculated using the steam tables.
 J 
h1  2827.9  103  
 kg 
 J 
h2  2675.5  103  
 kg 
(10 bar, 200 ºC)
(sat. H2O(v) at 100 kPa)
Therefore,

m
V2  552  
s
To solve for the area, the following relationship is used

AV2
m 
v̂2
From the steam tables
67
 m3 
ˆv2  1.6940 

 kg 
Now all but one variable is known.
 
A  3.07 10 3 m 2
68
2.38
First start with the energy balance around the nozzle. Assume that heat transfer and potential
energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance
reduces to
n 2 (h  ek ) 2  n1 (h  ek )1  0
The molar flow rates can be eliminated from the expression since they are equal. Realizing that
e K ,2  e K ,1 since the velocity of the exit stream is much larger than the velocity of the inlet
stream simplifies the energy balance to
h2  h1  ek ,2
Using Appendix A.2 and the definition of kinetic energy
T2



1
h2  h1  R  A  BT  CT 2  DT  2  ET 3 dT   ( MW ) C 3 H 8 V22
2
T1
From Table A.2.1
A  1.213
B  28.785  10  3
C  8.824  10  6
D0
E 0
It is also important that the units for the molecular weight and universal gas constant are
consistent. The following values were used
 J 
R  8.314 
 mol  K 
 kg 
( MW ) C3 H 8  0.0441 
 mol 
Integration of the above expression and then solving for T2 provides
T2  419.2 K 
69
2.39
First start an energy balance around the diffuser. Assume that heat transfer and potential energy
effects are negligible. The shaft work term is also zero. The energy balance reduces to
n 2 (h  ek ) 2  n1 (h  ek )1  0
A mass balance reveals
n1  n 2
The molar flow rates can be eliminated from the expression. Using the definitions of enthalpy
and the kinetic energy, the equation can be rewritten as

T2
 2 2
1
c
dT


(
MW
)
V
2  V1
P
air

2

T1
The temperature and velocity of the outlet stream are unknown, so another equation is needed to
solve this problem. From the conservation of mass,





P1V1 P1 A1V1
P2 A2V2


T1
T1
T2

where A2, the cross-sectional area of the diffuser outlet, is twice the area of the inlet. Therefore,

1  P  T  
V2   1  2 V1
2  P2  T1 
Using Appendix A.2 and the above expression, the energy balance becomes
T2

R  A  BT  CT  DT
T1
2
2
2

 
1  P1  T2   
1

 ET dT   ( MW ) air    V1   V12 
  2  P2  T1  

2


3

Substituting values from the problem statement provides an equation with one unknown:
T2  381 K
Therefore,
 1  1 bar  381 K 
m
300 m/s   111  
V2  

2  1.5 bar  343.15 K 
s
70
2.40
To find the minimum power required for the compressor, one must look at a situation where all
of the power is used to raise the internal energy of the air. None of the power is lost to the
surroundings and the potential and kinetic energy effects must be neglected. Therefore, the
energy balance becomes
0  n1h1  n 2 h2  W s
Performing a mass balance reveals
n1  n 2
The energy balance reduces to
W s  n1 h2  h1 
Using Equation 2.58 and Appendix A.2, the equation becomes
T2


W s  n1R  A  BT  CT 2  DT  2  ET 3 dT
T1
Table A.2.1 and the problem statement provide the following values
A  3.355
B  0.575  10 3
C 0
D  1600
E 0
 mol 
n1  50 
 s 
T1  300 K
To find the work, we still need T2. We need to pick a reasonable process to estimate T2. Since
the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston
situation. For this situation,
PV k  const.
Since we are assuming the air behaves ideally, we can rewrite the equation as
71
k
 n RT 
 n RT 
P1  1 1   P2  2 2 
 P1 
 P2 
k
P11 k T1k  P21 k T2k
Substituting values from the problem statement, we obtain
1 7 / 5 

7 / 5 1 bar 
T2  300 K 

10 bar 1 7 / 5 

5/7
 579 K
Substitute this value into the expression for the work and evaluate:
W s  417.4 kW 
72
2.41
(a)
Perform a mass balance:
n1  n 2  n out
Apply the ideal gas law:
P1V1 P2V2

 n out
RT1 RT2
Substitute values from the problem statement:
1105 5 10 3   2 105 2.5 10 3   nout
8.314373.15
8.314293.15
n out  0.366 mol/s
(b)
No work is done on the system, and we can neglect potential and kinetic energy effects. We will
assume the process is also adiabatic. The energy balance reduces to
0   nin hin   n out hout
in
out
n out hout  n1h1  n 2 h2   n1 hout  h1   n 2 hout  h2   0
We can calculate the enthalpy difference from the given ideal heat capacity:
 3.267  5.324  10
Tout
n1 R
373.15
3

 3.267  5.324  10
Tout
T dT  n 2 R
3

T dT  0
293.15
Again, we must calculate the molar flow rates from the ideal gas law. Upon substitution and
evaluation, we obtain
Tout  329 K 
73
2.42
(a)
Since the temperature, pressure, and volumetric flow rate are given, the molar flow rate is
constrained by the ideal gas law.
  
Note: 1.67  10 8 m 3 /s  1 cm 3 / min
n 



PV 1.0135  10 5 Pa 1.67  10 8 m 3 /s 

 7.45  10 7 mol/s
8.314 J/mol  K 273.15 K 
RT
To recap, we have shown
1 SCCM   7.45  10 7 mol/s
(b)
Assumptions:
N2 is an ideal gas
All power supplied by the power supply is transferred to the N2
Uniform temperature radially throughout sensor tube
Kinetic and potential energy effects negligible in energy balance
Let x represent the fraction of N2 diverted to the sensor tube, and n s represent the molar flow
rate through the sensor tube. Therefore, the total molar flow rate, ntotal , is
n
ntotal  s
x
We can use temperature and heat load information from the sensor tube to find the molar flow
rate through the sensor tube. First, perform an energy balance for the sensor tube:
H  n s hout  hin   Q
The enthalpy can be calculated with heat capacity data. Therefore,
n s 
Q
T2
 cP dT
T1
Now, we can calculate the total molar flow rate.
74
Q
ntotal 
T2
x  c P dT
T1
To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found
in Part (a)
Q
vtotal (SCCM ) 
T2
x  c P dT


1 SCCM 


 7.45 10  7 mol/s 


T1
(c)
To find the correction factor for SiH4, re-derive the expression for flow rate for SiH4 and then
divide it by the expression for N2 for the same power input, temperatures, and fraction of gas
diverted to the sensor tube.
Q
T2
Factor 
vtotal , SiH 4
vtotal , N 2
x  c P, SiH 4 dT

T1
Q
T2
x  c P, N 2 dT


1 SCCM 


 7.45  10  7 mol/s 




1 SCCM 


 7.45  10  7 mol/s 


T2
 c P, N

2
dT
T1
T2
 c P, SiH
4
dT
T1
T1
If we assume that heat capacities are constant, the conversion factor simplifies:
Factor 
c P, N 2
c P, SiH 4
Using the values in Appendix A.2.2 at 298 K, we get
Factor  0.67
75
2.43
(a)
It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both
cases the internal energy of the gas must be increased. In the constant pressure cylinder work,
Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is
not required with a constant volume.
(b)
As you perspire, sweat evaporates from your body. This process requires latent heat which cools
you. When the water content of the environment is greater, there is less evaporation; therefore,
this effect is diminished and you do not feel as comfortable.
76
2.44
From the steam tables at 10 kPa:
T(K)
h
323.15
2592.6
373.15
2687.5
423.15
2783
473.15
2879.5
h vs. T
6000
y = 0.0003x
5000
523.15
2977.3
573.15
3076.5
4000
673.15
3279.5
3000
773.15
3489
873.15
3705.4
973.15
3928.7
1073.15
4159.1
1173.15
4396.4
1273.15
4640.6
1373.15
4891.2
1473.15
5147.8
1573.15
5409.7
+ 1.6241x + 2035.7
2
R =1
Series1
2000
Poly. (Series1)
1000
0
0
 dh 
cP  
  1.6241  0.0006624T
 dT  P
2
500
1000
T
 kJ 
 J 
 kg  K   3.516  0.001434T R  mol  K 




Now compare the above values to those in Appendix A.2.
A
B
Steam Tables
3.516
0.001434
Appendix A
3.470
0.001450
% difference
1.3
1.4
77
1500
2000
2.45
For throttling devices, potential and kinetic energy effects can be neglected. Furthermore, the
process is adiabatic and no shaft work is performed. Therefore, the energy balance for one inlet
and one outlet is simplified to
n1 h1  n 2 h2
which is equivalent to
m 1 hˆ1  m 2 hˆ2
Since mass is conserved
hˆ1  m 2 hˆ2
From the steam tables:
 kJ 
hˆ1  3398.3  
 kg 
 kJ 
 hˆ2  3398.3  
 kg 
(8 MPa, 500 ºC)
Now that we know û 2 and P2 , T2 is constrained. Linear interpolation of steam table data gives
T2  457 º C
78
2.46
(a)
An expression for work in a reversible, isothermal process was developed in Section 2.7.
Equation 2.77 is
P 
W  nRT ln 2 
 P1 
Therefore,
P 
w  RT ln 2 
 P1 
Evaluating the expression with
 J 
R  8.314 
 mol  K 
T  300 K 
P2  100 kPa 
P1  500 kPa 
gives
 J 
w  4014  
 kg 
(b)
Equation 2.90 states
w
R
T2  T1 
k 1
Since the gas is monatomic
5
cP   R
2
3
cv    R
2
and
79
k
5
3
T2 can be calculated by applying the polytropic relation derived for adiabatic expansions. From
Equation 2.89
PV k  const
 Pv k  const
By application of the ideal gas law
RT
P
Since R is a constant, substitution of the expression for P into the polytropic relation results in
v
P 1 k T k  const
 P11 k T 1 k  P2 1 k T2 k
This relation can be used to solve for T2 .
T2  157.6 K 
Now that T2 is known, value of work can be solved.
 J 
w  1775.9  
 kg 
80
2.47
(a)
The change in internal energy and enthalpy can be calculated using
hˆ  hˆ l 1 atm, 100 º C   hˆ l 1 atm, 0 º C 
uˆ  uˆ l 1 atm, 100 º C   uˆ l 1 atm, 0 º C 
We would like to calculate these values using the steam tables; however, the appendices don’t
contain steam table data for liquid water at 0.0 ºC and 1 atm. However, information is provided
for water at 0.01 ºC and 0.6113 kPa. Since the enthalpy and internal energy of liquid water is
essentially independent of pressure in this pressure and temperature range, we use the steam
table in the following way
 kJ 
hˆ l 1 atm, 100 º C   419.02  
 kg 
 kJ 
hˆ l 1 atm, 0 º C   hˆ l 0.6113 kPa , 0.01 º C   0  
 kg 
 kJ 
uˆ l 1 atm, 100 º C   418.91  
 kg 
 kJ 
uˆ l 1 atm, 0 º C   uˆ l 0.6113 kPa , 0.01 º C   0  
 kg 
Therefore,
 kJ 
ĥ  419.02  
 kg 
 kJ 
û  418.91  
 kg 
(b)
The change in internal energy and enthalpy can be calculated using
hˆ  hˆ v 1 atm, 100 º C   hˆ l 1 atm, 100 º C 
uˆ  uˆ v 1 atm, 100 º C   uˆ l 1 atm, 100 º C 
From the steam tables
81
 kJ 
hˆ v  2676.0  
 kg 
 kJ 
hˆ l  419.02  
 kg 
 kJ 
uˆ v  2506.5  
 kg 
 kJ 
uˆ l  418.91  
 kg 
Therefore,
 kJ 
hˆ  2256.99  
 kg 
 kJ 
uˆ  2087.59  
 kg 
The change in internal energy for the process in Part (b) is 5.11 times greater than the change in
internal energy calculated in Part (a). The change in enthalpy in Part (b) is 5.39 times greater
than the change in enthalpy calculated in Part (a).
82
2.48
To calculate the heat capacity of Ar, O2, and NH3 the following expression, with tabulated values
in Table A.2.1, will be used,
cP
 A  BT  CT 2  DT  2  ET 3
R
where T is in Kelvin. From the problem statement
T  300 K 
and from Table 1.1
 J 
R  8.314 
 mol  K 
To find the A-E values, Table A.2.1 must be referred to.
Formula
Ar
O2
NH3
A
3.639
3.5778
B  103
0.506
3.02
C  10 6
0
0
D  10 5
-0.227
-0.186
E 109
0
0
The values are not listed for Ar since argon can be treated as a monatomic ideal gas with a heat
capacity independent of temperature. The expression for the heat capacity is
5
c P, Ar    R
2
Now that expressions exist for each heat capacity, evaluate the expressions for T  300 K  .
 J 
c P, Ar  20.785 
 mol  K 
 J 
c P, O2  29.420 
 mol  K 
 J 
c P, NH 3  35.560 
 mol  K 
By examining the heat capacity for each molecule, it should be clear that the magnitude of the
heat capacity is directly related to the structure of the molecule.
83
Ar
O2

Since argon is monatomic, translation is the only mode through which the atoms can
exhibit kinetic energy.

Translation, rotation, and vibration modes are present. Since oxygen molecules are
linear, the rotational mode of kinetic energy contributes RT per mol to the heat capacity.
NH3

Translation, rotation, and vibration modes are present. Ammonia molecules are nonlinear, so the rotation mode contributes 3RT/2 per mole to the heat capacity.
The vibration contributions can also be analyzed for oxygen and ammonia, which reveals that the
vibration contribution is greatest for ammonia. This is due to ammonia’s non-linearity.
84
2.49
For a constant pressure process where potential and kinetic energy effects are neglected, the
energy balance is given by Equation 2.57:
Q  H
The change in enthalpy can be written as follows
H  mh2  h1 
From the steam tables:
 kJ 
hˆ2  hsat. vapor at 10 kPa   2584.6  
 kg 
 kJ 
hˆ1  hsat. liquid at 10 kPa   191.81  
 kg 
Therefore,

 kJ  
 kJ 
Q  2 kg  2584.6    191.81   
 kg  
 kg 

Q  4785.6 kJ 
We can find the work from its definition:
Vf
W    PE dV
Vi
The pressure is constant, and the above equation can be rewritten as follows
W   PE mvˆ2  vˆ1 
From the steam tables:
 m3 
ˆv2  vˆsat. vapor at 10 kPa   14.674 

 kg 
 m3 
vˆ1  vˆsat. liquid at 10 kPa   0.00101 

 kg 
Therefore,
85

 m3  
 
W  10000 Pa 2 kg 14.674    0.00101 
   293.5 kJ

 kg 
 kg  

86
2.50
First, perform an energy balance on the system. Potential and kinetic energy effects can be
neglected. Since nothing is mentioned about work in the problem statement, W can be set to
zero. Therefore, the energy balance is
Q  U
Performing a mass balance reveals
m2  m1
where
m1  m1l  m1v
Now the energy balance can be written as

Q  m1 (uˆ 2  uˆ1 )  m1uˆ 2  m1l uˆ1l  m1v uˆ1v

Since two phases coexist initially (water is saturated) and P1 is known, state 1 is constrained.
From the saturated steam tables
 kJ 
uˆ1l  191.79  
 kg 
 kJ 
uˆ1v  2437.9  
(sat. H2O at 10 kPa)
 kg 
As heat is added to the system, the pressure does not remain constant, but saturation still exists.
One thermodynamic property is required to constrain the system. Enough information is known
about the initial state to find the volume of the container, which remains constant during heating,
and this can be used to calculate the specific volume of state 2.
l l
v v
V2 V1 m1 vˆ1  m1 vˆ1
v2 


m2 m1
m1l  m1v
From the saturated steam tables
87
 m3 
ˆv1l  0.001010 

 kg 
 m3 
vˆ1l  14.674 

 kg 
(sat. H2O at 10 kPa)
Therefore,
 m3 
vˆ2  1.335 

 kg 
The water vapor is now constrained. Interpolation of steam table data reveals
 kJ 
uˆ 2  2514.6  
 kg 
Now that all of the required variables are known, evaluation of the expression for Q is possible.
Q  11652 kJ 
88
2.51
Let the mixture of ice and water immediately after the ice has been added represent the system.
Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are
neglected, the energy balance reduces
H  0
We can split the system into two subsystems: the ice (subscript i) and the water (subscript w).
Therefore,
H  mi hi  mw hw  0
and
mi hi   mw hw
We can get the moles of water and ice.
 
Vw
0.0004 m 3

 0.399 kg 
vˆw 
 m3  
 0.001003   


 kg  

mw
0.399 kg   22.15 mol
nw 

MW H 2O 0.0180148  kg 
 mol 
mi
ni 
 5.55 mol
MW H 2O
mw 
Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0
ºC, the assumption is correct.) The following expression mathematically represents the change
in enthalpy.





ni c P, i 0   10 º C   h fus  c P, w T f  0 º C  nwc P, w T f  25 º C
Note: Assumed the heat capacities are independent of temperature to obtain this
expression.
From Appendix A.2.3
c P ,i  4.196 R
c P ,w  9.069 R
and
 kJ 
h fus  6.0 
 mol 
89
Substitution of values into the above energy balance allows calculation of Tf.
T f  3.12 º C
(Our assumption that all the ice melts is correct.)
(b)
To obtain the percentage of cooling achieved by latent heat, perform the following calculation
Fractionlatent 

ni  h fus


 nw c P, w T f  Tw, i

5.55 mol 6000 
J 

 mol  

Fractionlatent 
 0.911
 J 
3.12 º C  25 º C
 22.15 mol9.069  8.314 
 mol  K 
Percentlatent  91.1%
90
2.52
A mass balance shows
n2  n1l  n1v
To develop the energy balance, neglect kinetic and potential energy. Also, no shaft work is
performed, so the energy balance becomes
U  Q
The energy balance can be expanded to


Q  n1l  n1v u 2  n1l u1l  n1v u1v
If the reference state is set to be liquid propane at 0 ºC and 4.68 bar, the internal energies become
u1l  0
u1v  u vap 0 º C 
u 2  u vap 0 º C  
T2
 c P  R dT
273 K
Once the change in internal energy for vaporization and temperature of state 2 is determined, Q
can be solved. As the liquid evaporates, the pressure increases. At state 2, where saturated
propane vapor is present, the ideal gas law states
P2 
RT2
v2
To find v2 , assume that v1v  v1l . The volume of the rigid container is
V  n1v v1v  n1v
 m3 
RT1
 0.00485 

P1
 mol 
Therefore,
v2 
 m3 
 0.00243 

n1l  n1v
 mol 

V

91
Also, since the propane is saturated, P2 and T2 are not independent of each other. They are
related through the Antoine Equation,
 
B
ln P sat  A 
T
sat
C
where
P sat  P2 and T sat  T2
Substitution provides,
 RT 
B
ln 2   A 
T2  C
 v2 
 bar  m 3 
Using values from Table A.1.1 and R  8.314  10  5 

 mol  K 
T2  301.7 K
To find u vap , refer to the definition of enthalpy.
hvap  h v  h l  (u  Pv) v  u  Pv l
Since v v  v l , the above the change in internal energy of vaporization can be written as
u vap  h vap  Pv v  h vap  RT
Therefore,
 kJ 
u vap 0 º C   14.39 
 mol 
Evaluation of the following equation after the proper values have been substituted from Table
A.2.1


301.7 K


Q  n1l  n1v u vap 0 º C    c P  R dT   n1l 0  n1v uvap 0 º C 


273 K
92
gives
Q  18.1 kJ 
93
2.53
The equation used for calculating the heat of reaction is given in Equation 2.72. It states
 i

hrxn
  vi h f
This equation will be used for parts (a)-(e). Since the heat of reaction at 298 K is desired, values
from Appendix A.3 can be used.
(a)
First the stoichiometric coefficient must be determined for each species in the reaction.
vCH 4 ( g )  1
vO2 ( g )  1
vCO2 ( g )  1
v H 2 O( g )  1
From Tables A.3.1 and A.3.2
kJ 
hf ,298 CH ( g )  74.81  mol

kJ 
hf ,298 O ( g )  0  mol

kJ 
hf ,298 CO ( g )  393.51  mol

kJ 
hf ,298 H O( g )  241.82  mol

4
2
2
2
From Equation 2.72, the equation for the heat of reaction is













hrxn
,298  vCH 4 ( g ) h f ,298 CH ( g )  vO2 ( g ) h f ,298 O ( g )  vCO2 ( g ) h f ,298 CO ( g )  v H 2 O ( g ) h f ,298 H O ( g )
4
2
2
2








 kJ 
 kJ 
 kJ 
 kJ 

hrxn
   0 
    393.51 
    241.82 

, 298    74.81 



 mol     mol   
 mol   
 mol  

 kJ 

hrxn
, 298  560.52 
 mol 
Now that a sample calculation has been performed, only the answers will be given for the remaining
parts since the calculation process is the same.
94
(b)
 kJ 

hrxn
, 298  604.53 
 mol 
(c)
 kJ 

hrxn
, 298  206.12 
 mol 
(d)
 kJ 

hrxn
, 298  41.15 
 mol 
(e)
 kJ 

hrxn
, 298  905.38 
 mol 
95
2.54
The acetylene reacts according to the following equation
C2H2(g) + (5/2)O2(g)  2CO2(g) + H2O(g)
(a)
First, choose a basis for the calculations.
nC 2 H 2  1 mol
Calculate the heat of reaction at 298 K using Equation 2.72 and Appendix A.3
 i

hrxn
 h f 
C H

hrxn
  vi h f
2
 O  2hf CO  hf H O
 2.5 h f
2
2
2
2
 J 

hrxn
 1.255 10 6 
 mol 



H rxn,298  nC 2 H 2 hrxn
 1.255 10 6 J 
The required amount of oxygen is calculated as follows
nO 1  2.5(nC H
2
2
2
)1  2.5 mol
The compositions for both streams are
Streams
1 (Inlet)
2 (Outlet)
nC 2 H 2
nO2
nN 2
nCO2
nH 2O
1
0
2.5
0
0
0
0
2
0
1
D  10 5
-1.299
-0.227
-1.157
0.121
E  109
0
0
0
0
From Table A.2.2
Species
C2H2
O2
CO2
H2O
A
6.132
3.639
5.457
3.470
B  103
1.952
0.506
1.045
1.45
C  10 6
0
0
0
0
Integration of the following equation provides an algebraic expression where only T2 is
unknown.
96
T2
H rxn,298 
  ni 2 c P i dT  0
298 i
Substituting the proper values into the expression gives
T2  6169 K
(b)
The calculations follow the procedure used in Part (a), but now nitrogen is present. The basis is
nC 2 H 2  1 mol
The heat of reaction is the same as in Part (a), but the gas composition is different. Since
stoichiometric amount of air is used,
nO 1  2.5 mol
2
 yN 
 n N 2  nO2  2   9.40 mol
1
1 y

 O2  air
   
The composition of the streams are summarized below
Streams
nC 2 H 2
nO2
nN 2
nCO2
nH 2O
1
2
1
0
2.5
0
9.40
9.40
0
2
0
1
D  10 5
-1.299
-0.227
-1.157
0.121
0.04
E  109
0
0
0
0
0
From Appendix A.2
Species
A
C2H2
O2
CO2
H2O
N2
6.132
3.639
5.457
3.470
3.280
B  103
1.952
0.506
1.045
1.45
0.593
C  10 6
0
0
0
0
0
Therefore,
T2  2792 K
97
(c)
Now excess air is present, so not all of the oxygen reacts. The heat of reaction remains the same
because only 1 mole of acetylene reacts. Since the amount of air is twice the stoichiometric
amount
nO 1  5 mol
2
 yN 
 n N 2  nO2  2   18.80 mol
1
1 y

 O2  air
   
The compositions of the streams are summarized below
Streams
nC 2 H 2
nO2
nN 2
nCO2
nH 2O
1
2
1
0
5
2.5
18.8
18.8
0
2
0
1
The table of heat capacity data in Part (b) will be used for this calculation. Using the expression
shown in Part (a)
T2  1787 K
98
2.55
(a)
The combustion reaction for propane is
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
For all subsequent calculations, the basis is one mole of propane. The heat of reaction is
calculated as follows
 C H  5hf O  3hf CO  4hf H O

hrxn
  h f
3
8
2
2
2
 J 

hrxn
 2.044  10 6 
 mol 



H rxn,298  nC 3 H 8 hrxn
 2.044 10 6 J 
The required amount of oxygen for complete combustion of propane is
nO 1  5(nC H
2
3
8
)1  5 mol
 yN 
 n N 2  nO2  2   18.81 mol
1
1 y

 O2  air
The stream compositions are listed below
   
Streams
1 (Inlet)
2 (Outlet)
nC 3 H 8
nO2
nN 2
nCO2
nH 2O
1
0
5
0
18.8
18.8
0
3
0
4
D  10 5
0.04
-1.157
0.121
E  109
0
0
0
From Table (a)2.2
Species
N2
CO2
H2O
A
3.280
5.457
3.470
B  103
0.593
1.045
1.45
C  10 6
0
0
0
Now all of the necessary variables for the following equation are known, except T2 .
T2
H rxn,298 
  ni 2 c P i dT  0
298 i
Solving the resulting expression provides
T2  2374 K
99
(b)
The combustion reaction for butane is
C4H10(g) + (13/2)O2(g)  4CO2(g) + 5H2O(g)
For all subsequent calculations, the basis is one mole of butane. The heat of reaction is
calculated as shown in Part (a)
H rxn,298  2.657  10 6 J 
The moles of nitrogen and oxygen in the feed stream are calculated according to the method in
Part (a). The compositions are
Streams
1 (Inlet)
2 (Outlet)
nC 4 H 10
nO2
nN 2
nCO2
nH 2O
1
0
6.5
0
24.5
24.5
0
4
0
5
The c P data listed in Part (a) can also be used for this reaction since there is no remaining
butane.
T2  2376 K
(c)
The combustion reaction for pentane is
C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(g)
The basis is one mole of pentane. The heat of reaction is calculated as shown in Part (a).
H rxn,298  3.272  10 6 J 
The moles of nitrogen and oxygen in the feed stream are calculated according to the method in
Part (a). The compositions are listed below
Streams
1
2
nC 5 H 12
nO2
nN 2
nCO2
nH 2O
1
0
8
0
30.1
30.1
0
5
0
6
Substitution of the values into the expression used to find T2 and subsequent evaluation results
in
T2  2382 K
The adiabatic flame temperatures are nearly identical in all three cases.
100
2.56
The equation for the combustion of methane is
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Using Equation 2.72
 J 

hrxn
 8.02  105 
 mol 
The basis for this problem is
nCH 1  1 mol
4
Also, let  represent the fractional conversion of methane. Therefore, the composition of the
product gas leaving the reactor is
nCH 2  11    mol
nO 2  21    mol
nN 2  7.52 mol
nCO 2   mol
nH O 2  2 mol
4
2
2
2
2
Furthermore, the heat of reaction is calculated as follows


H rxn,298    8.02 105 J 
From Table A.2.2
Species
A
CH4
O2
CO2
H2O
N2
1.702
3.639
5.457
3.470
3.280
B  103
9.081
0.506
1.045
1.45
0.593
C  10 6
-2.164
0
0
0
0
D  10 5
0
-0.227
-1.157
0.121
0.04
E  109
0
0
0
0
0
After substitution of the outlet composition values, heat capacity data, and the heat of reaction
into the following equation
101
1273
H rxn, 298 
  n  c  dT  0
i 2
P i
298 i
integration provides an equation with one unknown:  . Solving the equation gives
 =0.42
Since the fractional conversion is 0.42, 58% of the methane passed through the reactor unburned.
102
2.57
For the entire cycle,
U11  U12  U 23  U 31  0
 U12  50 kJ 
From state 1 to state 2
U12  Q12  W12
 Q12  400 kJ 
From state 2 to state 3
U 23  Q23  W23
 Q23  0 kJ 
From state 3 to state 1
U 31  Q31  W31
W31  250 kJ 
Hence, the completed table is
Process
State 1 to 2
State 2 to 3
State 3 to 1
U kJ 
-50
800
-750
W kJ 
-400
800
-250
Q kJ 
350
0
-500
To determine if this is a power cycle or refrigeration cycle, look at the overall heat and work,
W11 and Q11 .
W11  W12  W23  W31  150 kJ 
Q11  Q12  Q23  Q31  150 kJ 
Since work is done on the system to obtain a negative value of heat, which means that heat is
leaving the system, this is a refrigeration cycle.
103
2.58
Refer to the graph of the Carnot cycle in Figure E2.20. From this graph and the description of
Carnot cycles in Section 2.9, it should be clear that state 3 has the lowest pressure of all 4 states,
and state 1 has the highest pressure. States 1 and 2 are at the higher temperature. States 3 and 4
have the lower temperature. Since both the temperature and pressure are known for states 1 and
3, the molar volume can be calculated using
v
RT
P
The table below summarizes the known thermodynamic properties.

State
T K 
P bar 
1
2
3
4
1073
1073
298
298
60
v m 3 /mol
0.00149
0.2
0.124

For each step of the process, potential and kinetic energy effects can be neglected. The step from
state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in
internal energy is 0, and the energy balance becomes
Q12  W12
From Equation 2.77,
P 
W12  nRT1 ln 2 
 P1 
where W12 is the work done from state 1 to state 2. The value of P2 is not known, but
recognizing that the process from state 2 to 3 is an adiabatic expansion provides an additional
equation. The polytropic relationship can be employed to find P2 . A slight modification of
Equation 2.89 provides
Pv k  const
From the ideal gas law
P
RT
v
Combining this result with the polytropic expression and noting that R is constant, allows the
expression to be written as
104
Tv k 1  const
Therefore,
1
T
 k 1
v2   3 v3k 1 
 T2

Substituting the appropriate values (k=1.4) gives
 m3 
v2  0.00504 

 mol 
Applying the ideal gas law
P2 
RT2
 17.7 bar 
v2
Now, W12 can be calculated.
W12  10.89 kJ 
Calculation of W34 follows a completely analogous routine as calculation for W12 . The
following equations were used to find the necessary properties
1
 m3 
T
 k 1
 0.0367 
v4   1 v1k 1 

 T4

 mol 
P4 
RT4
 0.675 bar 
v4
Now the following equation can be used
P 
W34  nRT3 ln 4 
 P3 
which gives
W34  3.01 kJ 
105
For an adiabatic, reversible process, Equation 2.90 states
W
nR
T2  T1 
k 1
This equation will be used to calculate the work for the remaining processes.
nR
T3  T2   16.11 kJ
k 1
nR
T1  T4   16.11 kJ 
W41 
k 1
W23 
To find the work produced for the overall process, the following equation is used
Wnet  W12  W23  W34  W41
Evaluating this expression with the values found above reveals
Wnet  7.88 kJ 
Therefore, 7.88 kilojoules of work is obtained from the cycle.
The efficiency of the process can be calculated using Equation 2.98:

Wnet
7.88

 0.72
QH 10.89
since QH  W12  10.89 kJ  . Alternatively, if we use Equation E2.20D.
  1
TC
TH
where TC  T3  T4 and TH  T1  T2 . Upon substitution of the appropriate values
  0.72
106
2.59
Since this is a refrigeration cycle, the direction of the cycle described in Figure 2.17 reverses.
Such a process is illustrated below:
Constant T H
QH
Isothermal
compression
Adiabatic
compression
State 2
Well - insulated
T2=T1
P2
T1
P1
State 3
State 1
T1
P1
Adiabatic
expansion
State 1
T3
P3
State 4
T4
P4
State 4
T4=T3
P4
State 2 State 3
T3
T2
P3
P2
Isothermal
expansion
QC
Constant T C
States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since
the both the temperature and pressure are known for states 2 and 4, the molar volume can be
calculated using
v
RT
P
The following table can be made
State
T K 
1
2
3
4
1073
1073
298
298

P bar 
v m 3 /mol
60
0.00149
0.2
0.124

For each step of the process, potential and kinetic energy effects can be neglected. The process
from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in
internal energy is 0, and the energy balance becomes
Q12  W12
107
From Equation 2.77,
P 
W12  nRTH ln 2 
 P1 
where W12 is the work done from state 1 to state 2. The value of P1 is not known, but
recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional
relation. The polytropic relationship can be employed to find P1 . A slight modification of
Equation 2.89 provides
Pv k  const
From the ideal gas law
P
RT
v
Combining this result with the polytropic expression and noting that R is constant allows the
expression to be written as
Tv k 1  const
Therefore,
1
T
 k 1
v1   4 v 4k 1 
 T1

Substituting the appropriate values (k=1.4) gives
1
k 1
T

 m3 
v1   4 v 4k 1   0.00504 

 mol 
 T1

RT
P1  1  17.7 bar 
v1
Now, W12 can be calculated.
W12  10.9 kJ 
Calculation of W34 follows a completely analogous routine as the calculation for W12 . The
following equations were used to find the necessary properties:
108
 m3 
v3  0.0367 

 mol 
Applying the ideal gas law
P3 
RT3
 0.675 bar 
v3
Now the following equation can be used
P 
W34  nRTC ln 4 
 P3 
which gives
W41  3.0 kJ 
Equation 2.90 can be used to determine the work for adiabatic, reversible processes. This
equation will be used to calculate the work for the remaining processes.
nR
TC  TH   16.11 kJ 
k 1
nR
TH  TC   16.11 kJ 
W41 
k 1
W23 
To find the work produced for the overall process, the following equation is used
Wnet  W12  W23  W34  W41
Evaluating this expression with the values found above reveals
Wnet  7.88 kJ 
Therefore, 7.88 kJ of work is obtained from the cycle. The coefficient of performance is defined
in Equation 2.99 as follows
COP 
QC
Wnet
where QC is the equal to Q34 . From the energy balance developed for the process from state 3
to state 4
109
Q34  W34  3.01 kJ 
Therefore,
COP 
3.01 kJ 
 0.382
7.88 kJ 
110
2.60
(a)
The Pv path is plotted on log scale so that the wide range of values fits (see Problem 1.13)
logP
100
0.075
4
1
3
2
log v
v
(b)
The work required to compress the liquid is the area under the Pv curve from state 3 to state 4.
Its sign is positive. The power obtained from the turbine is the area under the curve from state 1
to 2. Its sign is negative. The area under the latter curve is much larger (remember the log
scale); thus the net power is negative.
(c)
First, perform a mass balance for the entire system:
m 1  m 2  m 3  m 4  m
Since no work is done by or on the boiler, the energy balance for the boiler is
m hˆ1  m hˆ4  Q H
Similarly, the energy balance for the condenser is
m hˆ3  m hˆ2  Q C
To find the necessary enthalpies for the above energy balances, we can use the steam tables:
 kJ 
hˆ1  3424.5  
 kg 
(520 ºC, 100 bar)
111


 kJ  
 kJ  
 kJ 
hˆ2  0.10168.77     0.90 2574.8     2334.2  
 kg  
 kg  
 kg 


(sat. liq at 7.5 kPa) (sat. vap. at 7.5 kPa)
 kJ 
hˆ3  168.77  
 kg 
(sat. liquid at 0.075 bar)
 kJ 
hˆ4  342.81  
 kg 
(subcooled liquid at 80 ºC, 100 bar)
Now, we can calculate the heat loads:

 kJ  
 kJ 
Q H  100 kg/s  3424.5    342.81     308169 kW 
 kg  
 kg 


 kJ  
 kJ 
Q C  100 kg/s 168.77    2334.2     216543 kW 
 kg  
 kg 

(d)
Use Equation 2.96:
W net  Q net  0
From Part (c), we know
Q net  308169 kW   216543 kW   91626 kW 
Therefore,
W net  91626 kW 
(e)
Using the results from Parts (c) and (d):
η
91626 kW 
 0.297
308169 kW 
112
Chapter 3 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
3.1
Since entropy is a state function
s sys  s sys, step1  s sys, step 2
Step 1 is a constant volume process. Therefore, no work is done. After neglecting potential and
kinetic energy effects, the energy balance for a reversible process becomes
u  q
 du  q
cv dT  q
Using the definition of entropy,
final

s sys ,step1 
initial
q rev
T

T2
cv dT
 T
T1
Step 2 is an isothermal process. For and ideal gas u is zero and the energy balance is (PE and
KE neglected)
q  w
 dq  w
For an ideal gas, the following can be shown
wrev   Pdv  
RT
dv
v
Therefore,
final
s sys ,step 2 

initial
qrev
T

v2
R
 v2 

1 
 v dv  R ln v
v1
Combination of both steps yields
T2
s sys 

T1
v 
cv dT
 R ln 2 
T
 v1 
2
3.2
Equation 3.62 states
T2
s sys 
cP
T
T1
P 
dT  R ln 2 
 P1 
Substituting the equation for c P yields
T2
s sys 

T1
P 
A  BT  CT 2
dT  R ln 2 
T
 P1 


P 
T 
C
s sys  A ln 2   BT2  T1   T22  T12  R ln 2 
2
 P1 
 T1 
3
3.3
Equation 3.3 states
S univ  S sys  S surr
where
S sys  m( sˆ2  sˆ1 )
S surr 
Qsurr
Tsurr
(the temperature of the surroundings is constant)
First, we will calculate S sys . From the steam tables:
 kJ 
sˆ1  7.1228 

 kg  K 
(300 ºC, 10 bar)
Since the container is rigid and mass is conserved in the process,
vˆ2  vˆ1
 m3 
vˆ1  0.25794 

 kg 
(300 ºC, 10 bar)
 m3 
 vˆ2  0.25794 

 kg 
To find the number of phases present in state 2, compare the specific volume of state 2 to the
specific volume for saturated water and saturated water vapor at 1 bar. From the steam tables,
 m3 
vˆ2l , sat  0.001043 

 kg 
(sat. H2O(l) at 1 bar)
 m3 
vˆ2v, sat  1.6940 

 kg 
(sat. H2O(v) at 1 bar)
Since vˆ2l , sat  vˆ2  vˆ2v, sat , two phases are present. The quality of the water can be calculated as
follows
vˆ2  1  x vˆ2l , sat  xvˆ2v, sat
Therefore,
4
x  0.152
From the steam tables:
 kJ 
sˆ2l , sat  1.3025 

 kg  K 
 kJ 
sˆ2v, sat  7.3593 

 kg  K 
(sat. H2O(l) at 1 bar)
(sat. H2O(v) at 1 bar)
The entropy of state 2 can be calculated using the following equation:
sˆ2  1  x sˆ2l , sat  xsˆ2v, sat
 kJ 
sˆ2  2.2231 

 kg  K 
Now the entropy change of the system can be calculated.

 kJ 
 kJ  
kJ
  49.0  
S sys  10 kg  2.2231 
 7.1228 





K
 kg  K 
 kg  K  

To find the change in entropy of the surroundings, an energy balance will be useful. Since no
work is done, the energy balance is
U  Q
We also know
Qsurr  Q
To following expression is used to solve for the internal energy:

 
U  m 1  x uˆ 2l , sat  xu 2v, sat  uˆ1
From the steam tables
 kJ 
uˆ1  2793.2  
 kg 
 kJ 
uˆ 2l , sat  417.33  
 kg 
(300 ºC, 10 bar)
(sat. H2O(l) at 1 bar)
5
 kJ 
uˆ 2v, sat  2506.1  
 kg 
(sat. H2O(v) at 1 bar)
The expression for internal energy yields
U  20583.77 kJ 
Therefore,
Q  20583.77 kJ 
and
S surr 
Qsurr 20583.77 kJ 
 kJ 

 70.25  
Tsurr
293 K 
K
Now the change in entropy of the universe can be calculated
 kJ 
 kJ 
 kJ 
S univ  S sys  S surr  49.0    70.25    21.25  
K
K
K
Since Suniv > 0, this process is possible.
6
3.4
Entropy Balance:
S univ  S sys  S surr
Considering the copper block to be the system, no work is done on the system; thus, the energy
balance is
du  q
cv dT  q
(neglecting PE and KE)
This can be used in the following expression for entropy
T2
s sys 
q rev

