Chapter 2: Heat, Work, Internal Energy, Enthalpy, and the First
Law of Thermodynamics
Questions on Concepts
Q2.1) Electrical current is passed through a resistor immersed in a liquid in an adiabatic container. The temperature of the liquid is varied by 1°C. The system consists solely of the liquid.
Does heat or work flow across the boundary between the system and surroundings? Justify your
answer.
The work on the resistor is done by the surroundings. Heat flows across the boundary between
the surroundings and the system because of the temperature difference between them.
Q2.2) Explain how a mass of water in the surroundings can be used to determine q for a process.
Calculate q if the temperature of a 1.00-kg water bath in the surroundings increases by 1.25°C.
Assume that the surroundings are at a constant pressure.
If heat flows across the boundary between the system and surroundings, it will lead to a temperature change in the surroundings given by T 
q
. For the case of interest, we obtain:
Cp


q  q surroundings  -m C p T  1000 g   4.19 J g -1 K -1  1.15 K  5.24 103 J
Q2.3) Explain the relationship between the terms exact differential and state function.
In order for a function to be a state function f (x,y), it must be possible to express the function as
a total differential df as:
1
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
 f 
 f 
df    dx    dy
 x  y
 y  x
If that form exists, it is called the exact differential.
Q2.4) Why is it incorrect to speak of the heat or work associated with a system?
Heat and work exist only during the transitions of a given system between states. At system in a
particular state is not associated with heat or work.
Q2.5) Two ideal gas systems undergo reversible expansion starting from the same P and V. At
the end of the expansion, the two systems have the same volume. The pressure in the system that
has undergone adiabatic expansion is lower than in the system that has undergone isothermal expansion. Explain this result without using equations.
In the system undergoing an adiabatic expansion, all work that is done comes from lowering of
U, and therefore of the temperature. On the other hand, some of the work that is done in an isothermal expansion can come at the expense of the heat that has flown across the boundary between the system and surroundings.
Q2.6) A cup of water at 278 K (the system) is placed in a microwave oven and the oven is
turned on for 1 minute during which it begins to boil. Which of q, w, and U are positive, negative, or zero?
The heat, q, is positive since heat flows across the system-surrounding boundary into the system.
The work, w, is negative because the vaporized water does work on the surroundings. U is positive because the temperature increases an some of the liquid is vaporized.
Q2.7) What is wrong with the following statement?: because the well-insulated house stored a
2
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
lot of heat, the temperature didn’t fall much when the furnace failed. Rewrite the sentence to
convey the same information in a correct way.
Heat can not be stored since it only exists in transitions. The sentence should be rephrased to:
Since the house is well insulated, the walls are nearly adiabatic. Therefore, the temperature of the
house did not fall rapidly when in contact with the surroundings at lower temperature, as it would
have if the walls would be more or less diathermal.
Q2.8) What is wrong with the following statement?: burns caused by steam at 100°C can be
more severe than those caused by water at 100°C because steam contains more heat than water.
Rewrite the sentence to convey the same information in a correct way.
Heat is not a substance that can be stored. When steam gets into contact with your skin it condenses to the liquid phase. In doing so, energy is released that is absorbed by the skin. Hot water
does not release as much heat in the same situation since it does not undergo a phase transition.
Q2.9) Describe how reversible and irreversible expansions differ by discussing the degree to
which equilibrium is maintained between the system and the surroundings.
In a reversible expansion, the system and surroundings are always at equilibrium with one another. In an irreversible expansion, they are not at equilibrium with one another.
Q2.10) A chemical reaction occurs in a constant volume enclosure separated from the surroundings by diathermal walls. Can you say whether the temperature of the surroundings increases, decreases, or remains the same in this process? Explain.
No, the temperature will increase if the reaction is exothermic, decrease if the reaction is endothermic, and not change if no energy evolution or consumption takes place in the reaction.
3
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Problems
Problem numbers in RED indicate that the solution to the problem is given in the Student Solutions Manual.
P2.1) 3.00 moles of an ideal gas at 27.0°C expand isothermally from an initial volume of 20.0
dm3 to a final volume of 60.0 dm3. Calculate w for this process (a) for expansion against a constant external pressure of 1.00 105 Pa and (b) for a reversible expansion.
a) w for an expansion against a constant pressure is:

 

w  p ext  Vfinal  Vinitial   1.00  10 5 Pa  0.06 m 3  0.02 m 3  4.00  10 3 J
b) w for a reversible expansion is:
w  n R T






Vfinal
0.06 m 3
 3.00 mol   8.314472 J K 1 mol 1  300 K   ln
 - 8.22  10 3 J
3
Vinitial
0.02 m
P2.2) A major league pitcher throws a baseball at a speed of 150. km/h. If the baseball weighs
220. g and its heat capacity is 2.0 J g–1 K–1, calculate the temperature rise of the ball when it is
stopped by the catcher’s mitt. Assume no heat is transferred to the catcher’s mitt. Assume also
that the catcher’s arm does not recoil when he or she catches the ball.
The kinetic energy, E kin  12 m v 2 , of the ball is transferred into heat, q p  C p ΔT , in the mitt:
E kin  12 m v 2  q p  C p ΔT
Solving for T yields:
ΔT 
1
2
m v2

Cp
1
2
0.22 kg   41.667 2 m 2s 2 
2.0 J g -1K 1  220.0 g  0.43 K
P2.3) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using a constant external pressure of 5.00 atm. Calculate q, w, U, and H.
The work against a constant pressure:


w  p external V  5  101325 Pa   60  10 3 m 3  20  10 3 m 3   2.03  10 4 J
4
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
U  0 and H  0 since T  0
q  w  2.03  10 4 J
P2.4) A system consisting of 57.5 g of liquid water at 298 K is heated using an immersion heater
at a constant pressure of 1.00 bar. If a current of 1.50 A passes through the 10.0-ohm resistor for
150 s, what is the final temperature of the water? The heat capacity for water can be found in
Appendix B.
The heat produced by the resistor is:
q  I 2 R t  n C p,m R Tf  Ti 
Solving for Tf yields:
Tf
I

