ERT 206/4
Thermodynamics
CHAPTER 2
The First Law and Other Basic Concepts
Miss. Rahimah Bt. Othman
Email: rahimah@unimap.edu.my
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
Define, discuss, apply and analyze internal energy, first law,
energy balance-closed system, thermodynamic state and state
function, equilibrium, the Phase Rule, reversible process,
constant-V and constant-P processes, enthalpy and heat
capacity.
3. Chapter 3: Volumetric properties of pure fluids
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
Fundamental Laws of Thermodynamics
• First law of thermodynamics – energy can be
converted from one form to another, but cannot be
created or destroyed and the total quantity of
energy is constant.
(Eg: A stone is falling down from the highest place) - Potential
energy convert to kinetic energy.
• Second law of thermodynamics – energy
consist of quality and quantity where the real
processes normally occurred within the
decreasing quality of energy.
(Eg: A cup of hot coffee become a cool coffee after a few
minutes) – Energy with low temperature.
First Law of Thermodynamics
DEsystem + DEsurroundings = 0
(Eq 2.1)
or
DEsystem = -DEsurroundings
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
Energy Balance For Closed System
• Closed system does not allow mass transferred but,
allow heat transferred between the systems and
surroundings.
• Because of that, all energy exchange between a
closed system and its surrounding appears as heat
and work.
• Therefore the second term of Eq 2.1 can be stated as;
DE surrounding   Q  W
• Q and W always refer to the system, and the choice of
the signs depends on the way energy or work
transferred.
• If Q and W transfer into the system from the
surroundings, then, the sign is +ve.
• With that understanding, thus Qsurr can be written as –Q and
Wsurr can be written as –W
• So, energy of surrounding now becomes;
DE surrounding   Qsurr  Wsurr
 Q  W
• Equation 2.1 now becomes;
DE system  Q  W
(2.2)
• This can be reduced to Eq 2.3 where Ut is total Internal Energy of
the system;
DU  Q  W
t
(2.3)
• Equation 2.3 only applies to finite change, where as, for differential
change, it becomes as;
(2.4)
dU t  dQ  dW
Example 2.1
Water flows over a waterfall 100 mm in height. Take 1 kg of the water as
the system, and assume that it does not exchange energy with it
surroundings.
(a) What is the potential energy of the water at the top of the falls with
respect to the base of the falls?
(b) What is the kinetic energy of the water just before it strikes bottom?
(c) After the 1 kg of water enters the stream below the falls, what
change has occurred in its state?
Solution
The 1 kg of water exchanges no energy with the surroundings. Thus,
for each part of the process Eq. (2.1) reduces to;
D(Energy of the system )  DU  DEK  DEP  0
(a) From Eq. (1.7) with g equal to its standard value.
E P  mzg  1 kg x 100 m x 9.8066 ms 2
kgm 2
 980.66
 980.66 N m  980.66J
2
s
(b) During the free fall of the water no mechanism exists for conversion
of potential or kinetic energy into internal energy. Thus ∆U must be
zero:
DU  DEK  DEP  EK 2  E K1  E P2  E P1  0
As an excellent approximation, let
EK1  EP2  0 Then,
E K 2  E P1  980.66 J
(c) As the 1 kg of water strikes bottom and mixes with other falling water
to form a stream, the resulting turbulence has the effect of converting
kinetic energy into internal energy. During this process , ∆EP is
essentially zero, and Eq. (2.1) becomes;
DU  DEK  0
or
DU  E K 2  E K 3
However, the stream velocity is assumed small, making EK3 negligible.
Thus,
DU  EK 2  980.66 J
Example 2.1-cont’
The overall result of the process is the conversion of potential energy
of the water into internal of the water. This exchange in internal energy
is manifested by a temperature rise of the water. Because energy in
the amount of 4, 184 J kg-1 is required for a temperature rise of 1oC in
water., the temperature increase is;
980.66
T 
 0.234 o C
4184
Assuming
NO HEAT transfer with the surroundings.
Example 2.2
A gas is confined in a cylinder by a piston. The initial pressure of the
gas is 7 bar and the volume is 0.10 m3. The piston is held in place by
latches in the cylinder wall. The whole apparatus is placed in a total
vacuum. What is the energy change of the apparatus if the restraining
latches are removed so that the gas suddenly expands to double its
initial volume, the piston striking other latches at the end of the
process?
