Modern Engineering Thermodynamics

374 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.8
In Chapter 3, an equation of state developed in 1903 by Pierre Berthelot (1827–1907) was briefly discussed. Using this
equation of state, develop equations based on measurable properties for the changes in (a) specific internal energy, (b) specific
enthalpy, and (c) specific entropy for an isothermal process.
Solution
The Berthelot equation is given in Eq. (3.46) as
pv− ð b
where a and b are constants. Solving this equation for p gives
Þ = RT − av− ð bÞ/ Tv 2
p = RT/ ðv − bÞ− a/ Tv 2
a. The change in specific internal energy is given by Eq. (11.20), for which we need
∂p
= R/ ðv − bÞ+ a/ T 2 v 2
∂T
Then, for an isothermal process (T 1 = T 2 ), Eq. (11.20) gives
ðu 2 − u 1 Þ T =
Z v2
v1
v
RT/ ðv − bÞ+ a/ Tv 2 − RT/ðv − bÞ + a/ Tv 2 dv
= −ð2a/TÞð1/v 2 − 1/v 1 Þ = 2aðv 2 − v 1 Þ/ ðTv 1 v 2 Þ
b. To find the change in specific enthalpy, we could use Eq. (11.23). However, to evaluate this equation, we need to be
able to determine the relation ð∂v/∂TÞ p
: Since the Berthelot equation is not readily solvable for v = v (T, p), we choose
instead to use the simpler approach, utilizing only the definition of specific enthalpy, h = u + pv. Then,
ðh 2 − h 1 Þ T
= ðu 2 − u 1 Þ T
+ p 2 v 2 − p 1 v 1 = 2aðv 2 − v 1 Þ
+ p 2 v 2 − p 1 v 1
Tv 1
v 2
= 3aðv
2 − v 1 Þ v
+ RT 2
Tv 1 v 2 v 2 − b − v
1
v 1 − b
c. Finally, since we already evaluated the relation ð∂p/∂TÞ v , we choose to use Eq. (11.25) for the isothermal specific entropy relation:
Z v2
∂p
ðs 2 − s 1 Þ T = dv
∂T
v
=
v1
Z v2
v1
R/ðv − bÞ + a/ðT 2 v 2
Þ dv
= R ln ½ðv 2 − bÞ/ðv 1 − bÞŠ+ aðv 2 − v 1 Þ/ðT 2 v 1 v 2 Þ
Exercises
22. Setting a = 0 in the Berthelot equation of state used in Example 11.8 produces the Clausius equation of state, p(v – b) = RT
(see Eq. (3.43)). Determine equations for the change in specific internal energy, specific enthalpy, and specific entropy for a
Clausius gas undergoing an isothermal process. Answer:(u 2 – u 1 ) T = 0, (h 2 – h 1 ) T = RT [(v 2 /(v 2 – b) – v 1 /(v 1 – b)], and (s 2 – s 1 ) T = R
ln[(v 2 – b)/(v 1 – b)].
23. Evaluate the change in specific internal energy of water vapor when it is modeled as a Berthelot gas, with a = 4.30 MN·m 4·K/kg 2
and b = 4.50 × 10 –3 m 3 /kg, and undergoes an isothermal compression from a specific volume 40.0 m 3 /kg to a specific volume
of 5.00 m 3 /kg at a constant temperature of 100.°C. Answer: (u 2 – u 1 ) T = –4.03 kN·m/kg = –4.03 kJ/kg.
24. Evaluate the change in specific enthalpy of water vapor when it is modeled as a Berthelot gas, with a = 4.30 MN·m 4·K/kg 2 and
b = 4.50 × 10 –3 m 3 /kg, and undergoes an isothermal expansion from a specific volume of 10.0 × 10 –3 m 3 /kg to a specific volume
of 1.00 m 3 /kg at a constant temperature of 500.°C. Use R water = 461 N·m/(kg·K). Answer:(h 2 – h 1 ) T = 261 kN·m/kg = 261 kJ/kg.
Note that, for an ideal gas undergoing an isothermal process,
ðu 2 − u 1 Þ T = ðh 2 − h 1 Þ T = 0 and ðs 2 − s 1 Þ T = R lnðv 2 /v 1 Þ
Therefore, the equations developed in Example 11.8 can be considered to be Berthelot corrections to ideal gas
behavior.
Equation (11.24) has the same Mdx + Ndy form as Eq. (11.11), so that we can utilize Eq. (11.12) to produce
the property relation
∂ðc v /TÞ
∂v
T
= ∂
∂T
∂p
∂T
v
v