Chapter 8
8.3 Maxwell Relations
The fundamental property relations are written so that the left-hand side is the differential of
a state function as follow
dU = TdS − PdV
(8.3-1a)
dH = TdS + VdP
(8.3-1b)
dA = − SdT − PdV
(8.3-1c)
dG = − SdT + VdP
(8.3-1d)
The exact differential of a function z = z(x, y) is written as
dz =
∂z
∂x
∂z
∂y
dx +
y
dy = Mdx + Ndy
(8.3-2)
x
In this expression, we have
M=
∂z
∂x
and N =
y
∂z
∂y
x
A useful property of exact differential is that the order of partial differentiation does not
matter. That is
∂ ∂z
∂y ∂x
∂M
∂y
=
y
=
x
x
∂N
∂x
∂ ∂z
∂x ∂y
(8.3-2)
x
y
(8.3-3)
y
Applying equation (8.3-3) to equation (8.3-1a) we get
∂T
∂V
=−
S
∂P
∂S
(8.3-4a)
V
Similarly, from the other three fundamental property relationships we have
8-9
∂T
∂P
∂S
∂V
−
∂S
∂P
=
S
∂V
∂S
P
∂P
∂T
V
=
T
∂V
∂T
=
T
(8.3-4b)
(8.3-4c)
(8.3-4d)
P
The Maxwell relations can be used to calculate entropy from PVT data. The derivative
relations of equations (8.3-1a)−(8.3-1d) then enable us to calculate U, H, A, and G. The
Maxwell relations can also be obtained from the diagram
P
G
H
S
T
A
V
∂T
∂V
U
=−
S
∂P
∂S
V
The sign of the derivative is obtained from multiplication of the individual sign of each
variable in a group. For TVS the sign is (−)(−)(+) = +, for PSV the sign is (+)(+)(−) = −.
Similarly, for the other three Maxwell relations we have
P
G
H
S
T
A
V
G
P
=
S
∂V
∂S
P
∂P
∂T
V
H
S
T
A
∂T
∂P
U
V
U
8-10
∂S
∂V
=
T
P
G
H
S
T
A
V
U
−
∂S
∂P
=
T
∂V
∂T
P
Example 8.3-1.3 ---------------------------------------------------------------------------------One mole of propane gas is to be expanded from 0.001 m3 to 0.040 m3 while in contact with
a heating bath at 100oC. The expansion is not reversible. The heat extract from the bath is
600 J. Determine the work for the expansion using the Van der Waals equation of state.
P=
RT
a
− 2
v −b
v
Solution ----------------------------------------------------------------------------------------From the first law we have ∆u = q + w. Since the amount of heat transfer q is given (600
J/mol), we only need to evaluate ∆u to find w.
We write u as a function of the independent variables T and v since the values of these two
variables are given in the problem
u = u ( T, v )
du =
∂u
∂T
dT +
v
∂u
∂v
dv
T
Since the process occurs at constant T, dT = 0, we have
du =
We need to express
∂u
∂v
T
∂u
∂v
T
dv
in terms of measurable properties by using the diagram
G
P
T
A
3
V
H
G
S
T
U
A
P
H
S
V
U
Koresky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 222
8-11
∂u
∂s
du =
ds +
v
∂u
∂v
d v = Td s − P d v
s
Therefore
T∂s − P∂v
∂v
du =
∂s
∂v
dv = T
T
P
G
− P dv
T
H
S
T
A
V
U
∂S
∂V
=
T
∂P
∂T
V
∂s
∂P
=
, the change in internal energy can now
∂v T
∂T v
be expressed solely in terms of measurable properties:
Using the Maxwell relation we have
du =
T
∂P
∂T
− P dv
v
In terms of molar quantities
du =
T
∂P
∂T
− P dv
v
From the Van der Waals equation:
∂P
∂T
=
v
R
v −b
P=
T
∂P
∂T
RT
a
− 2
v −b
v
−P =
v
a
v2
Hence
∆u =
0.04 m3
0.001m3
∂s
T
∂v
a
a
dv = −
− P dv =
3
2
0.001m v
v
0.04 m3
T
8-12
0.04 m3
0.001m3
Using R = 8.314 J/(mol⋅oK) = 8.314×10-5 m3⋅bar/(mol⋅oK) and for propane, Tc = 369.9oK, Pc
= 42.46 bar, we have
a=
27 ( RTc ) 2
27 (8.314 × 10 −5 × 369.9) 2
=
= 9.36×10-6 m6⋅bar/mol
64 Pc
64
42.46
Therefore
∆ u = − 9.36×10-6
1
1
−
= 913×10-5 m3⋅bar/mol
0.04 0.001
∆ u = 913 J/mol
The required work for the expansion is
w = ∆ u − q = 913 − 600 = 313 J/mol
Example 8.3-2.4 ---------------------------------------------------------------------------------Propane at 350oC and 600 cm3/mol is expanded in an isentropic turbine. The exhaust
pressure is atmospheric. What is the exhaust temperature? PvT behavior has been fit to the
Van der Waals equation with a = 92×105 [(atm⋅cm6)/mol2] and b = 91 [cm3/mol]. Solve this
using T as v as the independent variables, that is s = s(T, v). The Van der Waals equation is
given as
P=
RT
a
− 2
v −b
v
The ideal gas heat capacity for propane is
cp
R
= 1.213 + 28.785×10-3T − 8.824×10-6T2
Solution ----------------------------------------------------------------------------------------Since propane is expanded in an isentropic turbine, we can construct a path such that the sum
of ∆s is zero.