2
R t  n C p,m Ti

n C p,m



57.5 g  
-1
1
 1.5 A 2  10 Ohm   150 s   

 18.02 g mol -1   75.291 J mol K  298 K 



  312 K


57.5 g    75.291 J mol -1 K 1

-1 
 18.02 g mol 


 


 


P2.5) Using the results from Problem P1.4, determine the average heat evolved by the oxidation
of foodstuffs in an average adult per hour per kilogram of body weight. Assume the weight of an
average adult is 70. kg. State any assumptions you make. Assume 420. kJ of heat are evolved per
mole of oxygen consumed as a result of the oxidation of foodstuffs.
From P1.4 we know that the number of moles of O2 inhaled are:
n O2  0.01966 mol min -1  1.1796 mol h -1
The average heat evolved by the oxidation of foodstuffs is then:
5
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
q avrg 
q O2 n O2
m

1.1796 mol h  420 10
-1
70 kg 
3
J mol -1
  7.08 kJ h
-1
kg -1
We assumed that all the oxygen that is inhaled is used in the oxidation of foodstuffs.
P2.6) Suppose an adult body were encased in a thermally insulating barrier. If as a result of this
barrier all the heat evolved by metabolism of foodstuffs were retained by the body, what would
the temperature of the body reach after 3 hours? Assume the heat capacity of the body is 4.18 J g–
1
K–1. Use the results of Problem P2.5 in your solution.
From P2.5 we know that the heat produced after 3 h would be:


q3 h   7.08 kJ h -1 kg -1  3 h   21.24 kJ kg -1
Then using q  C p T , we obtain:
q  Cp T 
 2.124 x 104J kg-1 
q 3 h 

 5.08 K
Cp
 4.18 J K -1 g-1   1000 g kg-1 
P2.7) For 1.00 mol of an ideal gas, Pexternal = P = 200.0  103 Pa. The temperature is changed
from 100.0°C to 25.0°C, and CV,m = 3/2R. Calculate q, w, U, and H.
U  n C V, m T 


3
 1.0 mol   8.314472 J K -1 mol -1  298 K - 373 K    935 J
2
H  n C p,m T  n C V, m  R  T 


5
 1.0 mol   8.314472 J K -1 mol -1  298 K - 373 K    1560 J
2
H  q p
w U -qp   935 J  -  1560 J   625 J
P2.8) Consider the isothermal expansion of 5.25 mol of an ideal gas at 450. K from an initial
6
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
pressure of 15.0 bar to a final pressure of 3.50 bar. Describe the process that will result in the
greatest amount of work being done by the system with Pexternal  3.50 bar and calculate w. Describe the process that will result in the least amount of work being done by the system with Pexternal
 3.50 bar and calculate w. What is the least amount of work done without restrictions on
the external pressure?
The most work is performed in a reversible process:
V
w reversible  n R T ln  f
 Vi


p
  n R T ln  f

 pi




 15.0 bar 
3
 5.25 mol   8.314472 J K -1 mol -1  450 K   ln 
   28.6  10 J
 3.50 bar 
The least amount of work is done in a single stage expansion at constant pressure with the external pressure equal to the final pressure:
 1 1
w  pexternal  Vf  Vi   -n R T p external   
 p f pi 
1
1


3
   5.25 mol    8.314472 J K -1 mol-1    450 K    3.50 bar   ln 

   15.110 J
15.0
bar
3.50
bar


The least amount of work done without restrictions on the pressure is zero, which occurs when
p external  0 .
P2.9) A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packing emergency rations that, if completely metabolized, will release 30. kJ of heat per gram of rations consumed. How much rations must the hiker consume to avoid a reduction in body temperature of 4.0 K as a result of heat loss? Assume the heat capacity of the body equals that of water.
Assume the hiker weighs 55 kg. State any additional assumptions.
7
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
We start by calculating the heat that corresponds to a temperature decrease of 4 K. Using
q  C p T , we obtain:
q 4 K  C p T  75.3 J K -1 mol -1  4.0 K  
1
-1
18.02  10 -3 kg mol -1   17000 J kg
We then determine the heat lost for a 55 kg person as:


q person  q 4 K m  17000 J kg -1  55 kg   9.2 105 J
And finally the mass of rations that needs to be consumed is given by:
m rations 
q person
q rations

9.2  10 J 
5
3.0  10
4
J g -1
  31.0 g
P2.10) A muscle fiber contracts by 2.0 cm and in doing so lifts a weight. Calculate the work performed by the fiber and the weight lifted. Assume the muscle fiber obeys Hooke’s law with a
constant of 800. N m–1.
The work is given by:
w  12 k d 2 
1
2
800.0 N m  0.02 m   0.16 J
1
2
P2.11) Calculate H and U for the transformation of 1.00 mol of an ideal gas from 27.0°C and
1.00 atm to 327°C and 17.0 atm if
C P,m  20.9  0.042
T
in units of J K 1 mol1
K
For an ideal gas, H is given by:
Tf
ΔH  n  Cp,m dT  n
Ti
T

 20.9  0.042  dT
K
300 K 
600 K


ΔH  20.9  600 K  300K  J  0.21 T 2

 


600 K
300 K
J
 6.27  103 J  56.7  103 J  63.0  103 J
8
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
ΔU  ΔH  ΔpV   ΔH  n R ΔT




 63.0 103 J  1 mol  8.314472 J mol -1 K -1  300 K   60.5 103 J
P2.12) Calculate w for the adiabatic expansion of 1.00 mol of an ideal gas at an initial pressure
of 2.00 bar from an initial temperature of 450. K to a final temperature of 300. K. Write an expression for the work done in the isothermal reversible expansion of the gas at 300. K from an
initial pressure of 2.00 bar. What value of the final pressure would give the same value of w as
the first part of this problem? Assume that CP,m = 5/2R.
 3

w ad  U  n C p,m  R  T    mol   8.314472 J K -1 mol -1  150 K    1.87 103 J
 2

p
w reversible  n R T ln  i
 pf
p
ln  i
 pf

p
 and ln  i

 pf
  w reversible
 
nRT



  w reversible
1.87 103 J
 

 0.7497
1 mol  8.314472 J K -1 mol -1  300 K 
nRT



pi
 2.12
pf
pf 
pi
 0.944
2.12
P2.13) In the adiabatic expansion of 1.00 mol of an ideal gas from an initial temperature of
25.0°C, the work done on the surroundings is 1200. J. If CV,m = 3/2R, calculate q, w, U, and H.
For an adiabatic expansion of an ideal gas:
q=0
w  ΔU  C V, m  Tfinal  Tinitial    1200 J
 ΔU
ΔH  C P,m  Tfinal  Tinitial   C V, m  R   
 C V, m
 3