T
T1
T2
s sys 
cv
T
dT
T1
From Table A.2.3
c P  2.723R
cv  c P  2.723R
(for liquids and solids)
Therefore,
T2
s sys 

T1
T 

2.723R
 J    280 K  
 ln
dT  2.723R ln 2   2.723 8.314 

T
 mol  K    373.15 K  

 T1 
 J 
s sys  6.50 
 mol  K 




10 kg 
  6.50  J    1024  J 

S sys 
 mol  K  
 K 

 kg  

0
.
063465

 mol  


Since the temperature of the lake remains constant, the change in entropy of the surroundings can
be calculated as follows
S surr 
Qsurr
 nCu c P T2  T1   2.723nCu RT2  T1 
Q



Tsurr Tsurr
Tsurr
Tsurr
7


 J 


280 K   373.15 K  
 2.723 8.314 


10 kg 

 mol  K  


S surr  




kg
280 K 




0.063546

 mol  




J
S surr  1184  
K 
From the definition of entropy:
J
J
J
S univ  S sys  S surr  1024    1184    160  
K 
K 
K 
8
3.5
(a)
Since the process is adiabatic and reversible,
S sys  0
(b)
The change in entropy is calculated by
liquid

S sys  n
qrev
T
vapor

 nh
Tb
vap
  kJ  
 1 mol 8.2 

 mol  
J


 73.9  
111 K
K 
The sign is negative because there is less randomness in the liquid phase.
(c)
For this situation
δq  c P dT
Therefore,
Tf
S sys  m 
Ti
q rev
T

 J    273.15 K 
cP



dT
18
.
0148
g
4
.
2

 g  K   ln 373.15 K 


T


 
373.15 K

273.15 K

J
S sys  23.6  
K 
The sign is negative because as the water cools, less translational energy states are occupied by
the molecules. Therefore, the randomness decreases.
(d)
First, calculate the temperature at which the blocks (block A and block B) equilibrate. The
energy balance for the process is
n Ac P T2  373.15 K   n B c P T2  473.15 K   0
Since the heat capacities number of moles of A and B are equal, we find that
T2  423.15 K
Now, calculate the change in entropy:
9
S sys  S sys, A  S sys, B
 T 
 T 
S sys  n Ac P ln 2   n B c P ln 2 
 T1, A 
 T1, B 




Substituting values, we obtain
J
S sys  0.337  
K 
The sign is positive because two objects at different temperatures will spontaneously equilibrate
to the same temperature when placed together.
10
3.6
The solution below compares problems 2.14 and 2.15, the calculation of 2.13 was erroneously
included in the problem statement of the first printing and is shown at the end of this problem.
Problem 2.14
Since the system is well-insulated no heat is transferred with the surroundings. Therefore, the
entropy change of the surroundings is zero and
S univ  S sys
The gas in the piston-cylinder system is ideal and c P is constant, so

T 
 P 
S sys  n c P ln 2   R ln 2 
 T1 
 P1 

From the problem statement, we know
P1  2 bar 
P2  1 bar 
The ideal gas law can be used to solve T1 .
PV
2 bar 10 L 
T1  1 1 
 240.6 K 
nR 
 L  bar  
 0.08314 
1 mol
 mol  K  

Solving for T2 is slightly more involved. The energy balance for this system where potential and
kinetic energy effects are neglected is
U  W
Conservation of mass requires
n1  n2
Let n  n1  n2
The energy balance can be rewritten as
T2
V2
T1
V1
n  cv dT    PE dV
11
Since cv and PE are constant
ncv T2  T1    PE V2  V1 
V2 and T1 can be rewritten using the ideal gas law
nRT2
P2
PV
T1  1 1
nR
V2 
Substituting these expressions into the energy balance, realizing that PE  P2 , and simplifying
the equation gives
5 

 P2  P1 V1
2 
T2  
7
nR
2
Using the following values
P1  2 bar 
P2  1 bar 
V1  10 L
n  1.0 mol
 L  bar 
R  0.08314 
 mol  K 
results in
T2  206 K 
Since both states are constrained, the entropy can be calculated from Equation 3.63:

 J    206 K  
 J   1 bar  
S univ  S sys  1.0 mol 3.5  8.314 
 ln
ln
  8.314 


 mol  K    240.6 K  
 mol  K   2 bar  

J
Suniv  S sys  1.24  
K 
12
Problem 2.15
Since the initial conditions in the piston-cylinder assembly are equal to the initial conditions of
Problem 2.14, T1 and P1 are known. Moreover, P2 is known, so we only need to find T2 in
order to calculate the entropy change. For adiabatic, reversible processes, the following
relationship (Equation 2.89) holds:
PV k  const
This can be used to find V2 .
1
k
P
V2   1 V1k 

 P2
c
7
Noting that k  P  and substituting the proper values results in
cv 5
V2  16.4 L
Th polytropic expression can also be manipulated to yield
TV k 1  const.
Therefore,
T2 
T1V1k 1
V2k 1
Substitution of the appropriate variables provides
T2  197.4 K 
Now the entropy can be calculated,

 J   1 bar  
 J    197.4 K  
S univ  S sys  1.0 mol 3.5  8.314 
ln

  8.314 
 ln

 mol  K   2 bar  
 mol  K    240.6 K  

J
S univ  S sys  0.004  
K 
The value of the entropy shown represents round-off error. Since the process is reversible and
adiabatic, we know from Table 3.1 and the related discussion in Section 3.3 that the entropy
changes of the system, surroundings, and universe will be zero.
13
Problem 2.13
Equation 3.3 states
S univ  S sys  S surr  m( sˆ2  sˆ1 ) 
Qsurr
Tsurr
where the temperature of the surroundings is constant. First, we will determine s 2 . The first law
can be applied to constrain state 2. With potential and kinetic energy effects neglected, the
energy balance becomes
U  Q  W
The value of the work will be used to obtain the final temperature. The definition of work
(Equation 2.7) is
V2
W    PE dV
V1
Since the piston expands at constant pressure, the above relationship becomes
W   PE V2  V1 
From the steam tables
 kJ 
sˆ1  6.2119 
 (10 MPa, 400 ºC)
 kg  K 
 m3 
vˆ1  0.02641 

 kg 
(10 MPa, 400 ºC)
 

 m3  
3
V1  m1vˆ1  (3 kg) 0.02641 
   0.07923 m


 kg  

Now V2 and v2 are found as follows
 
 748740 J
W
 0.4536 m3
 0.07923 m3 
6
PE
2.0  10 Pa
3
 m3 
V
0.4536 m
ˆv2  2 
 0.1512 

m2
3 kg 
 kg 
V2  V1 
 
14
Since v̂2 and P2 are known, state 2 is constrained. From the steam tables:
 kJ 
sˆ2  7.1270 

 kg  K 

 3 
 20 bar, 0.1512  m  

 kg  

Now U will be evaluated, which is necessary for calculating Qsurr . From the steam tables:
 kJ 
uˆ 2  2945.2  
 kg 

 3 
 20 bar, 0.1512  m  

 kg  

 kJ 
uˆ1  2832.4  
 kg 
100 bar, 400 º C

 kJ  
 kJ 
U  m1 uˆ 2  uˆ1   3 kg  2945.2    2832.4     338.4 kJ 
 kg  
 kg 

Substituting the values of U and W into the energy equation allows calculation of Q
Q  U  W
Q  338400 [J]   748740 J   1.09  10 6 J   Qsurr
so
S univ  m( sˆ2  sˆ1 ) 
Qsurr
Tsurr

 kJ   1.09  103 kJ 
 kJ 
 kJ 

 1.126  

 3 kg  7.1270 
6
.
2119




673.15 K 
K
 kg K  
 kg K 

Therefore, this process is irreversible.
15
3.7
(a)
From Equation 3.63:
T 
P 
s sys  c P ln 2   R ln 2 
 T1 
 P1 

 0.5 bar  
 J    7  500 K  
 J 
s sys   8.314 
  ln
  20.63 
  ln

 mol  K    2  300 K  
 mol  K 
 1 bar  

(b)
From Equation 3.65:
T 
v 
s sys  cv ln 2   R ln 2 
 T1 
 v1 


 3  
 0.025  m  




 J    5  500 K  
 mol    4.85  J 

ln
s sys   8.314 


ln







 mol  K 
 m 3  

 mol  K    2  300 K  

 
 0.05 


mol
 




(c)
First, we can find the molar volume of state 1 using the ideal gas law.
v1 
 m3 
RT1
 0.025 

P1
 mol 
Now, we can use Equation 3.65
T 
v 
s sys  cv ln 2   R ln 2 
 T1 
 v1 


 3  
 0.025  m  


 mol  

J 
 J    5  500 K  

  10.62 

ln
s sys   8.314 
ln








 m 3  

 mol  K    2  300 K  
 mol  K 

 
 0.025 


 mol  


16
3.8
(i). We wish to use the steam tables to calculate the entropy change of liquid water as it goes
from its freezing point to its boiling point. The steam tables in Appendices B.1 – B.5 do not
have data for subcooled water at 1 atm. However, there is data for saturated water at 0.01 ºC and
a pressure of 0.6113 kPa. If we believe that the entropy of water is weakly affected by pressure,
then we can say that the entropy of water at 0.01 ºC and 0.6113 kPa is approximately equal to the
entropy at 0 ºC and 1 atm. The molar volumes of most liquids do not change much with pressure
at constant temperature. Thus, the molecular configurations over space available to the water
molecules do not change, and the entropy essentially remains constant. We do not need to
consider the molecular configurations over energy since the temperature difference is so slight.
So, from the steam tables:
 kJ 
sˆ0 º C , 1 atm   sˆ0.01 º C , 0.6113 kPa   0 

 kg  K 
 kJ 
sˆ100 º C , 1 atm   1.3068 

 kg  K 
Therefore,
 kJ 
 kJ 
sˆ  sˆ100 º C , 1 atm   sˆ0 º C , 1 atm   1.3068 
0


 kg  K 
 kg  K 
 kJ 
ŝ  1.3068 

 kg  K 
(ii). From the steam tables:
 kJ 
sˆl 100 º C , 1 atm   1.3068 

 kg  K 
 kJ 
sˆ v 100 º C , 1 atm   7.3548 

 kg  K 
Therefore,
 kJ 
 kJ 
sˆ  sˆ v 100 º C , 1 atm   sˆ l 100 º C , 1 atm   7.3548 
 1.3068 


 kg  K 
 kg  K 
 kJ 
ŝ  6.048 

 kg  K 
The change in entropy for process (ii) is 4.63 times the change in entropy for process (i). There
are many ways to reconcile this difference, but think about it from a molecular point of view. In
process (i), the available molecular configurations over energy are increased as the temperature
increases. As the temperature increases, the molar volume also increases slightly, so the
17
available molecular configurations over space also increase. Now consider process (ii), where
the molecules are being vaporized and entering the vapor phase. The molecular configuration
over space contribution to entropy is drastically increased in this process. In the liquid state, the
molecules are linked to each other through intermolecular interactions and their motion is
limited. In the vapor state, the molecules can move freely. Refer to Section 3.10 for a discussion
of entropy from a molecular view.
18
3.9
Before calculating the change in entropy, we need to determine the final state of the system. Let
the mixture of ice and water immediately after the ice has been added represent the system.
Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are
neglected, the energy balance reduces
H  0
We can split the system into two subsystems: the ice (subscript i) and the water (subscript w).
Therefore,
H  mi hi  mw hw  0
and
mi hi   mw hw
We can get the moles of water and ice.
 
Vw
0.0004 m 3

 0.399 kg 
vˆw 
 m3  
 0.001003   


 kg  

mw
0.399 kg   22.15 mol

nw 
MW H 2O 0.0180148  kg 
 mol 
mi
 5.55 mol
ni 
MW H 2O
mw 
Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0
ºC, the assumption is correct.) The following expression mathematically represents the change
in internal energies (cP=cv).


ni c P ,i 0   10 º C   h fus  c P ,w T f  0 º C   nw c P ,w T f  25 º C 
Note: Assumed the heat capacities are independent of temperature to obtain this
expression.
From Appendix A.2.3
c P ,i  4.196 R
c P ,w  9.069 R
and
 kJ 
h fus  6.0 
 mol 
19
Substitution of values into the above energy balance allows calculation of Tf.
T f  3.12 º C
(Our assumption that all the ice melts is correct.)
Now, we can calculate the change in entropy. From Equation 3.3
suniv  s sys  s surr
Since the glass is considered adiabatic,
s surr  0
suniv  s sys
We will again break the system into two subsystems: the ice and the water. The change in
entropy of the universe can be calculated as follows
S univ  ni si  nw s w
The definition of change in entropy is
final
s 

initial
δqrev
T
Assuming the heat capacities of ice and water are independent of temperature, the expressions
for the change in entropy of the subsystems are
h fus
 276.27 K 
 273.15 K 
 c P, w ln
si  c P, i ln


 273.15 K 
 263.15 K  273.15 K
 276.27 K 
s w  c P, w ln

 298.15 K 
Therefore
h fus


 273.15 K 
 276.27 K 
 276.27 K 
S univ  ni c P, i ln
 c P, w ln

  n w c P, w ln

 263.15 K  273.15 K
 273.15 K 
 298.15 K 


Substituting the values used before, we obtain
20
 J 
Suniv  6.59 
 mol 
21
3.10
(a)
The maximum amount of work is obtained in a reversible process. We also know the entropy
change for the universe is zero for reversible processes. From the steam tables
 kJ 
sˆ1  6.2119 

 kg  K 
 kJ 
sˆ2  8.5434 

 kg  K 
(400 ºC, 100 bar)
(400 ºC, 1 bar)
Using the entropy criterion,
Suniv  0  msˆ2  sˆ1   S surr

 kJ 
 kJ  
S surr  msˆ2  sˆ1   0.5 kg  8.5434 
 6.2119 

 
 kg  K 
 kg  K  

 kJ 
S surr  1.1658  
K
Since the process is isothermal, we know that the temperature of the surroundings is constant at
400 ºC. Therefore,
Qsurr
 kJ 
 S surr  1.1658  
Tsurr
K
Qsurr  784.76 kJ 
To find the work obtained, perform an energy balance. An energy balance where potential and
kinetic energy effects are neglected is
U  Q  W
We can calculate the change in internal energy from the steam tables, and we also know that
Q  Qsurr . From the steam tables:
 kJ 
uˆ1  2832.4  
 kg 
 kJ 
uˆ 2  2967.8  
 kg 
(400 ºC, 100 bar)
(400 ºC, 1 bar)
Therefore,
22
W  muˆ 2  uˆ1   Q

 kJ  
 kJ 
 2832 . 4 
W  0 . 5 kg  2967 . 8 
   784 . 76 kJ 

 kg  
 kg 

W  717.06 kJ 
(b)
If steam is modeled as an ideal gas, the change in internal energy is zero. Therefore, the energy
balance is
Q  W
The work can be found using Equation 2.77 which is developed for reversible, isothermal
processes.
P 
W  nRT ln 2 
 P1 

0.5 kg 
 J 
 1 bar 
W
 8.3145 
673.15 K ln



 mol  K  
 100 bar 
 kg   
 0.0180148 

 mol  

W  715332 J   715.3 kJ 
Therefore,
Q  715.3 kJ 
and
Qsurr  715.3 kJ 
Since the temperature of the surroundings is constant,
 kJ 
S surr  1.06  
K
23
3.11
(a)
The initial pressure is calculated as follows
P1  Psurr 

25000 kg  9.81 m/s 2
0.05 m
2
  24.6 105 Pa
Similarly, the final pressure is
P2  Psurr 

35000 kg  9.81 m/s 2
0.05 m 2
  34.4  105 Pa
(b)
The temperature should rise, which can be understood by considering an energy balance.
Because the system is insulated, the work done by the mass being added to the piston is
transformed into molecular kinetic energy.
(c)
The final temperature can be calculated with an energy balance:
U  W
ncv T2  T1   W
Since the pressure of the surroundings is constant after the block is added to the piston, the work
is calculated as follows:
W   P2 V2  V1 
Assume ideal gas behavior:
V2 
nRT2
P2
V1 
nRT1
 0.00169 m 3
P1
Now, we can create one equation with one unknown:
 nRT2

n5 / 2 R T2  T1    P2 
 V1 
 P2

Substitute values and solve for T2:
T2  557 K
24
(d)
Since the system is well-insulated
s surr  0
Use Equation 3.65 to calculate the change in entropy:
T 
V 
T 
T P 
suniv  s sys  cv ln 2   R ln 2   5 / 2 R ln 2   R ln 2 1 
 T1 
 V1 
 T1 
 T1P2 
Substituting values, we obtain,
 J 
s sys  0.354 
 mol  K 
(e)
Because the change in entropy of the universe is equal to change in entropy of the system, which
is positive in this situation, the second law is not violated.
25
3.12
(a)
To calculate ssys, we need to take the gas from state 1 to state 2 using a reversible process. The
process in part a can be drawn as follows:
24.6 bar
500 K
State 1
34.4 bar
557 K
rev.
adiabatic
State 2
rev.
isothermal
PI
T I = 557 K
State I
We need to find the intermediate pressure, PI where we end up at the temperature in state 2, T2.
To find it, we can use the results from pages 78-79 of the text:
T1P11 k  / k  TI PI1 k  / k  T2 PI1 k  / k
Therefore,
k
 T  1k 
P1  35.9 [bar]
PI   1 
 T2 
(I)
If we draw the process on a PT diagram, we get:
40
P
I
2
Adiabatic compression
Isothermal
expansion
1
20
490
560
T
26
The change in entropy can be represented as follows:
s sys  s adiabatic  sisothermal
For the reversible adiabatic process, the change in entropy is zero. Therefore,
s sys  sisothermal
For an isothermal process,
q rev  wrev  
RT
dP
P
Therefore,
final

s sys 
q
initial
T
P2
 
PI
P 
R
 J 
dP   R ln 2   0.354 
P
 mol  K 
 PI 
If we substitute relation (I) in the expression above, we get the same expression in the book:
P
s sys   R ln 2
 P1

T 
T 
P 
k
 
R ln 2   c P ln 2   R ln 2 
 1  k 
 T1 
 T1 
 P1 
(b)
For this construction, we use the path diagrammed on the following PT diagram:
40
P
2
Isothermal
compression
Isobaric heating
1
20
490
560
T
We write
27
s sys  sisobar  sisothermal
First, find an expression for the isobaric heating
q rev  dh  c P dT
Therefore,
T2
sisobar 
T 
c
 TP dT  c P ln T12 
T1
Now, we need an expression for the isothermal, reversible expansion
q rev  wrev  Pdv
Therefore,
v 2  RT2 / P2
sisothermal 

v I  RT2 / P1
P
dv 
T
v 2  RT2 / P2

v1  RT2 / P1
P 
P 
R
dv  R ln 1    R ln 2 
v
 P2 
 P1 
Combine the two steps:
T 
P 
 J 
s sys  c P ln 2   R ln 2   0.354 
 mol  K 
 T1 
 P1 
(c)
For this construction, we use the path diagrammed on the following PT diagram:
28
40
P
2
Isothermal
compression
I
Isochoric heating
1
20
490
560
T
For isochoric heating followed by an isothermal expansion, the entropy can be expressed as
follows:
s sys  sisochoric  sisothermal
For the reversible, isothermal expansion, we obtain (refer to Parts (a) and (b) to see how this is
derived)
P 
sisothermal   R ln 2 
 PI 
However, we can relate the intermediate pressure to the initial pressure through the ideal gas law:
PI P1

T2 T1
or
PI  27.4 [bar]




 P2 
T 
P 
P2 

 sisothermal   R ln    R ln
 R ln 2   R ln 2 
 T2 
 PI 
 T1 
 P1 
P1 

 T1 
For the isochoric process:
q rev  cv dT
Therefore,
29
T2
sisochoric 
cv
T 
 T dT  cv ln T12 
T1
Combining these results:
T
s sys  cv  R  ln 2
 T1

P
  R ln 2

 P1

T 
P 
 J 
  c P ln 2   R ln 2   0.354 
 mol  K 

 T1 
 P1 
Using any of these three reversible paths, we get the same answer!
30
3.13
(a)
Since the process is reversible and adiabatic, the entropy change for the process is zero.
s sys  0
Furthermore, since the process is adiabatic, the changes in entropy of the surroundings and the
universe are also zero.
(b)
Since this process is isentropic (s=0), we can apply an expression for the entropy change of an
ideal gas.
T2
c
P
s  0   P dT  Rln 2
P1
T T
1
Insert the expression for cP from Appendix A
P 
3.639  0.506 10 3 T  0.227 10 6 T 2
dT  R ln 2 
T
 P1 
T1
T2
s  0  R 
which upon substituting
T1  250 K 
P1  1 bar 
P2  12.06 bar 
yields
T2  482 K 
(c)
An energy balance gives
T2
T2
T1
T1
w  u   cv dT 
 c
P
 1dT
31
 J 
w  R  2.639  0.506  10 3 T  0.227  10 6 T 2 dT  5390 
 mol 
T1
T2
(d)
Reversible processes represent the situation where the minimum amount of energy is required for
compression. If the process were irreversible, more work is required, and since the process is
adiabatic, the change in internal energy is greater. Since the change in internal energy is greater,
so too is the change in temperature. Therefore, the final temperature would be higher than the
temperature calculated in Part (b).
32
3.14
The problem statement states that the vessel is insulated, so we can assume that heat transfer to
the surroundings is negligible. Therefore, the expression for the entropy change of the universe
is
suniv  s sys
An energy balance will help us solve for the entropy change. Neglecting potential and kinetic
energy effects, the energy balance is
U  Q  W
Since the vessel is insulated, the heat term is zero. Furthermore, no work is done, so the energy
balance is
U  muˆ 2  uˆ1   0
 u1  u 2
The values for the initial pressure and temperature constrain the value of specific energy at state
1. From the steam tables,
 kJ 
uˆ1  2619.2  
 kg 
 kJ 
 uˆ 2  2619.2  
 kg 
(400 ºC, 200 bar)
The problem statement provides the pressure of state 2 (100 bar). Since we know the pressure
and internal energy at state 2, the entropy is constrained. Information given in the problem
statement also constrains state 1. From the steam tables,
 kJ 
sˆ1  5.5539 

 kg  K 
 kJ 
sˆ2  5.7754 

 kg  K 
(400 ºC, 200 bar)

 kJ  
100 bar, uˆ 2  2619.2   


 kg  

Therefore,
 kJ 
sˆuniv  sˆsys  sˆ2  sˆ1  0.2215 

 kg  K 
 kJ 
S univ  S sys  0.2215  
K
33
3.15
The subscript “2” refers to the final state of the system. “1” refers to the gas initially on one side
of the partition. If you take the system to be the entire tank (both sides of the partition), then no
net work is performed as the gas leaks through the hole. Furthermore, the tank is well insulated,
and kinetic and potential energy effects can be neglected. Thus, the energy balance is
u  0
and
T2  T1
(ideal gas)
Initially, the partition is divided into two equal parts. The gas fills the entire volume for the final
state ( V2  2V1 ). The ideal gas law can be used to calculate the moles of gas present.
n
 
P1V1
10 105 Pa 0.5 m 3   200.5 mol

RT1 
 J 
300 K 
 8.314 
 mol  K  

Now, the change in entropy can be solved using a slight modification of Equation 3.65:
3

 nv 
T 
 v 
T 
S sys  n cv ln 2   R ln 2   n  R ln 2   R ln 2 
 nv1 
 T1 
 v1 
 T1 
2


 J   3

S sys  200.5 mol 8.314 
  ln1  ln2

 mol  K    2


J
S sys  1155  
K 
34
3.16
(a) A schematic of the process is shown below:
1 bar, 298 K
N2
N2
N2
N2
P2, T2
1 bar, 298 K
O2
Mixing
Process
O2
N2
N2
N2
N2
N2
N2
N2
N2
O2
O2
N2
N2
N2
N2
The tank is insulated. The change in entropy of the universe can be rewritten as
S univ  S sys, N 2  S sys, O2
since the tank is well-insulated. After the partition ruptures, the pressure and temperature will
remain constant at 1 bar and 298 K, respectively. This can be shown by employing mass and
energy balances. Therefore, the partial pressures in the system are
p2, N 2  0.79 bar
p2,O2  0.21 bar
Use Equation 3.62 to determine the entropies:
298 K
s sys, N 2 
 P2 
cP

 J    0.79 bar 
    8.314 

 ln
ln
dT
R

 T
P1  N
mol  K    1 bar 



298 K
2
 J 
s sys, N 2  1.96 
 mol  K 
298 K
s sys, O2 
P 

cP
 J    0.21 bar 
dT  R ln 2    8.314 
 ln

T
P1 O
mol  K    1 bar 



298 K
2

 J 
s sys ,O2  12.98 
 mol  K 
Therefore,


 J 
 J 
S univ  0.79 mol1.96 
  0.21 mol12.98 


 mol  K  
 mol  K  


35
J
S univ  4.27  
K 
(b) A schematic of the process is shown below:
2 bar, 298 K
N2
N2
N2
N2
N2
N2
O2
N2
N2
N2
N2
N2
N2
N2
P2, T 2
1 bar, 298 K
N2
N2
Mixing
Process
N2
N2
N2
N2
N2
N2
N2
O2
N2
N2
O2
N2
N2
N2
N2
N2
O2
N2
N2
N2
Before calculating the change in entropy, we need to find how the temperature and pressure
change during the process. The energy balance simplifies to
U  0
which can be rewritten as
nO
2

 n N 2 cvT2  n N 2 cv 298 K   nO2 cv 298 K   0
Assuming the heat capacities are equal, we can show that
T2  298 K
We can find the pressure after the rupture by recognizing that the tank is rigid. Therefore,
VO2  V N 2  Vtot
By employing the ideal gas law, we get the following equation (it has been simplified):
nN 2
P1, N 2

nO2
P1, O2

nO
2
 nN 2

P2
Substitute values and solve to obtain
P2  1.65 bar
Now, use Equation 3.62 to calculate the entropies as was done in Part (a):
36
298 K
s sys, N 2 
P 

cP
 J    1.30 bar 
dT  R ln 2    8.314 
 ln

T
P1  N
mol  K    2 bar 



298 K
2

 J 
s sys, N 2  3.58 
 mol  K 
P 

cP
 J    0.347 bar 
dT  R ln 2    8.314 
 ln

T
 mol  K    1 bar 

 P1 O2
298 K
298 K
ssys ,O2 

 J 
s sys ,O2  8.8 
 mol  K 
Therefore,


 J 
 J 
S univ  0.79 mol 3.58 
  0.21mol 8.8 


 mol  K  
 mol  K  


J
Suniv  4.68  
K
37
3.17
First start with the energy balance for the throttle. Potential and kinetic energy effects can be
neglected. During the throttling process, no shaft work is performed and the rate of heat transfer
is negligible. Therefore,
m 2 hˆ2  m 1hˆ1
A mass balance allows the energy balance to be simplified to
hˆ2  hˆ1
From the steam tables:
 kJ 
hˆ1  3398.3  
 kg 
 kJ 
 hˆ2  3398.3  
 kg 
(500 ºC, 8 MPa)
Now calculate entropy:
sˆuniv  sˆsys  sˆsurr
Since the process is adiabatic,
 kJ 
sˆsurr  0 

 kg  K 
The steam tables can be used to calculate the change in entropy of the system.
 kJ 
sˆ2  8.7098 

 kg  K 
 kJ 
sˆ1  6.7239 

 kg  K 

 kJ  
100 kPa, hˆ2  3398.3   


 kg  

(8 MPa, 500 ºC)
Thus,
 kJ 
sˆuniv  sˆsys  sˆ2  sˆ1  1.9859 

 kg  K 
38
3.18
For this process to work, conservation of mass and the first and second laws of thermodynamics
must hold. The subscript “1” refers to the inlet stream, “2” refers to the cold outlet, and “3”
refers to the hot outlet. To test the conservation of mass, perform a mass balance
m 1  m 2  m 3
 kg 
 kg 
 kg 
2    0.5    1.5  
s 
 s 
s 
Clearly, the conservation of mass holds. Now test the first law of thermodynamics by writing an
energy balance. Since there is no heat transfer or work, the energy balance becomes
m 2 hˆ2  m 3hˆ3  m 1hˆ1  0
(PE and KE effects neglected)
Using the conservation of mass we can rewrite the mass flow rate of stream 1 in terms of streams
2 and 3:




m 2 hˆ2  hˆ1  m 3 hˆ3  hˆ1  0
If we assume that the heat capacity of the ideal gas is constant, the equation can be written as
follows:
m 2 cˆP T2  T1   m 3cˆP T3  T1   0
Therefore,
 kg 
 kg 
0.5   cˆP (60 K)  1.5   cˆP (20 K)  0
s 
s 
This proves that the first law holds for this system. For the second law to be valid, the rate of
change in entropy of the universe must greater than or equal to 0, i.e.,
dS 
0
dt univ
Assuming the process is adiabatic, we can write the rate of entropy change of the universe for
this steady-state process using Equations 3.48-3.50:
 dS 
   m 2 sˆ2  m 3 sˆ3  m 1sˆ1  m 2 sˆ2  sˆ1   m 3 sˆ3  sˆ1 
 dt univ
39
where a mass balance was used. For constant heat capacity, we can calculate the entropy
differences using Equation 3.63:
5  T 
 P 
P 
T 
s2  s1  cP ln 2   R ln 2   R  ln 2   ln 2 
 P1 
 P1 
 T1 
 2  T1 
5  T 
 P 
P 
T 
s3  s1  cP ln 3   R ln 3   R  ln 3   ln 3 
 P1 
 P1 
 T1 
 2  T1 
Applying these relations, we get:
1
 dS 
2.77 R  0.057cP 
  
 dt univ MW
Therefore, the second law holds if cP  48.6 R ; this value is far in excess of heat capacity for
gases, so this process is possible.
40
3.19
(a)
T1 = 640 oC
P2 = 100 kPa
P1 = 4 MPa
V1= 20 m/s
Nozzle
V2=?
(b)
Since the process is reversible and adiabatic,
J
S sys  0  
K
(c)
The steam tables can be used to determine the final temperature. From the interpolation of steam
table data
 kJ 
sˆ2  sˆ1 640 º C, 4 MPa   7.4692 

 kg  K 
Now two thermodynamic properties are known for state 2. From the steam tables:

 kJ  
 P2  0.1 MPa, sˆ2  7.4692 
 


kg
K



T2  121.4 º C
(d)
An energy balance is required to calculate the exit velocity. The energy balance for the nozzle is




0  m 2 hˆ  eˆK 2  m 1 hˆ  eˆ K 1
Realizing that m 1  m 2 allows the energy balance to be written as


eˆK ,2  hˆ  eˆ K 1  hˆ2
which is equivalent to




V2  2 hˆ  0.5V 2 1  hˆ2

Substituting the following values
41

m
V1  20  
s
 J 
hˆ2  2719100  
 kg 
 J 
hˆ1  3767000  
 kg 
(0.1 MPa, 121.4 ºC)
(4 MPa, 640 ºC)
yields

m
V2  1448  
s
Note: the velocity obtained is supersonic; however, the solution does not account for this type of
flow.
42
3.20
A schematic of the process follows:
Propane in
v1 = 600 cm3/mol
T1= 350 oC
Turbine
ws
Propane P = 1 atm
2
out
Since this process is isentropic (s=0), we can apply an expression for the entropy change of an
ideal gas. We must be careful, however, to select an expression which does not assume a
constant heat capacity.
T2
c
P
s  0   P dT  Rln 2
P1
T T
1
Insert the expression for cP from Appendix A
P 
1.213  28.785  10 3 T  8.824  10 6 T 2
s  0  R 
dT  R ln 2 
T
 P1 
T1
T2
We can also relate P1 to T1 and v1 (which are known) through the ideal gas law:
P1 
RT1
v1
This leaves us with 1 equation and 1 unknown (T2). Integrating:
T 
Pv 
s  0  1.213 ln 2   28.785  10 3 (T2  T1 )  4.412 10 6 T22  T12  ln 2 1 
 T1 
 RT1 

Using
T1 = 623 K,
v1 = 600 cm3/mol,
P2 = 1 atm
43

R=82.06 cm3 atm/mol K,
we get:
T2 = 454 K
To solve for
W S
we need a first law balance, with negligible ke and pe, the 1st law for a steady
n
state process becomes:
0  n h1  h2   Q  W S
If heat transfer is negligible (isentropic),
T2
T2
2
W S
 h2  h1   c p dT  R  1.213  28.785  10 3 T  8.824  10 6 T dT
n
T1
T1


integrating:




W S
 R 1.213T2  T1   14.393 10  3 T22  T12  2.94110  6 T23  T13
n
Substituting
 J 
R  8.314 
 mol  K 
T2  454 K
T1  623 K
results in
W S
 J 
 19860 
n
 mol 
44

3.21
A reversible process will require the minimum amount of work. For a reversible process,
suniv  0
 s sys  s surr
For this process, the change in entropy of the system can be calculated as follows
s sys  sO2  s N 2
where
sO2 
T2 c

P O2
T1
s N 2 
T2 c

T1
T
P 
dT  R ln 2 
 P1 
P N 2
T
O2
P 
dT  R ln 2 
 P1 
N2
Since the temperature of the streams do not change, the expressions reduce to
P 
sO2   R ln 2 
 P1  O2
P 
s N 2   R ln 2 
 P1  N 2
Therefore,
 
 s surr  R  ln 
 

P2
P1

P 

 ln  2 
O2
 P1  N 2




Since we are assuming the temperature of the surroundings remain constant at 20 ºC,
q
s surr  surr
Tsurr
where
 qsurr  q
45
Therefore,
 P 
 P2  
2



q surr  Tsurr s surr  RTsurr ln   ln  
  P1  O
 P1  N 2 
2

  1 bar 

 J 
 1 bar  





q surr   8.314 
293
.
15
K
ln
ln



 

 mol  K  
 0.79 bar  N 2 
  0.21 bar  O2

 J 
q surr  4378.2 
 mol 
and
 J 
q  4378.2 
 mol 
Energy Balance:
 n
h  nout hout  Q  W S  0
in in
in
out
Because the temperature of the oxygen and nitrogen doesn’t change in the process, the energy
balance per mole of feed becomes
wS   q
 J 
wS  4378.2 
 mol 
46
3.22
(a)
Take the entire container to be the system and assume no heat or work crosses the system
boundary.
Energy Balance:
U  0
After the oscillation cease, the temperature and pressure on both sides of the piston will be the
same assuming the metallic piston is very thin and the thermal conductivity coefficient is large.
Now, let’s rewrite the energy balance.
U  n1, A ugas A  n1,B ugas B  0
or
n1, A
ugas A  ugas B  0
n1,B
n1, A PA,1VA,1TB ,1
P T

 A,1 B ,1  2.41
n1,B PB ,1VB ,1TA,1 2 PB ,1TA,1
since
2V A,1  VB,1
so
T


2
3

2.41 RT2  773.15 K   R  5/2  1.5  10-3 T dT  0
2

373.15 K
T2  585 K
Mass Balance:
n A,1  n B,1  n A,2  n B,2
Using the ideal gas law, we get
47
PA,1VA,1 PB ,1VB ,1 PA, 2VA, 2 PB , 2VB , 2 P2Vtot




RTA,1
RTB ,1
RTA, 2
RTB , 2
RT2
since the pressure and temperature of state 2 are equal on both sides. But
Vtot  V A,1  VB,1  3V A,1
Using the volume relationships and simplifying, we get
P2 
T2  PA,1 2 PB ,1  585 K   10 bar 
1 bar  




2


3  TA,1 TB ,1 
3  773.15 K 
373.15 K  
P2  3.56 bar 
(b)
The container is well-insulated, so
suniv  s sys
The entropy change of the system can be split into two subsystems:
ssys 
n1, A
n1,B
s gas A 
s gas B
n1, A  n1,B
n1, A  n1,B
Using Equation 3.63 and realizing that c P  cv  R ,
sgas A 
sgas B
 P 
T 
2.41 
 J 
2.5 R ln 2   R ln 2   1.96 
3.41 
 mol  K 
 PA,1 
 TA,1 
T
 P 
1  2 3.5  1.5  10 3 T
 J 
R 

dT  R ln 2   1.51 
3.41  TB ,1
T
 mol  K 
 PB ,1 

Therefore,
 J 
 J 
 J 
suniv  ssys  1.96 
 1.51 
 3.47 


 mol  K 
 mol  K 
 mol  K 
The process is possible.
48
3.23
Equation 3.3 can be modified to show
S univ  S sys  S surr
Let’s first calculate S sys . Before this problem is solved, a few words must be said about the
notation used. The system was initially broken up into two parts: the constant volume container
and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant
volume container; “2” refers the piston-cylinder assembly. “i" denotes the initial state before the
valve is opened, and “f” denotes the final state.
First, the mass of water present in each part of the system will be calculated. The mass will be
conserved during the expansion process. Since the water in the rigid tank is saturated and is in
equilibrium with the constant temperature surroundings (200 ºC), the water’s entropy is
constrained. From the steam tables,
 kJ 
vˆ1l, i  0.001156  
 kg 
 kJ 
vˆ1v, i  0.12736  
 kg 
 kJ 
sˆ1l, i  2.3308  
 kg 
 kJ 
sˆ1v, i  6.4322  
 kg 
(Sat. water at 200 ºC)
P sat  1553.8 kPa 
Knowledge of the quality of the water and the overall volume of the rigid container can be used
to calculate the mass present in the container.
 