4
Koresky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 245
8-13
T2, v2
∆s2
Ideal gas
v
∆s1
∆s = 0
T1, v1
T
Choosing T and v as the independent variables the change in entropy is given by
∂s
∂T
ds =
Since ∂s =
∂Q
T
∂s
∂T
dT +
v
1
T
=
v
∂s
∂v
dv
T
∂Q
∂T
=
v
cv
T
Using Maxwell relation as shown on the following diagram, we have
∂s
∂v
=
T
∂P
∂T
v
P
G
S
T
A
Therefore
ds =
cv
∂P
dT +
T
∂T
From the Van der Waals equation:
H
V
U
∂S
∂V
=
T
∂P
∂T
V
dv
v
P=
RT
a
− 2
v−b
v
For an isentropic process, we have
8-14
∂P
∂T
=
v
R
v−b
∆s =
T2
T1
cv
dT +
T
v2
v1
R
dv = 0
v−b
c
cv
= p − 1 = 0.213 + 28.785×10-3T − 8.824×10-6T2
R
R
0.213ln(T2/T1) + 28.785×10-3(T2 − T1) − 4.412×10-6(T22 − T12) + ln
v2 − b
=0
v1 − b
(E-1)
v2 − b
v1 − b
(E-1)
In terms of molar quantities
0.213ln(T2/T1) + 28.785×10-3(T2 − T1) − 4.412×10-6(T22 − T12) + ln
Since R = 82.06 cm3⋅atm/mol⋅K, v2 =
=0
RT2
82.06T2
=
= 82.06T2
1
P2
Substituting the following numerical values: T1 = 350oC = 623.15oK, v1 = 600 cm3/mol, P2 =
1 atm, and b = 91 [cm3/mol] into equation (E-1), we have
0.213ln(T2/623.15) + 28.785×10-3(T2 −623.15) − 4.412×10-6(T22 − 623.152)
+ ln
82.06T2 − 91
=0
509
(E-2)
We can use the following Matlab statements to solve the nonlinear equation
>> fun=inline('0.213*log(x/623.15) + 28.785e-3*(x -623.15) - 4.412e-6*(x^2 623.15^2)+log((82.06*x-91)/509)');
>> T2=fsolve(fun,500,optimset('Display','off'))
T2 =
448.2512
>>
The temperature of the exhaust from the turbine is 448.3oK.
8-15
Example 8.3-3.5 ---------------------------------------------------------------------------------Develop a general relationship for the change in temperature with respect to pressure at
constant entropy
∂T
∂P
s
(a) Evaluate the expression for an ideal gas.
(b) From the result in part (a), show that for an ideal gas with constant cP, an isentropic
expansion from state 1 and state 2 yields equation Pv k = const.
(c) Evaluate the expression for a gas that obeys the Van der Walls equation of state.
Solution ----------------------------------------------------------------------------------------(a) Evaluate the expression for an ideal gas
The cyclic rule can be employed to give
∂T
∂T
∂s
=−
∂P s
∂s P ∂P T
Substituting the relation
∂s
∂T
=
P
cP
∂s
and
T
∂P
=−
T
∂v
∂T
yields
P
∂T
T ∂v
=
∂P s c P ∂T P
For an ideal gas: Pv = RT
∂v
R
=
∂T P P
Therefore,
∂T
RT 1
v
=
=
∂P s
P cP cP
(b) Show that for an ideal gas with constant cP, an isentropic expansion from state 1 and state
2 yields equation Pv k = const.
Separation of variables provides
∂T
R ∂P
=
T
cP P
5
Koresky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 245
8-16
Integration provides
R
T
P c
ln 2 = ln 2 P
T1
P1
This expression can be rewritten as
T2
P
= 2
T1
P1
R
cP
The ideal gas law is now employed
R
cP
P2 v 2
P
= 2
P1v1
P1
1−
P2
R
cP
1−
v 2 = P1
R
cP
v1
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) Evaluate the expression for a gas that obeys the Van der Walls equation of state.
In Part (a), we found
∂T
T ∂v
=
∂P s c P ∂T P
Using the derivative inversion rule, we find for the van der Waals equation
P=
RT
a
− 2
v−b
v
Rv 3 (v − b )
∂v
=
∂T P RTv 3 − 2a(v − b )2
Therefore,
∂T
1
RTv 3 (v − b )
=
∂P s c P RTv 3 − 2a (v − b )2
8-17