Modern Engineering Thermodynamics

11.2 Two New Properties: Helmholtz and Gibbs Functions 363
Solution
The constant volume specific heat is defined by Eq. (3.15) as
c v =
∂u
∂T
and the constant pressure specific heat is defined by Eq. (3.19) as
c p =
∂h
∂T p
Equation (11.1) is du = Tds – pdv, and if we it divide through by dT, we get
du
dT = T ds
dT − p dv
dT
If we now require the specific volume v to be constant during this operation, then this equation becomes
du
= T ds
− p dv
dT v dT v dT v
Now, a total derivative restrained with a constant parameter is just a partial derivative, or
du
=
∂u
= c v
dT v ∂T v
ds
=
∂s
dT v ∂T v
dv
=
∂v
= 0 ðsince v is to be held constant hereÞ
dT v ∂T v
Then, substituting these results into Eq. (a) gives one of the desired relations:
∂u
= c v = T ∂s
∂T v ∂T v
Similarly, beginning with Eq. (11.3), we have dh = Tds + vdp, and dividing this through by dT gives
dh
dT = T ds
dT + v dp
dT
Again, imposing the condition that p must be constant during this operation, we get
dh
=
∂h
= c p
dT p ∂T p
ds
=
∂s
dT p ∂T p
dp
=
∂p
= 0 ðsince p is to be held constant hereÞ
dT p ∂T
Then, substituting these results into Eq. (b) gives the other desired relation:
∂h
= c p = T ∂s
∂T p ∂T p
p
The following exercises are designed to strengthen your understanding of the thermodynamics and the mathematics of the
material presented in this part of the chapter.
Exercises
1. Use the results of Example 11.1 to show that a material that has an entropy function of the form s(T, v) = A + B(ln T) +
C(ln v), where A, B, and C are constants, has a constant volume specific heat given by c v = B. Hint: Substitute the given
function into the relation c v = T(∂s/∂T) v .
2. If the specific internal energy of a material is found to depend on its specific entropy and specific volume according to the
relation u(s, v) = A + Bs + Cv 2 + Ds/v, where A, B, C, and D are all constants, then determine an expression for p(s, v) for
this material. Answer: p(s, v) = –(2Cv – Ds/v 2 ).
3. If the specific enthalpy of a material depends on specific entropy and pressure according to h(s, p) = A + Bs + Cp 2 + Ds 3 p,
where A, B, C, andD are all constants, then determine an expression for T(s, p) for this material. Answer: T(s, p) = B + 3Ds 2 p.
v
(a)
(b)
We now introduce two new thermodynamic properties. The first is the total Helmholtz function F, named after
the German physicist and physiologist Hermann Ludwig Ferdinand von Helmholtz (1821–1894), defined as
F = U − TS
Dividing by the system mass gives the specific Helmholtz function f as
f = u − Ts