   2 R  R    ΔU   52  ΔU   2000 J
3
 R

2 

9
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.14) According to a story told by Lord Kelvin, one day when walking down from Chamonix
to commence a tour of Mt. Blanc, “whom should I meet walking up (the trail) but (James) Joule,
with a long thermometer in his hand, and a carriage with a lady in it not far off. He told me he
had been married since we parted from Oxford, and he was going to try for (the measurement of
the) elevation of temperature in waterfalls.” Suppose Joule encountered a waterfall 30. m in
height. Calculate the temperature difference between the top and bottom of this waterfall.
The decrease in potential energy (mgh) must equal the heat evolved as the water falls.
(mCp,mT). Note that the mass cancels. Heat is evolved so T increases, translating into a negative sign in front of heat term:
m g h  m C p,m ΔT
ΔT  


gh
9.81 m s 2  30.0 m 

 18.02  10 -3 kg mol -1   0.0704 K
-1
-1
C p,m
75.3 J m K




P2.15) An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to a final
state described by Pf, Vf, T in (a) a process at the constant external pressure Pf and (b) in a reversible process. Derive expressions for the largest mass that can be lifted through a height h in
the surroundings in these processes.
w  m g h  p f Vf  Vi 
m 
p f Vf  Vi 
gh
V 
w  m g h  n R T ln  f 
 Vi 
m 
n R T  Vf
ln 
gh
 Vi



P2.16) An automobile tire contains air at 320.  103 Pa at 20.0°C. The stem valve is removed
and the air is allowed to expand adiabatically against the constant external pressure of 100.  103
10
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Pa until P = Pexternal. For air, CV,m = 5/2R. Calculate the final temperature. Assume ideal gas behavior.
Since q = 0, we have U = w
n C V, m Tf  Ti   p ext Vf  Vi 
 n R Tf n R Ti
n C V, m Tf  Ti   p ext 

pi
 pf



The factors n cancel out. Rearranging the equation gives:

R p ext
 C V, m 
pf



R p ext
 Tf   C V, m 
pi



R p ext
 C V, m 
pi
Tf 

Ti 
R p ext
 C V, m 
pf


 Ti

 
 
 

 
 
8.314472 J K -1 mol -1  10 5 Pa
  2.5  8.314472 J K -1 mol -1 
3.20  10 5 Pa

 
8.314472 J K -1 mol -1  10 5 Pa
  2.5  8.314472 J K -1 mol -1 
10 5 Pa
 





 






Tf  0.804 Ti  235 K
P2.17) Count Rumford observed that using cannon-boring machinery, a single horse could heat
11.6 kg of water (T = 273 K) to T = 355 K. in 2.5 hours. Assuming the same rate of work, how
high could a horse raise a 150.-kg weight in 1 minute? Assume the heat capacity of water is 4.18
kJ K–1 kg–1.
The rate of work, L, for warming the water sample can be calculated as the ratio of work and
time:
L


w C m ΔT 4.18 kJ K -1 kg -1  11.6 kg   82 K 


 0.442 kJ s -1  442 J s -1
9000 s
t
t
The rate of work for lifting the weight depends on the potential energy:
11
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
L
w mgh

, where m, g, and h are mass, gravitational acceleration, and height, respectively.
t
t
Solving for h yields:


Lt
442 J s -1  60 s 
h

 18.03 m
m g 150.0 kg   9.80665 m s -2


P2.18) Count Rumford also observed that nine burning candles generate heat at the same rate
that a single horse-driven cannon-boring piece of equipment generates heat. James Watt observed
that a single horse can raise a 330.-lb. weight 100. feet in 1 minute. Using the observations of
Watt and Rumford, determine the rate at which a candle generates heat. (Note: 1.00 m = 3.281
ft.)
First we convert the units:
m = 330.0 lb = 149.685 kg
h = 100 ft = 30.479 m
Then using q = w, we obtain for the heat that horse generates:
q 9 candles  w 


m g h 149.685 kg   9.80665 m s -2  30.479 m

 44754.948 J min -1
1 min 
t
Since this heat is equivalent to the heat that nine candles generate, the heat evolved by one candle
is:
q1 candle


q 9 candles
44754.948 J min -1


 4972.8 J min -1
number of candles
9
P2.19) 3.50 moles of an ideal gas are expanded from 450. K and an initial pressure of 5.00 bar to
a final pressure of 1.00 bar, and CP,m = 5/2R. Calculate w for the following two cases:
a. The expansion is isothermal and reversible.
12
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
b. The expansion is adiabatic and reversible.
Without resorting to equations, explain why the result for part (b) is greater than or less than the
result for part (a).
a) Calculating the initial and final volumina:
Vi 


n R T 3.50 mol   8.314472 J K 1 mol 1  450 K 

 0.0262 m 3
pi
5.00  10 5 Pa




n R T 3.50 mol   8.314472 J K 1 mol 1  450 K 
Vf 

 0.1310 m 3
5
pf
1.00  10 Pa


w for an isothermal, reversible process is then given by:




V 
 0.1310 m 3 

w  n R T ln  final   3.50 mol   8.314472 J K 1 mol 1  450 K   ln 
3 
 0.0262 m 
 Vinitial 
w  -21076 J  -21.1 kJ


b) For an adiabatic, reversible process:
T 
V 
ln  final    - 1 ln  final  , where γ  C P,m /C V, m
 Tinitial 
 Vinitial 
T 
T

p
 - 1 ln  p initial 
ln  final    - 1 ln  final initial   
p


 Tinitial 
 Tinitial p final 
 final 
Therefore:
 γ - 1  p initial 

  ln Tinitial 
Tfinal  Exp 
ln 
γ
 p final 


With C P,m  52 R, and C V, m  32 R , the final temperature is:


 5 bar  
Tfinal  Exp  0.4  ln 
   ln  450 K   236 K
 1 bar  


And finally w for an adiabatic process and for 3.5 moles of gas:
w  CV ΔT  32   8.314472 J K 1 mol 1    450 K-236 K    3.5 mol   -9340.85 J  -9.34 kJ
Less work is done on the surroundings in part b) because in the adiabatic expansion, the temperature falls and therefore the final volume is less that that in part a).
13
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.20) An ideal gas described by Ti = 300. K, Pi = 1.00 bar, and Vi = 10.0 L is heated at constant
volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar.
It is then restored to its original state by the extraction of heat at constant pressure. Depict this
closed-cycle process in a P–V diagram. Calculate w for each step and for the total process. What
values for w would you calculate if the cycle were traversed in the opposite direction?
p 1 Vi
10
8
p [bar]
6
4
2
pi Vi
p i V2
0
10
20
30
40
50
60
70
80
90
100
110
V [L]
First we calculate the number of moles:
n
p i Vi
1.00 bar   10.0 L

 0.401 mol
R Ti
8.314472  10 -2 L bar K 1 mol 1  300 K 


The process in the diagram above is described by the steps:
step 1:
pi, Vi, Ti → p1,Vi, Ti
step 2:
p1, Vi, T1 → pi,V2, T1
step 3:
pi, V2, T1 → p1,Vi, Ti
In step 1 (pi, Vi, Ti → p1,Vi, Ti) w1 = 0 since V stays constant
14
120
130
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
In step 2 (p1, Vi, T1 → pi,V2, T1) we first calculate T1:
T1  Ti
p1
10.0 bar   3000 K
 300 K  
1.00 bar 
pi
Then the work is:
V
w 2  n R T ln  f
 Vi

p
  n R T ln  f

 pi





 10.0 bar 
3
 0.401 mol   8.314472 J K -1 mol -1  3000 K   ln 
   23.0  10 J
 1.00 bar 
In step 3 (pi, V2, T1 → p1,Vi, Ti) we first calculate V2:
p1Vi  p i V2 and V2 
p1 Vi
 10Vi  100L
pi
And the work:
w 3  p externmal V  1.00 bar  
10



Pa
10 3 m 3
 10 L  100 L 
 9.00  10 3 J
1 bar 
1 L
5
And for the entire circle:
w cycle  w1  w 2  w 3  0 J  23.0  10 3 J  9.00  10 3 J   14.0  10 3 J
P2.21) 3.00 mols of an ideal gas with CV,m=3/2R initially at a temperature Ti = 298 K and Pi =
1.00 bar are enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by
placing a 625-kg mass on the piston of diameter 20.0 cm. Calculate the work done in this process
and the distance that the piston travels. Assume that the mass of the piston is negligible.
The constant pressure during compression pconst, which is also the final pressure, pf, can be calculated using the gravitational acceleration. Pressure is force divided by area:
p f  p const 


625 kg   9.80667 m s 2  195084.2 Pa
F mg
mg

 π 2 
π
A
A
d
0.2 m2
4
4


15
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
For a adiabatic, non-reversible compression:
 n R Tf n R Ti
nC V, m Tf  Ti   p const Vf  Vi   p const 

pi
 pf



Solving for Tf and using C V, m  32 R yields:
Tf 
2  p const Ti 3  2  195084.2 Pa   298 K  3

 2 Ti  
 5 298 K   411.34 K
5  p i
5  1  10 5 Pa



The work performed by the gas is:


w  n C V, m Tf  Ti   3.0 mol   32  8.314472 J K 1 mol 1  411.34 K - 298 K   4.24  103 J To
calculate the distance the cylinder moved, we need the final volume of the piston:


n R Tf 3.0 mol   8.314472 J K 1 mol 1  411.34 K 
Vf 

 0.05259 m 3
195084.2 Pa 
pf
The height of the cylinder that the gas filled initially is then:
hi 




Vi
0.07433 m 3

 2.367 m
A
0.0314 m 2
The reduced height of the cylinder filled with gas after compression is:
hf 


h i Vf 2.367 m   0.0526 m 3

 1.675 m
Vi
0.07433 m 3


That means the piston moved the difference in height:
h  h i - h f  2.367 m - 1.675 m  0.69 m
P2.22) A bottle at 21.0°C contains an ideal gas at a pressure of 126.4  103 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against Pexternal = 101.9  103 Pa,
and some gas is expelled from the bottle in the process. When P = Pexternal, the stopper is quickly
replaced. The gas remaining in the bottle slowly warms up to 21.0°C. What is the final pressure
in the bottle for a monatomic gas, for which CV,m = 3/2R, and a diatomic gas, for which CV,m =
5/2R?
16
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
In an adiabatic expansion U = w:
n C V, m Tf  Ti   p ext Vf  Vi 
 n R Tf n R Ti
n C V, m Tf  Ti   p ext 

pi
 pf



The factors n cancel out. Rearranging the equation gives:

R p ext
 C V, m 
pf



R p ext
 Tf   C V, m 
pi



R p ext
 C V, m 
pi
Tf 

Ti 
R p ext
 C V, m 
pf


 Ti

 
 


Tf  0.923 Ti  271 K
Once the stopper is in place, the gas makes a transformation:
Ti  214 K , pi  101.9 103 Pa to Tf  294 K and pf.
p i Vi p f Vf

Ti
Tf
With Vi = Vf:
pf 
 
 
8.314472 J K -1 mol -1  101.9  10 3 Pa
  1.5  8.314472 J K -1 mol -1 
126.4  10 3 Pa

 
8.314472 J K -1 mol -1  101.9  10 3 Pa
-1
-1
  1.5  8.314472 J K mol 
101.9  10 3 Pa
 
294 K   101.9 103 Pa   110.5 103 Pa
Tf
pi 
271 K 
Ti
The same calculation carried out for C V, m 
5
R gives:
2
Tf
 0.945, Tf  278 K
Ti
17


 


 






Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
pf  107.8 103 Pa
P2.23) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a pressure of P = 1.00 bar and temperature of T = 298 K. What is the reaction that occurs? Calculate w
for the process.
The chemical equation for the process is:
Zn (s)  H 2 SO 4 (aq) 
 Zn 2 (aq)  SO 4 (aq)  H 2 (g)
2-
First we calculate the volume of H2 that is produced:
VH2 
10 g 
8.314472 J K mol    298 K   3.79 10