 
V1  0.05m1 vˆ1l, i  0.95m1 vˆ1v,i
 
Using the values from the steam table and V1  0.5 m 3 provides
m1  4.13 kg 
Using the water quality specification,
49
m1v  0.95m1  3.92 kg 
m1l  0.05m1  0.207 kg 
For the piston-cylinder assembly, both P and T are known. From the steam tables
 m3 
vˆ2, i  0.35202 

 kg 
 kJ 
sˆ2, i  6.9665 

 kg  K 
(600 kPa, 200 ºC)
Now enough information is available to calculate the mass of water in the piston assembly.
V
m2  2  0.284 kg 
vˆ2
Now the final state of the system must be determined. It helps to consider what physically
happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When
the valve is opened, the water will rush out of the rigid tank and into the cylinder until
equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the
surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of
the entire system will match the surrounding’s. In other words,
 kJ 
sˆ f  sˆ2, i  6.9665  
 kg 
(600 kPa, 200 ºC)
Thus, the change in entropy is given by
S sys  (m2  m1v, i  m1l, i ) sˆ f  m2 sˆ2, i  m1v, i sˆ1v, i  m1l, i sˆ1l, i
Substituting the appropriate values reveals
 kJ 
S sys  3.05  
K
Now we calculate the change in entropy of the surroundings. Since the temperature of the
surroundings is constant,
S surr 
Qsurr
Q

Tsurr Tsurr
After neglecting potential and kinetic energy effects, the energy balance becomes
50
U  Q  W
The change in internal energy and work will be calculated in order to solve for Q. The following
equation shows how the change in internal energy can be calculated.
U  (m2  m1v, i  m1l, i )uˆ f  m2 uˆ 2, i  m1v, i uˆ1v, i  m1l, i uˆ1l, i
From the steam tables
 kJ 
uˆ1l, i  850.64  
 kg 
 kJ 
uˆ1v, i  2595.3  
 kg 
 kJ 
uˆ f  uˆ 2, i  2638.9  
 kg 
(sat. H2O at 200 ºC)
(600 kPa, 200 ºC)
Using these values and the values of mass calculated above,
U  541.0 kJ 
Calculating the work is relatively easy since the gas is expanding against a constant pressure of
600 kPa (weight of the piston was assumed negligible). From Equation 2.7,
Vf
W   PE  dV   PE (V f  Vi )
Vi
where
PE  600000 Pa 
 
V f  (m2  m1v, i  m1l,i )vˆ2, i  1.55 m 3
   
 
Vi  0.1 m 3  0.5 m 3  0.6 m 3
Note: vˆ2, i was used to calculate V f because the temperature and pressure are the same
for the final state of the entire system and the initial state of the piston-cylinder assembly.
The value of W can now be evaluated.
W  570 kJ 
51
The energy balance can be used to obtain Q.
Q  U  W  541.0 kJ    570 kJ   1111 kJ 
Therefore,
S surr 
 1111 kJ 
 kJ 
 2.35  
473.15 K 
K
and
 kJ 
 kJ 
 kJ 
Suniv  S sys  S surr  3.05    2.35    0.70  
K
K
K
52
3.24
Let the subscript “1” represent the state of the system before the partition is removed. It has two
components: component a and component b. Subscript “2” represents the system after the
partition is removed.
Mass balance:
n1, a  n1, b  n2
Energy balance for the adiabatic process:
U sys  W






 n1, a cv, a T2  T1, a  n1, b cv, b T2  T1, b   P V2  V1, a  V1, b

The external pressure, P, for the above energy balance is equal to P1,a. To find the final
temperature, first find the volumes using the ideal gas law.
V1, a 
V2 
2 mol 8.314 
J 
300 K 
 mol  K  

 0.05 m 3
 1000 kg  9.8 m  s - 2 


0.098 m 2


 
  
 
4 mol 8.314 
J 
T2
 mol  K  

 3.3256  10  4 T2 m 3
2
 1000 kg  9.8 m  s 


0.098 m 2


 
  
 
Substitute the volumes into the energy balance and solve for T2 :
2 mol 3 R T2  300 K   2 mol 3 R T2  300 K   1 105 Pa 3.3256  10  4 T2  0.15 m 3 
2 
T2  360.4 K
2 
To calculate the change in entropy, we can use the following relationship


T 
 P 
T 
 P 
S sys  na c P ln 2   R ln 2   nb c P ln 2   R ln 2 
 T1 
 P1  b
 T1 
 P1  a


The only unknown in the above equation is P1,b, so we can calculate it with the ideal gas law:
53
P1, b 
J 
300 K 
 mol  K  
2 mol 8.314 

0 .1 m
3
 0.5  10 5 Pa
Substitute values into the expression for entropy and solve:
 5  360.4 K 
 5  360.4 K 
 1 bar 
 1 bar 
  ln
S sys  2 molR  ln

  2 molR  ln
  ln
 0.5 bar  b
 1 bar  a
 2  300 K 
 2  300 K 
J
S sys  3.72  
K 
54
3.25
First perform an energy balance on the process:
U sys  W
The change in internal energy can be written:
U  n A, f u A, f  nB , f u B , f  n A,i u
A mass balance gives:
n A,i  n A , f  nB , f
These two expressions can be substituted to give:
U  n A, f cv TA, f  Ti   nB , f cv TB , f  Ti   W
5
R , we obtain
2
5
5
5
PB, f VB, f  PA, f V A, f  PA, iV A, i  W
2
2
2
Using the ideal gas law and cv 
Now, find an expression for the work
Vf
W    PE dV
Vi
The pressure is not constant; its value is given by
k VB  VB ,i  mg
kx mg

 Patm 

A2
A
A
A
5
5
PE  1.06  10  8.284  10 VB Pa 
PE  Patm 
Integrating
W  1.06 105VB, f  4.142 105VB2, f
Substituting this expression into the energy balance gives:
5
5
5
PB, f VB, f  PA, f V A, f  PA,iV A, i  1.06  105VB, f  4.142  105VB2, f
2
2
2
55
This expression can be simplified by recognizing that PB, f  PA, f and V A, f  V A, i :


5
5
PB, f VB, f  V A, i  PA, iV A, i  1.06 105VB, f  4.142  105VB2, f
2
2
We still have two unknowns: PB, f and VB, f . We can eliminate PB, f by writing a force
balance for the final state:
PB, f  1.06 105  8.284 105VB, f Pa 
Substituting this into the energy balance, we obtain



5
5
1.06  105  8.284 105VB, f VB, f  V A, i  PA, iV A, i  1.06 105VB, f  4.142 105VB2, f
2
2
Solving for the final volume:
VB, f  0.268 m 3
and for the final pressure.
PB, f  PA, f  Pf  1.06  105  8.284  105 0.268  3.28  105 Pa
Since the gas in A has undergone an adiabatic, reversible expansion
S sys, A  0
Therefore,
T 
P 
0  c P ln 2   R ln 2 
 T1 
 P1 
 3.28  10 5 Pa 


 J    7  T2

0   8.314 
  ln
  ln
 7.0  10 5 Pa 
 mol  K    2  313.15 K 



Solve for T2:
T2  252 K
56
3.26
A schematic of the process is drawn:
Q surr=69.86 kW
Steam
in
P1 = 60 bar
T1 = 500 oC
W
Turbine
Steam
out
P2 = 10 bar
Tsurr=300 K
(a) and (b)
The maximum work occurs for a reversible process. Applying the second law, we get:
 dS 
 dS 
 dS 
      0
 dt  univ  dt  sys  dt  surr
or
m sˆ2  sˆ1  
Q surr
0
Tsurr
Calculate the change in entropy of the surroundings:
 kJ 
We can look up property values of state 1 from the steam tables sˆ1  6.8802 
 and
 kgK 
 kJ 
hˆ1  3,422.13   . Converting the units of mass flow rate gives:
 kg 

 kg   1 hr  
 kg 
m   4,500   
  1.25  
 hr   3,600 s 
 s 

so
sˆ2  sˆ1 
 kJ 
Q surr
 6.6939 

m Tsurr
 kgK 
We now know two properties of steam ( ŝ2 and P2) From the steam tables:
T2  200 oC
Using the steam tables for superheated steam at 1 MPa, we find that when water has this value of
entropy
57
 kJ 
hˆ2  2827.9  
 kg 
Energy balance:


m hˆ2  hˆ1  Q  W s
Calculate Ws:



 kJ 
 kJ  
W s  m hˆ2  hˆ1  Q  1.25 kg/s  2827.9    3422.1      69.86 kW 
 kg 
 kg  

W s  673 kW 
(c)
The isentropic efficiency is defined as follows:
η
W S actual
W S reversible
For this situation,
W S actual  0.665 673 kW   447.5 kW 
(d)
The real temperature should be higher since not as much energy is converted into work.
(e)
Use the energy balance:


m hˆ2  hˆ1  Q  W s
 kJ 
 kJ 
Q  W s ˆ  69.86 kW  -447.5 kW
 h1 
 3422.1    3008.2  
hˆ2 
1.25 kg/s
m
 kg 
 kg 
At 1 MPa, water has this value of enthalpy when
T2 actual  280.2 º C
58
3.27
Since the compressor is adiabatic, the energy balance after neglecting potential and kinetic
energy becomes
n h2  h1   W S
Using the ideal gas law and Appendix A.2, the above equation becomes


T
P1V1 2

WS 
A  BT  CT 2  DT  2  ET 3 dT
T1 
T1
Substituting the following values
P1  1 bar   1 105 Pa 
 m3 
V1  1 

 s 
T1  293.15 K 
T2  473.15 K 
A  3.355; B  0.575 10 3 ; C  0; D  0.016 105 ; E  0 (Table A.2.2)
gives
W S  218.8 kW 
This is the work of our real turbine (with 80% isentropic efficiency). We can use the isentropic
efficiency to calculate the work of an equivalent 100% efficient process.
W 
compressor   S reversible
WS 
compressor
W S reversible  W S compressor compressor  0.8218.8 kW 
W S reversible  175.4 kW 
Since the process is adiabatic, we can use the following equation again to calculate what the final
temperature would be in a reversible process.

T
P1V1 2

WS 
A  BT  CT 2  DT  2  ET 3 dT

T1
T1
59

P V
175400 W   1 1
T1
 A  BT  CT
T2
2
 DT  2  ET 3 dT

T1
Substituting the values from above and solving for T2 , we obtain
T2, rev  164.8 K
To obtain the pressure the final pressure for the real compressor, we can calculate the final
pressure for the reversible process because the final pressure is the same in both cases. For the
isentropic expansion
T2
s  0 
P 
dT  R ln 2 
T
 P1 
T1

cp
 A  BT  CT 2  DT  2  ET 3 
P 
s  0  R  
 dT  R ln 2 
T
 P1 

T1 
T2


 T2  438 K 

 P2 
3.355  0.575  10  3 T  1600T  2 
dT
ln
s  0  R 




 1 bar  
 

T




293
.
15
K
T

 1

Solve for P2:
P2  4.16 bar 
60
3.28
Isentropic efficiency for a turbine is defined as

ws actual
ws rev
If the rate of heat transfer is assumed negligible, the energy balance for this process is


0  m hˆ1  hˆ2  W s
For a reversible, adiabatic process,
s sys  0
sˆ2,rev  sˆ1 500 º C, 10 MPa 
From the steam tables
 kJ 
sˆ2,rev  sˆ1  6.5965 

 kg  K 
Since sˆ2l , sat  sˆ2  sˆ2v, sat , a mixture of liquid and vapor exists. The quality of the water can be
calculated as follows
sˆ2,rev  1  x sˆ2l ,sat  xsˆ2v ,sat

 kJ  
 kJ 
 kJ   
  x 7.3593 
6.5965 
 1  x 1.3025 
 


 
kg
K
kg
K




 kg  K 



 
x  0.874
Therefore,
hˆ2,rev  1  x hˆ2l ,sat  xhˆ2v ,sat


 kJ  
 kJ  
hˆ2,rev  1  0.874  417.44     0.874 2675.5   
 kg  
 kg  


 kJ 
hˆ2,rev  2391.0  
 kg 
Also, from the steam tables,
 kJ 
hˆ1  3373.6  
 kg 
(500 ºC, 10 MPa)
61
The reversible work is calculated as:
W S rev  hˆ
m 1
2
 kJ 
 kJ 
 kJ 
 hˆ1  2391.0    3373.6    982.6  
 kg 
 kg 
 kg 
and the actual work as
W S actual
m 1

W S rev  835  kJ 
m 1
 kg 
 
The exit temperature is calculated by determining the enthalpy of the actual exit state:

W S actual
 kJ 
ˆh
ˆ
 2538.4  
2, actual  h1 
m 1
 kg 
This state is still saturated (although the quality is higher), so the temperature is
T2 = 99.6 oC
62
3.29
The subscripts “2” and “3” represent the two outline streams, and “1” represents the inlet stream.
First, perform a mass balance:
n1  n 2  n3 
2
1
n1  n1
3
3
Now write the energy balance:
0  n1h1  n2 h2  n3 h3   Q  W S
Since the system is insulated and there is no shaft work, the energy balance can be rewritten as:
2
1
n1h2  n1h3  n1h1  0
3
3
2
h2  h1   1 h3  h1   0
3
3
Substituting expressions for heat capacity A.2.2, we obtain the following expression:
400 K


T


3


4000 
1
4000 
3
3
 3.280  0.593 10 T  T 2  dT  3 R  3.280  0.593  10 T  T 2  dT  0
300 K
300 K
Solving, we find
2
R
3
T3  100.6 K
Now set up an entropy balance for the process. The minimum pressure is required for a
reversible process.
2
1
n1s2  n1s3  n1s1  0
3
3
2
s2  s1   1 s3  s1   0
3
3
Assuming ideal gas behavior, we can express the changes in entropies using Equation 3.62:
400 K  3.280
 1bar 
4000 


ln
s 2  s1  R   
dT
 0.593  10  3 

3 
T
P

T


1


300 K
100.6 K  3.280
 1bar 
4000 


ln
s3  s1  R   
dT
 0.593 10  3 

3 
T
P

T


1


 300 K




63
Substitute these expressions into the entropy balance and solve for P1:
P1  1.85 bar
64
3.30
A schematic of the process is illustrated below:
reversible
Tsurr = 25 oC
Psurr = 1 bar
Pure H 2O
P2 = 20 bar
Pure H2O
T 2 = ? oC
P1 = 60 bar
T1 = 540 oC
m = 5 kg
m = 5 kg
well-insulated
(a)
Since the process is adiabatic and reversible,
suniv  0
s sys  0
s surr  0
(b)
We can obtain the final temperature using the steam tables.
 kJ 
sˆ1  6.999 

 kg  K 
 kJ 
 sˆ2  6.999 

 kg  K 
(540 ºC, 60 bar)
(20 bar)
The pressure and entropy of state 2 can be used to back out T2 from the steam tables.
T2  362.5 º C
(c)
To obtain the value of work, perform an energy balance. The process is adiabatic, and potential
and kinetic energy effects can be neglected. Therefore, the energy balance is
U  muˆ 2  uˆ1   W
From the steam tables,
 kJ 
uˆ2  2881.15  
 kg 
(362.5 ºC, 20 bar)
65
 kJ 
uˆ1  3156.12  
 kg 
(540 ºC, 60 bar)
Hence,

 kJ 
 kJ  
W  5 kg  2881.15    3156.12     1373 kJ 
 kg 
 kg  

(d)
The specific volume can be found in the steam tables
 m3 
vˆ2  0.14173 

 kg 
Therefore,
 

 m3  
3
V2  5 kg  0.14173 
   0.709 m


 kg  

66
3.31
A schematic of the process is shown below:
very tiny
leak hole
Tsurr = 25 oC
Psurr = 1 bar
Pure H2O
Pure H2O
P1 = 60 bar
P2 = 20 bar
T1 = 540 oC
T ' 2 = ? oC
m = 5 kg
well-insulated
In order to leave the system, the gas must do flow work on the surroundings. The initial state is
the same as for Problem 3.30 and the final pressures are the same. Since the water only expands
against 1 bar, the work is lower than that for the differential process described in Problem 3.30.
Thus, this adiabatic process looses less energy, leading to a higher final temperature.
Another way to view this argument is to look at this process as a closed system. This depiction
is the expansion analog of the compression process depicted for Example 2.5 in Figure E2.5B
(page 57). We can represent this process in terms of two latches, one that keeps the process in its
initial state at 60 bar and one that stops the expansion after the pressure has reduced to 20 bar.
The process is initiated by removal of the first latch and ends when the piston comes to rest
against the second latch. Such a process is depicted as “Problem 3.31’ below. The
corresponding reversible process of Problem 3.30 is shown next to it for comparison. Clearly the
process on the left does less work, resulting in a greater final temperature.
Problem 3.31
Problem 3.30
T surr = 25 oC
Psurr = 1 bar
latches
Pure H2O
P1 = 60 bar
T1 = 540 oC
irreversible
reversible
Tsurr = 25 oC
Psurr = 1 bar
Pure H2O
Pure H 2O
P2 = 20 bar
P2 = 20 bar
Pure H2O
T ' 2 = ? oC
T 2 = ? oC
P1 = 60 bar
m = 5 kg
m = 5 kg
T1 = 540 oC
m = 5 kg
m = 5 kg
well-insulated
67
3.32
(a)
Consider the tank as the system.
Mass balance
dm
 m in  m out  m in
dt
Separating variables and integrating:
m2
t
m1
0
 dm   m in dt
or
t
m2  m1   m in dt
0
Energy balance
Since the potential and kinetic energy effects can be neglected, the open system, unsteady state
energy balance is
 dU 

   m out hout   m in hin  Q  W s
dt

 sys out
in
The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream
and not outlet stream. Therefore, the energy balance simplifies to
 dU 

  m in hin
 dt  sys
The following math is performed
U2
t
U1 0
0
 dU   m
t
h dt  hin  m in dt
in in
0
U 2   m2uˆ 2  m2 hˆin
where the results of the mass balance were used. Thus,
uˆ 2  hˆin
68
From the steam tables,
 kJ 
hˆin  3456.5  
 kg 
(3 MPa, 773 K)
Now the water in the tank is constrained. From the steam tables:
 kJ 
sˆ2  7.743 

 kg  K 
3
m
vˆ2  0.14749 

 kg 

 kJ  
 3 MPa, uˆ 2  3456.5   


 kg  

Compute the change in entropy. An entropy balance gives:
 dS 
 dS 
      m in sin
 dt  univ  dt  sys
Integrating with sin constant
t
S univ  m2 sˆ2  sˆin  m in dt  m2 sˆ2  sˆin 
0
From the steam tables:
 kJ 
sˆin  7.2337 

 kg  K 
(3 MPa, 773 K)
Therefore,





 0.05 m 3
 kJ  
 kJ 
kJ
 7.743 
  0.173  
 7.2337 
S univ  







 m 3  
K
 kg  K  
 kg  K 
 0.14749 


kg
 



(b)
If it tank sits in storage for a long time and equilibrates to a final temperature of 20 ºC, some or
all of the vapor will condense and exchange heat with the surroundings. Let the subscript “3”
designate the final state of the water when it has reached a temperature of 20 ºC. The change in
entropy between state 2 and state 3 is given by the following equation
69
S univ  m2 sˆ3  sˆ2  
Qsurr
T
Let the subscript “3” designate the final state of the water when it has reached a temperature of
20 ºC. First, find quality of the water. From the steam tables:
 m3 
ˆv3sat
, l  0.001002  kg 


(sat. water at 20 ºC)
 m3 
vˆ3sat

57
.
79


,v
 kg 
Calculate the quality as follows:
ˆ sat
vˆ3  vˆ2  1  x vˆ3sat
, l  xv3, v

 m3 
 m3   
 m3  
0.14749    1  x  0.001002     x 57.79   
 kg 
 kg   
 kg  

x  0.0025
Now calculate the entropy of state 3
ˆ sat
sˆ3  1  x sˆ3sat
, l  xs3, v
Substitute values from the steam tables:

sˆ3  1  0.0025 0.2966

 kJ 
sˆ3  0.3175 

 kg  K 

 kJ  
 kJ  


0
.
0025
8
.
6671

 kg  K  
 kg  K  






Now calculate the amount of heat transferred to the surroundings:
Q surr  Q  m2 uˆ 3  uˆ 2 
Calculate the internal energy of state 3:
ˆ sat
uˆ 3  1  x uˆ 3sat
,l  xu 3,v
Substituting values from the steam tables:
70
 kJ 
uˆ 3  89.74  
 kg 
Therefore,





 0.05 m 3
 kJ  
 kJ 
 89.74    3456.5     1141 kJ 
Q surr  


 m 3  
 kg  
 kg 
0
.
14749





 kg  

Now calculate the change in entropy of the universe
S univ
S univ




3
 kJ 
 kJ   1141 kJ 

 0.05 m

0
.3175
7
.
743


 kg  K 
 kg  K    293 K
3 


m




 0.14749   


kg
 

 kJ 
 1.38  
K
The entropy change for both processes can be fount by adding together the entropy change from
Part (a) and Part (b):
S univ (a) and (b)  1.55 kJ 
K
71
3.33
A schematic is given below
valve maintains
pressure in system
constant
v
T1 = 200 oC
x1 = 0.4
V = 0.01 m3
l
Mass balance
dm
 m in  m out  m out
dt
Separating variables and integrating:
m2
t
m1
0
 dm   m
out
dt
or
t
m2  m1    m out dt
0
Entropy Balance:
Q
Q
 dS 
 dS 
Suniv     m out sout  surr     m out sout 
Tsurr  dt  sys
Tsurr
 dt  sys
Integrating with sout constant
Suniv  m2 s2  m1s1  m2  m1 sout 
Q
Tsurr
(1)
From steam tables:
72
 kJ 
s1  (1  x)u f  xu g  0.6  2.3309  0.4  6.4323  3.9715 

 kg K 
 kJ 
 kJ 
s 2  6.4323 
is the same as sout  6.4323 

  s2
 kg K 
 kg K 
Thus Equation 1 simplifies to
Suniv  m1 ( s 2  s1 ) 
Q
(2)
Tsurr
Energy Balance to find Q :
 dU 

  m out hˆout  Q
 dt  sys
Integrating
 dU   
m 2 uˆ 2
t
m1uˆ1
0

t
t
0
0
 m out hˆout  Q dt  hˆout  m out dt   Q dt
Substituting in the mass balance and solving for Q
Q  m 2 uˆ 2  m 1 uˆ 1  m 2  m 1 hˆ out
To find the mass in each state:
 m3 
v1  (1  x)v f  xv g  0.6  .001  0.4  0.1274  0.051 

 kg 
 m3 
v2  0.1274 

 kg 
 
V
0.01 m 3
 0.196 kg 
m1  1 
v1
 m3 
0.051 

 kg 
V
and m2  2 
v2
From the seam tables:
73
 
0.01 m 3
 0.0785 kg 
 m3 
0.1274 

 kg 
 kJ 
u1  (1  x)u f  xu g  0.6  850.64  0.4  2597.5  1549  
 kg 
 kJ 
u 2  2595.3  
 kg 
and
 kJ 
hout  2793.2  
 kg 
Plugging in:
Q  m2uˆ2  m1uˆ1  m2  m1 hˆout  228 kJ 
Back into Equation 2
 kJ 
228  
 kJ  

Q
 kg  
 s  0

S univ  m1 ( s 2  s1 ) 
  0.196    6.4323  3.9715 


Tsurr 
 s  
 kg K   473  K 
s
 
Note: Vaporization is reversible if Tsurr = Tsys.
74
3.34
The process can be represented as:
V1 = 1 m3
P1 = 10 bar
Process
V2 = 10 m3
P2 = ?
T1 = 1000 K
Work out
Q=0
T2 = ?
State 1
State 2
Solving for the number of moles:
PV
n  1 1  120.3 mol
RT1
The maximum work is given by a reversible process. Since it is also adiabatic, the entropy
change of the system is zero:
T 
P 
s  0  cP ln 2  R ln 2 
T1 
P1 
5
Since cP  R
2
 T2 
 
 T1 
5/ 2

P2 nRT2

P1 V2 P1
Solving for T2 gives
 nRT 5 / 2 
1
T2  

 V2 P1 
2/3
 215 K
An energy balance on this closed system gives:
T2
T2
T1
T1
U = Q  W  W  n  c v dT  n  cP  R dT
Solving for work, we get
3
6
W  n RT2  T1   1.18 10 J
2
75
3.35
The maximum efficiency is obtained from a Carnot cycle. From Equation 3.32
T
n  1 C
TH
where temperature is in Kelvin. Hence,
n  1
298.15 K
 0.614
773.15 K
76
3.36
The labeling described in Figure 3.8 will be used for this solution. First, a summary of known
variables is provided.
State
1
2
3
4
P (bar)
30
0.1
T (ºC)
500
From our knowledge of ideal Rankine cycles, the table can be expanded as follows
State
1
2
3
4
Thermodynamic
State
Superheated Steam
Sat. Liq. and Vapor
Sat. Liquid
Subcooled Liquid
P (bar)
T (ºC)
30
0.1
0.1
30
500
The saturation condition constrains state 3. First, start with the turbine. Since the rate of heat
transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the
following is known
sˆ1  sˆ2
and

W s  m hˆ2  hˆ1

From the steam tables:
 kJ 
sˆ1  7.2337 

 kg  K 
 kJ 
sˆ2l  0.6492 

 kg  K 
 kJ 
sˆ2v  8.1501 

 kg  K 
 kJ 
hˆ1  3456.5  
 kg 
 kJ 
hˆ2l  191.81  
 kg 
hˆ2v
 kJ 
 2584.6  
 kg 
(500 ºC, 30 bar)
(sat. H2O at 0.1 bar)
(500 ºC, 30 bar)
(sat. H2O at 0.1 bar)
77
The quality of the water can be calculated as follows
sˆ1  (1  x) sˆ2l  xsˆ2v
By substituting the steam table data, we find
x  0.878
Now the quality can be used for the following expression
hˆ2  (1  x)hˆ2l  xhˆ2v
Substituting the steam table data and the quality calculated above, we get
 kJ 
hˆ2  2292.7  
 kg 
Therefore,
 kJ 
 kJ  

 kg  
W s  100    2292.7    3456.5     116.38 MW 
 s  

 kg 
 kg  
At state 3,
 kJ 
hˆ3  191.81  
 kg 
 m3 
ˆv3  0.001010 

 kg 
(sat. H2O(l) at 0.1 bar)
Therefore,


Q C  m hˆ3  hˆ2  210.09 MW 
Since the molar volume of water does not change noticeably with pressure and the compressor is
adiabatic, the work required for compressing the fluid can be calculated as follows
W c  m vˆ3 P4  P3 
 m3  
 kg 
W c  100   0.00101    30  105  0.1  105 Pa   3.02  10 5 W   0.302 MW 
 s 
 kg  


78
Now we find the rate of heat transfer for the evaporator. For the entire Rankine cycle, Equation
3.84 gives
Q net  W net  Q C  Q H  W s  WC  0
We have
Q H  W s  WC  Q C
Using the values calculated above:
Q H  326.2 MW 
Equation 3.84:
W net  W S  WC
W net  116.08 MW 
The efficiency can be calculated with Equation 3.82
W
116.08 MW 
  net 
 0.356
326.02 MW 
QH
79
3.37
To make the Rankine cycle more efficient, we need to increase the area that represents the net
work in Figure 3.8. This can be done in a variety of ways:
1. Increase the degree of superheating of steam in the boiler. This process is sketched in the
upper left hand Ts diagram below. This change reduces the moisture content of steam
leaving the turbine. This effect is desirable since it will prolong the life of the turbine;
however, if the steam is heated too high, materials limitations of the turbine may need to be
considered.
2. Lower the condenser pressure. Lowering the pressure in the condenser will lower the
corresponding saturation temperature. This change will enlarge the area on the Ts diagram,
as shown on the upper right below. Thus, we may want to consider lowering the pressure
below atmospheric pressure. We can achieve this since the fluid operates in a closed loop.
However, we are limited by the low temperature off the heat sink that is available.
increase the superheating of steam in the boiler
T
lower the condenser pressure
T
1new
1
wnet
wnet
4new
4
2new
3
3new
2new
s
s
2 stage turbine with reheat
increase the boiler pressure
Turbine 1
Turbine 2
1
T
T
1
3
6
2
wnet
wnet
4
4
3
5
2
s
s
80
3. Increase the boiler pressure. This will increase the boiler temperature which will increase the
area as shown on the bottom left Ts diagram.
4. We can be more creative about how we use the energy available in the boiler. One way is to
divide the turbine into two stages, Turbine 1 and Turbine 2, and reheat the water in the boiler
between the two stages. This process is illustrated below.
Rankine cycle with reheat
Ws
1
Fuel
Air
Ws
Turbine
2
turbine
1
2
4
QH
3
Boiler
QC
cooling
water
6
5
condenser
Wc
compressor
The pressure in Turbine 1 will be higher than the pressure in turbine 2. This process is
schematically shown on the bottom right Ts diagram. This process leads to less moisture
content at the turbine exit (desirable) and limits the temperature of the superheat (desirable).
81
3.38
The labeling described in Figure 3.8 will be used for this solution. First, a summary of known
variables is provided.
State
1
2
3
4
P (bar)
100
1
T (ºC)
500
From our knowledge of ideal Rankine cycles, the table can be expanded as follows
State
1
2
3
4
Thermodynamic
State
Superheated Steam
Sat. Liq. and Vapor
Sat. Liquid
Subcooled Liquid
P (bar)
T (ºC)
100
1
1
100
500
The saturation condition constrains state 3. First start with the turbine (state 2). Since the rate of
heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the
following is known
sˆ1  sˆ2
From the steam tables,
 kJ 
sˆ1  6.5965 

 kg  K 
 kJ 
sˆ2l  1.3025 

 kg  K 
sˆ2v
 kJ 
 7.3593 

 kg  K 
(500 ºC, 100 bar)
(sat. H2O at 1 bar)
The quality of the water can be calculated as follows
sˆ1  (1  x) sˆ2l  xsˆ2v
By substituting the steam table data, we find
x  0.874
82
States 1, 2, and 3 are constrained; therefore, the enthalpies can be determined using the steam
tables. The enthalpy of state 4 can be calculated from Equation 3.80:
hˆ4  hˆ3  vˆl P4  P3 
 m3 
ˆ
ˆ
where vl  v3  0.001043 
 since specific volume of liquids are insensitive to pressure
 kg 
changes. From the steam tables:
State
1
 kJ 
3240.8  
 kg 
Note: hˆ2  (1  x)hˆ2l  xhˆ2v
ĥ
2
3
4
 kJ 
2391.0  
 kg 
 kJ 
417.44  
 kg 
 kJ 
427.8  
 kg 
Equation 3.84:
W net  Q H  Q C
Therefore,

 
 
 
W net  m hˆ1  hˆ4  m hˆ3  hˆ2  m hˆ1  hˆ4  hˆ3  hˆ2

and
m 
W net
hˆ1  hˆ4  hˆ3  hˆ2 

100 103 kW 

 kJ 
 kJ   
 kJ 
 kJ  
 3240.8    417.5      427.8    2391.0   

 kg 
 kg   
 kg 
 kg  
 kg 
m  116.3  
s 
Now consider the non-ideal turbine and compressor. Use the definition of isentropic
efficiencies:

W s actual
turbine  
Ws reversible
 W S actual  turbine W S reversible  turbine m hˆ2  hˆ1 reversible
83
WC reversible
WC actual

WC reversible m hˆ4  hˆ3 reversible


compressor 
 WC actual
compressor
compressor
Equation 3.84 is used to find the mass flow rate
W net  W s  WC



hˆ4  hˆ3 reversible 

W net  m turbine hˆ2  hˆ1

reversible
compressor 

Substituting the enthalpy values from the table shown above and the efficiency values gives
 kg 
m  147  
 s 
A higher flow rate is needed as compared to the reversible process.
84
3.39
The labeling shown in Figure 3.8 will be used for this solution. First, a summary of known
variables is provided.
State
Thermodynamic
State
Sat. Vapor
1
2
3
4
P (MPa)
1.7
0.7
From our knowledge of ideal Rankine cycles, the table can be expanded as follows
State
1
2
3
4
Thermodynamic
State
Sat. Vapor
Sat. Liq. And Vapor
Sat. Liquid
Subcooled Liquid
P (MPa)
1.7
0.7
0.7
1.7
The enthalpy of states 1 and 3 can be found using the NIST website. For states 2 and 4, use the
following expressions to find the enthalpies
hˆ2  (1  x)hˆ2l  xhˆ2v
hˆ4  hˆ3  vˆl P4  P3   hˆ3  vˆ3 P4  P3 
To find x, use the fact that the turbine is isentropic.
sˆ1  sˆ2
sˆ1  (1  x) sˆ2l  xsˆ2v
From the NIST website:
 J 
s1  146.38 
 mol  K 
 J 
s2l  90.45 
 mol  K 
 J 
s2v  150.07 
 mol  K 
(sat. vapor at 1.7 MPa)
(sat. mixture at 0.7 MPa)
By substituting the above values into the entropy relationship, we find
85
x  0.938
Also, from the NIST website:
 J 
h1  36046 
 mol 
 J 
h2l  18416 
 mol 
 J 
h2v  35353 
 mol 
 J 
h3  18416 
 mol 
 3
5 m
v3  6.954 10 

 mol 
(sat. vapor at 1.7 MPa)
(sat. mixture at 0.7 MPa)
(sat. liquid at 0.7 MPa)
Substituting these values into the expression for h4 and h2 yields
 J 
h2  34302.9 

 mol 
 J 
h4  18485.5 
 mol 
Equation 3.82 states
 rankine 
h2  h1  h4  h3 
h1  h4
 J 
 J  
 J 
 J 
 36046 
 18485.5 
 18416 

34302.9 



 mol 
 mol  
 mol 
 mol  
 rankine 
 J 
 J 
 18485.5 
36046 

 mol 
 mol 
 rankine  0.095
This efficiency is significantly lower than a conventional power plant.
86
3.40
(a)
Using the information in the problem statement and our knowledge of ideal vapor compression
cycles, the following table can be created
State
Thermodynamic
P (MPa)
State
1
Sat. Mixture
0.12
2
Sat. Vapor
0.12
3
Superheated Vapor
0.7
4
Sat. Liquid
0.7
Note: Refer to Figure 3.9 to review labeling convention.
Equation 3.85 states
Q C  n h2  h1 
which upon combination with Equation 3.89 gives
Q C  n h2  h4 
From the NIST website:
 kJ 
h2  39.295 
 mol 
 kJ 
h4  24.181 

 mol 
(sat. vapor at 0.12 MPa)
(sat. liquid at 0.7 MPa)
Therefore,

 mol  
 kJ 
 kJ  

Q C   0.5 
39
.
295
24
.
181




 mol 
 mol    7.557 kW 
 s  


(b)
The power input to the compressor can be calculated with Equation 3.86:
WC  n h3  h2 
Equation 3.87 states
s 2  s3
87
From the NIST website:
 J 
s3  s2 sat. vapor at 0.12 MPa   177.89 
 mol  K 
Now, state 3 is constrained.
 kJ 
h3  43.031 
 mol 

 J 
 0.7 MPa, s3  177.89 
 
mol

K



Therefore,

 mol  
 kJ 
 kJ  
 39.295 
WC   0.5 
 43.031 
  1.87 kW 


 s  
 mol 
 mol  

(c)
Equation 3.90 states:
 kJ 
 kJ 

24
.
181
39.295 
 mol 
h h
h h
 mol 
COP  2 1  2 4 
 4.05
h3  h2 h3  h2
 kJ 
 kJ 
 39.295 
43.031 
 mol 
 mol 
88
3.41
Using the information in the problem statement and our knowledge of ideal vapor compression
cycles, the following table can be created
State
Thermodynamic
P (MPa)
State
1
Sat. Mixture
0.12
2
Sat. Vapor
0.12
3
Superheated Vapor
0.7
4
Sat. Liquid
0.7
Note: Refer to Figure 3.9 to review labeling convention.
From Equation 3.90
COP 
Q C h2  h1

W C h3  h2
The enthalpy of state 2 can be found directly from the NIST website, but the enthalpies of states
1 and 3 require the use of additional information.
 kJ 
h2  39.295 
 mol 
(sat. vapor at 0.12 MPa)
For the process between state 2 and state 3,
s 2  s3
From the NIST website:
 J 
s3  s 2  177.89 
 mol  K 
(sat. vapor at 0.12 MPa)
Now, state 3 is constrained.
 kJ 
h3  43.031 
 mol 

 J 
 0.7 MPa, s3  177.89 
 
 mol  K  

The process between state 4 and state 1 is also isentropic.
 J 
s1  s 4  115.09 
 mol  K 
(sat. liquid at 0.7 MPa)
The quality of the R134a can be calculated as follows
89
s1  1  x s1l  xs1v
where
 J 
s1l  90.649 
 mol  K 
 J 
s1v  177.89 

 mol  K 
(sat. R134a(l) at 0.12 MPa)
(sat. R134a(v) at 0.12 MPa)
Therefore,

 J 
 J  
 J 
 1  x  90.649 
115.09 
  x177.89 



 mol  K 
 mol  K   
 mol  K  

 x  0.280
Using the quality of R134a, the enthalpy of state 1 can be calculated as follows
h1  1  x h1l  xh1v
From the NIST website:
 kJ 
h1l  17.412 
 mol 
 kJ 
h1v  39.295 

 mol 
(sat. R134a(l) at 0.12 MPa)
(sat. R134a(v) at 0.12 MPa)
Therefore,


 kJ  
 kJ  
h1  1  0.2817.412 
  0.28 39.295 


 mol  
 mol  


 kJ 
h1  20.74 
 mol 
Now, everything needed to calculate COP is available. Using Equation 3.90,
 kJ 
 kJ 
39.295 
 20.74 

h h
 mol 
 mol   4.97
COP  2 1 
h3  h2
 kJ 
 kJ 
43.031 
 39.295 

 mol 
 mol 
90
Is this modification practical?
No. An isentropic turbine adds significant level of complexity to the cycle. Turbines are
expensive and wear over time. Furthermore, the real turbine added to the cycle will not be 100%
efficient, so the COP will not increase as much. The cost of the turbine is not justified by the
increase in COP.
91
3.42
The subscript “h” refers to the hotter cycle, while “c” refers to the cooler cycle. The
(a)
The flow rate of the cooler cycle can be found by performing an energy balance on the
condenser/evaporator shared between the two cycles. An energy balance shows:
Q H , c  Q C , h  Q
Written in terms of enthalpy, the equation is
n c h8  h7    n h h2  h1 
We can find the enthalpies for positions 2 and 8 directly from the thermodynamic tables because
the fluid is saturated in at these positions. To find the other enthalpies we must use the following
relationships:
s 6  s7
h1  h4
Using the NIST website:
 J 
h1  24181 
 mol 
 J 
h2  40967 
 mol 
 J 
h7  41510 
 mol 
 J 
h8  21099 
 mol 
(sat. R134 liquid at 0.7 MPa)
(sat. R134 vapor at 0.35 MPa)
 J 
)
(R134 vapor at 0.35 MPa with s  177.89 
 mol  K 
(sat. R134 liquid at 0.35 MPa)
Now the flow rate can be calculated:
n c
nc

 J 
 J 
 0.5 mol/s 40967 
 24181 


 mol 
 mol  



 J 
 J 
 41510 
 21099 


 mol 
 mol  

 mol 
 0.411 
 s 
92
(b)
Performing an energy balance we find
Q C  n c h6  h5 
We can also use the following relationship
 J 
h5  h8  21099 
 mol 
From the NIST website:
 J 
h6  39295 
 mol 
(sat. R134 vapor at 0.12 MPa)
Therefore,

 mol  
 J 
 J 
Q C   0.411 
 21099 
 39295 
  7.48 kW 


 s  
 mol 
 mol  

(c)
The power input is calculated as follows:
WC , total  WC , c  WC , h
where
WC , c  nc h7  h6 
WC , h  n h h3  h2 
We have all of the required enthalpies except h3. State 3 is constrained because
 J 
s3  s 2  175.95
 mol  K 
P3  0.7 MPa
(sat. R134a vapor at 0.35 MPa)
From the NIST website:
 J 
h3  42428 
 mol 
 J 
)
(R134a vapor at 0.7 MPa with s  175.95
 mol  K 
Now compute the power for each unit:
93


 J 
 J 
 J 
 J 
 39295 
 40967 
WC , total  0.411 mol/s 41510 
  0.5 mol/s 42428 




 mol 
 mol  
 mol 
 mol  


W C , total  1.64 kW 
(d)
The coefficient of performance is calculated using the following equation:
COP 
Q C
WC , total