Solution
Because the question concerns the entire apparatus, the system
is taken as the gas, piston, and cylinder. No work is done during
the process, because no force internal to the system moves, and
no heat is transferred through the vacuum surrounding the
apparatus. Hence Q and W are zero, and the total energy of the
system does not change. Without further information we can say
nothing about the distribution of energy among the parts of the
system. This may well be different than the initial distribution.
Example 2.3
If the process described in Eg. 2.2 is repeated, not in a vacuum but in
air at atmospheric pressure of 101.3 kPa, what is the energy change of
the apparatus? Assume the rate of heat exchange between the
apparatus and the surrounding air is slow compared with the rate at
which the process occurs.
Solution 1
The system is chosen as before, but here work is done by the
system in pushing back the atmosphere. It is evaluated as the
product of the force of atmospheric pressure on the back side of the
piston F = Patm A and the displacement of the piston ∆l = ∆Vt/A.
Here, A is the area of the piston and ∆Vt is the volume change of
the gas. This is work done by the system on the surroundings, and
is a negative quantity; thus,
W   FDl   PatmDV t  (101.3)(0. 2  0.1) kPa m 3  10.13
W  10.13 kN m  10.13 kJ
kN 3
m
2
m
Example 2.3 cont’
Solution 2
Heat transfer between the system and surroundings is also possible
in this case, but the problem is worked for the instant after the
process has occurred and before appreciable heat transfer has had
time to take place. Thus Q is assumed to be zero in Eq. (2.2),
giving;
D(Energy of the system )  Q  W  0  10.13  10.13 kJ
The total energy of the system has decreased by an amount equal
to the work done on the surroundings.
Example 2.4
When a system is taken from a state a to state b as in the Fig 2.1 along
path acb, 100 J of heat flows into the system and the system does 40 J of
work
(a) How much heat flows into the system along path aeb if the work done
by the system is 20 J
(b) The system returns from b to a along bda. If the work done on the
system is 30 J, does the system absorb or liberate heat? How much?
Solution
Example 2.4 cont’
Assume that the system changes only in its internal energy and that
Eq. (2.3) is applicable. For path acb, and thus for any path leading
from a to b,
From first law of thermodynamics;
DEsystem + DEsurroundings = 0
DE system  Q  W
DU t  Q  W
DU
t
ab
 Qacb  Wacb
 100  40
 60 J
(2.3)
(a) For path aeb,
DU t ab  Qaeb  Waeb
DU t ab not depend on the route, so, its still 60 J
So,
DU t ab  60 J
Qaeb  ?
Waeb  20 J
60  Qaeb  20
Qaeb  60  20
 80 J
(b) For path bda,
DU t ab   DU t ba (reverse direction )
 60 J
So,
 DU t ba  60 J
Qbda  ?
Wbda  30 J
 60  Qbda  30 ( work done on the system)
Qbda  60  30
 90 J
therefore the heat transferred out from the
system to the surroundin g.
2.6 Equilibrium
- a static condition
- no tendency toward change
2.7 Phase Rule
F  2   N
• Phase rule as introduced by Gibbs;

= number of phases
N = number of chemical species
F = degree of freedom
• For multi phase equilibrium, the number of independent variables that must be
fixed to establish its intensive state is given by the formula above.
• The intensive state of a system in equilibrium is established when its
temperature, pressure and composition of all phases are fixed.
• The phase rule give the number of variables from this set which must be
specified to find others variables.
Example 2.5
How many degrees of
systems?
freedom has each of the following
(a) Liquid water in equilibrium with its vapor
(b) Liquid water in equilibrium with a mixture of water vapor and
nitrogen
(c) A liquid solution of alcohol in water in equilibrium with its vapor
Solution
(a) The system contains a single chemical species existing as two
phases (one liquid and one vapor). Thus,
F  2    N  2  2 1  1
This result is in agreement with the fact that for a given pressure
water has but one boiling point. Temperature or pressure, but not
both, may be specified for a system comprised of water in
equilibrium with its vapor.
Example 2.5 cont’
(b) In this case two chemical species are present. Again there are
two phases.