110 Pa 
-1
 65.39 g mol
-1
-1
5
-4
m3
Assuming that Vf  Vi   VH 2 , the work is:

 

w  p ext Vf  Vi    1  10 5 Pa  3.79  10 -3 m 3  379 J
P2.24) One mole of an ideal gas for which CV,m = 20.8 J K–1 mol–1 is heated from an initial temperature of 0.00°C to a final temperature of 275°C at constant volume. Calculate q, w, U, and
H for this process.
For a process with V = constant:
w=0


q  ΔU  C V ΔT  20.8 J K 1 mol 1  275 K   1.0 mol   5720.0 J  5.72 kJ
ΔH  C p ΔT  n R  C V  ΔT 
8.314472 J K
1




mol 1  1.0 mol   20.8 J K 1 mol 1  1.0 mol   273 K   8006.0 J  8.0 kJ
P2.25) One mole of an ideal gas, for which CV,m = 3/2R, initially at 20.0°C and 1.00  106 Pa
18
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
undergoes a two-stage transformation. For each of the stages described in the following list, calculate the final pressure, as well as q, w, U, and H. Also calculate q, w, U, and H for the
complete process.
a. The gas is expanded isothermally and reversibly until the volume doubles.
b. Beginning at the end of the first stage, the temperature is raised to 80.0°C at constant volume.
a) p 2 
p1 V1 p1

 0.500  10 6 Pa
V2
2
V
w  n R T ln  f
 Vi

  1.0 mol   8.314472 J K -1 mol -1  293.15 K   ln 2   1.69  10 3 J



U  H  0 because T  0
q  w  1.69  10 3 J
b)
T p
T1 T2
0.500  10 6 Pa  353 K   6.02  105 Pa
, and therefore p 2  2 1 

293 K 
T1
p1 p 2
U  n C p,m T  1.0 mol  


3
 8.314472 J K -1 mol -1  353 K - 293 K   748 J
2
w = 0 because V = 0
q = U = 748 J
H  n C p,m T  n C V, m  R T  1.0 mol  


5
 8.314472 J K -1 mol -1  353 K - 293 K   1.25  10 3 J
2
For the overall process:
19
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics


q  1.69  10 3 J  748 J   2.44  10 3 J


w   1.69  10 3 J  0 J    1.69  10 3 J
U  748 J   0 J   748 J


H  1.25  10 3 J  0 J   1.25  10 3 J
P2.26) One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00  105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 1.00  106
Pa. Calculate the final temperature of the gas. Calculate q, w, U, and H for this process.
Tf  Vf 
 
Ti  Vi 
1

 Tf

 Ti

p
   i

 pf
 Tf

 Ti
  pi
  
  pf
 Tf

 Ti





T
  f
 Ti






1
 pi

 pf



1
1
1 



  1.00  10 5 Pa 
  

6
  1.00  10 Pa 
 5
 1 
 3
5
 
3
 0.100 
 0.4
 2.51
Tf   2.51   298 K   749 K
q = 0 for an adiabatic process.
w  U  n C p,m T  1.0 mol  


3
 8.314472 J K -1 mol -1  749 K - 298 K   5.62  10 3 J
2
H  U    p V   U  R T   5.62 103 J   8.314472 J K -1 mol -1    749 K-298 K   9.37  103 J
20
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.27) The temperature of 1 mol of an ideal gas increases from 18.0° to 55.1°C as the gas is
compressed adiabatically. Calculate q, w, U, and H for this process assuming that CV,m = 3/2R.
q = 0 for an adiabatic process.
w  U  n C p,m T  1.0 mol  

 

3
 8.314472 J K -1 mol -1  55.1 C - 18.0  C  463 J
2

 

H  U  p V   U  R T  463 J   8.314472 J K -1 mol -1  55.1 C - 18.0  C  771 J
P2.28) A 1.00-mol sample of an ideal gas for which CV,m = 3/2R undergoes the following twostep process: (1) From an initial state of the gas described by T = 28.0°C and P = 2.00  104 Pa,
the gas undergoes an isothermal expansion against a constant external pressure of 1.00  104 Pa
until the volume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature falls to –40.5°C. Calculate q, w, U, and H for each step and for the overall process.
1) For an isothermal process:
q  w  p ext Vf - Vi   p ext 2 Vi - Vi   p ext Vi
The initial volume is:


nRT 1mol   8.314472 J K 1 mol 1  301.15K 
V

 0.1251 m 3
4
p
2.00  10 Pa


And:

 

q  w   0.1251 m 3  1.00  10 4 Pa   1251 J
ΔH  ΔU  0
2) For a process with V = constant:
w=0
q  ΔU  n C V ΔT 
3
2
8.314472 J K
1

mol 1  68.5 K   1.0 mol   854.3 J


ΔH  n C p ΔT  n R  32 n R  ΔT  52 n R  52  8.314472 J K 1 mol 1  1.0 mol   68.5 K   1423.9 J
For the overall process:
21
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
w tot  w 1  w 2   1251 J
q tot  q 1  q 2  1251 J  854.3 J   2.1  10 3 J
ΔU tot  ΔU 1  ΔU 2  854.3 J
ΔH tot  ΔH 1  ΔH 2  1423.9 J
P2.29) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless
adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with CV,m = 3/2R. Initially,
Ti = 298 K and Pi = 1.00 bar in each part. Heat is slowly introduced into the left part using an
electrical heater until the piston has moved sufficiently to the right to result in a final pressure Pf
= 7.50 bar in the right part. Consider the compression of the gas in the right part to be a reversible process.
a. Calculate the work done on the right part in this process and the final temperature in the
right part.
b. Calculate the final temperature in the left part and the amount of heat that flowed into this
part.
The number of moles in each part is given by:
n
p i Vi
1.00 bar   50.0 L 