7.48 kW
 4.56
1.64 kW
(e)
The COP for the cascade is 4.56, while the COP is 4.05 in Problem 3.40. The cascade system’s
COP is 12.6% greater.
94
3.43
One design follows; your design may differ:
In order to cool a system to -5 ºC, the temperature of the fluid must be colder than -5 ºC so that
heat transfer will occur. We arbitrarily specify that the working fluid evaporates at -10 ºC.
Similarly, in order to eject heat to the 20 ºC reservoir, the fluid must condense at a temperature
greater than 20 ºC. Arbitrarily, we choose 25 ºC. One possible refrigeration cycle is presented
below.
Ý
valve
3
T
1
QH
evaporator
refrigerated
unit at -5 C
4
4
QC
20 C
reservoir
QH
2
1
3
condenser
Wc
QC
2
s
compressor
A number of fluids will work sufficiently for this system, but the design process will be
illustrated using R134a. For states 3 and 4, the pressure is constant, and for states 1 and 2, the
pressure is constant at a different value. From the NIST website, we find
P1  P2  0.201 MPa
(Tsat = -10 ºC)
P3  P4  0.665 MPa
(Tsat = 25 ºC)
(Note: The temperatures of each state are not constant. The listed saturation temperatures
are the temperatures at which the fluid evaporates and condenses.)
Now that the pressures are known, we can compute the required flow rate required in order to
provide 20 kW of cooling.
Q C  n h2  h1 
We can get h2 from the thermodynamic tables for saturated R134a. In order to find h1, we can
use the following relationship:
h1  h4
From the NIST website:
 J 
h2  40064 
 mol 
(sat. R134 vapor at 0.201 MPa)
95
 J 
h4  23931 
 mol 
(sat. R134 liquid at 0.665 MPa)
Now, calculate the required flow rate of R134a.
Q C
20,000 J/s

h2  h1  
 J 
 J 
 23931 
 40064 


 mol 
 mol  

 mol 
n  1.24 
 s 
n 
96
3.44
The schematic of the process and the corresponding Ts is presented below
T = 5 oC
valve
Fridge
2
valve
6
T
QC,R
1
QH
1
evaporator
Wc
6
3
5 oC
T = -15 oC Freezer evaporator
QC,F
4
QH
5
high T
reservoir
-15 oC
QC,R
2
3
QC,F
4
5
condenser
Wc
s
compressor
A number of refrigerants will work for this system, but the design process will be illustrated for
R134a only. To define each state, we need to thermodynamically constrain each state. You
should note that the problem doesn’t state what temperature the fluid condenses at. Therefore,
we can assume the temperature is 25 ºC. By using information in the Ts diagram and from the
NIST website, we find
P1  P2  0.34966 MPa
P3  P4  0.16394 MPa
P5  P6  0.66538 MPa
(T1=T2=Tsat = 5 ºC)
(T3=T4=Tsat = -15 ºC)
(T6 = 25 ºC)
Before solving for additional pressures and temperatures, we will list the known temperatures
and pressures.
State
1
2
3
4
5
6
Temperature (ºC)
5
5
-15
-15
25
25
Pressure (MPa)
0.24334
0.24334
0.16394
0.16394
0.66538
0.66538
States 4, 5, and 6 are completely constrained as confirmed by Gibbs phase rule. Now, we need
to find the liquid and vapor compositions of states 1, 2, and 3 to completely constrain the states.
Since the heat duties are equal for the refrigerator and the freezer, we have the following
relationship:
Q C , F  Q C , R
97
which upon performing an energy balance becomes
h4  h3  h2  h1
Energy balances around the valves provide:
h1  h6
h2  h3
Substituting these relationships into the expression that equates the heat loads results in
h  h6
h2  h3  4
2
From the NIST website:
 J 
h4  39755 
 mol 
 J 
h6  23931 
 mol 
Now constrain states 1, 2, and 3 by determining the enthalpies:
 J 
h1  h6  23931 
 mol 
 J 
 J 
 23931 
39755 

 mol 
 mol   31843  J 
h2  h3 
 mol 
2
Technically, the states are now all constrained, but we would like also like to know the vapor and
liquid compositions of each state. The compositions of states 4, 5, and 6 are already known; we
can calculate the compositions of states 1, 2, and 3 as follows:
sat
h1  1  x1 h1sat
, l  x1h1, v
sat
h2  1  x2 h2sat
, l  x2 h2, v
sat
h3  1  x3 h3sat
, l  x3 h3, v
From the NIST website,
98
 J 
h1sat
, l  21095  mol 


 J 
h2sat
, l  21095  mol 


 J 
h3sat
, l  18380  mol 


 J 
h1sat
, v  40965  mol 


 J 
h2sat
, v  40965  mol 


 J 
h3sat
, v  39755 
 mol 
Now, we can solve the composition of vapor for each state:
x1  0.143
x2  0.541
x3  0.630
The following table presents a summary of our results:
State
Temperature
(ºC)
Pressure
(MPa)
1
5
0.24334
2
5
0.24334
3
-15
0.16394
4
-15
0.16394
5
25
6
25
Phases
Saturated Liquid
and Vapor
Saturated Liquid
and Vapor
Saturated Liquid
and Vapor
Liquid
Vapor
Composition Composition
0.857
0.143
0.459
0.541
0.37
0.630
Saturated Vapor
0
1
0.66538
Superheated
Vapor
0
1
0.66538
Saturated Liquid
1
0
99
3.45
In this case, the working material is a solid. The four states of the magnetic material are shown
of the sT diagram below. Note the axis are shifted from the usual manner.
3
2
QH
QC
1
4
(a) The heat expelled by the cold reservoir can be approximated by:
qC  TC s 2  s1 
where T is the average temperature between states 1 and 2, which is approximately 1 K. Taking
values of S/R from the sT diagram, we get:
 J 
qC  R1.9  1.3  5.0
 mol  K 
(b) Similarly, the heat absorbed by the hot reservoir can be approximated by:
 J 
q H  TH s 4  s 4   8.8R1.2  1.9  51
 mol  K 
(b) The coefficient of performance is given by:
COP 
qC
wC ,total

qC
 0.11
q H  qC
(d) The value of COP is much lower than a typical refrigeration process (COP= 4-6); in general,
refrigeration processes at these low temperatures are much less efficient.
F. Work is supplied to magnetize the material and to spin the wheel.
100
3.46
When the polymer is unstretched it is in a more entangled state. When stretched the polymer
chains tend to align. The alignment decreases the spatial configurations the polymer can have,
and therefore, reduces that component of entropy. If the process is adiabatic, the entropy of the
system cannot decrease. Consequently, its thermal entropy must increase. The only way this can
be accomplished is by increased temperature.
101
3.47
Assume the temperature is 298 K. The following data was taken from Table A.3.2:
Species
h ºf (kJ/mol)
g ºf (kJ/mol)
Cu2O (s)
O2 (g)
CuO (s)
-170.71
0
-156.06
-147.88
0
-128.29
The data listed above were used to create the next table using
s ºf 
h ºf  g ºf
T
Species
s ºf (kJ/mol K)
Cu2O (s)
O2 (g)
CuO (s)
-0.0766
0
-0.0932
Now the change in entropy of the reaction can be calculated in a method analogous to Equation
2.72
 

 kJ   
 kJ  
º
s rxn
  vi s ºf  2 - 0.0766 
  4 - 0.0932 


i
 mol  K   
 mol  K  

 kJ 
º
s rxn
 0.22 
 mol 
Does this violate the second law of thermodynamics? This problem shows that the entropy
change of the system is negative, but nothing has been said about the entropy change of the
universe. We must look at the change in entropy of the surroundings to determine if the second
law is violated. By looking at the enthalpies, we see that the reaction is exothermic, which
means that heat is transferred from the system to the surroundings. Therefore, the entropy of the
surroundings will increase during this reaction.
102
3.48
The temperature and pressure terms in the equation for entropy do not contribute anything to the
entropy change because they are constant. Therefore, the only remaining entropy contribution,
the randomness of the atomic arrangements, must be considered. The randomness does not
increase when CdTe forms. In pure crystals of Cd and Te, the location of each atom is known
because the crystal lattice constrains the atomic locations. In CdTe, the crystal lattice still
defines the location of each atom, so the randomness has not increased. Therefore, the change in
entropy is zero.
3.49
This argument is not scientifically sound. Morris is arguing that since evolution results in more
order, the second law of thermodynamics is violated, so evolution must be impossible. However,
the flaw in this argument is caused by ignoring the entropy change of the entire universe. The
second law states that the entropy of the universe will remain constant or increase for any
process. Morris’ argument was based on the entropy of the system undergoing evolution – not
the entropy of the entire universe. A system can decrease in entropy if the entropy of the
surroundings increases by at least that much.
103
3.50
Entropy is related to the number of configurations that a state can have. The greater the number
of configurations, the more probable the state is and the greater the entropy. We can
qualitatively relate this concept to the possible hands in a game of poker, but it is more
interesting to quantify the results using some basic concepts of probability.
We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards.. There
are a finite number of permutations in which we can arrange a 52 card deck in 5 cards. For the
first card in the hand, we select from 52 cards, the second card can be any other card so we select
from 51 cards, the third card has 50 cards, and so on. Thus the number of permutations of 5
cards is:
P  52  51 50  49  48  311,875,200
However, we do not care the order in which the cards are dealt, so we must divide this number
by the number of ways we can come up with the same hand. We do this math in a similar way.
For a given hand there are five cards we can pick first, four we can pick second, …. So the
number of ways we can make the same hand is:
N  5  4  3  2  1 120
The number of unique configurations can be found by dividing P by N . Thus, the cards can
display
C
P
 2,598,960
N
or 2,598,960 unique configurations. To find the “entropy” of a given hand, we need to find out
how many of these unique configurations belong to the hand
Consider a four of a kind. There are 13 different possible ranks of four of a kind, one for each
number A, 2, 3, 4, 5, 6, 7,8, 9, 10, J, Q, K. The fifth card in the hand could be any of the other
48 cards. Therefore, the number of combinations of four of a kind is: 13 x 48 = 624
The probability of a four of a kind is: 624 / 2598960 = 1 / 4165 which is very unlikely. Thus the
“entropy” of this hand is low. In contrast, there are 1,098,240 to have a hand that has one pair;
therefore the probability of getting this hand is much greater, 1/2.4, and its “entropy” is high.
The probability of having nothing is 1/2 which represents the most likely hand in poker, or the
hand with the highest “entropy”. In fact, the rules of poker are defined so the hand of lower
“entropy” always beats the hand of higher “entropy.”
104
Chapter 4 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
1
4.1.
(a)
Yes, the form of the equation is reasonable. This can be rewritten:
PA 
RT
a
 n
v A v A
It is equivalent to Pv  RT , except the pressure term has been corrected to account for the ions’
intermolecular forces. The coulombic forces between the gas molecules affect the system
pressure. This modification is similar to the van der Waals equation. Since we are limited to 1
parameter, we need to choose the most important interaction. Since net electric point charges
exert very strong forces, this effect will be more important than size.
(b)
The sign should be negative for a because the positively charged gas molecules repel each other
due to coulombic forces. Therefore, the system pressure increases, i.e.,:
PA  Pideal
Coulombic forces are much stronger than van der Waals interactions, so a will be large – much
larger than a values for the van der Waals EOS.
(c)
Coulombic repulsion is the primary intermolecular force present in the gas. Coulombic potential
energy is proportional to r-1. v is proportional to r3, so the coulombic potential goes as v-1/3.
Therefore,
n
1
3
a must have must have the following units to maintain dimensional homogeneity with the
pressure term:
 a  N  J 
 v1/ 3    m 2    m 3 
so
a  
J

1/3 
 m  mol 
2
2
4.2
The attractive interactions are described by van der Waals forces. For both O2 and propane, the
dipole moments are zero. Therefore, the expression for the interactions reduce to

3  i j
2 r6
 Ii I j

 Ii  I j





From Table 4.1:
I a  12.07 eV   1.933  10 11 erg 
I b  10.94 eV  1.753  10 11 erg
Now obtain the requested expressions:
   1.933 10

3 16  10  25 cm 3
aa  
2
r6
aa 
 3.71  10 59
r
ab  
ab 
6
cm
6
2
r
6
cm
6

   1.753 10

bb 
   1.933 10
 erg
3 62.9  10  25 cm 3
bb  
2
r6
 5.20  10 58
r
6
cm
6


 erg
3 16  10  25 cm 3 62.9  10  25 cm 3
2
r6
 1.39  10 58

erg 1.933  10 11 erg 

11
erg  1.933  10 11 erg 
1.933  10
 

11
2
 erg
11
11


erg 1.753  10 11 erg 

 1.933  10 11 erg  1.753  10 11 erg 


erg 1.753  10 11 erg 

1.753  10 11 erg  1.753  10 11 erg 

(b)
Calculation:





  5.20  10  58
  3.71  10  59
6

cm  erg 
cm 6  erg 
aa bb 



r6
r6



3
aa bb 
 1.39  10 58
cm
6

 erg
r
Note: Disregarded the positive value.
The value of
6
aa bb is equal to ab . The values are equal because the ionization energies are
similar.
(c)
An expression for the average intermolecular attraction in the mixture can be found using the
mixing rules
mix  y a2 aa  2 y a yb ab  yb2 bb
 3.71  10 59 y a2  1.39  10 58 y a yb  5.20  10 58 yb2
mix 
cm 6  erg
6
r


4

4.3
(a)
300 K, 10 atm. The intermolecular distance of molecules is greater at lower pressures.
Therefore, fewer intermolecular interactions exist, which cause less deviation from ideality.
(b)
1000 K, 20 atm. At higher temperatures, the kinetic energy of the molecules (speed) is greater.
The molecules interact less; thus, the compressibility factor is closer to unity.
(c)
Let subscript “1” denote BClH2 and “2” denote H2. For the mixture, we calculate the
compressibility factor as follows
B
z  1  mix
v
where
Bmix  y12 B1  2 y1 y2 B12  y 22 B2
Since BClH2 is polar and H2 is non-polar and small
B1  Bmix  B2
Therefore, the plot may look like the following.
quadradic, not
linear
zH2 close to 1
1
z
0.9
zBClH2 lowest
0.8
0.7
0.6
0
0.2
0.4
0.6
yBClH2
5
0.8
1
4.4
(a)
The intermolecular attractions and volume occupied by the styrene monomers will contribute to
the deviations from ideality. Since styrene monomers are essentially non-polar, the order of
importance is as follows
dispersion  dipole - dipole  induction
The van der Waals EOS is appropriate since it accounts for the occupied molar volume and
intermolecular forces, but it should be noted that more modern EOSs will give more accurate
results.
(b)
Use critical data to calculate the a and b parameters:
2


J 
  8.314 
647.15 K 

2

3
 mol  K  
27 RTc 
27  
  3.13  J  m 

a


2
64 Pc
64
39  105 Pa 
 mol 
b
RTc

8 Pc

 J 
647.15 K 
 8.314 
 mol  K  



8 39  105 Pa 
 m3 
 1.72  10  4 

 mol 
Now we can use the van der Waals EOS to solve for temperature.
P
RT
a

v  b v2

 J 
 8.314 
T
 mol  K  

5
10  10 Pa  
3
4  m 
3.0  10 
 1.72  10 4

 mol 
T  550.8 K  277.65 º C
 m3 
 mol 



 J  m3 
3.13 
2 
 mol 

 3.0  10 4


 m3  
 mol  


2
The styrene will not decompose at this temperature.
(c)
The a parameter is related to attractive intermolecular forces. Dispersion is the controlling
intermolecular force in this system, and its magnitude is directly related to the size of the
molecules (polarizability component of dispersion). For the 5-monomer long polymer chain, a is
5 times the a value in Part (b). The b parameter is also related to the size of the molecule since it
accounts for the volume occupied by the molecules. Again, the b parameter for the reduced
polymer chain is 5 times the b value from Part (b).
6
 J  m3 
a  15.65 
2 
 mol 
 m3 
b  8.6  10  4 

 mol 
(d)
We must realize that if we initially believed there were 100 moles of styrene in the reactor, then
there can only be 20 moles of the 5-monomer long polymer chain. Therefore,
 m3 
v  1.5  10  3 

 mol 
and
 J  m3 

 J 
15
.
65
 8.314 
T
 mol 2 
 mol  K  



5

10  10 Pa  
2
3
3
3




m
m


4
3  m 


1.5  10 3 
8
.
6
10

 mol  1.5  10 
 
 mol 

 
 mol  
T  612.41K  339.26 º C
Decomposition will occur.
7
4.5
There are many ways to solve this problem, and the level of complexity varies for each method.
To illustrate the principles in Chapter 4, two of the simplest solution methods will be illustrated.
Method 1. Polarizability of each atom
The polarizability of a molecule scales with the number of atoms; the polarizabilities of
individual atoms are additive. Using the first two molecules, solution of the following system of
equations
1 C  4 H  1 CH 4
2 C  6 H  1 C2 H 6
gives
 
cm 
 C  11.4  10 25 cm 3
 H  3.65  10  25
3
Now, the polarizability of the chlorine atom can be found. For chloroform,
1 C  4 Cl  1 CCl 4
Therefore,
 
 Cl  23.4  10 25 cm 3
The values of C3 H 8 , CH 3Cl , CH 2Cl2 , and CHCl3 are calculated with the polarizabilities found
above as follows, and compared to the values given.
Species
C3H8
CH3Cl
CH2Cl2
CHCl3
 calculated
(x 1025 cm3)
63.4
45.8
65.5
85.3
 reported
(x 1025 cm3)
62.9
45.6
64.8
82.3
% Difference
0.8
0.3
1.1
3.6
All agree reasonably well. Now, the polarizabilities of C 4 H10 and C 2 H 5Cl will be calculated.
C
4
H 10

  
 
 
 4 11.4  10  25 cm 3  10 3.65  10  25 cm 3  82.1  10  25 cm 3
8
C
2
H 5 Cl
  

  
 
 
 2 11.4  10  25 cm 3  5 3.65  10  25 cm 3  1 23.4  10  25 cm 3  64.45  10  25 cm 3
Method 2. Bond Polarizability
For this method, we calculate the molecules’ polarizabilities by adding the polarizability of each
bond, instead of the atoms. For the methane molecule
4 C  H  1 CH 4
 
 
26  10  25 cm 3
 6.5  10  25 cm 3
4
 C  H 
To calculate the polarizability of a C-C bond, use ethane as follows:
6 C  H   C  C  1 C 2 H 6
  
 
 
  C  C  44.7  10  25 cm 3  6 6.5  10  25 cm 3  5.7  10  25 cm 3
The C-C and C-H polarizability calculated above predict the given polarizability of propane well.
The polarizability of C-Cl bonds is calculable with the polarizability of chloroform.
4 C  Cl  1 CCl 4
  C  Cl 
 
 
105  10  25 cm 3
 26.25  10  25 cm 3
4
The values of C3 H 8 , CH 3Cl , CH 2Cl2 , and CHCl3 are calculated with the polarizabilities found
above as follows, and compared to the values given.
Species
C3H8
CH3Cl
CH2Cl2
CHCl3
 calculated
(x 1025 cm3)
63.4
45.8
65.5
85.3
 reported
(x 1025 cm3)
62.9
45.6
64.8
82.3
% Difference
0.8
0.3
1.1
3.6
All agree reasonably well. This value predicts the polarizabilities of the other species in the table
reasonably well. Now, the polarizabilities of C 4 H10 and C 2 H 5Cl will be calculated.
1 C 4 H 10  3 C  C  10 C  H
C
4
H 10

  
 
 
 3 5.7  10  25 cm 3  10 6.5  10  25 cm 3  82.1  10  25 cm 3
9
1 C 2 H 5 Cl  1 C  C  5 C  H  1 C  Cl
C
2
H 5 Cl

  
  
 
 
 1 5.7  10  25 cm 3  5 6.5  10  25 cm 3  1 26.25  10  25 cm 3  64.45  10  25 cm 3
Note both the atom method and the bond method yield identical results. More accurate values
for the polarizabilities can be calculated using more of the data given in the problem.
10
4.6
(a)
 increases with molecular size. Therefore,
 I  S  O
2
2
2
 scales with magnitude of van der Waals interactions. Because these are non-polar, diatomic
molecules, only dispersion forces are present. Dispersion forces depend on the first ionization
potential and polarizability. Ionization energy is approximately equal for each molecule. The
polarizability scales with molecular size, so
 I   S  O
2
2
2
(b)
 increases with molecular size. Diethylether and n-butanol have the same atomic formula and
similar spatial conformations. Therefore, they should be about equal in size. Methyl ethyl
ketone has fewer atoms, but has two exposed electron pairs on the double-bonded oxygen. The
size of the molecular electron orbital of methyl ethyl ketone is approximately equal to the sizes
of diethyl ether and n-butanol, so
 n - butanol   diethyl ether   methyl ethyl ketone
Methyl ethyl ketone and n-butanol are much more polar than diethyl ether due to their greater
asymmetry, so their  values are greater than diethyl ether’s. Now we must determine if there is
greater attraction in n-butanol or methyl ethyl ketone.  scales with the magnitude of van der
Waals interactions. Since induction and dispersion forces are similar in these molecules, we
must consider the strength of dipole-dipole forces. There is greater charge separation in the
double bond of ketone, so
 methyl ethyl ketone   n - butanol   diethyl ether
11
4.7
(a)
At 30 bar, the water molecules are in closer proximity than they are at 20 bar. Intermolecular
attractions are greater, so the magnitude of molecular potential energy is greater. The potential
energy has a negative value for attractive interactions. The molecular kinetic energy is identical
since the temperature is the same. Hence, the internal energy, the sum of kinetic and potential
energies, is less at 30 bar.
(b)
The key to this phenomenon is hydrogen bonding. At 300 K and 30 bar, isopropanol and npentane are both liquids. The hydrogen bonding and dipole-dipole interactions are present in
isopropanol, and dispersion is present in n-pentane. The intermolecular forces are greater in the
isopropanol, so the compressibility factor is smaller for isopropanol.
At 500 K and 30 bar, both species are gases. In the gas phase, hydrogen bonding does not play a
significant role. The dispersion forces in n-pentane are stronger than the dipole-dipole forces of
isopropanol. Therefore, the compressibility factor is smaller for n-pentane.
12
4.8
(a)
Ideal: For ideal NH3, the compressibility factor is equal to one. For real NH3, the strong
intermolecular forces (dipole-dipole and dispersion) cause the molar volume to decrease. They
outweigh the volume displaced by the physical size of NH3; thus, z will be less than one.
(b)
Internal energy will be greater for the ideal gas. In the real gas, intermolecular attractions are
present. Internal energy value is the sum of potential and kinetic energies of the molecules. The
absolute values of the kinetic energies are identical at identical temperature: however, the
potential energy decreases for real NH3 due to attractive interactions - so the internal energy is
less for the real NH3.
(c)
The entropy will be greater for the ideal gas. Entropy is a measure of possible molecular
configurations or “randomness”. Ammonia has an electric dipole in which positive and negative
charge are separated. The intermolecular forces in the real gas cause the molecules to align so
that the positive charge in one molecule is adjacent to a negative charge in a neighboring
molecule to reduce potential energy. Therefore, fewer possible configurations exist, which
creates less randomness and lower entropy.
13
4.9
(a)
We can determine which case has the higher compressibility factor by comparing the molar
volumes at constant T and P. With Ne, very weak intermolecular attractions are present, so
volume displacement becomes important. The compressibility factor will be slightly greater than
one. In NH3, the strong intermolecular attractions decrease the molar volume, so z is less than
one. The compressibility factor is greater for Ne.
(b)
Entropy is a measure of randomness. Both species are gases at these conditions. The
intermolecular attractions present in NH3 reduce the number of possible configurations. The
weak forces present in Ne have a much smaller effect. However, NH3 is asymmetrical, while Ne
is symmetrical. The asymmetry of NH3 results in more possible configurations that NH3 can
have. Therefore, it is difficult to qualitatively show for which case the entropy is greater. Since
both species are gases, intermolecular interactions are relatively weak, and we can guess that
entropy is greater in NH3.
14
4.10
(a)
To find the average distance between the two atoms of Ar, we can find the volume that each
atom occupies. The molar volume can be found from the compressibility factor. At 300 K and
25 bar,
Tr  1.99
Pr  0.513
From the generalized compressibility charts:
z  0.9859
Therefore,
v  0.9859

 J 
 8.314 
300 K 
 mol  K  

25  105 Pa 
 9.84  10
4
 m3 


 mol 
 m3 
v  1.63 10  27 

 atom 
The second number was found by dividing the molar volume by Avogadro's number. To
estimate the distance between each atom, we note that the distance between molecules can be
related to the volume by:
r3  v
Consider the geometry shown below:
3
v
r
Ar
Ar
A rough estimate of r is

r  1.63  10  27 m 3
1 / 3  1.18 109 m
15
r  11.8 A 
(b)
The potential energy due to gravity can be calculated as follows
G 
 G  m 2Ar
r

 m2  
where G is the gravitation constant  G  6.67  10 11 
  and mAr is the mass of an argon
2 

 kg  s  

atom. The mass of an argon atom is calculated as follows
 1 kg  
 39.948 g Ar 
1 mol
 kg 

  6.633  10 - 26 
m Ar  


23


 atom 
 1 mol Ar   6.023  10 atoms  1000 g  
Using the distance calculated in Part (a), we get
G  2.49 10 -52 J 
(c)
Equation 4.13 quantifies the potential energy due to London interactions
Ar  Ar  
3  Ar Ar
2 r6
 I Ar I Ar

 I Ar  I Ar



From Table 4.1
I Ar  15.76 eV  2.52  10 -4 erg
 
 Ar  16.6  10 25 cm 3


Using the distance calculated in Part (a) r  1.18  10  7 cm , we get
Ar  Ar  1.93  10 10 erg  1.93  10 17 J 
(d)
The potential energy due to London interactions is around 10 35 times greater than the potential
energy due to gravity. Clearly, London interactions are much more important, and gravitational
effects can be neglected.
16
4.11
We want to compare the Lennard-Jones potential to one with an exponential repulsion term. As
provided in the text, Equation 4.19, the Lennard-Jones potential is
  12    6 
  4      
 r  
 r 
To simplify further analysis, we can rewrite the equation in dimensionless quantities:
* 
  1
1 



4  r *12 r *6 
r* 
r
σ
where
We want to compare this to an exponential repulsive function
*
exp


 c2 
1 

 c1 exp  

 r *  r * 12 


 
We have two adjustable parameters, c1 and c2, to match the first (LJ) potential to the second
(exp) potential. We need to choose reasonable criteria to specify. For this solution we choose
equal well depths and equal values at =1. Other choices may be just as valid; you should
realize that you have two parameters to fit and so must specify two features.
Using the above criteria, we iterate on a spreadsheet, to get the solution:
c1 = 143,000 and
c2 = 11.8
This solution is shown at two magnifications in the plots on the following page:
17
Potential functions
0.3
0.2
Lennard-Jones
exponential
0.1
0
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
-0.1
-0.2
-0.3
Potential functions
20
Lennard-Jones
15
exponential
10
5
0
0.7
1
1.3
1.6
1.9
-5
We draw the following conclusions:
1. The most stable configuration (the bottom of the well) occurs at a greater separation for
the exp model.
2. The Lennard-Jones potential increases more steeply at small radii, i.e., it behaves more
like the hard-sphere potential.
3. The two models are in reasonable qualitative agreement
18
4.12
(a)
The bond strength of a sodium ion can be viewed as the amount of energy it would take to
remove the sodium ion from the crystal lattice. The interaction between the chlorine and sodium
ions is Coulombic attraction.



QNa QCl
4.803  10 10 esu  4.803  10 10 esu
6
 5.01  10 11 erg 
r
2.76  10 -8 cm
  31.3 eV 
6


(b)
6
QNa QCl
Q Q
 12 Na Na
r
r
  5.01  10 11 erg  12
4.803  10
  2.089  10 11 erg  13.0 eV
esu 4.803  10 10 esu 
3.90  10 -8 cm
10
(c)
QNa QCl
Q Q
Q Q
 12 Na Na  8 Na Cl
r
r
r
4.803  10 10 esu  4.803  10 10 esu
  2.089  10 11 erg  8
4.78  10 -8 cm
  1.77  10 11 erg  11.1 eV
6


(d)
QNa QCl
Q Q
Q Q
Q Q
 12 Na Na  8 Na Cl  6 Na Na
r
r
r
r
10
4.803  10 esu 4.803  10 10 esu
11
  1.77  10 erg   6
5.52  10 -8 cm
  7.37  10 12 erg  4.60 eV
6


19


4.13
The boiling points of the halides depend upon the strength of intermolecular attractions. The
stronger the intermolecular attraction, the higher the boiling point. Dispersion and dipole-dipole
interactions are present in all five species listed. The magnitude of the dipole-dipole interactions
is similar so the pertinent intermolecular force in these molecules is dispersion. The molecular
size increases from left to right. Polarizabilities are greater in larger molecules, which manifests
in larger dispersion forces. Therefore, the boiling point increases from left to right.
20
4.14
van der Waals forces hold the Xe atoms together in a molecule of Xe2. The potential energy can
be quantified with the Lennard-Jones potential function. The bond length is the r value where
the potential is a minimum. From Table 4.2:

k
and
 229 K
  4 .1 A
We start with Equation 4.19: to get
  12    6 
  4      
 r  
 r 
Differentiation with respect to r yields
 12   12 6    6 
d
 4         0
dr
r  r  
 r  r 
where we set this derivative equal to zero to find the minimum. Solving gives
1
 
  
2
r
6
or
r  6 2  1.12  4.60 A
21
4.15
The data in the following table were taken from Table A.1.1
Species
Tc K 
He
CH4
NH3
H2O
5.19
190.6
405.6
647.3
Pc  10 -5 Pa 
2.27
46.00
112.77
220.48
The van der Waals a parameter can be calculated using Equation 4.39.
27 RTc 2
a
64 Pc
The van der Waals b parameter can be calculated using Equation 4.40.
b
RTc
8 Pc
Using the data table and equations listed above, the following table was created; the values of
dipole moment and polarizability reported in Table 4.1 are also included.
Species
He
CH4
NH3
H2O
 J  m3 
a

2
 mol 
0.00346
0.230
0.425
0.554
 m3 
b  105 

 mol 
2.38
4.31
3.74
3.05

[D]

[cm x 1025]
0
0
1.47
1.85
2.1
26
22.2
14.8
3
The values of a for helium is two orders of magnitude less than the other species since it only has
weak dispersion forces (small atom, small The values of a for methane, ammonia, and water
are of the same magnitude because the sums of the intermolecular attractions for each molecule
are similar. All three molecules have comparable dispersion forces; although slightly weaker in
the ammonia and water. However, unlike methane, these two molecules also have dipole-dipole
and induction forces. In fact, the strong dipole in water gives it the largest value
The values of b are all of the same magnitude, as expected since b scales with size according to
the number of atoms in the molecule.
Size of Molecules:
 Value of b:
CH 4  NH 3  H 2O  He
bCH 4  bNH 3  bH 2O  bHe
22
4.16
The van der Waals b parameter can be calculated using Equation 4.40.
b
RTc
8 Pc
Critical point data can be found in Table A.1.1. The following table was made:
Tc K 
Species
Pc  10 -5 Pa 
CH4
C6H6
CH3OH
190.6
46.00
562.1
48.94
512.6
80.96
 J 
Note: Used R  8.314 
for calculation of a.
 mol  K 
 m3 
b  10 5 

 mol 
4.306
11.94
6.58
The equation from page 186 can be rewritten to calculate  .
 3
3b
2N A
The table listed below was created using this equation, and data from Table 4.2 are included
along with the percent difference.
Species
  1010 m
CH4
C6H6
CH3OH
3.24
4.56
3.74
Table 4.2 Value
  1010 m
3.8
5.27
3.6
23
Percent
Difference
14.7
13.5
3.8
4.17
The van der Waals a parameter can be calculated using Equation 4.39. The values for the above
equation were taken from Table A.1.1, and the following table was made:
Tc K 
Species
 J  m3 
a

2
 mol 
0.2303
1.883
0.9464
Pc  10 -5 Pa 
CH4
C6H6
CH3OH
190.6
46.00
562.1
48.94
512.6
80.96
 J 
Note: Used R  8.314 
for calculation of a.
 mol  K 
The equation from page 187 can be used to find C6.
a
2N a2C6
3 3
3a 3
 C6 
2N a2
The  values can be found in Table 4.2. The following table can now be created
Species
 1010 m
CH4
C6H6
CH3OH
3.8
5.27
3.6

C 6  10 77 J  m 6
1.66
36.3
5.82

To compare the values obtained from Equation 4.13, first calculate the potential energy with
Equation 4.13:

 3  i j
2 r6
 Ii I j

 Ii  I j





 II
3
 i j  i j
2
 Ii  I j




Therefore
C6 
We obtain the following values using this equation, and the corresponding percent differences
were calculated.
24
Species
CH4
C6H6
CH3OH


C 6  10 77 J  m 6
Page 187
Equation 4.13
1.66
1.021
36.3
12.0
5.82
1.36
Percent Difference
63
202
328
The values for the van der Waals a constant have the correct qualitative trends and order of
magnitude; however, those values predicted from basic potential theory vary significantly from
corresponding states. The basic potential result presented in the text assumes that the species are
evenly distributed throughout the volume. It does not take into account the structure given to the
fluid through intermolecular forces. In fact, a more careful development includes a radial
distribution function, which describes how the molecular density of the fluid varies with r. The
radial distribution function depends on pressure and temperature of the fluid.
25
4.18
P
RT
a

v  b v2
Redlich-Kwong
P
RT
a
 1/2
v  b T v(v  b)
Peng-Robinson:
P
RT
a(T )

v  b v(v  b)  b(v  b)
van der Waals:
As you may have discovered, these equations are largely empirical with no theoretical
justification. They simply represent experimental data better. We can use our knowledge of
intermolecular forces, however, to explain why these may work better.
If we compare these equations to the van der Waals equation, we note that the first term
on the right hand side of all three equation is identical; we accounted for this term as a correction
for finite molecular size (or alternatively repulsive interaction due to the Pauli Exclusion
Principle). This form represents a hard sphere model.
The second term, that which deals with intermolecular attraction, is different in all three
models. Both of the later equations include a temperature dependence in this term. We have
seen that if attractive forces depend on orientation (dipole-dipole), they fall off with T as a result
of the averaging process (recall discussion of Equation 4.11).
Another explanation goes as follows: as T increases, the molecules move faster, reducing
the effect of intermolecular forces. If we say that the potential energy between two molecules
depends on the amount of time that they spend close to each other, then it would be inversely
related to velocity (The faster molecules are moving, the less time they spend in the vicinity of
other species). In this case, the correction term would go as V-1, where V is the molecular
velocity. If we relate molecular velocity to temperature
2
1
2 mV
 32 kT
Therefore the correction term goes as T-1/2, as shown in the Redlich-Kwong equation.
The inclusion of a "b" term in the second term may help relax van der Waal's "hard
sphere" model with a more realistic potential function, i.e., something closer to a Lennard Jones
potential (Figure 4.8) than the Sutherland potential (Figure 4.7) upon which the van der Waals
equation is based. It makes sense that this should be included in the force correction since this is
taking into account repulsive forces. One example of a more detailed explanation follows:
If we look at the Redlich-Kwong equation, it says that if we have 2 species with the same
attractive strength (same a -> same magnitude of van der Waals forces), the larger species will
have less of an effect on P. The following sketch illustrates how 2 species could have the same
van der Waals attractive forces:
26
- +
London
= London
+
Dipole
Species 2
larger and non-polar
(larger b, same a)
Species 1
smaller and polar
(smaller b, same a)
In the case above, when the two species are the same distance apart, they have the same
attractive force; however, the smaller species (1) can get closer before its electron cloud
overlaps. Thus it has more opportunity for attractive interactions than the larger species. The
Peng-Robinson equation exhibits the most complicated form in an attempt to better fit
experimental data.
27
4.19
(a)
Work is defined as follows
W    PdV
Substitution of the ideal gas law for P yields
W  
nRT
dV
V

 1 L 
 J 
W  2.0 mol 8.314 
1000 K ln


 mol  K  
 10 L 

W  38.29 kJ
(b)
Instead of substituting the ideal gas law into the definition of work, the Redlich-Kwong equation
is used:
v2
W  n 
v1
RT
a
dv

1
/
2
v  b T v (v  b)
Substituting
 J 
R  8.314 
 mol  K 
T  1000 K 
n  2 mol
 J  K1/2  m 3 
a  14.24 

2
 mol

 3
5 m
b  2.1110 

 mol 
 m3 
v1  0.0005 

 mol 
 m3 
v2  0.005 

 mol 
28
and evaluating the resulting formula gives
W  37.35 kJ
(c)
Energy balance:
u  q  w
Since the process is reversible
q  Ts
and
w  u  Ts
To use the steam tables conveniently, we need the initial and final pressures. We can calculate
these with the Redlich-Kwong EOS:
P1  1.65 MPa
P2  15.6 MPa
From the steam tables:
 kJ 
uˆ1  3522.6  
 kg 
 kJ 
uˆ 2  3462.2  
 kg 
 kJ 
sˆ1  8.101 

 kg  K 
 kJ 
sˆ2  7.00 

 kg  K 
Therefore,


 kJ  
 kJ 
 kJ  
 kJ 
ŵ   3462.2    3522.6     1000 K  7.00 
 8.101 

 
kg
kg
kg

K
kg

K











 kJ 
ŵ  1040  
 kg 
W  37.5 kJ 
The answers from the three parts agree very well. Part (a) is not as accurate as Part (b) and Part
(c) because water is not an ideal vapor. The value from Part (b) is 1.1 % smaller than the value
from Part (c). Clearly, the Redlich-Kwong EOS or steam tables are appropriate for this
calculation.
29
4.20
First, we can obtain an expression for the compressibility factor with the van der Waals equation.
1
a
Pv
v
a




b RTv
RT v  b RTv
1
v
To put this equation in virial form, we can utilize a series expansion:
1
 1  x  x 2  x 3  ...
1 x
Therefore,
2
1
b b
 1        ...
b
v v
1  
v
and
a 

b 

2
Pv
RT   b 

 1
    ...
v
RT
v
This expression can also be expanded in pressure. From Equation 4.58, we know that
B' 
C'
B
RT
C  B2
RT 2
where
Pv
 1  B' P  C ' P 2  ....
RT
Substituting B and C found above, we get
B' 
bRT   a
RT 2
and C ' 
2abRT   a 2
RT 4
Therefore,
30
2

 bRT   a 
Pv
 P   2abRT   a  P 2  ....
 1 
 RT 2 

RT
RT 4 



31
4.21
Using the virial expansion, the pressure can be written as follows

1 B C
P  RT  

 ...

 v v 2 v3
The virial expansion in pressure is
P


RT
1  B' P  C ' P 2  ...
v
If we substitute the first expression into the second, we obtain
2

 1 B C   
1 B C
 RT 
1 B C 

 ... 

