Modern Engineering Thermodynamics

11.6 Determining u, h, and s from p, v, and T 377
and, from Table C.1b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
Then, from Eq. (11.30), we have
c p − c v = Tβ2 v
κ
v = v f ð20:0°CÞ = 0:001002 m 3 /kg
= ð293 KÞð0:207 × 10−6 K −1 Þ 2 ð0:001002 m 3 /kgÞ
45:9 × 10 −11 m ⋅ s 2 /kg
= 2:74 × 10 −5 J/ðkg ⋅ KÞ = 2:74 × 10 −8 kJ/ðkg ⋅ KÞ
In most applications, this difference is clearly negligible, since the value of c p for liquid water at standard temperature and
pressure is 4.18 kJ/(kg·K).
Exercises
28. Show that, for an ideal gas, defined by pv = RT, the isobaric coefficient of volume expansion is β = 1/T and the
isothermal coefficient of compressibility is κ = 1/p. Then, show that, for an ideal gas, Eq. (11.30) gives c p – c v = R.
29. Using the methods of Example 11.10, determine the difference between the constant pressure specific heat c p and
the constant volume specific heat c v for liquid mercury at 20.0°C. For liquid mercury, v = 7.4 × 10 –5 m 3 /kg. Answer:
(c p – c v ) mercury = 1.79 × 10 –5 J/(kg·K).
30. Rework Example 11.10 for liquid benzene at 20°C. For liquid benzene, v = 1.15 × 10 –3 m 3 /kg. Answer: (c p – c v ) benzene =
5.46 × 10 –4 J/(kg·K).
Now we need to develop a strategy for the integration of Eqs. (11.19), (11.22), (11.24), and (11.26) for
arbitrary states. Since u, h, and s are point functions, the integration results are independent of the actual integration
path, so we should pick a path that is easy to evaluate. In addition, since u and h lack well-defined absolute
zero values, the integration path must begin at an arbitrary reference state.
Since all equations of state should reduce to the ideal gas equation of state at low pressures, we can postulate
that they must all be reducible to the following form:
pv = RT + fðv, TÞ
where the function f(v, T) must be on the order of 1/v so that it vanishes as p → 0 and v → ∞. Consequently, the
reference state is usually taken to be at some arbitrary reference temperature T 0 and at essentially zero pressure p 0 =0,
and zero density or infinite specific volume, v = ∞.
To generate a numerical value from Eqs. (11.19), (11.22), (11.24), and (11.26) for u, h, and s, we start the integration
process at this reference state. Now, the choice of the easiest integration path from the reference state to
the actual state depends on the form of the arguments in the integrals. For example, in evaluating Eq. (11.19)
for the specific internal energy, the easiest path to follow is a constant specific volume line for the first integral
(since c v is defined for a constant volume process) and to follow a constant temperature line for the second integral.
Since the first integration line is at zero pressure, the constant volume specific heat along this path is c 0 v
(the superscript zero indicates a zero pressure constant volume specific heat). Note that the integration path must
be only piecewise continuous; therefore, it can have “kinks.”
Since all of our equations of state must have the form pv = RT + f(v, T),
and
then,
so that Eq. (11.19) now gives
u − u 0 =
T
Z T
∂p
∂T
T 0
c 0 v dT + Z v
− p = RT
v + T v
v
v 0 =∞
p = RT
v
+
f ðv, TÞ
v
∂p
= R ∂T v + 1
∂f
v
v
∂T
v
T
∂p
= R T ∂T v + T
∂f
v ∂T
T
v
∂p
∂T
∂f
−
∂T
v
− p
v
dv =
v
RT
v + f
v
Z T
= T v
T 0
c 0 v dT + Z v
∂f
− f ∂T v
v
v 0 =∞
T
v
∂f
− f ∂T v
v
dv