 2.02 mol
R Ti
8.314472  10 -2 L bar K 1 mol 1  298 K 


a) We first calculate the final temperature in the right side:
Tf  Vf

Ti  Vi



1
T
  f
 Ti



1
 pi

 pf



1
22
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

 Tf

 Ti

p 
   i 

 pf 
 Tf

 Ti
  pi
  
  pf
 Tf

 Ti



1
1 

  1.00 bar  
  



7
.
50
bar



 5
 1 
 3
5
 
3
 2.24
Tf   2.24   298 K   667 K


3
w  ΔU  n C V ΔT  2.02 mol    8.314472 J K 1 mol 1  667 K - 298 K   9.30  10 3 J
2
b) First we calculate the volume of the right part:
-2
1
1
n R Trf  2.02 mol    8.314472 10 bar L K mol    667 K 
Vrf 

 14.9 L
p rf
 7.50 bar 
Therefore Vlf  1000 L  14.9 L  85.1 L , and
Tlf 
 7.50 bar   85.1 L 
plf Vlf

 3800 K
nR
 2.02 mol   8.314472 10-2 bar L K 1 mol 1 


3
ΔU  n C V ΔT  2.02 mol    8.314472 J K 1 mol 1  3800 K - 298 K   88.2  10 3 J
2
From part a) w = 9.30 103 J
q = U – w = 88.2  103 J + 9.30 103 J  97.5  10 3 J
P2.30) A vessel containing 1.00 mol of an ideal gas with Pi = 1.00 bar and CP,m = 5/2R is in
thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equi-
23
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
librium, initially at 298 K, and as separated by adiabatic walls from the rest of the universe. The
vessel, gas, and water bath have an average heat capacity of CP = 7500. J K–1. The gas is compressed reversibly to Pf = 10.5 bar. What is the temperature of the system after thermal equilibrium has been established?
From P2.19 for a reversible, adiabatic process:
 γ - 1  p initial 

  ln Tinitial 
Tfinal  Exp 
ln 
γ
 p final 




 1 bar 
Tfinal  Exp  0.4  ln 
  ln 298 K   763.29 K
 10.5 bar 


The heat transferred to the bath in the adiabatic process is:
q  C V, m gas  ΔT 
5
2
8.314472 J K
1

mol 1  763.3 K - 298 K  1.0 mol   9671.6 J
The temperature of the bath increases to:
Tf 
9671.6 J   298 K   299.3 K
q
 Ti 
7500 J K 1
C p system 


P2.31) DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of length L is bent such that it lies on the arc of a circle of radius Rc. The reversible work
involved in bending DNA without twisting is
wbend 
BL
2R 2 c
where B is the bending force constant. The DNA
in a nucleosome particle is about 680 Å in length. Nucleosomal DNA is bent around a protein
complex called the histone octamer into a circle of radius 55 Å. Calculate the reversible work
involved in bending the DNA around the histone octamer if the force constant B = 2.00  10–28 J
m–1.
The bending work is given by:
24
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
w bend

 

BL
2.00  10 -28 J m -1  680  10 -10 m


 2.25  10 -19 J m -2
2
2
10
2 Rc
2  55  10 m


P2.32) Compare the energy of DNA bending calculated in Problem 2.31 to the thermal energy kB
T, where kB is Boltzmann’s constant. Assume T = 310. K. Propose a source for the excess energy
required to bend the DNA in Problem 2.31.
The thermal energy at 310 K is:


E thermal  k B T  1.38  10 -23 J K -1  310 K   4.278  10 -21 J
The ratio between the energy from P2.31 and the thermal energy is:




E bend
2.25  10 -19 J

 52.6
E thermal
4.278  10 -21 J
The energy source could be the formation of intermolecular interactions.
P2.33) The reversible work involved in twisting a short DNA molecule depends quadratically on
f/L, the angle of twist per unit length, where the twist angle is expressed in units of radians. The
expression for the reversible work is:
2
wtwist
CL   

. A DNA oligomer 20 base pairs in length
2  L 
undergoes a twisting deformation of 36 degrees in order to bind to a protein. Calculate the reversible work involved in this twisting deformation. Assume C = 2.5  10–28 J m. Assume also
that each base pair is 3.4 Å in length (1 Å = 10–10 m).
25
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
2
w twist



CL 
2.5  10 -28 J m  20  3.4  10 -10

  
2 L
2
  36   Pi rad 
 


m   180


-10


20

3
.
4

10
m





2




- 21
  7.26  10 J



P2.34) The formalism of Young’s modulus is sometimes used to calculate the reversible work
involved in extending or compressing an elastic material. Assume a force F is applied to an elastic rod of cross-sectional area A0 and length L0. As a result of this force, the rod changes in length
by L. Young’s modulus E is defined as
E
FL0
tensile stress F / A0


tensile strain L/ L0 A0 L
a. Derive Hooke’s law from the Young’s modulus expression just given.
b. Using your result from part (a), show that the reversible work involved in changing by L
the length L0 of an elastic cylinder of cross-sectional area A0
2
1  L 
is w  
EA0 L0 .
2  L0 
a) Hooke’s law says that the force with which a mass on a spring opposes displacement is directly proportional to the displacement:
Fkd
Young’s modulus shows the same dependency of the force:
E
F L0
A 0 L
26
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
F
E L A 0
 k  L ,
L0
where k is a constant given by:
k 
E A0
L0
b) The work is given by the integral:
L
w twist
L
 L 
1
1 E A0
1


  F ds  k  L ds  k  L2 
L2  E A 0 L 0 
2
2
L
2
L
0
 0
0
0
2
P2.35) The heat capacity of solid lead oxide is given by
C P,m  44.35 1.47 103
T
in units of J K1 mol1
K
Calculate the change in enthalpy of 1 mol of PbO(s) if it is cooled from 500 to 300 K at constant
pressure.