 C '  RT  
   
RT  
1  B ' RT  
2
v 
v v 2 v 3 
v
v
v3   
 v v 2 v3






If we combine like terms on the right side and set them equal to terms on the left, we find
B  B' RT
C  BB' RT  C ' RT 2
Substitute the expression for B’ into the expression for C and solve for C’:
C' 
CB
RT 2
and
B' 
B
RT
32
4.22
At the critical point we have:
Pc 
RTc
a

vc  b Tc vc2
(1)
RTc
2a
 P 

  0
2
 v Tc
vc  b  Tc vc3
(2)
 2 
2 RTc
6a
 P

 2  0
3
Tc vc4
 v 

vc  b 
Tc

(3)
If we multiply Equation 2 by 2 and Equation 3 by (v-b) and add them together:
0
4a
vc3

6avc  b 
vc4
this can be solved to give:
vc  3b
If we plug this back into Equation 2 and solve for a, we get:
a
9
vc RTc2
8
Finally, if we plug this back into Equation 1 we can solve for the Berthelot constants in terms of
the critical temperature and the critical pressure:
a
27 R 2Tc2
64 Pc
and
b
RTc
8Pc
33
4.23
Before we can find the reduced form, we need to find expressions for a and b in terms of critical
point data (Problem 4.22). We find
vc  3b
(1)
27 R 2Tc3
a
64 Pc
RTc
b
8 Pc
(2)
(3)
The Berthelot equation is
P
RT
a

v  b Tv 2
First, substitute Equation 1 into the first term on the right-hand side and rearrange to get
RT
a
P b 
3vr  1 Tv 2
Now, substitute Equation 3:
P
8 PcTr
a

3v r  1 Tv 2
Substitute Equation 2:
P
8 PcTr

3v r  1
27 R 2Tc3
64 Pc
Tv 2
From Equation 3:
R 2Tc2  64 Pc2 b 2
Substituting this result, we get
P
8PcTr 27 PcTc b 2

3v r  1
Tv 2
34
Finally, substitute Equation 1:
P
8PcTr 3PcTc vc2

3v r  1
Tv 2
This can be rewritten as
Pr 
8Tr
3

3vr  1 Tr vr2
35
4.24
(a)
The virial equation is:
z
Pv
D
B C
 1   2  3 ...
RT
v v
v
(1)
We can rewrite this equation:
C
 Pv 
 1v  B  ...
RT
v
z  1v  
(2)
PvT data from the steam tables are given in the steam tables and corresponding values of (z-1)v
and 1/v can be calculated. An illustrative plot of (z-1)v vs. 1/v for T = 300 oC is shown below.
Plot to determine the second virial coeff at 300 C
-115
(z -1)v (cm3 / mol)
-120
-125
(z -1)v
-130
2
1.5
1
0.5
0
-135
1/v (mol/l )
We see that at pressures above 1.5 MPa (15 bar) we see that the plot reaches a constant value of
around -118 cm3/mol. We may choose to report this value as B. A more careful examination of
Equation 2 suggests another possibility.
If we plot (z-1)v vs. 1/v, we should get a straight line at low to moderate pressures with an
intercept equal to the second virial coefficient, B, and a slope equal to the third virial coefficient,
C. At very low pressures, the curve is indeed linear and gives an intercept of -140 cm3/mol, as
shown on the next page. The slope of this region would yield the third virial coefficient, C.
Can you calculate C?
36
The first value uses much more data while the second method uses limited data in a better range.
What value would you be more apt to use? Water exhibits this “odd” general behavior at all
temperatures.
Plot to determine the second virial coeff at 300 C
-115
(z -1)v (cm3 / mol)
-120
y = 177.058x - 140.373
-125
(z -1)v
-130
lin
2
1.5
1
0.5
0
-135
1/v (mol/l )
If we do this at other values of T, we can compile a set of second virial coefficients vs.
temperature and get an idea of the differences in the two approaches. Temperatures of 200, 250,
300, 350, 400, 500, 600, 700, 800, 900, 1000, and 1200 oC analyzed in this manner. The results
are reported in the table and figure below. Also reported were the average value and the value at
the lowest pressure used. Note that the average value is indicative of the 1st method above while
the value at the lowest pressure used is indicative of the second value used.
T
200
250
300
350
400
500
600
700
800
900
1000
1200
B (Level)
-214
-156
-118
-89
-72
-49
-34
-24
-17
-13
-9
-3
B (Linear)
-226
-173
-140
-102
-103
-83
-73
-68
-66
-51
-50
-51
37
B (AVG)
-215
-157
-120
-92
-76
-54
-39
-30
-24
-19
-16
-11
B(1st value)
-220
-168
-134
-99
-95
-75
-63
-57
-54
-52
-52
-52
Second Virial Coefficient of Water by 4 Methods
0
B (water) cm 3 /mol
-50
-100
B (average)
B (lowest P)
-150
B (Level)
-200
B (Linear)
CRC
1500
1250
1000
750
500
250
-250
T (K)
For comparison, values of -142.2 cm3/mol and -7160 cm6/mol2 are reported for B and C,
respectively, at 260 oC1. Values of B reported in the CRC are also shown on the summary plot.
They agree most closely with the first (level) method.
(b)
There are several alternatives how to solve this problem, each of which comes up with a slightly
different result:
Alternative 1:
Rewrite the virial equation:

 1 BH 2 O
P  RT  
 ...
v2

v
Take the derivative:


1 2 BH 2 O 
 P 

   0  RTc  2 
 v
 v Tc
v 3c 

 c
so
 cm3 
v
BH 2O   c  28 

2
 mol 
1
J.H. Dymond and E.B. Smith, The Virial Coefficients of Pure Gases and Mixtures, A Critical Compilation, Oxford
University Press, Oxford, 1980.
38
Alternative 2:
From the virial equation:
 cm3 
Pv

BH 2O   c c  1vc  43.2 

 mol 
 RTc

Alternative 3:

 1 BH 2 O C H 2 O
P  RT  

 ...
v2
v3

v
so


1 2 BH 2O 3C H 2O 
 P 


   0  RTc  2 
3
 v
 v Tc
v
v 4c 
c

 c
(1)

 2P 
6B
12C H 2O 


  0  RTc  2  H 2O 
4
5
 v 2 
 v3
vc
vc 

Tc

 c
(2)
and
Multiply Equation 1 by (4/vc) and add to Equation 2 to get:

2
v 3c

2 BH 2 O
v 4c
0
so
 cm3 
BH 2O  vc  56 

 mol 
39
4.25
A trial-and-error is the easiest method for solving this problem. The general method is as
follows
1. Guess P sat
2. Calculate values of molar volumes that result at the chosen P sat : vlow , vmid , vhigh
3. Find areas under the curves with the following expressions
 P
v mid
sat

 f v  dv
v low
  f v   P
v high
sat
dv
v mid
[ f v  represents the particular EOS implicit in molar volume being used.]
4. If the values of the expressions are not equal, repeat the process until they are.
(a)
1. Guess P sat :
P sat  6 bar 
2. Calculate molar volume solutions for:
RT
a

P sat 
1
/
2
v  b T v (v  b )
 J  K1/2  m 3 

 J 
41
.
851


 8.314 
2
 363.15 K 
mol

mol
K





6  105 Pa   

 m3 

 3 
v  0.0001 
363.15 K 1 / 2 v v  0.0001  m  

 mol 
 mol  

 m3 
 m3 
 m3 

0
.
000551

0
.
00459
vlow  0.000152 
v
v


 high
 mid

 mol 
 mol 
 mol 
3.
0.000551 

a
sat  RT
dv  1157.19
P



 
 v  b T 1 / 2 v (v  b )  


0.000152 
0.00459 


a
 RT 
  P sat dv  1325.78

  v  b T 1 / 2 v (v  b ) 


0.000551 
40
4. Repeat this process until the areas are equal. This occurs approximately at
P sat  6.38 bar 
(b)
1. Guess Psat :
Psat  6 bar 
2. Calculate molar volume solutions for
P sat 
RT
a T 

v  b v(v  b)  bv  b 
using
 m3 
 J  m3 
a  2.066 
, b  9  10  6 

 ,  T   1.188
2
 mol 
 mol 
 m3 
 m3 
 m3 

0
.
000551

0
.
00459
vlow  0.000152 
v
v


 high
 mid

 mol 
 mol 
 mol 
3.
 sat

RT
a T 
 P  v  b  v(v  b)  bv  b  dv  1568.48

0.000152 
0.000551

 RT
a T 
sat 
 v  b  v(v  b)  bv  b   P dv  1100.19

0.000551 
0.00459

4. Repeat this process until the areas are equal. This occurs approximately at
P sat  4.945 bar 
The value calculated with the Redlich-Kwong EOS is 11.9% greater than the measured value of
5.7 bar. The Peng-Robinson EOS results in a value that is 13.25% less than the measured value.
41
4.26
First, calculate a , b , and  T  . From Table A.1.1:
Pc  48.84 Pa 
Tc  305.4 K 
w  0.099
Now we can calculate the required parameters.
a


 J 
305.4 K 
0.45724 8.314 

 mol  K  


48.84  105 Pa 
2
 J  m3 
 0.6036 

2
 mol 

 J 
0.07780 8.314 
305.4 K 
 mol  K  

 4.045  10  5
b
5
48.84  10 Pa 


 T   1  0.37464  1.542260.099  0.269920.0992

 T   1.12
 m3 


 mol 


243.15 K 

1 


3
05.4
K


2
Now, we can find the three solutions to the Peng-Robinson EOS:
P sat 
RT
a T 

v  b v(v  b)  bv  b 
Using the values calculated above and
P sat  10.6  105 Pa 
 J 
R  8.314 
 mol  K 
we get
 m3 
4
v  7.199  10  5 
 , v  2.167 10
 mol 
 m3 
 3
3 m


,
v
1
.
578
10




 mol 
 mol 
The molar volume of saturated ethane liquid is the smallest value from the list above, while the
molar volume of saturated ethane vapor is the largest value. Therefore,
42
 g 
36.0694 
3

MW ethane
 mol   1 m
  0.501  g 

 liq 
 3
liq
3
3

 3   100  cm 
v
 cm 
5 m
7.199  10 

 mol 
 g 
36.0694 
3

MW ethane
 mol   1 m
  0.0229  g 

 vap 
 3
 3   100 3 cm 3 
v vap
 cm 
3 m
1.578 10 

 mol 
Both of the densities calculated with the Peng-Robinson EOS are larger than the reported values.
The liquid density is 7.05% larger, and the vapor density is 18.7% greater.
43
4.27
(a)
We will use the Redlich-Kwong EOS in order to obtain an accurate estimate. First, calculate the
a and b parameters:
a
 J  m 3  K1 / 2 
0.42748R 2Tc2.5
 1.56 

Pc
mol 2


 m3 


 mol 
(Note: Critical data for nitrogen obtained in Table A.1.1.)
0.08664 RTc
 2.69  10  5
b
Pc
Now use the EOS to find the molar volume:
P
RT
a

1
/
2
v  b T vv  b 
Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a
solver function:
 m3 
v  1.97 10  4 

 mol 
Now calculate the total volume of gas in the 30,000 units:
Vtotal  30000 units 43 liter/unit   1.29  10 6 liters  1290 m 3
Therefore, the number of moles is:
n
1290 m 3
 6.55  10 6 mol
3
m 
1.97  10  4 

 mol 
and the mass


m  6.55  10 6 mol 0.02801 kg/mol  1.83  105 kg
Now, calculate the value of the gas:


Value  $6.1 / kg  1.83  105 kg  $1,116,300
44
If we use the ideal gas law, we find:

 J 
295 K 
 8.314 
RT 
 mol  K  
 1.98  10  4

v
P
12400000 Pa
 m3 


 mol 
Following the steps above,
Value  $1,113,200
The value calculated using the ideal gas law is $3100 less than the value calculated using the
Redlich-Kwong EOS.
(b)
First, calculate the a and b parameters:
 J  m 3  K1 / 2 
0.42748R 2Tc2.5
 1.74 
a

Pc
mol 2


 m3 


 mol 
(Note: Critical data for oxygen obtained in Table A.1.1.)
b
0.08664 RTc
 2.21  10  5
Pc
Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a
solver function:
 m3 
v  1.53  10 4 

 mol 
Following the steps presented in Part (a), we find that
Value  $2,428,000
With the ideal gas law, we find
 m3 
v  1.64  10 4 

 mol 
which provides
Value  $2,265,000
45
The value found with the ideal gas law is $163,000 less than the value found using the RedlichKwong EOS.
46
4.28
First, rewrite the Redlich-Kwong equation in cubic form.
v3 
RT 2  a
RT
ab

v 

b  b 2 v 
0
P
P
 PT 1 / 2
 PT 0.5
or at the critical point
v3 

RTc 2  a
RT
ab
v 
 c b  b 2 v 
0
0
.
5
0.5


Pc
P
P
T
P
T
c
c c
 c c

Now expand v  vc 3  0
v 3  3v 2 vc  3vvc2  vc3  0
Setting the coefficients equal we get the following expressions.
3vc 
RTc
Pc
3vc2 
a
vc3 
(1)
PcTc0.5
ab

RTc
b  b2
Pc
(2)
(3)
Pc Tc0.5
Using Equation 2, find an expression for a:


RT
a   3vc2  C  b 2  Pc Tc0.5
PC


Substitute the above expression into Equation 3 to get
b 3  3vc b 2  3vc2b  vc3  0
The one real root to this equation is


b  21 / 3  1 vc
(4)
Substitute Equation 4 into Equation 1:
47
b
0.08664 RTc
Pc
(Equation 4.48)
Now, substitute Equation 1 and Equation 4.48 into the expression for a to get
a
0.42748RTc2.5
Pc
(Equation 4.47)
Pv
To verify Equation 4.50, substitute Equation 1 into z c  c c
RTc
 RT
Pc  c
3Pc
zc  
RTc


1
3
(Equation 4.50)
The Redlich-Kwong equation is
P
RT
a

v  b T 0.5vv  b 
We can use Equation 4 to get
0.2599 RT
a
b
P

vr  0.2599 T 0.5vvc vr  0.2599 
Now, substitute Equation 4.48
P
3Tr Pc
a

0
.
5
v r  0.2599 T vvc v r  0.2599 
Replacing a with Equation 4.47 yields
P
3Tr Pc
0.42748R 2Tc2.5

v r  0.2599 PC T 0.5 vvc v r  0.2599
From Equation 4.48
R 2Tc2 
Pc2 b 2
0.086642
48
which upon substitution yields
P
3Tr Pc
Pc b 2
0.42748

v r  0.2599 0.08664 2 Tr0.5 vvc v r  0.2599
Now, use Equation 4:
P
3Tr Pc
Pc
0.42748  0.2599 2

v r  0.2599
Tr0.5 v r v r  0.2599
0.086642
Therefore,
Pr 
3Tr
1

0
.
5
vr  0.2599 0.2599Tr vr vr  0.2599
49
(Equation 4.49)
4.29
Since we are concerned with liquid water, we can base all our calculations on saturated liquid
water since the molar volume of liquids are weakly dependent on pressure. Therefore, our
results are also applicable to sub-cooled water (the pressure is greater than the saturation
pressure). To find the thermal expansion coefficient, we will use the following approximation
 vˆ sat T  5 º C   vˆ sat T  º C  
1  vˆ sat T  5 º C   vˆ sat T  º C  





10 K
vˆ sat T   T  5 º C   T  5 º C   vˆ sat T  

This approximation is valid even though the saturation pressure changes because molar volume
is weakly dependent on pressure. From the steam tables

1
 m3 
vˆ sat 15 º C   0.001001 

 kg 
 m3 
ˆv sat 20 º C   0.001002 

 kg 
 m3 
vˆ sat 95 º C   0.001040 

 kg 
 m3 
ˆv sat 100 º C   0.001044 

 kg 
 m3 
vˆ sat 25 º C   0.001003 

 kg 
 m3 
vˆ sat 105 º C   0.001047 

 kg 
Using these values, we can calculate the thermal expansion coefficients

 m3 
 m3  


0
.
001003
0
.
001001




1

kg
kg
 m3   



 



 20 º C    0.001002    


10 K
 kg   







 m3 
 m3  
  0.001040 

1  0.001047 

kg 
kg  
 m3   




 100 º C    0.001044    


10 K
 kg   






 
 100 º C   6.71  10 K 
 20 º C   2.0  10  4 K -1
4
-1
The accuracy of these values may be limited since they are based on the small differences
between liquid volumes.
To calculate the isothermal compressibility, a similar approximation will be used. It is
50

 dvˆ sat

vˆ sat T   dP
1



T
The derivatives are determined from the following graph.
Specific Volume of Liquid Water vs. Pressure
v = -5E-07*P + 0.001044
R2 = 0.9949
0.001040
3
Specific Volume [m /kg]
0.001050
0.001030
T = 20ºC
T = 100 ºC
0.001020
Linear (T = 100 ºC)
Linear (T = 20ºC)
0.001010
v = -5E-07*P + 0.001002
R2 = 0.9979
0.001000
0.000990
0
5
10
15
20
Pressure [MPa]
It is clear that
 dvˆ sat

 dP


 dvˆ sat 



 5  10  7

 dP 
T  20 0 º C 
T 100 0 º C
 m3 


 kg  MPa 
Therefore,

 m3  
 20 º C    0.001002   


 kg  

1

 5  10  7


 m3  




kg
MPa

 
 1 
1
 4.99  10 10  

 Pa 
 MPa 
 20 º C   4.99  10  4 

 m3  
 100 º C    0.001044   

 kg  

1

 5 10  7


 m3  



kg
MPa

 
 1 
1
 4.79 10 10  
 100 º C   4.79 10  4 

 MPa 
 Pa 
51
25
4.30
(a)
Substitute critical data into the Rackett equation:
vcalc 
8.314190.6 0.29056  0.087750.0081 1111 / 190  
5
2/7
46  10
 cm 3 
vcalc  38 

 mol 
Now, calculate the error.
 v
calc  vexp
Error  

vexp

Error  0.8%


  100 %




(b)
 cm 3 
vcalc  54.2 

 mol 
Error  1.1%
 cm 3 
vcalc  161.2 

 mol 
Error  10.8%
 cm 3 
vcalc  20.5 

 mol 
Error  13.5%
 cm 3 
vcalc  68.5 

 mol 
Error  19.7%
(c)
(d)
(e)
1-hexanol had the largest percent error, while ethane and methane had the smallest percent
errors. The alkenes have low percentile errors due to their nonpolar nature, which the alcohols
and acids had large percentile errors due to being polar substances and exhibiting hydrogen
bonds.
52
4.31
(a)
Since we are given the temperature and pressure, we can make use of the generalized
compressibility charts. Using data from Table A.1.1, the required quantities can be found.
343.15 K 
T

 1.12
305.4 K 
Tc
30 bar 
P

 0.616
Pr 
Pc 48.74 bar 
w  0.099
Tr 
By double interpolation of the charts
z 0   0.8376
z 1  0.0168
and
z  z 0   wz 1  0.8376  0.099 0.0168  0.8393
Therefore,
V
znRT

P

30 kg 
 J 
 8.314 
343.15 K 
 mol  K  
 0.03007 kg/mol 
0.8393
 
30 10
5
Pa 
V  0.796 m 3
(b)
The Redlich-Kwong EOS should give reasonably accurate results. Room temperature was
assumed to be 25 ºC. The molar volume is required for the calculations, so
v V 
  
V
0.1 m 3

 7.518  10  5
n 

40 kg 




0
.
03007
kg/mol


 m3 


 mol 
Using data from Table A.1.1 and Equations 4.47 and 4.48,
 J  m 3  K1 / 2 
a  9.88 

2
 mol

53
 3
5 m


b 4.51 10 

 mol 
Substituting these values into the Redlich-Kwong EOS and evaluating gives
P  191.3 bar 
54
4.32
(a)
For propane
MW  propane  0.0441  kg 
 mol 
n 
50 kg 
 1130 mol
 kg 
0.0441 
 mol 
Now calculate the volume:
V 
nRT

P
1130 mol 8.314 

 
J 
50  273.15 K 
 mol  K  
35  105 Pa 
V  0.870 m 3
(b)
The Redlich-Kwong EOS is
P
RT
a

v  b T 1 / 2 vv  b 
a
 J  m 3  K1 / 2 
0.42748R 2Tc2.5
 18.33 

Pc
mol 2


where
 m3 


 mol 
(Note: Critical data for propane obtained in Table A.1.1.)
b
0.08664 RTc
 6.28  10  5
Pc
Now that these values are known, there is only one unknown in the Redlich-Kwong EOS: v .
Using a numerical technique, e.g., the solver function on a graphing calculator
 m3 
v  0.000109 

 mol 
 
V  0.124 m 3
55
(c)
The Peng-Robinson EOS is
P
RT
a T 

v  b vv  b   bv  b 
where
 
  1   1  Tr
Tr 
2
50  273.15 K  0.873
T

370 K
Tc
  0.37464  1.542260.152  0.269920.11522  0.605
  1.081
a
 J  m3 
0.45724RTc 2
 1.02 

Pc
 mol 
0.07780 RTc
b
 5.64  10  5
Pc
 m3 


 mol 
Now every variable in the Peng-Robinson EOS is known, except v.
 m3 
v  0.0000942 

 mol 
 
V  0.107 m 3
(d)
We must calculate the reduce temperature and pressure to use the compressibility charts:
323.15 K
T

 0.873
370 K
Tc
35 bar
P

 0.825
Pr 
Pc 42.44 bar
Tr 
By double-interpolation on the compressibility charts (Appendix C),
z 0   0.1349
z 1  0.052
Therefore,
56
Pv
 z  z 0   wz 0   0.1349  0.152 0.052  0.127
RT
 m3 
0.127 RT
 0.00009749 
v

P
 mol 
V  0.111 m 3
(e)
From ThermoSolver
Using the Peng-Robinson equation,
 m3 
v  0.00009429 

 mol 
 
V  0.107 m 3
Using the generalized compressibility charts:
 m3 
v  0.0000971 

 mol 
Therefore,
 
V  0.110 m 3
57
4.33
Note: Multiple possibilities exist for which substance to use in the vial. The solution using one
possibility is illustrated below.
(a)
The substance must have a critical temperature above room temperature but below the
temperature of one’s hand. From the Appendix A.1.2, we that for carbon dioxide
Tc  304.2 K  31.1 º C
Clearly, CO2 is a suitable substance, and it is safe to use.
(b)
The vial must be able to withstand the pressure of the substance at its critical point. Therefore,
the vial must withstand 73.76 bar (the critical pressure of carbon dioxide).
(c)
Since the substance passes through its critical point, the molar volume at that state is constrained
by the critical temperature and pressure. To estimate the molar volume, we can use the PengRobinson EOS. Calculate the necessary parameters for the EOS:
2
a  0.45724

 J 
 8.314 
 304.2 K 2

 mol  K  

73.76  10 5 Pa 
 J  m3 
 0.40 

2
 mol 

 J 
 8.314 
304.2 K 
 mol  K  

b  0.07780
 3.4  10  4
5
73.76  10 Pa 
α 1
 m3 


 mol 
Substitute the above parameters and solve for the molar volume:
 cm 3 
 m3 
vc  1.18  10  4 
118




 mol 
 mol 
Now, we can calculate the required amount of CO2.
n
 
100 cm 3
V

 0.847 mol
vc
 cm 3 
118 

 mol 
58
(d)
If the vial contains less substance than needed, the molar volume will be greater. Therefore, the
substance will not pass through the critical point. As the substance is heated, it will go through a
transition from a saturated liquid and vapor mixture to superheated vapor. Then, it will become a
supercritical fluid once the critical temperature is exceeded.
P
critical
point
liquid
part c
part d
not enough
CO2
vapor
liquid - vapor
v
59
4.34
Methane:
The following quantities are required to calculate the molar volume with the Peng-Robinson
equation.
 
 T   1   1  Tr
 T   0.9626
2
a
 J  m3 
0.45724RTc 2
 0.25 

Pc
 mol 
b
0.07780 RTc
 2.68  10  5
Pc
T  Tr Tc  209.66 K 
 m3 


 mol 
P  Pr Pc  55.2  10 5 Pa 
The molar volume is the only unknown in the following equation.
P
RT
a T 

v  b vv  b   bv  b 
v  1.83  10
4
 m3 


 mol 
Therefore,



 3 
4 m

55.2  10 Pa  1.83  10 



mol

 

z
 0.58

 J 
 8.314 
209.66 K 
 mol  K  

5
From the compressibility charts,
z 0   0.5984
z 1  0.0897
Therefore,
z  z 0   wz 1  0.5984  0.0080.0897
z  0.599
60
Methanol:
Following the procedure outlined above, the Peng-Robinson equation produces
 m3 
v  3.14 10  4 

 mol 
 z  0.651
Using the compressibility charts:
z 0   0.5984
z 1  0.0897
From Table A.1.1,
w  0.559
Therefore,
z  z 0   wz 1  0.5984  0.559 0.0897
z  0.649
Summary:
Methane
z  0.58 (Peng-Robinson)
z  0.599 (Compressibility charts)
The value from the Peng-Robinson EOS is 3.2% smaller than the value
from the charts.
Methanol
z  0.651 (Peng-Robinson)
z  0.649 (Compressibility charts)
The value from the Peng-Robinson EOS is 0.31% smaller than the value
from the charts.
61
4.35
From Table A.1.1:
Tc  405.6 K
Pc  112.77 bar
w  0.25
Calculate reduced temperature and pressure:
92  273.15 K  0.9
T

405.6 K
Tc
306.5 bar
P

 2.718
Pr 
Pc 112.77 bar
Tr 
From interpolation of the compressibility charts:
z 0   0.4133
z 1  0.1351
Therefore,
Pv
 z  z 0   wz 1  0.38
RT
and
v
0.38RT
 3.76  10  5
P
 m3 


 mol 
Since the temperature of the ammonia in this system is below the critical temperature, we know
the ammonia is not a supercritical fluid. The pressure is greater than the critical pressure, so the
ammonia is a liquid.
62
4.36
Let the subscript “ace” represent acetylene and “but” represent n-butane. First, convert the given
quantities of acetylene and n-butane into moles.
nace 
nbut 
30 kg 
26.038  10
-3
50 kg 
58.123 10
-3
 kg 
 mol 
 kg 
 mol 
 1152.2 mol
 860.2 mol
Therefore,
y ace  0.573
ybut  0.427
We can also calculate a and b parameters for pure species using the following equations
0.42748 R 2Tc2.5
a
Pc
b
0.08664 RTc
Pc
Substituting critical data from Table A.1.1 gives
 J  K1/2  m3 
aace  8.03 

2
 mol

 J  K1/2  m 3 
abut  29.07 

2

 mol
 m3 
bace  3.62 10  5 

 mol 
 m3 
bbut  8.08110  5 

 mol 
Now, we can use mixing rules to calculate the parameters for the mixture.
amix  y12 a1  2 y1 y 2 a12  y 22 a2
 J  K1/2  m3 
a12  8.03  29.07 1  0.092  13.87 

2
 mol

 J  K1/2  m 3 
 amix  14.72 

2
 mol

63
bmix  y1b1  y 2b2  5.53  10
5
 m3 


 mol 
Substitution of the mixture parameters into the Redlich-Kwong EOS results in an equation with
one unknown.
 m3 
v  0.00111 

 mol 

 m3  

V  nace  nbut v  1152.2 mol  860.2 mol 0.00111 



mol

 

 
V  2.23 m 3
64
4.37
First, we need to calculate the a and b parameters for species (1) and (2). From Table A.1.1 and
A.1.2:
CO2 (1):
Toluene (2):
Tc  304.2 K 
Pc  73.76 bar 
a1 
 J  m 3  K1/2 
0.42748R 2Tc2.5
 6.466 

Pc
mol


b1 
0.08664 RTc
 2.97  10  5
Pc
 m3 


 mol 
Tc  591.7 K 
Pc  41.14 bar 
 J  m 3  K1/2 
0.42748R 2Tc2.5
a2 
 61.17 

Pc
mol


b1 
0.08664 RTc
 1.036  10  4
Pc
 m3 


 mol 
We can use the mixing rules to calculate bmix.
bmix  y1b1  y 2b2
 2 mol 
5
bmix  
 2.97  10

5
mol


 3
5 m
bmix  7.40  10 

 mol 
 m 3    3 mol 
4

  
1.036  10


mol
5
mol


  
 m3  



mol

 
Now we can substitute the known values into the Redlich-Kwong EOS. Note that the molar
volume can be calculated as follows
v
 
 m3 
0.01 m 3
V

 0.002 

5 mol
ntot
 mol 
We can then solve for amix.
 J  m 3  K1/2 
a mix  31.0 

mol


65
From the mixing rules, we know
 J  m3  K1/2 
2
2
amix  31.0 
  y1 a1  2 y1 y2 a12  y2 a2
mol


Therefore,
 J  m 3  K1/2 
a12  16.55 

mol


Equation 4.79 states
a12  a1a 2 1  k12 
Substitution of the values provides:
k12  0.17
66
4.38
(a)
We can match the species by looking at the b parameter alone. The b parameter is directly
related to the size of the molecule. Because sizeC 2 H 6  size H 2 O  size H 2 ,
bC 2 H 6  bH 2 O  bH 2 . Therefore,
 J  m3 
a

 mol 
0.564
 m3 
b

 mol 
Species
6.38  10 5
C2H6 (3)
0.025
2.66  10 5
H2 (1)
0.561
3.05 10 5
H2O (2)
These values are also consistent with what we would expect for the magnitude of van der Waals
interactions given by a.
(b)
The van der Waals parameters for the mixture can be calculated according to the mixing rules.
From Equations 4.81 and 4.82
a mix  y12 a1  y1 y 2 a12  y1 y3 a13  y 2 y1 a21  y 22 a 2  y 2 y3 a 23  y3 y1 a31  y3 y 2 a32  y32 a3
a mix  y12 a1  2 y1 y 2 a12  2 y1 y3 a13  y 22 a2  2 y 2 y3 a 23  y32 a3
bmix  y1b1  y 2b2  y3b3
Calculate mole fractions:
n1
5 mol

 0.5
n1  n2  n3 10 mol
y 2  0 .4
y3  0.1
y1 
Since binary interaction parameters are not available, we must use Equation 4.78 to calculate
a12 , a13 , a23 .
 J  m3 
a12  a1a2  0.118 

 mol 
 J  m3 
a13  0.119 

 mol 
67
 J  m3 
a23  0.562 

 mol 
Now we can find the numerical values for the van der Waals parameters.
 J  m3 
amix  0.206 

 mol 
 m3 
bmix  3.19  105 

 mol 
(c)
For the mixture, the van der Waals equation is
P
a
RT
 mix
v  bmix
v2
The molar volume is calculated as follows
v
 
 m3 
0.0125 m 3
V

 0.00125 

10 mol
n1  n2  n3
 mol 
Therefore,
 J  m3 

 J 
0
.
206


 8.314 
 300 K 
mol

mol
K






P

2

 3 
 3
 m3  
 0.00125  m   3.19  10  5  m   

0.00125 




 mol   
 mol 

 mol  
P  1.92 MPa
68
4.39
(a)
To calculate the molar volume of the mixture, we will use the virial expansion in pressure since
it is more accurate at moderate pressures. But first, we must calculate the second virial
coefficient for the mixture.
Bmix  y12 B1  2 y1 y 2 B12  y 22 B2



 cm 3  
 cm 3  
 cm 3  
2








0
.
75
110
2
0
.
25
0
.
75
153
Bmix  0.252   625 















 mol  
 mol  
 mol  



 cm 3 
Bmix  158.3 

 mol 
Therefore,
RT  Bmix  RT
P 
 Bmix
1 
P 
RT  P
 cm 3 
 m3 
v  2445.6 
  0.002446 

 mol 
 mol 
v
(b)
We can estimate k12 using an EOS where the a parameter is the only unknown. b can be
calculated as follows
bmix  y1b1  y 2b2
Using the Redlich-Kwong EOS,

bmix  0.25 5.83 10  5



 m3  
5

   0.75 2.97 10


mol

 

 m3  
 3
5 m

   3.69  10 


 mol  
 mol 
Substitute the known properties into the Redlich-Kwong EOS and solve for a.
 J  K1/2  m 3 
amix  8.693 

2
 mol

Now we can calculate k12 as follows
a mix  y12 a1  2 y1 y 2 a12  y 22 a2
69
 J  K1/2  m 3 
a1  28.03 

2
 mol

 J  K1/2  m 3 
a2  6.466 

2
 mol

 J  K1/2  m 3 
 a12  8.81 

2
 mol

(Calculated using Equation 4.47)
From Equation 4.79
a12  a1a 2 1  k12 
8.81
k12  1 
 0.35
6.466  28.03
70
4.40
During this process the gas in Tank A undergoes an expansion from 7 bar to 3.28 bar. During
this isentropic process, the gas cools. If attractive forces are present, they will manifest
themselves more in state 1 at the higher pressure than in state 2. (While T has some effect to
counter this trend, we expect it will be secondary to the effect of pressure). Thus, in the nonideal case, additional energy must be supplied to “pull” the molecules apart. Since the tank is
insulated, the only place that this can come from is the kinetic energy of the molecules. Thus,
they slow down even more than in the ideal gas case, and the final temperature is lower. If we
had an equation of state that appropriately described the non-ideal behavior, we could apply the
concepts of the thermodynamic web that we have learned in this chapter to solve for the final
temperature.
71
4.41
Select the “Equation of State Solver” from the main menu. Enter the pressure and temperature
and solve for the molar volume using the Lee-Kesler EOS. The program provides
 m3 
v  1.07  10  4 

 mol 
To calculate the volume occupied by 10 kilograms of butane, we need to know the number of
moles present in 10 kg.
n
10 kg 
 172.1 mol
0.05812 kg/mol
Therefore,

V  172.1 mol1.07  10  4


 
 m3  
3

   0.018 m

 mol  
The answers provided in Example 4.9 agree well with this answer. The answer found using the
Redlich-Kwong EOS is 16.7% larger than the answer from ThermoSolver, but the answer from
the compressibility charts is only 5.56% larger.
72
4.42
Using ThermoSolver should be straightforward; thus, only the answers are provided.
(a)
Tc  305.4 K 
Pc  48.84 bar 
 kJ 
h f ,298  84.68 

 mol 
(b)
Tsat  299.42 K 
(c)
The percent difference is calculated as follows
 v LK  v PR

z LK  z PR
% Difference  
or
 100 %
v LK
z LK


(i).
Quantity
v
z
Lee Kessler EOS
 m3 
8.34  10  5 

 mol 
0.1383
Peng Robinson EOS
 m3 
8.70  10  5 

 mol 
0.1444
% Difference
Lee Kessler EOS
 m3 
3.68  10  4 

 mol 
0.5859
Peng Robinson EOS
 m3 
3.58  10  4 

 mol 
0.5711
% Difference
4.3
4.4
(i).
Quantity
v
z
73
2.7
2.5
Chapter 5 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
5.1
(a)
Following the example given by Equation 5.5a in the text
 u 
 u 
du  
 dT    dP
 P T
 T  P
(b)
 u 
 u 
du  
 dT    ds
 T  s
 s T
(c)
 u 
 u 
du    dh    ds
 s  h
 h  s
2
5.2.
The internal energy can be written as follows
 u 
 u 
du  
 dT    dv
 T  v
 v T
Substituting Equations 5.38 and 5.40
  P 

 u 
 u 
  P

  cv and    T 
 v T   T  v
 T  v

into the above expression yields
  P 

du  cv dT  T 
  P  dv
  T  v

From the ideal gas law, we have
R
 P 

 
 T  v v
Therefore,
 RT

du  cv dT  
 P  dv
 v

which upon noting that P 
RT
for an ideal gas, becomes
v
du  cv dT
3
5.3
The heat capacity at constant pressure can be defined mathematically as follows
 h 
 u   Pv  
 u 
 v 
cP  
 
 
  P

T
 T  P 
 P  T  v
 T  P
For an ideal gas:
R
 v 

 
 T  P P
Therefore,
 u 
cP  
 R
 T  v
One mathematical definition of du is
 u 
 u 
du  
 dT    dP
 T  P
 P T
 u 
We can now rewrite 
 :
 T  v
 u 
 u   T   u   P 

 
 
   
  cv
 T  v  T  P  T  v  P T  T  v
For an ideal gas:
 u 
  0
 P  T
so
 u 
cv  

 T  P
Substituting this result into our expression for c P gives
c P  cv  R
4
5.4
In terms of P, v, and T, the cyclic equation is
 P   T   v 
1  
 
  
 T  v  v  P  P T
For the ideal gas law:
Pv  RT
so the derivatives become:
R
 P 

 
 T  v v
P
 T 

 
 v  P R
 RT  v
 v 
   2 
P
 P T
P
Therefore,
 P   T   v 
 R  P   v 

 
      
  1
 T  v  v  P  P T  v  R  P 
The ideal gas law follows the cyclic rule.
5
5.5
For a pure species two independent, intensive properties constrains the state of the system. If we
specify these variables, all other properties are fixed. Thus, if we hold T and P constant h cannot
change, i.e.,
 h 
0
 
 v T , P
6
5.6
Expansion of the enthalpy term in the numerator results in
 Ts  vP 
 h 

 

T
 T  s 
s
 P 
 h 


  v
 T  s
 T  s
Using a Maxwell relation
 s 
 h 
  v 

 v  P
 T  s
 s   T 
 h 


 
  v
 T  P  v  P
 T  s
We can show that
c
 s 
  P

 T  P T
(use thermodynamic web)
1  RT
2a 2ab 
 T 




  
 v  P R  v  b v 2 v 3 
(differentiate van der Waals EOS)
Therefore,
 v
vc  RT 2a 2ab 
2a  b  
 h 
  c P 



1   
  P 

2
3
 T  s RT  v  b v
v 
 v  b vRT  v 
7
5.7
 h 
  :
 P T
 s 
 Ts  vP 
 h 
  T   v
  
P
 P T
T
 P T 

R
 v 
 s 
2
   1  B' P  C ' P
   
P
 T  P
 P T



RT
 h 
1  B ' P  C ' P 2  v  v  v
  
P
P

 T
 h 
  0
 P T
 h 
  :
 P  s
 h   Ts  vP 
 v
  
P
s
 P  s 
RT
 h 
1  B' P  C ' P 2
  
P
 P  s


 h 
 :

 T  P
 h 
  cP

 T  P
(Definition of cP)
 h 
 :

 T  s
 P 
 Ts  vP 
 h 

  v
 

T
 T  s
s
 T  s 
c  T 
c
P
 s   P 
 P 
  P
    P
  

T  v  P T R 1  B' P  C ' P 2
 T  P  s T
 T  s

8



Pv
1
1  B' P  C ' P 2
 h 

c

c


P
P
RT 1  B' P  C ' P 2
 T  s
1  B' P  C ' P 2
 h 
  cP

 T  s


9


5.8
(a)
A sketch of the process is provided below
well
insulated
m
s = 0
T1
P1
T2
P2
The diagram shows an infinitesimal amount of mass being placed on top of the piston of a
piston-cylinder assembly. The increase in mass causes the gas in the piston to be compressed.
Because the mass increases infinitesimally and the piston is well insulated, the compression is
reversible and adiabatic. For a reversible, adiabatic process the change in entropy is zero.
Therefore, the compression changes the internal energy of the gas at constant entropy as the
pressure increases.
(b)
To determine the sign of the relation, consider an energy balance on the piston. Neglecting
potential and kinetic energy changes, we obtain
U  Q  W
Since the process is adiabatic, the energy balance reduces to
U  W
As the pressure increases on the piston, the piston compresses. Positive work is done on the
system; hence, the change in internal energy is positive. We have justified the statement
 u 
  0
 P  s
10
5.9
(a)
By definition:
1  v 

v  T  P
 
and
1  v 

v  P  T
  
Dividing, we get:
 v 
 

 v   P 
 T  P

    

 v 
 T  P  v  T
 
 P  T
where derivative inversion was used. Applying the cyclic rule:
 v   P   T 
1       
 T  P  v  T  P  v
Hence,
  P 
 
  T  v
(b)
If we write T = T(v,P), we get:
 T 
 T 
dT    dv    dP
 P  v
 v  P
(1)
From Equations 5.33 and 5.36
ds 
cv
c
 P 
 v 
dT    dv  P dT    dP
T
T
 T  v
 T  P
We can solve for dT to get:
11
dT 
T  P 
T  v 
  dP
  dv 
c P  cv  T  v
c P  cv  T  P
(2)
For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence,
T
 T 
  
 v  P c P  cv
 P 
 
 T  v
or
  v 
 P   v 
c P  cv  T      T  
  T  P
 T  v  T  P
where the result from part a was used. Applying the definition of the thermal expansion
coefficient:
Tv
 P   v 
c P  cv  T     