F  2   N  2  2  2  2
The addition of an inert gas to a system of water in equilibrium with its vapor
changes the characteristics of the system. Now temperature and pressure may be
independently varied, but once they are fixed the system described can exist in
equilibrium only at a particular composition of the vapor phase. (If nitrogen is
considered negligibly soluble in water, the liquid phase is pure water).
(c) Here N = 2, and π = 2, and
F  2   N  2  2  2  2
The phase-rule variables are temperature, pressure, and the phase compositions.
The composition variables are either the mass or mole fractions of the species in a
phase, and they must sum to unity for each phase. Thus fixing the mole fraction of
the water in the liquid phase automatically fixes the mole fraction of the alcohol.
These compositions cannot both be arbitrary specified.
2.8 Reversible Process
- DEFINITION: A process where its direction can be reversed at any point by
infinitesimal change in external condition.
The piston (Fig. 2.2) confines the gas at a pressure just specified to balance
the weight of the piston and all that it supports.
Reversible Chemical Reaction
CaCO3  CaO  CO2
Increased the weight
CO2 pressure rises
CO2 combines with CaO to form
CaCO3 (exothermic process)
Temperature rise in the
bath
Example 2.6
A horizontal piston/cylinder arrangement is placed in a constant-temperature
bath. The piston slides in the cylinder with negligible friction, and an external
force holds it in place against an initial gas pressure of 14 bar. The initial gas
volume is 0.03 m3. The external force on the piston is reduce gradually, and
the gas expands isothermally as its volume doubles. If the volume of the gas
is related to its pressure so that the product PVt is constant, what is the work
done by the gas in moving the external force?
How much work would be done if the external force were suddenly
reduced to half its initial value instead of being gradually reduced?
Solution
The process, carried out as first described, is mechanically reversible, and
Eq. (1.3) is applicable. If PVt = k, then P = k/Vt, and
V2t
V2t
V1
V1
W    t PdV t  k  t
dV t
V2t
 k ln t
t
V
V1
With;
V1t  0.03 m3
V2t  0.06 m3
Example 2.6 cont’
and;
k  PV t  P1V1  (14 x 105 )(0.03)  42000 J
t
W  42000 ln 2  29112 J
The final pressure is;
k 42 000
P2 t 
 700 000 Pa  7 bar
V2
0.06
In the second case, reduction of the initial force by half is followed by sudden
expansion of the gas against a constant force equivalent to a pressure of 7 bar.
Eventually, heat transfer returns the system to the same final equilibrium state as in
the reversible process. Thus ∆ Vt is the same as before, but the work accomplished
is not given by Eq. (1.3). Rather, the work done against the external force equals
the equivalent external pressure times the volume change:
W  (7 x 105 )(0.06  0.03)  21 000 J
This process is clearly irreversible, and compared with the reversible
process is said to have an effieciency of:
21 000 /29 112  0.721
or
72.1 %
Example 2.7
• See Example 2.7 (page 36)
2.9 CONSTANT-V and CONSTANT-P PROCESSES
For n moles of homogeneous fluid in a closed system;
d (nU )  dQ  dW
(2.6)
W meanwhile can be replaced by;
dW   Pd (nV )
Will be explained by
Carnot group next
week
Combined both equations;
d (nU )  dQ  Pd (nV )
(2.8)
Equation 2.8 is a general equation for n moles of homogeneous fluid in a
closed system experienced a mechanically reversible process
(1.2)
CONSTANT-V PROCESS
For constant-V process, the work is equal to zero
d (nU )  dQ  dW
(2.6)
W meanwhile can be replaced by;
0
d (nU )  dQ  dW
dQ  d (nU )
(2.9)
Q  nDU
(2.10)
The integration of 2.9 yield
For the constant-V, closed system process, the heat transfer is equal to
internal energy change of the system
CONSTANT-P PROCESS
For constant-P process,
d (nU )  dQ  Pd (nV )
(2.8)
Solved for dQ;
dQ  d (nU )  Pd (nV )
Since P now is a constant, therefore
dQ  d (nU )  d (nPV )
 d [n(U  PV )]
The term (U+PV) is known as enthalpy and has only mathematical definition,
which is
H  U  PV
(2.11)
To be continue…
THANK YOU