ΔH  n C p ΔT  44.35  1.47  10 -3  300 J K 1 mol 1  1.0 mol   300 K - 500 K   - 8.96  103 J
P2.36) Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with CV,m =
3/2R. The initial state is described by P = 3.25 bar and T = 300. K.
a. Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a
final pressure of P = 1.00 bar.
b. Calculate the final temperature if the same gas undergoes an adiabatic expansion against an
external pressure of P = 1.00 bar to a final pressure of P = 1.00 bar.
c. Explain the difference in your results for parts (a) and (b).
27
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
a) The final temperature for the reversible adiabatic expansion is:
Tf  Vf 
 
Ti  Vi 
1
T
  f
 Ti

 Tf

 Ti

p 
   i 

 pf 
 Tf

 Ti
  pi
  
  pf
 Tf

 Ti






1
 pi

 pf



1
1
1 

  3.25 bar  
  

  1.00 bar  
 5
 1 
 3
5
 
3
 0.626
Tf  0.626  300 K  188 K
b) The final temperature for the expansion against an external pressure can be obtained by using
the fact that in an adiabatic expansion U = w:
n C V, m Tf  Ti   p ext Vf  Vi 
 n R Tf n R Ti
n C V, m Tf  Ti   p ext 

pi
 pf



The factors n cancel out. Rearranging the equation gives:

R p ext
 C V, m 
pf



R p ext
 Tf   C V, m 
pi



 Ti

28
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

R p ext
 C V, m 
pi
Tf 

Ti 
R p ext
 C V, m 
pf

 
 

 
8.314472 J K -1 mol -1  1.00 bar  
  1.5  8.314472 J K -1 mol -1 

3.25 bar 


-1
-1
 
8.314472 J K mol  1.00 bar  
  1.5  8.314472 J K -1 mol -1 

1.00 bar 

 



Tf  0.723 Ti  217 K
P2.37) The relationship between Young’s modulus and the bending force constant for a deformable cylinder is B = EI, where, I = R4/4 and R is the radius of the cylinder.
a. Calculate the Young’s modulus associated with a DNA of radius 10. Å (1 Å = 10–10 m).
Assume the value of B given in Problem 2.31.
b. Suppose a DNA molecule 100 base pairs in length is extended by 10. Å. Calculate the reversible work assuming the DNA can be treated as a deformable rod.
c. Compare this work to the thermal energy. Assume T = 310. K.
a) Young’s modulus is given by:
E


B
4B
4  2.00  10 -28 J m1


 2.55  10 8 J m -3
4
4
10
I πR
π  10  10 m


b) Assuming 3  10 -10 m per base pair for the DNA, the work is given by:

 
 
E A 0 L2   10  10 -10 m  2.55  10 8 J m -3  10  10 -10 m
w

2 L0
2  300  10 -10 m
2



2
 4.45  10 -22 J
c) The thermal energy at 310 K is:


E thermal  k T  1.38  10 -23 J K -1  310 K   4.28  10 -21 J
P2.38) The Young’s modulus of muscle fiber is approximately 3.12  107 Pa. If a muscle fiber
2.00 cm in length and 0.100 cm in diameter is suspended with a weight M hanging at its end, calculate the weight M required to extend the length of the fiber by 10%.
We use the result from P2.34 a:
29
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
F
E L A 0
 k  L ,
L0
We set this force equal to the force exerted by the weight in the earth’s gravitational field
F  g m , and solve for the mass, m:
E L A 0
gm
L0
2


d
 0.001 m 
E L   
3.12  10 7 Pa  0.1  0.02 m     

2
E L A 0 E L  r
2
2 


m



L0 g
L0 g
L0 g
0.02 m   9.81 m s 2


2
 0.25 kg
P2.39) An ideal gas undergoes a single-stage expansion against a constant external pressure Pexternal
at constant temperature from T, Pi, Vi to T, Pf, Vf.
a. What is the largest mass m that can be lifted through the height h in this expansion?
b. The system is restored to its initial state in a single-state compression. What is the smallest
mass m that must fall through the height h to restore the system to its initial state?
c. If h = 10.0 cm, Pi = 1.00  106 Pa, Pf = 0.500  106 Pa, T = 300. K, and n = 1.00 mol, calculate the values of the masses in parts (a) and (b).
a) For the expansion:
w  m g h  p ext Vf  Vi 
m
 p ext Vf  Vi 
gh
For a final volume of Vf, the external pressure can not be larger than pf:
30
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
m max 
 p f Vf  Vi 
gh
b) For the compression:
w  m g h  p ext Vi  Vf 
m
 p ext Vi  Vf 
gh
For a final volume of Vi, the external pressure can not be larger than pi:
m min 
 p i Vi  Vf 
gh
c) Vi 
n R Ti 1.0 mol   8.314472 J K 1 mol 1  300 K 

 2.49  10 -3 m 3
pi
1.00  10 6 Pa




p i Vi  p f Vf
Vf 

m max 
m min
 

p i Vi
1.00  10 6 Pa  2.49  10 -3 m 3

 4.98  10 -3 m 3
6
pf
0.500  10 Pa



 
 
  1.27 10
 p f Vf  Vi   0.500  10 6 Pa  4.98  10 -3 m 3  2.49  10 -3 m 3

gh
9.81 m s -2  0.100 m




6
-3
3
-3
3
pf  Vf  Vi   1.00 10 Pa    4.98 10 m    2.49 10 m 


gh
 9.81 m s-2    0.100 m 
3
kg
 2.54 103 kg
P2.40) Calculate q, w, U, and H if 1.00 mol of an ideal gas with CV,m = 3/2R undergoes a reversible adiabatic expansion from an initial volume Vi = 5.25 m3 to a final volume Vf = 25.5 m3.
The initial temperature is 300. K.
For an adiabatic process:
31
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
q =0
From P2.19 for a reversible, adiabatic process:
C p,m nR  C V, m  nR  32 R 


V 



Tfinal  Exp  γ  1 ln  final   ln Tinitial  , γ 
3
C V, m
C V, m
 Vinitial 
2R


Tfinal
5
3
 5

 25.5 m 3 



 Exp   3  1  ln 

ln
300
K
  104.6 K
3 
 5.25 m 




w  U  C V, m ΔT  32  8.314472 J K 1 mol 1  300 K - 104.6 K   1.0 mol   2437.0 J  2.44 kJ


H  C p,m ΔT  52  8.314472 J K 1 mol 1  300 K - 104.6 K   1.0 mol   4061.6 J  4.06 kJ
P2.41) A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and
temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K, respectively.
The final volume of the air in the tire is Vf = 1.00 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire. Assume that CV,m = 5/2R
Tf  Vf 
 