 T  v  T  P
2
12
5.10
We need data for acetone, benzene, and copper. A table of values for the molar volume, thermal
expansion coefficient and isothermal compressibility are taken from Table 4.4:
 
 m3 
v  10 6 

 mol 
73.33
86.89
7.11
Species
Acetone
Benzene
Copper
 
  103 K -1
  1010 Pa -1
1.49
1.24
0.0486
12.7
9.4
0.091
We can calculate the difference in heat capacity use the result from Problem 5.9b:
c P  cv 
vT 2

or

 73.33  10  6

c p  cv  
Species
Acetone
Benzene
Copper
 

 m3  
3
-1 2

 293 K  1.49  10 K

 mol  
12.7  10
10
Pa 
-1
 J 
c p  cv 
 mol  K 
37.6
41.6
0.5
 J 
cp 
 mol  K 
125.6
135.6
22.6
 J 
 37.6 
 mol  K 
% difference
30%
31%
2%
We can compare values to that of the heat capacity given in Appendix A2.2. While we often
assume that cP and cv are equal for condensed phases, this may not be the case.
13
5.11
We know from Equations 4.71 and 4.72
1  v 

v  T  P
 
and
1  v 

v  P T
  
Maxwell relation:
 P 
 s 

  
 v T  T  v
Employing the cyclic rule gives
 P   v 
 P 

    

 v T  T  P
 T  v
which can be rewritten as
1  v 


v  T  P
 P 
 s 
  
 
 v T  T  v  1  P 
 
v  v T
Therefore,

 s 
  
 v T 
Maxwell Relation:
 s 
 v 
   

 P T
 T  P
From Equation 4.71:
 v 

  v
 T  P
Therefore,
 s 
    v
 P T
14
5.12
(a)
An isochor on a Mollier diagram can be represented mathematically as
 h 
 
 s  v
This can be rewritten:
 h 
 Ts  vP 
 P 
  
 T  
s
 s  v 
v
 s  v
Employing the appropriate Maxwell relation and cyclic rule results in
 h 
 T   s 
   T  v
  
 s  v  v  T
 s  v
We know
T
 T 
 s 
 P 
and    

 

 s  v c v
 v  T  T  v
For an ideal gas:
R
 s 
 P 
  
 
 v  T  T  v v
Therefore,

T R
R
 h 
 T 1  
  T  v
cv v
cv 
 s  v

(b)
In Part (a), we found
T
 h 
  T  v
cv
 s  v
 P 


 T  v
For a van der Waals gas:
15
R
 P 

 
 T  v v  b
Therefore,
RT
 h 
  T 
cv
 s  v
 v 


v b
16
5.13
(a)
The cyclic rule can be employed to give
 T 
 T   s 

  
  
 s  P  P T
 P  s
Substitution of Equations 5.19 and 5.31 yields
T
 T 

 
 P  s c P
 v 


 T  P
For an ideal gas:
R
 v 

 
 T  P P
Therefore,
RT 1
v
 T 

 

P cP cP
 P  s
(b)
Separation of variables provides
T
R P

T
cP P
Integration provides
T
ln 2
 T1

P
  ln 2

 P1
R
 cP


which can be rewritten as
R
 P2  cP
 
T2

T1  P1 
The ideal gas law is now employed
17
R
P2 v 2  P2  cP
 
P1v1  P1 
 R
1
c
P2 P


v
2

 R
1
c
P1 P


v
1
where
1
R c P  R cv 1



cP
cP
cP k
If we raise both sides of the equation by a power of k, we find
P2 v 2k  P1v1k
 Pv k  const.
(c)
In Part (a), we found
T  v 
 T 

 


 P  s c P  T  P
Using the derivative inversion rule, we find for the van der Waals equation
Rv 3 v  b 
 v 

 
 T  P RTv 3  2av  b 2
Therefore,
1
RTv 3 v  b 
 T 

 
 P  s c P RTv 3  2av  b 2
18
5.14
The development of Equation 5.48 is analogous to the development of Equation E5.3D. We
want to know how the heat capacity changes with pressure, so consider
 c P

 P


T
which can be rewritten as
 c P

 P
   h  
   h  

  
     
T  P  T  P  T  T  P T  P
 h 
Consider the   term:
 P T
 h 
 Ts  vP 
 s 
 v 
  T    v  T 
 v
  
P
 P T
 T  P
 P T 
T
 c 
Substitution of this expression back into the equation for  P  results in
 P T
    v 

 c P 
     T 

  v 
 P T  T   T  P
 P
  2v 
T  v 
 c P 

   v 



T




2


T  T  P
 P T
 T  P  T  P
  2v 
 c P 

  T  2 
 T 
 P T

P
Therefore,
c Preal
Preal
c Pideal
Pideal
 dc P 

   2v  
  dP
 T 
2

  T  
P

and
ideal

c Preal  c P
Preal 
  2v
T 
  2
Pideal   T

 
  dP
 
P 
19
5.15
In order to solve this problem we need to relate the change in entropy from 10 to 12 bar to the
change in molar volume (for which we have complete data). First, we can rewrite the change in
entropy as
12 bar
s  s 2  s1 
 s 
  dP
 P T
10 bar

Applying a Maxwell relation, we can relate the above equation to the change in molar volume:
12 bar
s 2  s1 
12 bar
 s 
 v 
  P T dP  s1     T  P dP
10 bar
10 bar
As 10 bar:
 v 
 v 
4

 
  5.60  10
 T  P  T  P
 m3 


 kg  K 
At 12 bar:
 v 
 v 
4

 
  4.80 10
 T  P  T  P
 m3 


 kg  K 
 v 
To integrate the above entropy equation, we need an expression that relates 
 to pressure.
 T  P
Thus, we will fit a line to the data. We obtain

 v 
10

    4.0  10
 T  P 
 m3  
 3 
4 m

  P  9.6  10 


 kg  K  Pa  
 kg  K 
Now integrate the equation to find the entropy:
 4.0 10
1.210 6 Pa
s 2  s1 
1.010 6 Pa
10
P  9.6 10 4 dP 5.4960  0.104  kgkJ K   5.392  kgkJ K 
20
5.16
A schematic of the process follows:
Propane in
v1 = 600 cm3/mol
T1= 350 oC
Turbine
ws
Propane P = 1 atm
2
out
We also know the ideal gas heat capacity from Table A.2.1:
cP
 1.213  28.785 10  3 T  8.824 10  6 T 2
R
Since this process is isentropic (s=0), we can construct a path such that the sum of s is zero.
(a) T, v as independent variables
Choosing T and v as the independent variables, (and changing T under ideal gas conditions), we
get:
 s=0
s1
step 1
 s2
volume
v2 ,T2
Ideal
step 2 Gas
v1 ,T 1
Temperature
or in mathematical terms:
 s 
 s 
ds    dT    dv  0
 T  v
 v T
However, From Equation 5.33:
21
c
 P 
ds  v dT 
dv
T v
T
To get s1
P
a
RT
 2
vb v
so
R
 P 

T v v  b
and
v
 dv  2 R dv  R ln v2  b 



 
v1  b 
v1 T v
v1 v  b
v 2 P
s1   ds  
or, using the ideal gas law, we can put s1 in terms of T2:
RT2  b 
P

s1  R ln  2
v1  b 



For step 2
T2
T2
T1
623.15 K
c
s 2   v dT  R
T

0.213  28.785  10  3 T  8.824  10  6 T 2
dT
T
Now add both steps
s  s1  s 2  0
 RT2

 P  b
T2 
8.824  10  6 2
3
  0.213 ln


 ln  2

28
.
785

10
T

623
.
15
K

T2  623.15 K 2

2
2
 v1  b 
 623.15 





Substitute
T1  623.15 K

v1  600 cm 3 /mol
P2  1 atm

 cm 3  atm 
R  82.06 

 mol  K 
22

and solve for T2:
T2  448.3 K 
(b) T, P as independent variables
Choosing T and P as the independent variables, (and changing T under ideal gas conditions), we
get:
P1,T1
step 1
Pressure
s=0
step 2
s1
Ideal
Gas
s2
P2,T2
Temperature
Mathematically, the entropy is defined as follows
 s 
 s 
ds  
 dT    dP  0
 P T
 T  P
Using the appropriate relationships, the expression can be rewritten as
c
 v 
ds  P dT  
 dP  0
T
 T  P
For the van der Waals equation
 R 
 v  b  


 v 

 
 T  P 
RT
2a 
 

2
v 3 
 v  b 
Therefore,
23
s 
T2

T1
cP
dT 
T
P2

 R 
 v  b  


2a 
P1 
 

 v  b 2 v 3 
dP  0
RT
We can’t integrate the second term of the expression as it is, so we need to rewrite dP in terms
of the other variables. For the van der Waals equation at constant temperature:
 2a
RT 
dP   
 dv
3
v  b 2 
 v
Substituting this into the entropy expression, we get
1.213  28.785 10  3 T  8.824  10  6 T 2
s  
dT 
T
T2
v2
T1
v1
Upon substituting
T1  623.15 K
v1  600 cm 3
v2 
RT2
(gas acts ideally at 1 atm)
P2
 cm 3 
b  91 

 mol 
 cm 3  atm 
R  82.06 

 mol  K 
we obtain one equation for one unknown. Solving, we get
T2  448.3 K
24
R
 v  b dv  0
5.17
(a)
Attractive forces dominate. If we examine the expression for z, we see that at any absolute
temperature and pressure, z  1. The intermolecular attractions cause the molar volume to
deviate negatively from ideality and are stronger than the repulsive interactions.
(b)
Energy balance:
h2  h1  q
Alternative 1: path through ideal gas state
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
P [bar]
P,T1
P,T2
q = h
50
step 3
step 1
h1
h
3
h2 step 2
0
ideal gas
300
500
Choosing T and P as the independent properties:
 h 
 h 
dh  
 dT    dP
 T  P
 P T
or using Equation 5.46
  v 

dh  c P dT   T 
  v  dP
  T  P 
The given EOS can be rewritten as
1

v  R  aT 1 / 2 
P

25
T [K]
Taking the derivative gives:
R
 v 
 0.5

   0.5aRT
 T  P P
so


dh  cP dT  0.5aRT 0.5 dP
For step 1
 0.5aRT1
0
0.5
h1 
50 bar
dP  0.5aRT10.5 P  252  molJ 
For step 2
 3.58  3.02 10
500 K
h2  R
3
300 K

 J 
T  0.875T  0.5 dT  7961 

 mol 
For step 3:
 0.5aRT2
50 bar
h3 
0.5
0
dP  0.5aRT20.5P  323  molJ 
Finally summing up the three terms, we get,
 J 
q  h1  h2  h3  7888 
 mol 
Alternative 2: real heat capacity
For a real gas
h  c Preal
From Equation 5.48:
  2v
T
  T 2
ideak

P
P real
c Preal  c Pideal 

 dP

P
For the given EOS
26

1
v  R  aT 1 / 2 
P

Therefore,
  2v 

  0.25aRT 1.5
 T 2 

P
and
  2v
T
  T 2
ideak

P
P real

 
P 50 bar

 dP 
 0.25aRT 0.5 dP  0.875 K 1/2 RT 0.5


P
P ideak 0 bar
real
We can combine this result with the expression for c Preal and find the enthalpy change.
 3.58  3.02 10
500 K
h  R
3

T  0.875T  0.5 dT
300 K
 J 
q  h  7888 
 mol 
The answers is equivalent to that calculated in alternative 1
27
5.18
(a)
Calculate the temperature of the gas using the van der Waals equation. The van der Waals
equation is given by:
P
RT
a
 2
v b v
First, we need to find the molar volume and pressure of state 1.
  
 m3 
V
Al 0.1 m 2 0.4 m 

 0.00016 
v1  1 

n
n
250 mol
 mol 
P1 
mg
 Patm 
A
10000 kg 9.81  m2  

0.1 m 2
 
s  
 1.01325 10 5 Pa   1.08 10 6 Pa 
Substituting these equations into the van der Waals equation above gives
 J  m3 

 J 
0
.5


 8.314 
T1
 mol 
 mol  K  

6
1.08  10 Pa  

2
3

 m3 
3 


5 m
m
0.00016 
  0.00016 
  4  10 


mol
 mol  
 mol 
 

T1  297.5 K
Since the process is isothermal, the following path can be used to calculate internal energy:
v2 ,T2 = T 1
v
u
s
v1 ,T1
T
Thus, we can write the change in internal energy as:
28
 u 
 u 
 u 
du  
 dT    dv    dv
 v T
 v T
 T  v
Using Equation 5.40
v2
u 
  P 

 T  T  v  P  dv
v1
For the van der Waals EOS:
P
RT
a
 2
v b v
so
R
 P 

 
 T  v v  b
Therefore,
v2
u 
a
 v 2 dv
v1
We can assume the gas in state 2 is an ideal gas since the final pressure is atmospheric.
Therefore, we calculate v 2 ,
v2 
 m3 
RT2
 0.0244 

P2
 mol 
and
 J  m3 
0
.
5


0.0244
 mol 
 J 
dv  3104.5 
u  
 mol 
v2
0.00016
or
U  776.1 kJ 
29
(b)
From the definition of entropy:
suniv  s sys  s surr
First, let’s solve for s sys using the thermodynamic web.
 s 
 s 
ds sys  
 dT    dv
 T  v
 v T
Since the process is isothermal,
 s 
ds sys    dv
 v T
v2
 P 
 s sys   
 dv
T  v

v
1
Again, for the van der Waals equation,
R
 P 

 
 T  v v  b
Substitution of this expression into the equation for entropy yields
v2
s sys 
R
 v  b dv
v1
 J 
8.314 
 mol  K  dv  44.17  J 
s sys  
 mol  K 
 3
0.00016 v  4  10  5 m


 mol 
J
S sys  11042.5  
K
0.0244
The change in entropy of the surroundings will be calculated as follows
s surr 
Qsurr
Tsurr
where
30
Qsurr  Q
(Q is the heat transfer for the system)
Application of the first law provides
Q  U  W
We know the change in internal energy from part a, so let’s calculate W using
v2
W  n  Pdv
v1
Since the external pressure is constant,
m
 250 mol1.01325  10 5 Pa  0.0244 


W
W  614030 J 

 m3  
  0.00016 


mol
mol



 


3
Now calculate heat transfer.
Q  776100 J    614030 J   1.39  10 6 J 
Therefore,
S surr 
 1.39  10 6 J 
J
 4672  
297.5 K 
K 
and the entropy change of the universe is:
J
J
J
S univ  11042.5    4672    6370.5  
K 
K 
K 
31
5.19
First, calculate the initial and final pressure of the system.
Pi  10  10
5
P f  10  10

20000 kg 9.81 m/s 2 
Pa  
 4.92  10 6 Pa 
2
 
Pa   30000 kg9.81 m/s   6.89  10
0.05 m 
0.05 m
5
2
6
2
Pa 
To find the final temperature, we can perform an energy balance. Since the system is wellinsulated, all of the work done by adding the third block is converted into internal energy. The
energy balance is
u  w
To find the work, we need the initial and final molar volumes, which we can obtain from the
given EOS:

vi  8.37 10 4 m 3 /mol
vf 

8.314T f



25 
6.89  10 6 1 
 Tf 



 3.2  10 -5 m 3 /mol

Now, calculate the work






8.314T f
w   Pf v f  vi   6.89 10 6 Pa 
 3.2  10 -5  8.37  10 4 
 25 


6

6
.
89
10
1



 Tf 






We also need to find an expression for the change in internal energy with only one variable: Tf.
To find the change in internal energy, we can create a hypothetical path shown below:





step 2
v
ideal gas
step 3
step 1

vi ,Ti
u = -P (v -v )
f f i
vf ,Tf
500
Tf
32
T
For step 1, we calculate the change in internal energy as follows
v  RT / Plow

u1 
vi
u1 
 u 
  dv 
 v T
v  RT / Plow

vi
  P 

  P dv
T 
  T  v

v  RT / Plow 
2 
2
 aRTi  dv  aRTi ln RTi / Plow  b 

 T  a 2  v  b  T  a 2 
vi  b

i
 i


vi
Similarly, for step 3:
vf
v
f
  P 

 u 
dv
T
P
u3 







  v T
   T  v dv
v  RT / Plow
v  RT / Plow
 aRT 2  dv


aRT f 2
vf b
f




ln
  T  a 2  v  b T  a 2  RT f / Plow  b 


f
v  RT / Plow  f

vf
u3 




Insert the expression for the final molar volume into the equation for u3 :

8.314T f
u3 
ln
T f  a 2  6.89  10 6 1  25 / T f RT f / Plow  b

aRT f 2









Since the pressure is low (molar volume is big) during the second step, we can use the ideal heat
capacity to calculate the change in internal energy.
u 2 
Tf
Tf
Ti  500 K
Ti  500 K
 cv dT 

 20  R  0.05T dT


u 2  11.686 T f  500  0.025 T f2  500 2

If we set the sum of the three steps in the internal energy calculation equal to the work and
choose an arbitrary value for Plow, 100 Pa for example, we obtain one equation with one
unknown:
33


 RT / P  b 
  11.686 T f  500  0.025 T f2  500 2 
ln i low
2
Ti  a   vi  b 

aRTi 2



8.314T f

ln

2 
6
Tf  a
 6.89  10 1  25 / T f RT f / Plow  b 






8.314T f
 6.89  10 6 Pa 
 3.2  10 -5  8.37  10  4 



25 
6

 6.89  10 1 



 Tf 



aRT f 2









Solving for Tf we get
T f  536.2 K
The piston-cylinder assembly is well-insulated, so
suniv  s sys
Since the gas in the cylinder is not ideal, we must construct a hypothetical path, such as one
shown below, to calculate the change in entropy during this process.
P
Pf ,Tf
6.9
step 1
step 3
Pi ,Ti
4.9
ideal gas
Plow
step 2
536
500
For steps 1 and 3
Plow
Plow
 s 
s1     dP  
 P T
Pi
Pi
  v  
  dP
 
  T  P 
34
T
Pf
P
f
  v  
 s 
s3     dP    
  dP
P T
T



P 
Plow
Plow 
We can differentiate the given EOS as required:
Plow 
 RTi 2a  Ti  
 RTi 2a  Ti   Plow
dP 
ln
s1 


2
2




a
T
P
a
T




 Pi
i
i

Pi 

s3 

  RT 2a  T
f
f
  a  T 2 P
f
Plow 
Pf





 dP   RT f 2a  T f  ln Pf 

a  T f 2  Plow 

For step 2
Tf
Tf
Tf
c
 s 
 20

s 2   
 dT   P dT     0.05  dT
T
 T  P
T

Ti
Ti
Ti
Tf 
  0.05 T f  Ti
s2  20 ln

T
 i 


Sum all of the steps to obtain the change in entropy for the entire process
suniv  s sys  s1  s 2  s3


RT f 2a  T f
Tf 
 Pf 
 RTi 2a  Ti   Plow 
  0.05 T f  Ti 


  20 ln
ln
ln
T 
P 
2
a  Ti 2

a
T
i
low
 Pi 




f
Arbitrarily choose Plow (try 100 Pa), substitute numerical values, and evaluate:

suniv 
 J 
suniv  0.388 
 mol  K 

 J 
J
S univ  2 mol 0.388 
  0.766  

 mol  K  
K 

35



5.20
A schematic of the process is given by:
well
insulated
V=1L
V=2L
V=1L
T = 500 K
T=?
Vacuum
nCO =1 mole
nCO=1 mole
State i
State f
(a)
The following equation was developed in Chapter 5:
  2P  
 T  T 2  dv
v 
v ideal  
v
cvreal  cvideal 
For the van der Waals EOS
 2P 

0
 T 2 


Therefore,
cvreal  cvideal
From Appendix A.2:

3100 
 J 
  R  22.0 
cvreal  R 3.376  5.57  10  4 500 
 mol  K 
500 2 

(b)
As the diaphragm ruptures, the total internal energy of the system remains constant. Because the
volume available to the molecules increases, the average distance between molecules also
increases. Due to the increase in intermolecular distances, the potential energies increase. Since
the total internal energy does not change, the kinetic energy must compensate by decreasing.
Therefore, the temperature, which is a manifestation of molecular kinetic energy, decreases.
36
(c)
Because the heat capacity is ideal under these circumstances we can create a two-step
hypothetical path to connect the initial and final states. One hypothetical path is shown below:
v
vf ,Tf
step 2
u
2
s2
step 1
vi ,Ti
u1, s1
Tf
500
For the first section of the path, we have
Tf
Tf
u1 

cvreal dT
  cvideal dT
Ti
Ti

Tf


3100 
4
2.376  5.57 10 T  2  dT
T 
500 K i 
25773.4
u1  2.32 10  3 T f2  19.75T f 
 10507.4
Tf
u1  R



For the second step, we can use the following equation
vf
u 2 
 u 
  v T dv
vi
If we apply Equation 5.40, we can rewrite the above equation as
vf
u 2 
  P 


T
P



   T  v  dv
vi
For the van der Waals EOS, P 
RT
a
 2 ,
v b v
37
T
R
 P 

 
 T  v v  b
Therefore,
v f  0.002
v f  0.002
a
 RT

 J 
u1 
  v  b  P dv   v 2 dv  73.7  mol 
v i  0.001
v i  0.001
Now set the sum of the two internal energies equal to zero and solve for Tf:


u1  u 2  2.32  10  3 T f2  19.75T f 
25773.4
 10507.4  73.7  0
Tf
T f  497 K
(d)
Since the system is well-insulated
suniv  s sys
To solve for the change in entropy use the following development:
 s 
 s 
ds sys  
 dT    dv
 v T
 T  v
Using the thermodynamic web, the following relationships can be proven
cv
 s 
 

 T  v T
 P 
 s 

  
 v T  T  v
For the van der Waals EOS
R
 P 

 
 T  v v  b 
Now we can combine everything and calculate the change in entropy
38
497 K
s sys 

500 K
cv
dT 
T
0.002
R
 v  b  dv
0.001


0.002
497 K  2.376
dv
3100 
dT

s sys  R   
 5.57  10  4 


5
500 K  T
T3 
0.001 v  3.95  10
 J 
suniv  s sys  5.80 

 mol  K 

39




5.21
A schematic of the process is given by:
well
insulated
V = 0.1 m3
V = 0.1 m3
T = 300 K
V = 0.2 m3
T=?
Vacuum
nA =400 moles
nA=400 moles
State f
State i
Energy balance:
u  0
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
step 2
v [m 3/mol]
ideal gas
step 1
step 3
0.2/400
vf ,Tf
u
0.1/400
vi ,Ti
Tf
300
For the first section of the path, we have
v 
u1 
 u 
  dv
 v T
vi

If we apply Equation 5.40, we can rewrite the above equation as
v 
u1 
  P 

  P  dv
T 
 T  v

vi 

40
T [K]
For the van der Waals EOS
R
a
 P 


 
2
 T  v v  b T v 2
Therefore,
v 
u1 
v 
 RT

a
2a
 J 
P



 4  v  b Tv 2  dv   4 T v 2 dv  1120  mol 
i
vi  2.510
vi  2.510
Similarly for step 3:
v f  510 4
u3 

v
 RT

a
 v  b  2  P  dv 
Tv


v f  510 4

v 
2a
Tf v
2
dv 
 168000  J 
 mol  K 
Tf
For step 2, the molar volume is infinite, so we can use the ideal heat capacity given in the
problem statement to calculate the change in internal energy:
u 2 

3
R T f  300 K
2

If we set sum of the changes in internal energy for each step, we obtain one equation for one
unknown:


 168000
 J  3
 J 
u1  u 2  u3  1120 
  8.314 
 T f  300 K 
0


Tf
 mol  K  
 mol  2 
Solve for Tf:
T f  261.6 K
41
5.22
A schematic of the process is shown below:
Ethane 3 MPa; 500K
T surr = 293 K
Initially:
vacuum
(a)
Consider the tank as the system. Since kinetic and potential energy effects are negligible, the
open system, unsteady-state energy balance (Equation 2.47) is
 dU 

   nin hin   n out hout  Q  W s
dt

 sys in
out
The process is adiabatic and no shaft work is done. Furthermore, there is one inlet stream and no
outlet stream. The energy balance reduces to
 dU 
  nin hin

 dt  sys
Integration must now be performed
U2
t
U1
0
 dU   n
h dt
in in
t
n2 u 2  n1u1  hin  n in dt  nin hin  (n2  n1 )hin
0
Since the tank is initially a vacuum, n1=0, and the relation reduces to:
u 2  hin
42
As is typical for problems involving the thermodynamic web, this problem can be solved in
several possible ways. To illustrate we present two alternatives below:
Alternative 1: path through ideal gas state
Substituting the definition of enthalpy:
u 2  uin  Pin vin
or
u2 at 3 MPa, T   uin at 3 MPa, 500 K   Pin vin
(1)
From the equation of state:



 J 
Pin vin  RT 1  B' P    8.314 
552 K  1  2.8  10 8 3  10 6 Pa  3,800

mol
K





 J 
 mol  (2)
The change in internal energy can be found from the following path:
P
step 1
Plow
u1
u3
step 3
3 MPa
u2
step 2 ideal gas
T2
500 K
T
For steps 1 and 3, we need to determine how the internal energy changes with pressure at
constant temperature: From the fundamental property relation and the appropriate Maxwell
relation:
 u 
 s 
 v 
 v 
 v 
   T    P   T 
  P 
 P T
 P T
 P T
 T  P
 P T
From the equation of state
RT
 u 
1  B'P   P  RT2    B' RT
  
P
 P T
 P 
43
So for step 1:
 u 
u1     dP    B ' RTdP B ' RT Pin  349 [J/mol]
P  T
Pin 
Pin
0
0
(3)
and for step 3:
 u 
u 3     dP    B ' RTdP   B ' RT P2  0.7T
P  T
0
0
P2
P2
(4)
For step 2
T
T
 h   Pv  
 u 
u 2   
 dT   c P  R dT
 
 dT   
T  P  T  P 
T  P
500 
500 
500
T
or
u 2  R
 0.131  19.225 10
T2
3

T  5.56110 6 T 2 dT
T1 500 K
Substituting Equations 2, 3, 4, and 5 into 1 and solving for T gives:
T2  552 K
Alternative 2: real heat capacity
Starting with:
u 2  hin
The above equation is equivalent to
h2  P2 v2  hin
 h2  hin  P2 v2
To calculate the enthalpy difference, we can use the real heat capacity
ideal
c Preal  c P

P   2
 v
T 
2


Pideal   T

 
 dP
 
P 
For the truncated viral equation,
44
(5)
  2v 

 0
 T 2 

P
Therefore,
ideal
c Preal  c P
Now, we can calculate the change in enthalpy and equate it to the flow work term.
T2
 cP
ideal
dT  P2 v2
T1  500 K
 1.131  19.225 10
T2
R
3

T  5.561 10 6 T 2 dT  P2 v2  RT2 1  B' P2 
T1 500 K
Integrate and solve for T2:
T2  552 K
(b)
In order to solve the problem, we will need to find the final pressure. To do so, first we need to
calculate the molar volume. Using the information from Part (a) and the truncated virial
equation to do this
v
RT
1  B' P  
P

 J 
552 K 
 8.314 
 mol  K  

6
3  10 Pa
1  2.8 10 3 10 Pa 
8
6
 m3 
v  0.0014 

 mol 
This quantity will not change as the tank cools, so now we can calculate the final pressure.

 m3  
P2  0.0014 


mol

 

 1  2.8 10 8 P2

 J 
293 K 
 8.314 
 mol  K  



Solve for P2 :
45
P2  1.66  10 6 Pa
The entropy change of the universe can be expressed as follows:
S univ  S sys  S surr
To solve for the change in entropy of the system start with the following relationship:
 s 
 s 
ds sys  
 dT    dP
 T  P
 P T
Alternative 1: path through ideal gas state
Using the proper relationships, the above equation can be rewritten as
ds sys 
cP
 v 
dT  
 dP
T
 T  P
We can then use the following solution path:
P
s3
1.66 MPa
step 1
s 1
s 2
Plow
step 2 ideal gas
500 K
Choosing a value of 1 Pa for Plow, for step 1:
R
 v 
'
s1   
 dP    1  B P dP
T  P
P
1.66 MPa 
1.66 MPa
1 Pa
1 Pa
For step 3,
3 MPa
s1 

1 Pa
 v 

 dP 
 T  P
3 MPa

1 Pa
step 3
3 MPa



R
1  B ' P dP
P
46
T2
T
For step 2:
293 K
cP
1.131

dT
R

 19.225  10 3  5.561 10 6 T  dT



T
T

552 K
552 K 
293 K
s 2 
Adding together steps 1, 2 and 3:
 J 
s sys  46.9 
 mol  K 
______________________________________________________________________________
Alternative 2: real heat capacity
Using the proper relationships, the above equation can be rewritten as
c real
 v 
ds sys  P dT  
 dP
T
 T  P
For the truncated virial equation
 v 
1


  R  B ' 
 T  P
P

Now, substitute the proper values into the expression for entropy and integrate:
293 K
s sys  R

552 K
1.6610 6 Pa
1.131
3
6 
 T  19.225  10  5.561  10 T  dT  R
1

  B' dP
P

310 ^ Pa

 J 
s sys  46.9 
 mol  K 
______________________________________________________________________________
In order to calculate the change in entropy of the surroundings, first perform an energy balance.
u  q
Rewrite the above equation as follows
h  Pv   q
47
Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change,
we obtain the following equation
Tf
 cP
ideal


dT  v P f  Pi  q
Ti
T f  293K
R

1.131  19.225 10
Ti  552 K
3


 J 
q  15845 
 mol 
Therefore,
 J 
q surr  15845 
 mol 
and
 J 
s surr  54.08 
 mol  K 
Before combining the two entropies to obtain the entropy change of the universe, find the
number of moles in the tank.
n


 m3  
6
6
T  5.561  10  6 T 2 dT   0.0014 
  1.66  10 Pa - 3  10 Pa  q


mol

 

 
0.05 m 3
 75.7 mol
 m3 
0.0014 

 mol 
Now, calculate the entropy change of the universe.

 J 
 J 
S univ  75.7 mol 54.08 
 46.9 


 mol  K 
 mol  K  

J
S univ  544  
K
48
5.23
First, focus on the numerator of the second term of the expression given in the problem
statement. We can rewrite the numerator as follows:


uTr , v r  uTideal
 uTr , v r  uTideal
 uTideal
 uTideal
r ,vr
r ,vr  
r ,vr
r ,vr  

For an ideal gas, we know
uTideal
 uTideal
0
r ,vr
r ,vr  
Therefore,
uTr , v r  uTideal
 uTr , v r  uTideal
,vr
,vr 
r
r
Substitute this relationship into the expression given in the problem statement:
uTdep
,v
r
r
RTc
uTr , v r  uTideal
uTr , v r  uTideal
,vr
,vr 
r
r


RTc
RTc
Now, we need to find an expression for uTr , vr  uTideal
. Note that the temperature is constant.
r ,vr  
Equation 5.41 reduces to the following at constant temperature:
  P 

duT  T 
  P  dv
  T  v

The pressure can be written as
P
zRT
v
and substituted into the expression for the differential internal energy
  RT  z 
 RT 2  z  
RT 
zRT 





duT  T 
dv




  dv

v 
v 
v
T

 v 
  v  T  v

v

Applying the Principle of Corresponding States
49
 T 2  z
  r 
RTc  vr  Tr

duTr
 
  dvr
 v r 
If we integrate the above expression, we obtain
vr  2
uTr , v r  uTideal
T  z 
r, vv  

   r 
 RTc 
RTc
 v r  Tr  v
v 
v  
r
v
duTr

 dv r


Therefore,
uTdep
,v
r
RTc
r
vr  2
uTr , v r  uTideal
uTr , v r  uTideal
T  z 
v
,
,vr 
r
r
r



   r 
RTc
RTc
 v r  Tr  v
v  
r
50

 dv r


5.24
We write enthalpy in terms of the independent variables T and v:
 h 
 h 
dh  
 dT    dv
 T v
 v T
using the fundamental property relation:
dh  Tds  vdP
At constant temperature, we get:
  P 
 P  
dhT  T 
  v   dv
 v T 
  T  v
For the Redlich-Kwong EOS
1
R
a
 P 


 
3
/
2
 T  v v  b 2 T
vv  b 
 RT
a
a
 P 


  
 v T v  b 2 T 1 / 2 v 2 v  b  T 1 / 2 vv  b 2
Therefore,
 RT

3
RTv
a
a



dhT  
 dv
 v  b v  b 2 2 T 1 / 2 vv  b  T 1 / 2 v  b 2 
To find the enthalpy departure function, we can integrate as follows
v
h
dep

 dhT 
v 
 RT

RTv
a
a
3




  v  b v  b 2 2 T 1 / 2 vv  b  T 1 / 2 v  b 2  dv

v  
v
Since temperature is constant, we obtain
h dep 
3a
RTb
a
 v 


ln

v  b 2bT 1 / 2  v  b  T 1 / 2 v  b 
To calculate the entropy departure we need to be careful. From Equation 5.64, we have:


gas
gas
ideal gas
gas
sT , P  sTideal
 sT , P  sTideal
 sTideal
,P
, P  0  sT , P
, P 0
51

However, since we have a P explicit equation of state, we want to put this equation in terms of v.
Let’s look at converting each state. The first two states are straight -forward
sT , P  sT , v
and
gas
ideal gas
sTideal
, P  0  sT , v  
For the third state, however, we must realize that the ideal gas volume v’ at the T and P of the
system is different from the volume of the system, v. In order to see this we can compare the
equation of state for an ideal gas at T and P
P
RT
v'
to a real gas at T and P
P
RT
a

vb
T vv  b 
The volume calculated by the ideal gas equation, v’, is clearly different from the volume, v,
calculated by the Redlich-Kwong equation. Hence:
gas
gas  ideal gas
gas 
sTideal
 s ideal' gas  sTideal
s '
 sTideal

,P
,v
,
v
,
T ,v
T
v


Thus,



gas
gas
ideal gas
gas  ideal gas
gas 
 sT , v  sTideal
 sTideal
 sTideal
sT , P  sTideal

,P
, v    sT , v
, v     sT , v '
,v

Using a Maxwell relation:
 ds 
 P 
  

 dv T  T  v
Therefore,
 P 
dsT  
 dv
 T  v
52
For the Redlich-Kwong EOS
1
R
a
 P 


 
3
/
2
 T  v v  b 2 T
vv  b 
so
s

ideal gas
T , v  sT , v   
v 
R

a
1


 dv
3/ 2

2
v
b



T
v
v
b


v  

For an ideal gas
R
 P 

 
 T  v v
so


gas
gas
 sTideal
sTideal
,v
,v  
v
R
 v  dv
v 

Finally:
v'
dv
v'
RT
ideal gas
ideal gas
s '
 sT , v
R
 R ln  R ln
T ,v
v
v
Pv
v

Integrating and adding together the three terms gives:
s dep  R ln
v  b  
v
RT
 v 
ln
 R ln

Pv
2bT 3 / 2  v  b 
a
53
5.25
Calculate the reduced temperature and pressure:
Tc  647.3 K 
Pc  220.48 bar 
(Table A.1.2)
w  0.344
300 bar 
 1.36
220.48 bar 
673.15 K
Tr 
 1.04
647.3 K
Pr 
By double interpolation of data from Tables C.3 and C.4
 h dep
 Tr , Pr
 RTc





( 0)
 2.921
 h dep
 Tr , Pr
 RTc





(1)
 s dep
 Tr , Pr
 R





(1)
 h dep
Tr , Pr
 w
 RTc





 1.459
From Tables C.5 and C.6:
 s dep
 Tr , Pr
 R





( 0)
 2.292
 1.405
Now we can calculate the departure functions
  dep
 hTr , Pr
h dep  RTc  

  RTc
 




( 0)
(1) 





 J 
 J 
h dep   8.314 
647.3  2.921  0.344 1.459  18421 

 mol  K  
 mol 

  dep
 sTr , Pr
s dep  R  

 R
 




( 0)
 s dep
Tr , Pr
 w
 R





(1) 





 J 
 J 
s dep   8.314 
 2.292  0.344 1.405  23.07 

 mol  K  
 mol  K 

54
To use the steam tables for calculating the departure functions, we can use the following
relationships.
h dep  hT , P  hTideal
,P
s dep  sT , P  sTideal
,P
From the steam tables
 kJ 
 kJ 
hT , P  2151.0   and sT , P  4.4728 

 kg 
 kg  K 
We need to calculate the ideal enthalpies and entropies using the steam tables’ reference state.
vap
0 .01º C  
hTideal
, P  h
673.15 K
c ideal
dT
p
273.16 K

 kJ 
from the steam tables and heat capacity data from Table A.2.2.
We can get h vap  45.1 
 mol 
Using this information, we obtain
 kJ  
 kJ  
hTideal
  0.008314 

, P  45.1 

mol
mol  K  

 
673.15 K 
 3.47  1.45  10
 273.16 K 

 kJ 
hTideal
, P  59.14 
 mol 
Now, calculate the ideal entropy.
vap
0 .01º C 
sTideal
, P  s
673.15 K c ideal
p

273.16 K
T
P 
dT  R ln 2 
 P1 
From the steam tables:
 kJ 
s vap 0.01 º C   0.165 
 mol  K 
Substitute values into the entropy expression:
55
3
0.121  10 5 
T
 dT
T2

673.15 K 
5
kJ   
3.47
30


 3 0.121  10

dT





1
.
45
10
ln






 mol  K   273.16 K  T
 0.000613 
T3



sTideal
, P  0.165   0.008314 

 J 
sTideal
, P  107  mol  K 


Now, calculate the departure functions:
 kJ 
 kJ 
 kJ 
h dep  2151.0   0.0180148 kg/mol  59.14 
 20.4 

 mol 
 mol 
 kg 
 kJ 
0.0180148 kg/mol  0.107  kJ   0.0264  kJ 
s dep  4.4728 

 mol  K 
 mol  K 
 kg  K 
Table of Results
Generalized
Percent Difference
Steam Tables
Tables
(Based on steam tables)
 kJ 
h dep 
 mol 
 kJ 
s dep 
 mol  K 
-18.62
-20.4
9.9
-0.0231
-0.0264
12.5
56
5.26
State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature and
pressures are
30 bar 
 0.616
48.74 bar 
300 K
T1, r 
 0.982
305.4 K
P1, r 
50 bar 
 1.026
48.74 bar 
400 K
T2,r 
 1.31
305.4 K
P2,r 
and
  0.099
By double interpolation of data in Tables C.3 and C.4
( 0)
 h dep
 T1, r , P1, r
 RT
c





 h dep
 T2 , r , P2 , r
 RT
c





(1)
 0.825
 h dep
 T1, r , P1, r
 RT
c





 0.711
 h dep
 T2 , r , P2 , r
 RT
c





( 0)
 0.799
(1)
 0.196
Therefore,
 h dep 
 T1, r , P1, r 
 RT   0.825  0.099 0.799   0.904
c 


 h dep 
 T2 ,r , P2 ,r   0.711  0.099 0.196  0.730
 RTc 


The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cP data from Table
A.2.1.
400 K
hTideal
R
T
1
2
1.131  19.225 10
3
T  5.56110  6T 2 dT  717.39 R
300 K
The total entropy change is
57
dep
dep
h  hT , P  hTideal
 hT , P
T2
1, r
1, r
1
2,r
2,r
h  R  0.904TC  717.39  0.730TC 

 J 
h   8.314 
 0.904305.4 K   717.39 K  0.730305.4 K 
 mol  K  

 J 
h  6406.2 

 mol 
Using the data in Table C.5 and C.6
 s dep
 T1, r , P1, r

R

 s dep
 T2 , r , P2 , r

R


  0.601  0.099 0.756   0.676




  0.394  0.099 0.224   0.416

Substituting heat capacity data into Equation 3.62, we get
400 K 1.131  19.225  10 3 T  5.561  10 6 T 2
 50 bar 
dT  ln
s ideal  R  

T
 30 bar 
300 K
s ideal  1.542 R
Therefore,
s  sT , P  s ideal  sT , P  R0.676  1.542  0.416 
1, r
1, r
2,r
2,r
dep
dep
 J 
s  14.98 

 mol  K 
58
5.27
The turbine is isentropic. Therefore, we know the following
dep
dep
s  sT , P  s ideal  sT , P  0
1, r
1, r
2,r
2,r
Using the van der Waals EOS, we can find P1,r, which leaves one unknown in the above
equation: T2.
P1 


