Modern Engineering Thermodynamics

376 CHAPTER 11: More Thermodynamic Relations
Finally, if Eqs. (11.24) and (11.26) are set equal to each other,
c v
dT +
∂p
dv =
c
p
dT −
∂v
dp
T ∂T T ∂T p
and, if we solve for dT,
dT =
T
c p − c v
v
∂v
∂T
T ∂p
dp +
p
c p − c v ∂T
Then, writing the general relation T = T(p, v) and differentiating it, we get
dT =
∂T
dp +
∂T
dv
∂p ∂v p
and, by comparing coefficients of dp and dv in these two equations, it is clear that
∂T
=
T ∂v
∂p c p − c v ∂T p
and
or
Using Eq. (3.3), we can write
v
∂T
∂v p
c p − c v = T ∂v
∂T p
∂p
∂T
v
v
T ∂p
= c p − c v ∂T
∂p
∂T
v
= − ∂v
∂T
which, when substituted into the previous equation, yields
c p − c v = −T ∂v
∂T
v
= T ∂p
∂T
p
2
p
∂p
∂v
T
∂p
∂v
In Chapter 3, we define the isobaric coefficient of volume expansion β as
β = 1
∂v
v ∂T p
and the isothermal coefficient of compressibility κ as
κ = − 1
∂v
v ∂p
T
Substituting these two relations into the previous equation gives the final result:
T
v
∂v
∂T
dv
v
p
(3.5)
(3.6)
c p − c v = Tβ 2 v/k (11.30)
This equation reveals several important results. First of all, c p = c v for all simple substances at absolute zero
temperature. Second, since β 2 /κ = 0 for incompressible materials, then c p = c v for all incompressible materials. In
this case, the p and v subscripts are normally dropped and we write c p = c v = c for all incompressible materials.
Finally, since T, β, v, and κ are always positive, then c p ≥ c v for all simple substances.
EXAMPLE 11.10
Using the data in Table 3.2, determine the difference between c p and c v for saturated liquid water at 20.0°C.
Solution
From Table 3.2, for water, we find that
β = 0:207 × 10 −6 K −1
κ = 45:9 × 10 −11 m 2 /N