Modern Engineering Thermodynamics

11.6 Determining u, h, and s from p, v, and T 375
or
∂c v
= T ∂2 p
∂v T ∂T 2
v
Similarly, Eq. (11.26) has the same Mdx + Ndy form and application of Eq. (11.12), so it gives
"
∂ c p /T #
= − ∂
∂v
∂p ∂T ∂T p
or
T
∂c p
= −T ∂2 v
∂p
T
∂T 2
p
Both Eqs. (11.28) and (11.29) give specific heat information from measurable p, v, and T properties.
p
(11.28)
(11.29)
EXAMPLE 11.9
Using the Berthelot equation of state given in Example 11.8, determine an equation for the isothermal variation in the
constant volume specific heat with volume change.
Solution
From Eq. (11.28) we have what we seek,
∂c v
= T ∂2 p
∂v T ∂T 2
v
and from Example 11.8, the Berthelot equation of state can be written as
so that
and
then,
p =
RT
v − b − a
Tv 2
∂p
= R
∂T
v
v − b + a
T 2 v 2
∂ 2
p
∂T 2 = − 2a
v
T 3 v 2
∂c v
= − 2a
∂v T T 2 v 2
and, to find an explicit c v = c v (T, v ) equation, the preceding equation can be integrated from a reference state specific
volume v o to give
c v = − 2a
Z v
dv
T 2 v 2
v0
= 2a ð v 0 − vÞ/ T 2 v 0 v + f ðTÞ
where f (T) is a function of integration. Note that c v is independent of v only in the case where a = 0 in the Berthelot equation
of state.
Exercises
25. The Clausius equation of state, p(v – b) = RT (see Eq. (3.43)), can be obtained by setting a = 0 in the Berthelot equation
of state. Rework Example 11.9 to determine how the constant volume specific heat of a Clausius gas undergoing an
isothermal process depends on the specific volume of the gas. Answer: For a Clausius gas undergoing an isothermal
process, c υ does not depend on the specific volume of the gas.
26. For the Berthelot equation of state used in Example 11.9, determine an expression for the mixed partial derivative
∂ ∂c v
= ?
∂T ∂v T
Answer: 4a/(T 3 v 2 ).
27. Evaluate the constant volume specific heat relation developed in Example 11.9 for a material in which a = 2.30 MN·m 4·K/kg 2 .
Use v 0 = 0.100 m 3 /kg and v = 0.0200 m 3 /kg. Answer: c v = 1.321 kN·m/kg = 1.321 kJ/kg.
v