 3
 82.06  cm  atm  623.15 K 


 mol  K  


 3 
 3
 600  cm   91  cm  


 mol  
 mol 


 atm  cm 3 
91  10 5 

2

 mol

 3 
 600  cm  


 mol  

P1  75.19 atm  76.19 bar 
2
Calculate reduced temperature and pressures using data from Table A.1.1
76.19 bar 
 1.8
42.44 bar 
623.15 K
T1, r 
 1.68
370.0 K
P1, r 
P2,r 
1.013 bar 
 0.024
42.44 bar 
Also,
  0.152
From Tables C.5 and C.6:
 s dep
 T1, r , P1, r

R


  0.327  0.152 0.102   0.343


Substituting heat capacity data into Equation 3.62, we get
s
ideal
 T2 1.213  28.785  10  3 T  8.824  10  6 T 2

 1 atm 

R
dT  ln

75.19 atm 
T
 

623.15 K

Therefore,
59

 s dep
T2
1.213  28.785  10 3 T  8.824  10 6 T 2

 T2 , r , P2 , r
R 0.343 
dT

4
.
32



T
R

623.15 K


We can solve this using a guess-and-check method
T2  600 K : T2,r  1.62
 J 
s  33.84 

 mol  K 
T2  450 K : T2,r  1.22
 J 
s  0.77 

 mol  K 
T2  446.6 K : T2,r  1.21
 J 
s  0 
 mol  K 
Therefore,
T2  446.6 K
60


   s
 
5.28
A reversible process requires the minimum amount of work. Since the process is reversible and
adiabatic
s  0
which can be rewritten as
dep
dep
s  sT , P  s ideal  sT , P  0
1, r
1, r
2,r
2,r
Calculate reduced temperature and pressures using data from Table A.1.1
1 bar 
 0.0217
46.0 bar 
300 K
T1, r 
 1.57
190.6 K
P1, r 
P2,r 
10 bar 
 0.217
46.0 bar 
From Tables C.5 and C.6:
 s dep
 T1, r , P1, r

R



  0.00457  0.008 0.0028  0.0046

Substituting heat capacity data into Equation 3.62, we get
s
ideal
 T2 1.702  9.081 103T  2.164  106 T 2

 10 bar 
 R 
dT  ln

T
300 K
 1 bar 


Therefore,

 s dep
T2
3
6 2




T
T
1
.
213
28
.
785
10
8
.
824
10

 T2 , r , P2 , r
s  R 0.0046  
dT  2.303  
T
R

300 K


We can solve using a guess-and-check method
T2  400 K : T2,r  2.10
 J 
s  4.98 

 mol  K 
T2  385 K :
T2,r  2.02
61



 
 J 
s  1.42 

 mol  K 
T2  379 K :
T2,r  1.99
 J 
s  0.018 

 mol  K 
Therefore,
T2  379 K
An energy balance reveals that
h2  h1  h  ws
We can calculate the enthalpy using departure functions. From Tables C.3 and C.4:
 h dep 
 T1, r , P1, r 
 RT
  0.0965  0.008 0.011  0.0966
c


 h dep 
 T2 , r , P2 , r 
 RT
  0.0614  0.00890.015  0.0613
c


Ideal heat capacity data can be used to determine the ideal change in enthalpy
379 K

h ideal  R   1.702  9.081  10  3 T  2.164  10  6 T 2 dT 
300 K

Therefore,
379 K


 J  
3
6 2





190
.
6
K
0
.
0966
0
.
0613
1
.
702
9
.
081
10
T
2
.
164
10
T
dT
h   8.314 









 mol  K   


300 K
 J 
w s  h  3034.2 

 mol 
and

 mol  
 J 


3034
.
2
W S  1 / 30 



 mol    101.1 W
 s  



62
5.29
Equation 4.71 states
1  v 
  
v  T  P
 v 
 v  

 T  P
This can be substituted into Equation 5.75 to give
 JT 
vT  1
cP
63
5.30
For an ideal gas
R
 v 

 
 T  P P
Therefore,
 RT

 v

P
  v  v   0
 JT  
cP
cP
This result could also be reasoned from a physical argument.
64
5.31
The van der Waals equation is given by:
P
RT
a

v  b v2
(1)
The thermal expansion coefficient is given by:
1  v 
 1   T 
    
v  T  P  v   v  P
 
(2)
Solving Equation 1 for T:
a  v  b 

T   P  2 

v  R 

Differentiating by applying the chain rule,

a  1  v  b  2a Pv 3  av  2ab
 T 


    P  2   
 v  P 
v  R  R  v3
Rv 3
Substitution into Equation 2 gives

Rv 2
Pv 3  av  2ab
Substituting Equation 1 for P gives b in terms of R, T, v, a , and b:

Rv 2 v  b 
2
RTv 3  2av  b 
The isothermal compressibility is given by:
1  v 
 1   P 
     
v  P  T
 v   v  T
  
From the van der Waals equation:
RT
2a  RTv 3  2av  b 
 P 


  
2
v  b 2 v 3
v 3 v  b 
 v T
2
so
65
(3)

v 2 v  b 
2
RTv 3  2av  b 
2
For the Joule-Thomson coefficient, we can use Equation 5.75:
 JT 
c Pideal
  v 

T    v 
  T  P

Preal
2
  v 
  T  2   dP
Pideal 
  T  P 
Substituting the van der Waals equation into Equation 3 gives
RTv 3  2av  b 
T
2av  b 
 T 


  
3
v  b Rv
v  b  Rv 3
 v  P
2
(4)
Thus, the second derivative becomes:
  2T 
1  T 
2a 6av  b 
T
 2   

   3
2
Rv 4
v  b  v  b  v  P Rv
 v  P
or simplifying using Equation 4,
  2T 
2a v  3b 
 2  
Rv 4
 v  P
(5)
Substituting Equations 5 and 4 into Equation 5.75 gives:
 bRTv 3  2avv  b 
2
RTv 3  2av  b 

Preal
 RTv 4 
dP
c Pideal   
2av  3b  
Pideal 
2
 JT
At a given temperature the integral in pressure can be rewritten in terms of volume using the van
der Waals equation to give:
66
 bRTv 3  2avv  b 
2
RTv 3  2av  b 
vreal
 RTv  RTv 3  2av  b 2
dv
  
2av  3b  
v  b 2
videal 
2
 JT 
c Pideal
67
5.32
We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation
5.75. The following approximation can be made
 vˆ 
 vˆ 

 

 T  P  T  P
At 300 ºC,
vˆ350 º C,1MPa   vˆ250 º C,1MPa 
 vˆ 

 
350  250 º C
 T  P
 m3 
 m3 
0.28247 

  0.23268 
 vˆ 
 kg 
 kg 

 
350  250 º C
 T  P
 m3 
 m3 
 vˆ 

  0.0005 

  0.0005 
 T  P
 kg  º C 
 kg  K 
A similar process was followed to find cP.
 ˆ
 hˆ 
   h 
cˆ P  



 dT  P  T  P
At 300 ºC,
 hˆ 
hˆ350 º C,1MPa   hˆ250 º C,1MPa 

 
 T 
350  250 º C

P
 kJ 
 kJ 
3157.7    2942.6  
 kg 
 kg 

350  250 º C
 hˆ 


 T 

P
 hˆ 
 ˆ




   h   2.15  kJ   2.15  kJ 
cˆ P  



 kg  º C 
 kg  K 
 dT  P  T  P
Now,  JT can be found.
68
3 
3 


  vˆ 
 573.15 K  0.0005  m    0.25794  m  




T 



  T   vˆ  

kg
kg
K



 




P

 JT  
cˆ P
 kJ 
2.15 

 kg  K 
 m3  K 

 kJ 
 JT  0.0133 
69
5.33
At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.75:
 JT 
c Pideal

  v 
T    v 

  T  P
0
Preal
2
  v 
  T  2   dP
  T  P 
Pideal 
This is true when the numerator is zero, i.e.,
  v 

T    v   0
  T  P

For the van der Waals equation, we have
P
RT
a
 2
v b v
Solving for T:
a  v  b 

T   P  2 

v  R 

so
a  1  v  b  2a Pv 3  av  2ab
 T 


   P  2  

v  R  R  v3
Rv 3
 v  P 
Substituting for P:
RTv 3  2av  b 
 T 
  
v  b Rv 3
 v  P
2
Hence,
 bRTv 3  2avv  b 
 v 
T   v  0 
2
RTv 3  2av  b 
 T  P
2
Solving for T:
2avv  b 
bRv 3
2
T
(1)
70
Substituting this value of T back into the van der Waals equation gives
P
2avv  b  a a2v  3b 
 2 
bv 3
v
bv 2
(2)
We can solve Equations 1 and 2 by picking a value of v and solving for T and P. For N2, the
critical temperature and pressure are given by Tc = 126.2 [K] and Pc = 33.84 [bar], respectively.
Thus, we can find the van der Waals constants a and b:
2

RT 
 Jm 3 
a  27  c  = 0.137 
2
64 Pc
 mol 
b
 m3 
RTc
= 3.88  10 -5 

8 Pc
 mol 
Using these values in Equations (1) and (2), we get the following plot:
Joule-Thomson inversion line
1000
800
600
400
200
0
0
100
200
T [K]
71
300
400
5.34
We can solve this problem using departure functions, so first find the reduced temperatures and
pressures.
50 bar 
 0.99
50.36 bar 
273.15 K
T1, r 
 0.967
282.4 K
P1, r 
P2,r 
10 bar 
 0.2
50.36 bar 
Since the ethylene is in two-phase equilibrium when it leaves the throttling device, the
temperature is constrained. From the vapor-liquid dome in Figure 5.5:
T2,r  0.76
T2  214.6
The process is isenthalpic, so the following expression holds
dep
dep
h  hT , P  hTideal
 hT , P  0
T2
1, r
1, r
1
2,r
2,r
Therefore,
dep
dep
hT , P  hT , P  hTideal
T2
2,r
2,r
1, r
1, r
1
From Table A.2.1:
 1.424  14.394 10
214.6 K
hTideal
1  T2
R
3

T  4.392  10  6 T 2 dT
273.15 K
From Tables C.3 and C.4   0.085 :
 h dep
 T1,r , P1,r
 RT
c


  3.678  0.085 3.51  3.976


Now we can solve for the enthalpy departure at state 2.
72
hTdep, P
2,r
2,r
RTc
hTdep, P
2,r
2,r
RTc
214.6 K


1
3
6 2

1
.
424
14
.
394
10
T
4
.
392
10
T
dT
  3.976 





282.4 K


273.15 K


 3.01
We can calculate the quality of the water using the following relation
hTdep, P
2,r
2,r
RTc
 1  x 
hTdep,, liq
P
2,r
2,r
RTc
x
hTdep,, vap
P
2,r
2,r
RTc
where x represents the quality. From Figures 5.5 and 5.6:
hTdep,, liq
P
2,r
2,r
RTc
hTdep,, vap
P
2,r
2,r
RTc
 4.6  0.085 5.5  5.068
 0.4  0.0859 0.75  0.464
Thus,
x  0.447
55.3% of the inlet stream is liquefied.
73
5.35
Density is calculated from molar volume as follows:

MW
v
2
Substitute the above into the expression for Vsound
:


P
2
Vsound  
  MW
 
  v


  1  P 
MW   1 / v   s


 s
The following can be shown using differentials:
v
1
   
v
v2
Therefore,
 P 
v 2  P 
2
Vsound
    
 
MW  v  s
   s
74
5.36
From Problem 5.35:
2
Vsound
 P 
v 2  P 
    
 
MW  v  s
   s
The thermodynamic web gives:
 P   s 
 T   P 
 T   s   P   s 
 P 

        
 
  
     
 v  s
 s  v  v  P  v  s  T  s
 s  v  v T  s T  T  P
T
 P 
   
 v  s
 cv
c
 s   P   c P 
    
   P
 cv
 v T  s T  T 
 P   T 

 

 T  v  v  P
If we treat air as an ideal gas consisting of diatomic molecules only
R
 P 

 
 T  v v
cP 7

cv 5
P
 T 

 
 v  P R
Therefore,
 P 
 7  P 
     
 v  s
 5  v 
and
Vsound 
v 2  7  P 
   
MW  5  v 
7  RT 


5  MW 
Vsound  343 m/s
The lightening bolt is 1360 m away.
75
5.37
From Problem 5.35:
2
Vsound
 P 
v 2  P 
    
 
MW  v  s
   s
The thermodynamic web gives:
 P   s 
 T   P 
 T   s   P   s 
 P 

        
 
  
     
 v  s
 s  v  v  P  v  s  T  s
 s  v  v T  s T  T  P
so
 T  P   T   c P 
 P 
  
    
 
 v  s
 cv  T  v  v  P  T 
For liquids c P  cv water at 20 ºC, so
 P 
 P   T 
   
 

 v  s
 T  v  v  P
However, the cyclic rule gives:
 P   T   v 
1  
 
  
 T v  v  P  P T
So
 P 
 P 
   
 v  s  v T
From the steam tables, for saturated water at 20 oC:
 m3 
ˆ

.
001002
P = 2.34 kPa and v
 
 kg 
For subcooled water at 20 oC:
P = 5 MPa and vˆ  . 0009995
m3 


kg


76
So
 kg kPa 
5000  2.34
 P 
 P 
 P 
    



 vˆ  s  vˆ T  vˆ  .0009995  .001002  m 3 
and
 P 
Vsound  vˆ 2    1414 m/s
 vˆ  s
77
5.38
(a)
The fundamental property relation for internal energy is
dU  Qrev  Wrev
Substituting the proper relationships for work and heat, we obtain
dU  TdS  Fdz
The fundamental property relation for the Helmholtz energy is
dA  dU  d TS   dU  TdS  SdT
Substitute the expression for the internal energy differential:
dA  Fdz  SdT
(b)
First, relate the entropy differential to temperature and length.
 S 
 S 
dS  
 dT    dz
 T  z
 Z T
Now we need to find expressions for the partial derivatives.
 u 
 TS  Fz 
 S 
nc z  n
 
  T

T
 T  z 
z
 T  z
Therefore,
nc
 S 
a

  z  n  b 

T
 T  z
T

The following statement is true mathematically (order of differentiation does not matter):

Z
 A  
  A  
  

  
 T  z  T T  Z T  z
Furthermore,
78

Z
 A  
 S 
    

 Z T
 T  z T

T
 A  
 F 

    
 Z T  z  T  z
 S 
 F 
   
  k z  z0 
 Z T
 T  z
Substituting the expressions for the partial derivatives into the expression for the entropy
differential, we obtain
a

dS  n  b dT  k z  z 0 dz
T

(c)
First, start with an expression for the internal energy differential:
 U 
 U 
dU  
 dT  
 dz
 T  z
 Z T
From information given in the problem statement:
 U 

  na  bT 
 T  z
Using the expression for internal energy developed in Part (a) and information from Part (b)
 U 
 S 

  T    F  T  k  z  z 0   kT  z  z 0   0
 Z T
 z T
Therefore,
dU  na  bT dT  0dz  na  bT dT
(d)
We showed in Part (c) that
 U 
FU  
 0
 z T
79
 S 
Using the expression for   developed in Part B, we obtain
 z T
 S 
FS  T    kT  z  z0 
 z T
(e)
First, perform an energy balance for the adiabatic process.
dU  W
Substitute expressions for internal energy and work.
na  bT dT  Fdz  kT z  z 0 dz
Rearrangement gives
dT kT  z  z 0 

dz na  bT 
The right-hand side of the above equation is always positive, so the temperature increases as the
rubber is stretched.
80
5.39
The second law states that for a process to be possible,
suniv  0
To see if this condition is satisfied, we must add the entropy change of the system to the entropy
change of the surroundings. For this isothermal process, the entropy change can be written
ds 
cv
 P 
 P 
dT    dv    dv
T
 T  v
 T  v
Applying the van der Waals equation:
ds 
R
dv
v b
Integrating
s sys  R ln
v2
 J 
 11.5 
v1
 mol K 
For the entropy change of the surroundings, we use the value of heat given in Example 5.2:
 J 
q  q surr  600 
 mol 
Hence the entropy change of the surroundings is:
s surr 
q surr  600
 J 

 1.6 
373
Tsurr
 mol K 
and
 J 
suniv  s sys  ssurr  9.9 
 mol K 
Since the entropy change of the universe is positive we say this process is possible and that it is
irreversible.
Under these conditions propane exhibits attractive intermolecular forces (dispersion). The closer
they are together, on average, the lower the energy. That we need to put work into this system
says that the work needed to separate the propane molecules is greater than the work we get out
during the irreversible expansion.
81
5.40
A schematic of the process is given by:
Gas A in
Pi = 100 bar
Turbine
ws
Ti = 600 K
Gas A
out
Pf = 20 bar
Tf = 445 K
The energy balance for this process is provided below:
h  wS
Because the gas is not ideal under these conditions, we have to create a hypothetical path that
connects the initial and final states through three steps. One hypothetical path is shown below:
Pi ,Ti
100
ideal gas
P [bar]
step 1
h
Pf ,Tf
Plow
step 3
20
step 2
445
600
For the first section of the path, we have
P 0
h1 
 h 
  dP
 P T
P1

If we apply Equation 5.45 we can rewrite the above equation as
P 0
h1 


 v 
  v  dP
 Ti 
 T  P

Pi 

For the given EOS:
82
T [K]
R aP
 v 

   2
 T  P P Ti
Therefore,
P 0
v 0
 2aP



aP
J 

dP  2467 


P
v
dP
b
h1 





 5


Ti
T
 mol 


Pi 10010
Pi 10010 5  i
Similarly for step 3
Pf  2010 5

h3 
P 0
 2aP

 J 

 b dP  250 

 Tf

 mol 


For step 2, the pressure is zero, so we can use the ideal heat capacity given in the problem
statement to calculate the enthalpy change.
h2 
Tf
445 K
Ti
600 K
 J 
 c P dT   30  0.02T dT  6270  mol 
Now sum each part to find the total change in enthalpy:
 J 
h  h1  h2  h3  8487 
 mol 
 J 
ws  8487 
 mol 
In other words, for every mole of gas that flows through the turbine, 8487 joules of work are
produced.
83
Chapter 6 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
koretsm@engr.orst.edu
6.1
(a)
The Clausius-Clapeyron equation:
 P sat  h vap 1
1 
ln i
  i  
R T 373 [K] 

101 kPa
vap
dPisat hi dT
2
sat 
RT
Pi
or
so
Pi
sat
 h vap 1
1 
 101 kPaexp i  

 R T 373 [K]
(b) and (c)
Using
 kJ 
h vap  40.626 
 mol 
we obtain the following table
T [K]
Eqn 6.24
[kPa]
Steam
Tables [kPa]
%
Difference
273.156
278.15
283.15
288.15
293.15
298.15
303.15
308.15
313.15
318.15
323.15
328.15
333.15
338.15
343.15
348.15
353.15
358.15
363.15
368.15
373.15
0.84
1.16
1.58
2.13
2.84
3.76
4.93
6.40
8.24
10.54
13.36
16.82
21.04
26.13
32.26
39.58
48.28
58.56
70.67
84.84
101.35
0.6113
0.87
1.23
1.71
2.34
3.17
4.25
5.63
7.38
9.59
12.35
15.76
19.94
25.03
31.19
38.58
47.39
57.84
70.14
84.55
101.35
37.30%
33.02%
28.30%
24.51%
21.51%
18.62%
15.94%
13.68%
11.71%
9.86%
8.19%
6.75%
5.50%
4.40%
3.42%
2.58%
1.87%
1.25%
0.75%
0.34%
0.00%
2
The logarithmic trend is well-represented. However, at lower temperatures the Clausius kJ 
Clapeyron equation is up to 37% off. The actual heat of vaporization changes from 2501.3  
 kg 
 kJ 
at 0.01 oC to 2257.0   at 100 oC, a difference of around 10%.
 kg 
100.00
10.00
Eqn 6.23
Tables
1.00
0.10
273
283
293
303
313
323
333
343
353
363
373
Temperature [K]
(d)
For 100 ºC to 200 ºC, we obtain the following table:
T [K]
Eqn 6.24
[kPa]
Steam Tables
[kPa]
%
Difference
373.15
378.15
383.15
388.15
393.15
398.15
403.15
408.15
413.15
418.15
423.15
428.15
433.15
438.15
443.15
448.15
453.15
458.15
463.15
101.35
120.51
142.64
168.11
197.30
230.63
268.55
311.54
360.11
414.82
476.24
545.00
621.74
707.16
801.99
906.98
1022.93
1150.68
1291.10
101.35
120.82
143.28
169.06
198.53
232.1
270.1
313
361.3
415.5
475.9
543.1
617.8
700.5
791.7
892
1002.2
1122.7
1254.4
0.00%
0.26%
0.44%
0.56%
0.62%
0.63%
0.57%
0.47%
0.33%
0.16%
0.07%
0.35%
0.64%
0.95%
1.30%
1.68%
2.07%
2.49%
2.93%
3
468.15
473.15
1445.10
1613.62
1397.8
1553.8
3.38%
3.85%
10000.00
Eqn 6.23
1000.00
Tables
100.00
373
383
393
403
413
423
433
443
453
463
473
Temperature [K]
Over this range the Clausius-Clapeyron equation represents the data well and is no more than 4
% off. The actual heat of vaporization changes from 2257.0 kJ/kg  at 100 oC to 1940.7 kJ/kg 
at 200 oC, a difference of around 15%.
(e)
The heat of vaporization can be corrected for temperature as follows
hvap T  
Tb

T
l
cP
dT
 hvap Tb    c Pv dT
Tb
T
We can acquire heat capacity data from Appendix A.2, but to simplify the analysis, we will use
an average heat capacity for the vapor.
hvap T   75.4373.15  T   40626  34.13T  373.15
hvap T   56026  41.27T
Substitute this expression into the Clausius-Clapeyron equation
dPisat
Pisat

56026  41.27T  dT
RT 2
Integrate:
4
1 
1 
1
 T  
Pisat  101.35 kPa  exp   56026 
  41.27 ln
 
 T 373.15 
 373.15  
R 
Now plot the data as before from 0.01 ºC to 200 ºC.
T [K]
Eqn 6.24
[kPa]
Steam Tables
[kPa]
%
Difference
273.16
278.15
283.15
288.15
293.15
298.15
303.15
308.15
313.15
318.15
323.15
328.15
333.15
338.15
343.15
348.15
353.15
358.15
363.15
368.15
373.15
378.15
383.15
388.15
393.15
398.15
403.15
408.15
413.15
418.15
423.15
428.15
433.15
438.15
443.15
448.15
453.15
458.15
463.15
0.64
0.91
1.28
1.78
2.43
3.29
4.39
5.81
7.60
9.86
12.66
16.12
20.35
25.49
31.68
39.10
47.91
58.32
70.54
84.80
101.35
120.46
142.40
167.48
196.00
228.29
264.70
305.56
351.26
402.16
458.64
521.10
589.92
665.51
748.27
838.59
936.89
1043.55
1158.98
0.6113
0.87
1.23
1.71
2.34
3.17
4.25
5.63
7.38
9.59
12.35
15.76
19.94
25.03
31.19
38.58
47.39
57.84
70.14
84.55
101.35
120.82
143.28
169.06
198.53
232.1
270.1
313
361.3
415.5
475.9
543.1
617.8
700.5
791.7
892
1002.2
1122.7
1254.4
4.95
4.96
4.24
3.88
3.86
3.65
3.35
3.17
3.03
2.78
2.51
2.28
2.06
1.84
1.58
1.34
1.09
0.82
0.57
0.29
0.00
0.30
0.61
0.94
1.28
1.64
2.00
2.38
2.78
3.21
3.63
4.05
4.51
4.99
5.49
5.99
6.52
7.05
7.61
5
468.15
473.15
1283.56
1417.67
1397.8
1553.8
8.17
8.76
Water Saturation Pressure as a Function of Temperature
10000.00
Equation
Steam Tables
Psat (kPa)
1000.00
100.00
10.00
1.00
250
300
350
0.10
400
450
500
T (K)
The agreement between the two values at lower temperatures improves significantly at lower
temperatures, but actually worsens at higher temperatures. The agreement could potentially be
improved by not averaging the heat capacity.
6
6.2
We can find the required pressure by applying the Clapeyron equation:
dP
hl  h s

dT
vl  v s T


We can find the molar volume of water ice from any number of reference books. At 0 ºC and 1
bar:
 kg 
  1000  
 m3 
l
l
 v  1.80  10
5
 m3 


 mol 
 m3 
 v s  1.97  10  5 

 mol 
 kg 

 m3 
 s  917 
Also at 0 ºC and 1 bar,
 J 
h l  h s  6010 
 mol 




If we assume that h l  h s and v l  v s are independent of temperature and pressure, we can
separate variables in the Clapeyron equation and integrate.
P2  P1 
T 
ln 2 
v v
 T1 
hl  h s

l
s

P2  1  10 5 Pa  
so
1.80 10
6010 J/mol
5
 1.97  10
P2  66.1 bar 
7
5
 268.15 K 

ln
273
.
15
K
m / mol 


3

6.3
(a)
At 1 bar, the gas will act as an ideal gas.

 J 
300 K 
 8.314 
 m3 
RT 
 mol  K  
 0.0249 
v


P
1  10 5 Pa 
 mol 
The number of moles of vapor are found as follows (neglect molar volume of liquid)
V
n  
v
v
 
0.001 m 3
 m3 
0.0249 

 mol 
n v  0.0402 mol
(b)
At 21 bar, the gas will not behave ideally. Since we are assuming that the molar volume of
liquid is negligible and the heat of vaporization is independent of temperature, the Clapeyron
equation becomes
dP h vap
 v
dT
vT
The molar volume using pressure expansion of the virial equation is
vv 


RT 1  B ' P
 1

 RT 
 B'
P
P


Substituting this expression into the Clapeyron equation yields
dP

dT
h vap
1

R  B ' T 2
P

Separation of variables yields
P2  21 105 Par

P1 1 105 Pa
h vap
1
'
  B dP 
R
P

T2
dT
2
T1 300 K T

and integration results in
8
P
ln 2
 P1

 h vap
  B ' P2  P1  
R

1 1
  
 T2 T1 
We can substitute values for given quantities and constants to solve for T2.
T2  523.3 K
(c)
Using the virial equation,
3


1
1
 
 J 
7  m 







v v  RT   B '    8.314 
523
.
3
K
1
10
  

5


P
 
 mol  K  
 J 
 21  10 Pa
 m3 
v v  0.00164 

 mol 
We can assume the volume occupied by the liquid is negligible. Therefore,
nv 
V

vv
 
0.001 m 3
 m3 
0.00164 

 mol 
n v  0.61 mol
9
6.4
We can use the following computational path to solve for pressure at which graphite and
diamond are in equilibrium at 25 oC.
graphite
P = 1 [atm]
g  2866
g1 
diamond
 J 
mol
P
 v graphdP
g 3 
1
1
 v diam dP
P
P
graphite
g 2  0
diamond
Summing together the three steps we get:
 J 
g 1[atm]  2866 
 g1  g 2  g 3
 mol 
To find the change in Gibbs energy with pressure, we apply the fundamental property relation,
Equation 5.9. At constant temperature:
0
dgi  v idP  sidT
If the solid is assumed incompressible, we can integrate to get
g i   vi dP  vi P
Thus the sum of Gibbs energy becomes
P
1


 J 
g 1[atm]  2866 
   v graph dP  0   v diam dP  v graph  v diam P  1
 mol  1
P
Solving



1   cm 3    g   1  m 3  
 J    1
5

 2866 



 3  P  1.01  10 Pa 
 12 






6


 mol    2.26 3.51   g    mol   10  cm  

or
P = 1,514 [MPa] = 15,143 [bar]
10
6.5
From the Clausius-Clapeyron equation:
s
hTfus
h lAl  h Al
dP


s
s
dT
v lAl  v Al
T
v lAl  v Al
T

 

(I)
where h Tfus is the enthalpy of fusion at temperature T. We can get the molar volumes from the
densities:
 kg 
0.027 
MW
 mol   1.17  10  5
v lAl 

 kg 
ρl
2,300 

 m3 
 m3 


 mol 
 kg 
0.027 
MW
 mol   1.00  10  5
s
v Al


 kg 
ρs
2,700 

 m3 
 m3 


 mol 
and
so
 m3 
s
v lAl  v Al
 1.7  10 6 

 mol 
We can use the following path to calculate for h Tfus .
solid
liquid
T
h Tfus
933
 cP dT
T
T
933
 cPl dT
s
T = 933 [K]
solid
h fus  10,711
 J 
mol 
liquid
11
hTfus 
T
933.45
 J 
s
l
c P dT  10,711 
cP
dT 

 mol 
T
933.45


 J 
 J 
l
Using c Ps  20.608  0.0138T 
 and c P  31.748 
 , we get:
 mol K 
 mol K 
h Tfus  5,819.9  11.68T  0.0069T 2
Back into Equation (I) gives:
 5,819.9  11.68T  0.0069T 2 
dP  
 dT
1.7  10  6 T




Integrating:
100 [bar]
 5,819.9  11.68T  0.0069T 2 
 dT
 dP   
6

1
.
7
10
T


1
933.45 [K]
T


or
100  1105 1.7 10  6   5819.9 ln 933T.45   11.68T  993.45  0.00345T 2  993.452 
solving for T gives
T = 934.91 [K]
12
6.6
We can assume that silver acts as an ideal gas at 1500 K. We can also assume the molar volume
of the vapor is much greater than the molar volume of liquid. Therefore, we can use Equation
6.22
dP sat
P sat

h vap dT
RT 2
This can be rearranged to show
dP sat h vap P sat

dT
RT 2
This relation assumes
1. vv>>vl
2. Silver acts as an ideal gas
We can differentiate the expression for pressure in the problem statement to obtain
 14260 0.458 
  14260



 0.458 ln T   12.23 
 exp
2
dT
T 
 T
 T

sat
dP
 14260 0.458  sat


P
2
dT
T 
 T
dP sat
Therefore,
vap
 14260 0.458  h



2
T  RT 2
 T
Therefore,
h vap  R14260  0.458T 
At 1500 K,

 J 
14260  0.4581500 K 
h vap   8.314 
 mol  K  

 kJ 
h vap  112.8 
 mol 
13
6.7
For a single component system:
  Gi  g i  g
From the fundamental property relation given by Equation 5.9:
dg   sdT  vdP
We can identify a phase transition from the vertical line of the g vs. T plot, as indicated below.
Since this transition is vertical, i.e., the temperature is constant, the pressure must also be
constant. Thus, we can differentiate the Gibbs energy with respect to temperature at constant
pressure to get:
 g 

  s
 T  P
Hence the slope of a plot of g (or ) vs. T at any temperature must be the negative of the value of
entropy on the plot for s vs. T. The resulting curve is sketched below.

s
phase transition;
vertical line indicates
P = const

T




Thick line denotes
lowest value of 
Slope = the negative
value of s at the same
T on the curve above

T*
T
14
6.8
The ferrite phase has stronger bonds. At room temperature, iron is in the ferrite phase. The
heating to 912 ºC has the effect of increasing the entropy contribution to the Gibbs energy. At a
high enough temperature, the austenite phase becomes stable, so that its entropy must be greater
than the ferrite phase. If the entropy of the austenite phase is greater, the enthalpy of the ferrite
phase must be greater or else the austenite phase would be stable over the entire temperature
range. Hence, the ferrite phase has stronger bonds.
15
6.9
Since the pressures are low, we can assume ideal gas behavior. We can also assume that the
molar volume of the vapor is much greater than the molar volume of liquid and the heat of
vaporization is independent of temperature. Therefore, we can rearrange Equation 6.24 to obtain
h vap
h vap

 J    760 torr 
  8.314 
 ln

 mol  K    400 torr 


1
1





 353.25 K 333.75 K 
 kJ 
 32.3 
 mol 
This value is 7.7% smaller than the reported value.
16
6.10
(a)
The freezing point occurs where there is a discontinuity in the g vs. T plot, as indicated below.
The liquid is at a temperature higher than the freezing point and the solid at lower temperature.
These are demarked below. The melting temperature is 250 K, which occurs at a value g = 3,000
[J/mol]
Pure species "a"
6000
Gibb's Energy, g (J/mol)
1 unit
up
New Freezing
point
4,000
solid
Freezing
point
2,000
liquid
100
200
1.2
units
up
300
Temperature, T (K)
(b)
At constant pressure, the entropy can be found from Equation 5.14. For the solid we have:
g 1,000
 g 
 J 

 10 
s  
 

T
10
 T  P
 mol K 
And for the liquid, we get:
g 2,000
 g 
 J 

 40 
s  
 
T
50
 T  P
 mol K 
(c)
As we change pressure, we can see how the Gibbs energy changes at any given temperature by
Equation 5.14:
17
 g 
  v
 P  T
Assuming the molar volumes of the liquid and vapor stay constant over the temperature range
around the melting point, we see that the Gibbs energy of the liquid increases by 1.2 times the
Gibbs energy of the solid, since the molar volume of the liquid is 20% larger. The Gibbs energy
of the new freezing point at higher pressure is schematically drawn on the plot above. For
convenience, we choose the solid to increase by 1 unit on the plot. Thus, the liquid increases by
1.2 units. As the sketch shows, the freezing point, where the two lines intersect, will shift to
higher temperature.
18
6.11
For a single component system the fundamental property relation, Equation 5.9, gives:
dg   sdT  vdP
We can identify a phase transition from the vertical line of the g vs. P plot, as indicated below.
Since this transition is vertical, i.e., the pressure is constant, the temperature must also be
constant. Thus, we can differentiate the Gibbs energy with respect to pressure at constant
temperature to get:
 g 
  v
 P  T
Hence the slope of a plot of g vs. P must have a slope that matches the plot for v vs. T. Since the
molar volume of phase  is about twice the value of phase , its slope should be twice as big.
The resulting curve is sketched below.
v
phase transition;
vertical line indicates
T = const

v
v

P
g
 Slope of top line is
Straight line since
v is constant


Thick line denotes
lowest value of g

P*
P
19
about twice as big as
the slope of bottom
line
6.12
The saturation pressure can be found using the Clausius-Clapeyron equation with the assumption
that the heat of vaporization is independent of temperature. First, we need to use the given data
for the 63.5 ºC and 78.4 ºC to find the heat of vaporization.
 P sat
 R ln 2
 P sat
 1



 J    760 torr 
 ln

   8.314 
 mol  K    400 torr 


vap

h

1
1


1
1
 351.55 K  336.65 K 
  
 T2 T1 
 kJ 
h vap  42.39 
 mol 
Now we can calculate the vapor pressure at 100 ºC.
  h vap
sat
sat
P3  P2 exp

R



 kJ 
  42.39 



1

1 
1
1
mol  




     760 torr  exp
 kJ   373.15 K 351.55 K  
 T3 T2  
 0.008314 

 mol  K 


P3sat  1760 torr  2.32 atm
In comparison, ThermoSolver gives a value of 2.23 atm, using the Antoine equation.
20
6.13
We can show using the Chain Rule that
  gi
 
 T
 T


  Tg i  g i T  

 

 
T2
   1  g i   g i
   





T
T  T  P T 2



P 
P
Using fundamental property relations, Equation 5.14 states
 g i

 T

   si
P
Therefore,
  gi
 
 T
 T




    Tsi  g i   Tsi  hi  Tsi    hi

T2
T2
T2

P
21
6.14
Let T1  922 K, T2  1,300 K
g 2  h2  T2 s 2
dh  c P dT
1,300 K
h2  h1 
 c P dT
922 K
 J 
h2  h1  c P T2  T1   39,116 
 mol 
c
ds  P dT
T
T 
 J 
s2  s1  c P ln 2   85.10 
 mol K 
 T1 
 J 
g 2  h2  T2 s2  71,500 
 mol 
Alternative solution using the result from Problem 6.13:
d g
T

 dT

g
 
 T 2

   hT   h1  c P T  T1  We must leave h as a function of T

T2
T2

 h  c T  T  
g
 d  T      1 PT 2 1  dT
922 K
g
1300 K
 
 T 1
1 1
T 
g
g
 J 
      h1  c PT1     c P ln 2   55.01 
 T  2  T 1
 mol K 
 T2 T1 
 T1 
 J 
g 2  71,500 
 mol 
22
6.15
A possible hypothetical solution path is presented below:
monoclinic
orthorhombic
T = 298 [K]
g
1
g
3
T = 368.3 [K]
monoclinic
g = 0
2
orthorhombic
From the diagram, we see that the Gibbs energy for steps one and three can be calculated as
follows:
 g m
g1   
T
298 K 
368 K

 dT
P
and
 g o 
  T  dT
368 K 
P
respectively. We can apply Equation 5.14 from the thermodynamic web
298 K
g 3 
 g 

  s
 T  P
At 368 K, sulfur undergoes a phase transition, so
g 3m68o
K 0
mo
Using these above relationships, the expression for g 298
K becomes
mo
g 298
K 
368 K
m
s
dT  0 
298 K
298 K
o
s
368 K
dT  
368 K
298 K
298 K
368 K
 13.8  0.066T dT   11  0.071T dT
 J 
m o
g 298
K  79.5 
 mol 
Therefore, the transition from the monoclinic to orthorhombic state occurs spontaneously. The
orthorhombic state is more stable.
23
6.16
At the phase transition, the following is true
g
g
 
 
 T  Sr ( s )  T  Sr (l )
Using the thermodynamic web, the following can be shown (see Problem 6.13)
h
 g / T  

  2
 T  P T
The enthalpies can be written as follows
T
h l T   49179 
 35.146 dT  35.146T  3540
h s T   20285 
1500 K
T
 37.656 dT  37.656T  16305.4
900 K
g
can also be calculated at 900 K for solid Sr and 1500 K for liquid Sr.
T
 gl

T


 J 

 83.85 

 mol  K 
 ref
 gs

 T


 J 

 68.68 

 mol  K 
 ref
g
We can find   at any temperature using the differential equation as follows
T 
g
h
 g / T  
 dT    2 dT
T  P
T
 d  T    
Substituting our expressions, we get
g l /T
T
 35.146T  3540 
g
dT
 d  T     
2
T

 83.85
1500 K
24
 gl

T


  35.146 ln(T )  3540  176.04

T

g s /T
T
 37.656T  16305.4 
g
dT
 d  T     
2
T

 68.68
900 K
 gs

 T

 gl
Set 
T


  37.646 ln(T )  16305.4  205.5

T

  gl

 T
 

 and solve for T:


T melt  1059.8 K
The enthalpy of melting is defined as

 
h fus  h s T melt  h l T melt

Using the expressions developed above
 kJ 
 kJ 
 kJ 
 7.41 
 33.71 
h fus  26.30 


 mol 
 mol 
 mol 
25
6.17
At the phase transition, the temperature and Gibb’s energy of both phases must be equal.
Mathematically, this is equivalent to
g
g
 
 
 T  SiO2 ( s )  T  SiO 2 (l )
Using the thermodynamic web, the following can be shown (see Problem 6.13)
h
 g / T  

  2
 T  P T
Using the definition of enthalpy, we can write the following
hT 
T
href
Tref
 dh   c P dT
 hT   href 
T
 c P dT
Tref
The enthalpies can be written as follows
h T   738440 
l
h s T   856840 
T
 85.772 dT
2500 K
T
 53.466  0.02706T  1.27  10
5 2
1100 K
g
can also be calculated at 1100 K for solid SiO2 and 2500 K for liquid SiO2.
T
 gl

T


 J 

 487.3 

 mol  K 
 ref
 gs

 T


 J 

 903.5 


 mol  K 
 ref
26

T  2.19  10  9 T 3 dT
We can substitute our expressions for
g
and hT  into the above differential equation and
T
separate variables to obtain
T






dT
738440
85
.
772
l

g /T
T


g
2500 K


d



dT
 T   
2
T

 487.3
2500 K






T


5 2
9 3









856840
5
3.466
0
.
02706
1
.
27
10
2
.
19
10
T
T
T
dT
s

g /T
T


g
1100 K


d



dT
 T   
2
T

 903.5
1100 K




Integration provides
g l  9.5268  10 9  85.772 ln(T )  1052.4T

 487.5
T
T
g s  1.82  10 10 T 4  2.12  10 6 T 3  0.01353T 2  927190.9  53.466T ln(T )  1230T

 903.5
T
T
 gl
If we plot 
T

  gs

  T
 

 vs. T, we obtain the following:


27
There are three solutions, but only the solution between 1100 K and 2500 K is physically
meaningful. If we magnify the plot near the middle solution, we find
T  1983 K
The enthalpy of fusion is defined as

 
h fus  h s T melt  h l T melt

Using the expressions developed above
 kJ 
 kJ 
 kJ 
 9.72 
h fus  792.5 
  782.78 


 mol 
 mol 
 mol 
28
6.18
From the Clausius-Clapeyron equation
sat
dPCS

2
dT
h vap

T vv  vl

Assuming:
v v  v l
we get
sat
dPCS

2
dT
h vap
(I)
Tv v
The saturation pressure is given by:
sat
ln PCS
 62.7839 
2
4.7063  10 3
 6.7794 ln T  8.0194  10  3 T
T
(II)
sat
5
 4.48 10 Pa. Taking the derivative of Equation II
At T = 373 K, PCS
2
sat
d ln PCS
2
dT

sat
dPCS
1
2
sat
PCS
dT

4.7063  10 3
T2

6.7794
 8.0194  10  3 (III)
T
2
Plugging Equation III into Equation I,
 sat h vap
 4.7063  10 3 6.7794


 8.0194  10  3  PCS

2
T
T2
Tv v


 kJ 
vap
Solving for vv using hCS
gives:
 24.050 
2
 mol 

h vap  4.7063  10 3 6.7794
v 

 8.0194  10 3 

sat
T
TPCS
T2


v
2
z
Pv
B
 0.878  1 
RT
v
29
1
 m3 
 6.08  10 3 

 mol 
or
 m3 
 cm3 
B  7.4  10  4 
  740 

 mol 
 mol 
This value is about 50% higher than the reported value.
Alternative solution:
Following similar development as Problem 6.3:
sat
dPCS
dT
2
h vap



1
RT 2 
 B '
sat

PCS 2


h vap
1
sat
 sat  B' dPCS

2 dT
2
P
RT
 CS2

We must be careful about the limits of integration. We need to pick a value of T close so
enthalpy of vaporization is not too different, but far enough away to avoid round off error. If we
sat
5
choose T = 378 K, Equation I gives PCS
 5.04 10 Pa. Integrating:
2
4.4810 5 Pa

373
h vap
sat
 sat  B dPCS  

2 dT
2
P
RT
5


378
5.04 10 Pa CS2
1
'
4.48  105  '
h vap  1
1 
5
5

B
ln
4.48

10

5.04

10



5 
R 373 378 
5.04  10 


Solving for B’ gives:
B'  2.55  107 Pa
or
m 3 
B  B ' RT   7.9  10 4  
mol
30
6.19
Calculate vA, vB, v, VA, VB, and V from the ideal gas law:
RT
 0.05 m 3 / mol
P
RT
vB 
 0.05 m 3 / mol
P
RT
v 
 0.05 m 3 / mol
P
vA 
V A  n A v A  0.1 m 3
V B  n B v B  0.15 m 3
V  ntot v  0.25 m 3
We calculate the partial molar volumes as follows
 V 
 

n A  n B  RT   RT  0.05 m 3 / mol

V A  

P 
P
 n A T , P, n B n A 
 V
V B  
 n B

 

n A  n B  RT   RT  0.05 m 3 / mol


P 
P
T , P, n A n B 
To find the remaining quantities, we can apply Equations 6.44 and 6.46
Vmix  n A V A  v A   n B V B  v B 
Vmix  20.05  0.05  30.05  0.05  0
vmix  x A V A  v A   x B V B  v B 
vmix  0
31
6.20
(a)
For a pure species property
va  v y a  1
Substitution yields
 cm 3 
va  1001  800  2.510   100 

 mol 
(b)
From Equation 6.29
 V
Va  
 na


 n b ,T , P
We can find V by multiplying the given expression for molar volume by the total number of
moles.

