Modern Engineering Thermodynamics

752 CHAPTER 18: Introduction to Statistical Thermodynamics
where V is the total volume of the system, ħ and k are Planck’s constant and Boltzmann’s constant, and
Z rot = Z vib = 0. Then, Eqs. (18.41), (18.42), and (18.44) give
u = 3 2 RT
9
ð18:46aÞ
h = u+RT = 5 2 RT
c v = 3 2 R
>=
For monatomic gases only
ð18:46bÞ
ð18:46cÞ
c p = 5 2 R
n h
i
s = R ln ð2πm/ħ 2 Þ 3/2 ðkTÞ 5/2 /p + 5 o
>;
2
ð18:46dÞ
ð18:46eÞ
Theseformulaeforu, h, c v ,andc p are the same as those obtained from the kinetic theory of gases discussed
earlier in the chapter. The equation for entropy, on the other hand, is more complex and therefore one would
expect it to be more accurate. We now progress to the next most complex geometric molecular structure,
diatomic gases.
EXAMPLE 18.8
In a fiendish plan to incapacitate Superman, the arch villain Dorkmann proposes to compress 3.50 kg of krypton gas from
1.00 atmosphere, 20.0°C to 10.0 MPa, producing a concentrated and possibly toxic concentration of kryptonite. The foolish,
unschooled fiend intends to try to carry out the compression process adiabatically using only 100. kJ of work. But you, as
the ever-present hero Thermoperson, seeker of truth and wisdom, hold the power to foil the plan by computing
a. The final temperature of the krypton gas after compression.
b. The entropy production of the compression process.
Solution
a. The final temperature of the gas can be found from an energy balance and Eq. (18.46a) as
1Q 2 − 1 W 2 = mðu 2 − u 1 Þ = m 3 2 R ðT 2 − T 1 Þ
where R krypton = R/M krypton = 8.3143/83.80 = 0.0992 kJ/kg·K. Since 1 Q 2 = 0 for an adiabatic compression, the energy
balance equation can then be solved for T 2 as
T 2 = T 1 − 1 W 2
−100: kJ
= ð20:0 + 273:15 KÞ −
3mR/2 3ð3:50 kgÞð0:0992 kJ/kg⋅KÞ/2 = 458 K
b. An entropy balance using Eq. (18.46e) produces
1Q 2
T b
+ 1 ðS P Þ 2
= ms ð 2 − s 1 Þ
( "
= m R ln ð2πm/ħ2 Þ 3/2 5/2
#)
( " # )!
ðkT 2 Þ
+ 5 − R ln ð2πm/ħ2 Þ 3/2 ðkT 1 Þ 5/2
+ 5 p 2 2
p 1
2
Solving this equation for 1 (S P ) 2 gives
" 5/2 #
T
1ðS P Þ 2
= mR ln 2 p 1
T 1 p 2
= ð3:50 kgÞð0:0992 kJ/kg ⋅ KÞln
485 K 5/2
0:101325 MPa
293:15 K 10:0 MPa
= −1:16 kJ
kg.K
Since 1 (S P ) 2 is less than zero here, it violates the second law of thermodynamics and this process can not possibly occur.
Therefore, Superman has nothing to worry about.
Exercises
22. If Dorkmann in Example 18.8 increases the work input to the process from 100. kJ to 1000. kJ with all the other
variables unchanged, would the process work then? Answer: Possibly, because now 1 (S P ) 2 = 0.160 kJ/kg·K, which, being
positive, does not violate the second law of thermodynamics.