Ti  Vi 
1

T
  f
 Ti
 Tf

 Ti

p 
   i 

 pf 
 Tf

 Ti
  pi
  
  pf
 Tf

 Ti






1
 pi

 pf



1
1
1 

  1.00 bar  
  

  5.00 bar  
 7
 1 
 5
7
 
5
 1.58
Tf  1.58  298 K  472 K
32
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.42) One mole of an ideal gas with CV,m = 3/2R is expanded adiabatically against a constant
external pressure of 1.00 bar. The initial temperature and pressure are Ti = 300. K and Pi = 25.0
bar, respectively. The final pressure is Pf = 1.00 bar. Calculate q, w, U, and H for the process.
For an adiabatic, irreversible process:
q=0
However, dw = - pdV, where p is the pressure of the system. Rather we have dw = - pf dV, where
pf is the constant, external pressure that is producing the work. Since we are still assuming an
ideal gas in this case, we again have dU = nCV,mdT. Equating these expressions for dw and dU,
we obtain:
n C V, m dT  p f dV
Integrating on both sides yields:
n C V, m Tf - Ti   p f Vf - Vi 
C V, m Tf - Ti   


R Tf
Vf - Vi   R Tf  Vi  1
Vf
 Vf

Solving for Tf results in:
Tf 




C V, m Ti
R Ti p f
8.314472 J K 1 mol 1  300 K   1  10 5 Pa 3


 5  300 K   184.8 K
p i C V, m  R  C V, m  R 
25  10 5 Pa 52  8.314472 J K 1 mol 1



For an adiabatic process:


w  U  C V, m ΔT  32  8.314472 J K 1 mol 1  300 K - 184.8 K   1.0 mol   1436.7 J  1.44 kJ


H  C p,m ΔT  52  8.314472 J K 1 mol 1  300 K - 184.8 K   1.0 mol   2394.6 J  2.39 kJ
P2.43) One mole of N2 in a state defined by Ti = 300. K and Vi = 2.50 L undergoes an isothermal
reversible expansion until Vf = 23.0 L. Calculate w assuming (a) that the gas is described by the
ideal gas law and (b) that the gas is described by the van der Waals equation of state. What is the
percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals
parameters for N2 are listed in Table 1.3.
a) For an ideal gas:
33
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
V
w reversible  n R T ln  f
 Vi

 23.0 L 
3
  1.0 mol   8.314472 J K -1 mol -1  300 K   ln 
   5.54  10 J
 2.50 L 



b) For a van der Waals gas:
Vf
w    p external
Vi
 RT
a
dV    

V b V 2
m
Vi 
m
Vf
Vf
Vf





 dV    R T dV    a dV
V  Vm  b  V  V 2 

m 

i
i
The first integral can be solved by substituting y = Vm - b:
f
f
 RT 
RT
dV    
dy   R Tln Vf  b   ln Vi  b 
  
V b
y 
Vi  m
yi 
V
y
Therefore, the work is given by:
 V  b  
1
1 
  a   
w reversible  n R T ln  f
 Vi  b  
 Vi Vf 
 23.0 L  0.0380 L  

 1.0 mol   8.314472 J K -1 mol -1  300 K   ln 
 2.50 L  0.0380 L  



Pa 
 10
 10
1 bar 
 1.366 L2 bar 
5
6

1
1
m 6 L2  

-3
3
23.0  10-3 m 3
 2.50  10 m
 
 


   5.52  103 J

And the percent error is:
 5.52 10 J    5.54 10 J 
percent error  100 
3
3
 5.52 10 J 
3
  0.4%
P2.44) One mole of an ideal gas, for which CV,m = 3/2R, is subjected to two successive changes
in state: (1) From 25.0°C and 100.  103 Pa, the gas is expanded isothermally against a constant
pressure of 20.0  103 Pa to twice the initial volume. (2) At the end of the previous process, the
gas is cooled at constant volume from 25.0° to –25.0°C. Calculate q, w, U, and H for each of
the stages. Also calculate q, w, U, and H for the complete process.
34
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
1) For an isothermic expansion against a constant pressure:
w  p ext  Vfinal  Vinitial   p ext

nRT
pi
 1.0 mol   8.314472 J K
100  10
w   100  10 3 Pa 
1
3

mol 1  298.15 K 
 - 495.8 J
Pa

ΔH = ΔU = 0, since process is isothermal
q = -w = 495.8 J
2) For cooling at constant volume:
w = 0, since V = constant
q  ΔU  C V ΔT 
3
2
8.314472 J K
1

mol 1  50.0 K   1.0 mol   - 623.6 J


ΔH  C p ΔT  n R  32 n R  ΔT  52 n R  52  8.314472 J K 1 mol 1  1.0 mol   50.0 K   - 1039.3 J
For the overall process:
w tot  w 1  w 2   495.8 J
q tot  q1  q 2  495.8 J  623.6 J   127.8 J
ΔU tot  ΔU1  ΔU 2  623.6 J
ΔH tot  ΔH1  ΔH 2  - 1039.3 J
P2.45) The adhesion of leukocytes (white blood cells) to target cells is a crucial aspect of the
body’s immune system. The force required to detach leukocytes from a substrate can be measured with an atomic force microscope (AFM). The work of de-adhesion is proportional to the
cell–substrate contact area Ac, which in turn is related to Young’s modulus, the force F applied to
 RF 
the cell by the AFM, and the radius R of the cell by Ac    
 E 
2/3
. Assuming R = 5.00 
10–6 m and F = 2.00  10–7 N, calculate the change in contact area when Young’s modulus is reduced from 1.40 to 0.30 kPa. By what amount does the work of de-adhesion change when
Young’s modulus is changed from 1.40 to 0.30 kPa? Explain this effect.
The difference in contact area is:
35
Chapter 2/Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
2
 R F
 R F
   

ΔA c   
 E1 
 E2 

3
2
3

 

2

 

 5.00  10 6 m  2.00  10 7 N 
 5.00  106 m  2.00  107 N 
   

  
1400 Pa 
300 Pa 




 
3

 2.51 10 10 m 2  7.01 1010 m 2  4.50  10 10 m 2
Since the work is proportional the cell-substrate area, the work ratio is the ratio of the two areas:
10
2
F2  7.0110 m 

 2.78
F1  2.511010 m 2 
36
2
3