V  na  nb 100 ya  80 yb  2.5

y a yb 
n n
  100na  80nb  2.5 a b
y a yb 
na  nb
Differentiating with respect to na we get,
Va 

na

n n 
nb
na nb
100na  80nb  2.5 a b   100  2.5
 2.5
na  nb  n
na  nb
na  nb 2

b
so
Va  100  2.5 yb 1  ya   100  2.5 yb2
To find the molar volume at infinite dilution, we can use the following relation
Va  lim Va
ya 0
 cm 3 
Va  102.5 

 mol 
32
(c)
Since species A contributes more to a mixture than to a pure species,
v mix  0
Note: The Gibbs-Duhem equation says that species B also contributes more.
33
6.21
Calculate mole fractions:
n1
1 mol

 0.2
ntot 5 mol
y1 
y3  0.4
y 2  0.4
Calculate v.
Obtain an expression for v:
v
RT
P

B 
2 A
1  P  RT  y1  y 2   RT  



Substitute values:
v


 3
 82.06  cm  atm  500 K 

 mol  K  

50 atm
1  50   9.0 10
2
5

0.2  0.4  3.0  10  5 
 cm 3 
v  919.0 

 mol 
Calculate V.
V  ntot v

 cm 3  
V  5 mol 919.0 


mol
 


 
V  4595 cm 3
Calculate v1.
The value of v1 can be found by substituting y1=1 into the expression for v1.
v1 


 3
 82.06  cm  atm  500 K 

 mol  K  

50 atm 
1  50   9.0 10
2
 cm 3 
v1  698 

 mol 
34
5

1  0  3.0  10  5 
Calculate v2.
v2 


 3
 82.06  cm  atm  500 K 


 mol  K  

50 atm 
1  50   9.0 10
2
5

0  1  3.0  10  5 
 cm 3 
v2  1067 

 mol 
Calculate v3.
v3 


 3
 82.06  cm  atm  500 K 

 mol  K  

50 atm
1  50  3.0 10 
2
5
 cm 3 
v3  882 

 mol 
Calculate V1 .
From Equation 6.29:
  nv  
 V 


V1  
 
 n1  n 2 , n 3 ,T , P  n1  n 2 , n 3 ,T , P
We can substitute the expression for V into this derivative and use the fact that
ntot  n1  n2  n3 to obtain
V1 
  RT 
B
  
2 A














n
n
n
P
n
n
n
n
n
1
2
3
1
2
1
2
3

 RT
  
n1  P 
RT
  n 2 , n 3 ,T , P
Differentiating we get
V1 
RT
P

B 
2 A
1  P  RT  RT  



Substitute values:
35
V1 


 3
 82.06  cm  atm  500 K 

 mol  K  

50 atm
1  50   9 10
2
 cm 3 
V1  697.5 

 mol 
36
5
 3  10  5

6.22
(a)
By definition:
 H
H a  
 na


T , P , n b , n c
h  5,000 xa  3,000 xb  2,200 xc  500 xa xb xc J/mol
n  na  nb  nc
H  nh   5,000na  3,000nb  2,200nc  500
na nbnc
na  nb  nc 2

H 
nbn c
2na nbnc 
 
 5,000  500


na T ,P,n ,n
na  n b  nc 2 na  nb  n c 3 


b c
H 
Ha   
 5,000  500xb x c 1 2xa  J/mol
na T ,P,n ,n
b c
(b)
xa  xb  xc 
1
3
Ha  5,018.5 J/mol
(c)
xa  1 , xb  xc  0
Ha  5,000 J/mol
(d)
xb  1 , xa  xc  0
Hb  hb  3,000 J/mol
37
6.23
Let the subscript “1” designate CO2, and “2” designate propane. To calculate the partial molar
volumes, the following formulas will be used:
V1  v  y 2
dv
dy 2
v  y1V1  y 2V2
Expressions can’t be obtained for the molar volume with the van der Waals EOS; therefore, the
problem will be solved graphically. First, obtain an expression for the pressure that contains the
mole fractions of CO2 and propane:
a mix  y12 a1  2 y1 y 2 a1a 2  y 22 a 2
bmix  y1b1  y 2 b2
y12 a1  2 y1 y 2 a1a2  y 22 a2
RT
P

v   y1b1  y 2 b2 
v2
Solve for a and b using data from the appendices.
 J  m3 
a1  0.366 

 mol 2 
 J  m3 
a 2  0.941 

2
 mol 
 mol 
b1  4.29  10  5 

 m3 
 mol 
b2  9.06  10  5 

 m3 
Now we can create a spreadsheet with the following headings:
y1
y2
amix
bmix
v
The last column contains the molar volumes obtained by solving the van der Waals equation
with the spreadsheet’s solver function. After the table is completed, we create the following
graph.
38
v vs. y 2
1.48E-03
1.46E-03
v (m 3 mol -1)
1.44E-03
1.42E-03
1.40E-03
1.38E-03
1.36E-03
y = -0.00008x 2 - 0.00008x + 0.00147
R2 = 0.99946
1.34E-03
1.32E-03
1.30E-03
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y2 (Mole Fraction)
From the line of best fit, we find
v  8  10  5 y 22  8  10  5 y 2  0.00147
Therefore,
dv
 1.6  10  4 y 2  8  10  5
dy 2
and

V1  v  y 2  1.6  10 4 y 2  8  10 5

We can find the partial molar volume of propane from the following relationship
v  y1V1  y 2V2
V2 
v  y1V1
y2
Tabulate the values of the partial molar volumes in the spreadsheet and create the following
graph
39
Partial Molar Volumes as a Function of Carbon Dioxide Mole
Fraction
Partial Molar Volume (m
3
mol -1)
0.0016
V1
V2
0.00155
0.0015
0.00145
0.0014
0.00135
0.0013
0.00125
0
0.1
0.2
0.3
0.4
0.5
0.6
y1 (CO 2 mole fraction)
40
0.7
0.8
0.9
1
6.24
(a)
ga  40
kJ
mol
Ga  g  x b
dg
dx b
dg
 40  60  RT  ln xa  1  ln x b  1  5x a  5xb
dx b
Ga  40xa  60x b  RT xa ln x a  x b ln xb   5xa x b
 x b40  60  RT  ln xa  1  ln x b  1  5x a  5xb 
Ga  40xa  xb   RT xa  xb  ln x a  5xb2
Ga   40  RT ln x a  5 xb2
Ga   40 
kJ
kJ
8.314(300)
ln 0.2  5(0.64)
  40.8
mol
mol
1000
Ga  
Gmix  ng  x a ga  xb gb 
g  xa ga  x bgb   RT x a ln xa  xb ln x b   5 x a xb
Gmix  nT RT x a ln xa  xb ln x b   5xa x b   2.2 kJ
(b)
gmix  hmix  Tsmix
Assume the entropy of mixing is ideal:
hmix  5x a xb  0
so
h  hmix  hsensibleheat  0
so
hsensibleheat  0 and T goes down
41
6.25
To find V1 and V2 , we can read values directly from the graphs. Calculate mole fractions
x1 
1
 0.2
5
At x1  0.2 ,
 cm 3 
V1  46.5 

 mol 
 cm 3 
V2  69.8 

 mol 
The following relationships are employed to calculate the molar volumes of pure species
v1  lim V1
x1 1
v 2  lim V2  lim V2
x 2 1
x1  0
From the graph
 cm 3 
v1  50 

 mol 
 cm 3 
v2  70 

 mol 
Therefore,
 
 280 cm 
V1  50 cm 3
V2
3
To calculate the total volume, we can use
V  n1V1  n2V2
Substituting the values, we find
 
V  146.5  469.8  325.7 cm3
42
Therefore
v
V
325.7

ntot
1 4
 cm 3 
v  65.14 

 mol 
Using Equation 6.43 we can calculate the change in volume.
Vmix  V  V1  V2   325.7  50  280
 
Vmix  4.3 cm 3
43
6.26
(a)
Expression for hmix :
hmix   X i H i  X i hi X Cd H Cd  X Sn H Sn    X Cd hCd  X Sn hSn 
Multiply both sides by the total number of moles
H mix  nCd H Cd  n Sn H Sn   nCd hCd  n Sn hSn 
Therefore,
H mix Cd
 H mix
 
 nCd


 H Cd  hCd
 n Sn ,T , P
(b)
We can show by repeating Part (a) for Sn that
H mix Sn  H Sn  hSn
Equation 6.65:
dhmix
dX Sn
H mix Sn  hmix  X Cd dhmix
dX Cd
H mix Cd
 hmix  X Sn
Since,
hmix  13000 X Cd X Sn
We get,
dhmix
d
13000 X Cd X Sn   13000 X Cd  X Sn 

dX Sn
dX Sn
dhmix
d
13000 X Cd X Sn   13000 X Sn  X Cd 

dX Cd
dX Cd
Therefore, for 3 moles of cadmium and 2 moles of tin at 500 ºC:
44
2
H Cd  hCd  H mix Cd  13000 X Cd X Sn  X Sn13000 X Cd  X Sn   13000 X Sn
 J 
H Cd  hCd  H mix Cd  2080 
 mol 
and
 J 
2
H Sn  hSn  H mix Sn  13000 X Cd
 4680 
 mol 
(c)
Equation 6.37:
nCd d H mix Cd  n Sn d H mix Sn  0
Differentiate with respect to XCd:
nCd
d H mix Cd
dxCd
 n Sn
d H mix Sn
dxCd
0
where
d H mix Cd
dX Cd
d H mix Sn
dX Sn
 26000 X Sn  26000  26000 X Cd
 26000 X Cd
Therefore,
nCd
d H mix Cd
dxCd
 nSn
d H mix Sn
dxCd
 xCd ntot  26000  26000 xCd   1  xCd ntot 26000 xCd
Inspection of the above expression reveals that
nCd
d H mix Cd
dxCd
 n Sn
d H mix Sn
dxCd
0
(d)
A graphical solution can be found using the tangent-slope method discussed on pages 285-287:
A plot of a line tangent to the enthalpy of mixing curve at XCd = 0.6, is given below:
45
Heat of mixin g in cadmium (Cd )-Tin (Sn) s ystem
6000
5400


J


mol
4800
4200
3600
data
 hmix
3000
2400
1800
1200
600
0
0
0 .2
fit to:
  J 
 hmix  13, 000 xCd x Sn  

mol
0 .4
0 .6
0 .8
1
xCd
The intercepts give the respective partial molar quantities as follows:
 J 
H Cd  hCd  2050 
 mol 
 J 
H Sn  hSn  4800 
 mol 
The values using the graphical method are reasonably close to the analytical method.
46
6.27
The following can be shown with the Gibbs-Duhem equation
0  x1V1  x 2V2
Differentiation with respect to x1:
0  x1
dV1
dV
 x2 2
dx1
dx1
If the partial molar volume of species 1 is constant, the Gibbs-Duhem equation simplifies to
0
dV2
dx1
Therefore, the partial molar volume of species 2 is also constant.
Note that in this case, since the partial molar volume of species 1 is constant:
V1  v1
and similarly for species 2:
`
V2  v2
Hence, the molar volume can be written:
v  x1V1  x2V2  x1v1  x2v2
This is known as Amagat’s law.
47
6.28
(a)
Let species 1 represent HCl and species 2 represent H2O. An expression for the enthalpy of the
solution is
h  x1h1  x 2 h2  hmix
which can be written
~
h  x1h1  x2 h2  x1hs
Therefore,
~
H  n1h1  n2 h2  n1hs
To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2
consistent with the convention used in the table. As seen in Example 6.6,
x1 
n1
1

n1  n2  1  n
For this problem
x1  0.2
Therefore,
n1  1
n2  n  4
Now we can find expressions for the partial molar enthalpies.
 H 

H H 2 O  H 2  

n
 2  n1 , T , P
~
dhs
H H 2 O  H 2  h2  n1
dn2
~
dhs
 H H 2 O  hH 2 O  H 2  h2  n1
dn2
Using the data in Table 6.1 for n  4 ,
48
~
J


hs  61,204 
 mol solute 
J
J




 64049 
  56852 
~
~
~

dhs hs n  5  hs n  3
 mol solute 
 mol solute 


5 mol  3 mol
5 mol  3 mol
dn2
~
dhs
J


 3598.5 
dn2
 mol  mol solute 
Therefore,

J


 J 
H H 2 O  hH 2 O  1 mol solute  3598.5 
  3599 

 mol  mol solute  
 mol 

We can calculate H HCl  hHCl using Equation 6.46
H HCl  hHCl 

hmix  x H 2 O H H 2 O  hH 2 O
x HCl
  xHCl h~s  xH O H H O  hH O 
2
2
2
x HCl
 

 J 
 J  
  0.8  3599 

0.2  61204 

 mol  
 mol  
 J 



 H HCl  hHCl 
 46808 
0.2
 mol 
(b)
For n1  2 and n2  80 ,
 n  n2 
82
  1 
n   1
 1  40
2
 n1 
The new values for the number of moles consistent with Table 6.1
n1  1 mol
n2  40 mol
Using the data in the table for n  40 ,
~
~
~
dhs hs n  50   hs n  30

50 mol  30 mol
dn2
Interpolating the data in Table 6.1
49
~
J


hs n  30  72428 
 mol solute 
Therefore,
J
J




 73729 
  72428 
~

dhs
J

 mol solute 
 mol solute   65.05 
and


50 mol H 2O  30 mol H 2O
dn2
 mol  mol solute 



J
J


H H 2 O  hH 2 O  1 mol solute  65.05 
  65.05 


 mol  mol solute  

 mol H 2 O 
50
6.29
First perform an energy balance on the mixing process.
hmix  q
We can calculate hmix using data from Table 6.1. Referring to Equation E6.7A, we find
~
hmix  x HCl hs
Calculate x HCl :
x HCl
wHCl
0.30
MW HCl
36.46


 0.175
0
.
30
0.70
w
wHCl
H 2O


MW HCl MW H 2O 36.46 18.0148
Heats of data are tabulated for a solution containing one mole of the solute for various amounts
of water. Thus, we need to calculate how many moles of water must be added to HCl to obtain
the above mole fraction.
1 mol HCl
; where n is the number of moles of H2O
1 mol HCl  n
n  4.71 mol H 2 O 
x HCl 
By interpolation of data from Table 6.1, we get
~
 J 
hs  63224 
 mol 
(for n  4.71 )
Therefore,

 J 
 J 
hmix  0.175  63224 
  11064 

 mol  
 mol 

and
 J 
q  hmix  11064 
 mol 
51
6.30
To calculate the enthalpy of mixing from Table 6.1, we must use the following expression
~
hmix  x H 2 SO 4 hs
The mole fraction of sulfuric acid is
x H 2 SO4 
1
1 n
where n is the number of moles of water.
Equation 6.47 states

hmix  74.4 x H 2 SO 4 x H 2 O 1  0.561x H 2 SO
4

For n  1 , x H 2 SO 4  0.5 and x H 2 O  0.5
Table 6.1:
~
 J 
hs  31087 
 mol 

 J 
 J 
  15543.5 
 hmix  0.5  31087 

 mol  
 mol 

Equation 6.47: hmix  74.40.50.51  0.5610.5
 J 
hmix  13383 
 mol 
The following table was made
n [mol H2O]
x H 2 SO 4
1
2
3
4
5
10
20
50
100
0.5
0.333333
0.25
0.2
0.166667
0.090909
0.047619
0.019608
0.009901
hmix [kJ/mol]
hmix [kJ/mol]
(Table 6.1)
(Eq. 6.47)
-15543.5
-14978.7
-13001.8
-11414
-10174.2
-6367.27
-3548.43
-1497.22
-762.238
-13382.7
-13441.6
-11993.5
-10568.4
-9367.17
-5835.17
-3284.01
-1414.49
-725.289
52
% Difference
14.94
10.82
8.07
7.69
8.26
8.72
7.74
5.68
4.97
As you can see, the percent difference between the two methods decreases as the mole fraction
of sulfuric acid decreases. Although Equation 6.47 fit data at 21 ºC, while the Table 6.1
tabulates data taken at 25 ºC, we do not expect the temperature dependence to account for all the
observed difference. The table and equation come from different experimental data sets, and
also represent measurement uncertainty. Nevertheless, the agreement is reasonable.
53
6.31
To calculate the enthalpy of mixing from Table 6.1, we must use the following expression
~
hmix  x HCl hs
The mole fraction of HCl is
x HCl 
1
1 n
where n is the number of moles of water. The following table was made using these two
equations.
n [mol H2O]
x1
~ 
J

hs 
 mol HCl 
 J 
hmix 
 mol 
1
2
3
4
5
10
20
50
0.5
0.333
0.25
0.2
0.167
0.091
0.048
0.020
-26225
-48819
-56852
-61204
-64049
-69488
-71777
-73729
-13112.5
-16273
-14213
-12240.8
-10674.8
-6317.09
-3417.95
-1445.67
100
0.0099
-73848
-731.168
54
6.32
A schematic for the process is given below. The inlet streams are labeled “1” and “2” and the
exit stream “3”.
q
50 wt% NaOH
10 wt% NaOH
50 wt% H2O
90 wt% H2O
Stream 1
Stream 3
H 2O
Stream 2
The energy balance for this process reduces to
Q  H 3  H 2  H1
We first convert from weight percentage to mole fraction. For stream 1,
x NaOH ,1
wNaOH
0.50
MW NaOH
40


 0.311
0.50
0.50
w H 2O
wNaOH


MW NaOH MW H 2O 40 18.0148
and for stream 3,
x NaOH ,3
wNaOH
0.10
MW NaOH
40


 0.048
0.10
0.90
wH 2 O
wNaOH


MW NaOH MW H 2O 40 18.0148
We now calculate the moles of water per mole of NaOH so that we can use Table 6.1:
x NaOH 
1
1  nH 2 O
Therefore, for every mole of NaOH
n H 2 O,1  2.21
55
n H 2 O,3  19.8
Since enthalpy is a state function, we can choose any hypothetical path to calculate the change in
enthalpy. One such path is shown below. The box in our original schematic is depicted with
dashed lines below. We pick a basis of 1 mole NaOH. In step A, the inlet stream is separated
into its pure components. In step B, 17.6 additional moles of water are added to the pure water
stream. Finally the H2O and NaOH streams are remixed
1 mol NaOH
Step C
Step A
1 mol NaOH
2.2 mol H2O
1 mol NaOH
19.8 mol H2O
Step B
hmix=0
2.2 mol H2O
19.8 mol H2O
17.6 mol H2O
The enthalpy change is found by adding each step
H 3  H 2  H 1  H A  H B  H C
Since H B represents the mixing of water with water, H B  0 .
The enthalpies of mixing for steps A and C can be related to enthalpy of solution data from Table
6.1:
~
J


hs ,1  23906 
 mol NaOH 
~
J


hs ,3  42858 
 mol NaOH 
Note: The enthalpy of solution for Stream 1 is calculated by extrapolation. Generally,
extrapolation should be avoided, but it is necessary to complete this problem, and we are
not extrapolating very far.
For step A, we need the negative value of the heat of solution of stream 1. Thus for a basis of 1
mole NaOH:
H A   23906  J 
56
while for step C:
H C  42858 J 
Now adding the enthalpies of each step per 1 mole of NaOH:
H  23906  0  42858  18952 J 
To get the total heat that must be removed per mole of product solution, we divide by the number
of moles of product per mol of NaOH:
q
H
 J 
 910 
n NaOH
 mol 
57
6.33
The partial molar property can be written as follows:
 n K 
K1   T 
 n1 T , P ,n2 ,n3
Applying the chain rule to the above relationship:
 n
K1  k  T
 n1

 k 


 nT 
 T , P ,n2 ,n3
 n1  T , P ,n2 ,n3
 k 

K1  k  nT 
 n1  T , P ,n2 ,n3
(1)
 k 

. At constant T and P, we can write,
Now focus on nT 

n
 1 T , P ,n2 ,n3
 k 
 k 
 k 



dx3
dx2  
dx1  
dk  
 x1 T , P, x2 , x3
 x2 T , P, x1 , x3
 x3 T , P, x1 , x2
Therefore,
 k 
 k 
 x1 
 k 
 x 2 








 
 
 n1  T , P ,n2 ,n3  x1  T , P , x2 , x3  n1  T , P ,n2 ,n3  x2 T , P , x1 , x3  n1  T , P ,n2 ,n3
 k 
 x3 



 
 x3  T , P , x1 , x2  n1  T , P ,n2 ,n3
but,
x1 
n1
n1  n2  n3
so
 x1 
n1
1
1




 1  x1 
2
nT
 n1 T , P ,n2 ,n3 n1  n2  n3 n1  n2  n3 
Similarly,
58
(3)
(2)
 x 2 
n2
x



 2
2
nT
n1  n2  n3 
 n1  T , P ,n2 ,n3
(4)
 x3 
n3
x



 3
2
nT
n1  n2  n3 
 n1 T , P ,n2 ,n3
(5)
and
Substituting Equations 2, 3, 4, and 5 into the expression 1 for K1 and simplifying:


 k 



1  x1    k 
 x2    k 
 x3 
K1  k  
 x1 T , P, x 2 , x 3
 x 2 T , P, x1 , x 3
 x3 T , P, x1 , x 2
Utilize the fact that x1  x 2  x3  1


 k 

 k 
 k 

 (6)
  x3  k 




 
K1  k  x 2 

 x1 T , P, x , x  x3 T , P, x , x 
 x1 T , P, x , x  x 2 T , P, x , x 
2
3
1
3 
2
3

1
2 

When we hold species 3 constant:
 k 
 k 


dk  
dx1  
dx 2
 x 2 T , P, x1 , x3
 x1 T , P, x2 , x3
 k 
 k 
 dx1   k 
 dx2 




  



 
 x 2  T , P , x3  x1  T , P , x2 , x3  dx2   x2  T , P , x1 , x3  dx2 
Thus,
 k 
 k 
 k 




 
 
 x2 T ,P , x3
 x1 T ,P , x2 , x3  x2 T ,P , x1 , x3
(7)
When we hold species 2 constant, a similar analysis shows:
 k 
 k 
 k 




 
 
 x1 T , P , x2 , x3  x3 T , P , x1 , x2
 x3 T , P , x2
Substituting Equations 7 and 8 into Equation 6 gives
59
(8)
 k 
 k 


K1  k  x 2 
 x3 
 x2  T , P , x3
 x3  T , P , x2
Furthermore, the above analysis can be extended to m components. In general,
 k 

K i  k   xm 
m i
 xm T , P , x ji ,m
60
6.34
The expression can be found by employing the Gibbs-Duhem equation:
0  n1V1  n2V2
Differentiate with respect to x1 and then divide by the total number of moles:
0  x1
dV
dV
dV
dV1
 x 2 2  x1 1  1  x1  2
dx1
dx1
dx1
dx1
Differentiate the expression given in the problem statement.
dV1
 5.28  5.28 x1
dx1
Substitution of this result into the Gibbs-Duhem equation and rearrangement yields
dV2 5.28 x12  5.28 x1

 5.28 x1
dx1
x1  1
Integrate:
 cm 3 
V2  2.64 x12  C 

 mol 
To determine C, we can use the density information given in the problem statement.
V2  x 2  1  V2  x1  0  C  v 2
where
C  v2 
1
 2 


 MW2 

 cm 3 
 109.58 

mol 
 0.768 g/cm 3 




 84.16 g/mol 


1


Therefore,
 cm 3 
V2  2.64 x12  109.58 

 mol 
61
6.35
An expression for the enthalpy of the solution is
h  x1h1  x 2 h2  hmix
which is equivalent to
~
h  x1h1  x 2 h2  x1hs
Multiplication by the total number of moles yields
~
H  n1h1  n2 h2  n1hs
To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2
consistent with the convention used in the table. As seen in Example 6.6,
x1 
n1
1

n1  n2  1  n
For this problem
x1  0.33
Therefore,
n1  1
n2  n  2
Now we can find expressions for the partial molar enthalpies.
 H 

H H 2 O  H 2  
 n2  n1 ,T , P
~
dhs
H H 2 O  H 2  h2  n1
dn2
~
dhs
 H H 2 O  hH 2 O  H 2  h2  n1
dn2
Using the data in Table 6.1 for n  2 ,
62
J
J




 2,787 
  812 
~
~
~

dhs hs n  3  hs n  1
 mol solute 
 mol solute 


3 mol  1 mol
3 mol H 2O  1 mol H 2O
dn2
~
dhs
J


 987.5 
dn2
 mol  mol solute 
Therefore,

J


 J 
H H 2 O  hH 2 O  1 mol solute  987.5 
  987.5 

 mol  mol solute  
 mol 

Calculate the partial molar enthalpy:
 J 
H H 2O  hH 2 O  987.5 
 mol 
From the saturated steam tables at 25 ºC:
 kJ 
hˆH 2 O  104.87  
 kg 
 kJ 
 hH 2 O  1.89 
 mol 
Now we can find the partial molar enthalpy






kJ
J
J
H H 2 O  1.89 
  0.988 
  0.90 

 mol H 2 O 
 mol H 2 O 
 mol H 2 O 
63
6.36
(a)
Calculate the mole fraction of sulfuric acid
x1 
w1
MW 1
w1
w2

MW 1 MW 2
0.20
98.078

 0.044
0.20
0.80

98.078 18.0148
Calculate n to use in Table 6.1:
1
 0.044
1 n
n  21.7 mol H 2 O
x1 
Interpolating from Table 6.1
~
 J 
hs  74621 
 mol 
Now calculate the heat transfer
~

 J 

q  hmix  x1hs  0.044  74621 
 mol  

 J 
q  3283 
 mol 
(b)
Calculate the mole fraction of pure sulfuric acid. Consider a mixture of 20 kg of 18 M sulfuric
acid and 80 kg of water. Find the mass of sulfuric acid present.
VH 2 SO4 
mH 2 SO4

 H 2 SO4
20 kg
 10.9 L
1.84 kg/L
nH 2 SO4  VM  10.9 L 18 mol/L  196.2 mol


m H 2 SO 4  n H 2 SO4 MW H 2 SO 4  196.2 mol0.098 kg/mol  19.2 kg
Since both the initial (i) and final (f) states contain mixtures, to get the enthalpy of mixing, we
need to calculate the relative differences follows:
64
~
~
q  hmix  x1, f hs, f  x1,i hs ,i
calculate the mole fraction in the final state
w1
MW 1
0.192
98.078

 0.042
x1, f 
w1
w2
0.192
0.808


MW 1 MW 2 98.078 18.0148
Calculate n to use in Table 6.1:
1
 0.042
1 n
n  22.8 mol H 2 O
x1, f 
Interpolating from Table 6.1
~
 J 
hs , f  74689 
 mol 
For the initial 18 M sulfuric acid:
w1
MW 1
0.192
98.078

 0.81
x1,i 
w1
w2
0.192
0.008


MW 1 MW 2 98.078 18.0148
Calculate n to use in Table 6.1:
1
 0.81
1 n
n  0.23 mol H 2 O
x1,i 
We must extrapolate from Table 6.1. To do this we wish to extend the trend at low water
concentration. A plot of the data in Table 6.1 is useful. A semi-log plot follows:
65
0
3200 [J/mol]
enthalp of solution [J/mol solute]
-10000
-20000
-30000
-40000
-50000
-60000
-70000
-80000
-90000
0.1
1
10
100
n
~
 J 
hs ,i  3200 
 mol 
Now calculate the heat transfer
~
~


 J 
 J 
q  hmix  x1, f hs , f  x1,i hs,i  0.042  74689 
  0.81  3200 


 mol  
 mol  


 J 
q  710 
 mol 
(c)
Calculate the mole fraction of sodium hydroxide
x1 
w1
MW 1
w1
w2

MW 1 MW 2
0.20
40

 0.101
0.20
0.80

40 18.0148
Calculate the n value to use in Table 6.1:
1
 0.101
1 n
n  8.9 mol H 2 O
x1 
Interpolating from Table 6.1
~
 J 
hs  41458 
 mol 
66
Now calculate the heat transfer
~

 J 

q  hmix  x1hs  0.101  41458 
 mol  

 J 
q  4187 
 mol 
(d)
Calculate the mole fraction of ammonia
x1 
w1
MW 1
w1
w2

MW 1 MW 2
0.20
17.03

 0.209
0.20
0.80

17.03 18.0148
Calculate the n value to use in Table 6.1:
1
 0.209
1 n
n  3.78 mol H 2 O
x1 
Interpolating from Table 6.1
~
 J 
hs  33153 
 mol 
Now calculate the heat transfer
~

 J 

q  hmix  x1hs  0.209  33153 
 mol  

 J 
q  6929 
 mol 
67
6.37
Let species 1 designate ethanol and species 2 designate water. We need to obtain an expression
for the molar volume, so first, convert the given the mass fractions and densities to mole
fractions and molar volumes.
x1 
Mole fractions:
Specific molar volumes:
w1
MW 1
w1
w2

MW 1 MW 2
v  vˆMW mixture
where MW mixture  x1 MW 1  x 2 MW 2
Using this set of equations, the following table was made.
Mole Frac. EtOH
0.000
0.042
0.089
0.144
0.207
0.281
0.370
0.477
0.610
0.779
1.000
Mole Frac. H2O v (ml/mol)
1.000
18.05
0.958
19.54
0.911
21.18
0.856
23.11
0.793
25.47
0.719
28.34
0.630
31.85
0.523
36.19
0.390
41.65
0.221
48.73
0.000
58.36
The following graph plots the data. The trendline relates v to x1.
60.00
v (ml/mol)
50.00
40.00
2
v = 4.5491x1 + 35.918x1 + 17.957
30.00
R2 = 1
20.00
10.00
0.00
0.000
0.200
0.400
0.600
Mole Fraction EtOH (x1)
68
0.800
1.000
Now, we can calculate V1 .
 V 
  nv  


 
V1  
 n1  n 2 , n 3 ,T , P  n1  n 2 , n 3 ,T , P
We can substitute the trendline for V into this derivative and use the fact that ntot  n1  n2 to
obtain

n12
 
V1 
4.5491
 35.918n1  17.957n1  n2 

n1  n2 
n1 

n 2 ,T , P
Differentiating we get
 2n n  n   n 2 
2
1  53.875  ml 
V1  4.5491 1 1

 mol 


n1  n2 2 

 ml 
V1  4.5491 2 x1  x12  53.875 
 mol 


Calculate V2 :
V2 

n2
2


 4.5491 n1
 35.918n1  17.957n1  n2 


n1  n2 
 n1 ,T , P

Differentiating we get
  n2 
 ml 
1
V2  4.5491
  17.957 

 mol 
 n1  n2 2 
 ml 
V2  4.5491x12  17.957 
 mol 
Plotting V1 and V2 vs. x1 we obtain
69
60
59
20
19
58
57
56
55
18
17
16
15
54
53
52
14
13
12
0
0.2
0.4
0.6
0.8
H2O Partial Molar
Volume (ml/mol)
EtOH Partial Molar
Volume (ml/mol)
Partial Molar Volumes vs. EtOH Mole Fractions
EtOH
H2O
1
Mole Fraction EtOH (x1)
(b)
v mix can be calculated using Equation 6.46


v mix  x H 2 O V H 2 O  v H 2 O  x EtOH V EtOH  v EtOH 
From the data table in Part (a),
 ml 
v EtOH  v x1 1  58.36 
 mol 
 ml 
v H 2 O  v x1  0  18.05 
 mol 
Using the expressions for partial molar volumes


 ml 
V1  4.5491 20.5  0.52  53.875  57.29 
 mol 
 ml 
V2  4.54910.52  17.957  16.82 

 mol 
Therefore,


 ml  
 ml  

  0.5 57.29  58.36 
v mix  0.516.82  18.05 

 mol  
 mol  


 ml 
vmix  1.15 

 mol 
70
6.38
We can use the density data given in the problem statement to determine the pure species
properties. For pure ethanol x1  1 :
46 g/mol
 cm 3 

 58.55 
v1 

1
0.7857 g/cm 3
 mol 
MW1


 

 cm 3  
3
V1  n1v1  3 mol 58.55 
   175.7 cm


 mol  

For pure formamide  x1  0  :
45 g/mol
 cm 3 
 39.77 

2
1.1314 g/cm 3
 mol 

 cm 3  
3
V2  n2 v 2  1 mol 39.77 
   39.77 cm


mol
 


v2 
MW2



 
To calculate V and v, interpolate in the data table to obtain the density of the mixture
when x1  0.75 :

  0.8550 g/cm 3

Therefore,
v
where
MW

MW  0.7546 g/mol  0.2545 g/mol  45.75 g/mol
Substitute numerical values:
v
45.75 g/mol

0.8550 g/cm
3


 53.51 cm 3 /mol

 

 cm 3  
3
V  nv  4 mol 53.51 
   214.0 cm


 mol  

Now, we can calculate the volume change of mixing by using Equation 6.43:
71
 
 
 
Vmix  V  V1  V2  214 cm 3  175.7 cm 3  39.77 cm 3
 
Vmix  1.47 cm 3
The intensive volume change of mixing:
 

Vmix  1.47 cm 3
v mix 

 0.368 cm 3 / mol
n
4

We can use Equation 6.65 to determine the partial molar volume of formamide:
V2  v  x1
dv
v
 v  x1
dx1
x1
From the provided data table
45.8
45.7

v
 0.8401 0.8701  19.50 cm 3 / mol
x1 0.8009  0.6986


Therefore,


 


V2  53.51 cm 3 / mol  0.75 19.5 cm 3 / mol  38.89 cm 3 / mol

Now calculate the partial molar volume of ethanol:
V  n1V1  n2V2
 




214.0 cm 3  1 mol 38.89 cm 3 /mol
V1 
 58.37 cm 3 / mol
3 mol
72

6.39
Using the definition of G, the Gibbs energy of mixing of an ideal gas can be rewritten in terms of
the enthalpy of mixing and the entropy of mixing:
ideal gas
ideal gas
ideal gas
g mix
 hmix
 Tsmix
Since an ideal gas exerts no intermolecular interactions,
ideal gas
hmix
0
From Equation 6.48:
ideal gas
smix
  R xa ln xa  xb ln xb 
so
ideal gas
g mix
 RT  xa ln xa  xb ln xb 
To find the partial molar Gibbs energy of mixing of species a, we apply Equation 6.29:
Gmix a    nng mix 

a
 T , P , nb
Applying the expression above

nb 
na
  RT na ln na  nb ln nb  na  nb lnna  nb 
ng mix  RT  na ln
 nb ln
na  nb
na  nb 

where the mathematical relation of logarithms was used. Thus,

n  nb   RT ln x
n
 RT ln na  a  0  ln na  nb   a
a
na  nb 
na
a

 T , P , nb

at infinite dilution xa goes to zero, and the ln term blows up,
Gmix a    nng mix 
Gmix a  a  g a  
As chemical engineers, we are often interested in the limiting case of infinite dilution. We see
that even for ideal gas mixtures the chemical potential in this limit is not mathematically well.behaved, In Chapter 7, we will develop a different function, the fugacity, which behaves better.
73
6.40
At equilibrium
l
H
2
O
v
 H
2O
or
G Hl O  G Hv O
2
2
Since the water in the liquid phase is pure
GHl 2O  g Hl 2O  hHl 2O  Ts Hl 2O
The enthalpy and entropy of liquids are not sensitive to pressure changes. We can use data from
the saturated steam tables at 25 ºC to determine the Gibbs energy.
l
gˆ H
 104.87 kJ/kg   298.15 K 0.3673 kJ/kg  K 
2O
l
gˆ H
 4.640 kJ/kg 
2O



l
l
gH
 MWH 2 O gˆ H
 0.0180148 kg/mol 4.640 kJ/kg   0.0836 kJ/mol
2O
2O
Therefore,
v
H
2O
l
 gH
 83.6 J/mol
2O
74