Modern Engineering Thermodynamics

368 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.4 (Continued )
and
∂N
= ∂ðv2 Þ
∂x y ∂v
p
= 2v ≠ ∂M
∂y
x
Since Eq. (11.12) is not satisfied here, then z cannot be a thermodynamic property.
Exercises
10. Suppose the expression experimentally discovered in Example 11.4 is dz = p 2 dv + v 2 dp, would z be the thermodynamic
property? Answer: No, because 2p ≠ 2v.
11. If the expression experimentally discovered in Example 11.4 is dz = pdv–vdp,wouldz be a thermodynamic property? Answer:No,1≠ –1.
12. If the expression reported in Example 11.4 is dz = pdv+ vdp,wouldz be a thermodynamic property? Answer: Yes,dz = d(pv).
If we now look at our four basic property relationships as differential equations of the form of Eq. (11.11),
du = Tds− pdv (11.1)
dh = Tds+ vdp (11.3)
df = − pdv − sdT (11.5)
dg = vdp − sdT (11.8)
then, Eq. (11.12) must be valid for these equations, since we already know that all of these functions are thermodynamic
properties. Applying Eq. (11.12) to each of these equations yields a new set of equations, known as
the Maxwell thermodynamic equations:
Maxwell thermodynamic equations
∂T
= − ∂p
(11.13)
∂v s ∂s
v
∂T
= ∂v
(11.14)
∂p
s
∂s p
∂p
= ∂s
(11.15)
∂T ∂v T
v
∂v
∂T p
= − ∂s
∂p
T
(11.16)
While the Maxwell thermodynamic equations provide additional information about u, h, ands in terms of p, v,
and T, they cannot be solved to produce the direct functional relations between these properties that we seek.
However, these relations are used a little later in this chapter in conjunction with other material to provide the
desired u, h, and s relations from experimental p, v, T data.
EXAMPLE 11.5
Suppose we have the ideal gas equation of state, pv = RT, but know nothing about the entropy of this type of gas. Use the appropriate
Maxwell equations to determine a mathematical relation for the entropy of an ideal gas during an isothermal process.
Solution
The ideal gas equation of state is pv = RT, and so we know a p, v, T relation for our material. Perusing the Maxwell equations,
we see two, Eqs. (11.15) and (11.16), that involve only p, v, T variables on one side of the equation. We can choose
either of these equations to satisfy the problem statement, so we select Eq. (11.16):
∂v
= − ∂s
∂T p ∂p
Solving for v from the ideal gas equation of state gives v = RT/p, so the partial derivative we need is
∂v
= R/p
∂T p
T