5.10 The Examples of Reversibly Controlled “Free
Expansion” and Reversible Mixing of Ideal Gases:
Why Kelvin’s Second General Conclusion Is Not True?
We now consider the same oxygen and nitrogen systems at the same 300 K temperature but kept in different containers (V O 2 ¼ 0:5 m
3 and V N 2 ¼ 1 m
3 ). Correspondingly, the pressures are 300 kPa and 180 kPa, respectively. These initial
conditions are chosen so that when the two gases are mixed to become a mixture in
a container of 1.5 m
3 , the final state of the mixture is the same final state in the
above example: 300 K, 220 kPa (as it can be readily confirmed by the application
of Dalton’s law).
The entropy of the mixture is the same as in the example above. The entropy of
the initial oxygen and nitrogen systems is different, thus, the entropy gain of this
mixing process will be different, which can be readily calculated by using the Gibbs
theorem and Eq. (68A) or (69A)
S mixture À S O 2 þ S N 2
½
Š initial ¼ S O 2 300 K; 1:5 m
3
À
Á À S O 2 300 K; 0:5 m
3
À
Á
Â
Ã
þ S N 2 300 K; 1:5 m
3
À
Á À S N 2 300 K; 1 m
3
À
Á
Â
Ã
Recall Eq. (68A)
S B À S A ¼ Nc V ln
T B
T A
þ NRln
V B
V A
Thus, as the mole numbers of the two gases are determined in the aforementioned
example, and there is no change in temperature
S mixture À S O 2 þ S N 2
½
Š initial ¼ S O 2 300 K; 1:5 m
3
À
Á À S O 2 300 K; 0:5 m
3
À
Á
Â
Ã
þ S N 2 300 K; 1:5 m
3
À
Á À S N 2 300 K; 1 m
3
À
Á
Â
Ã
¼ 0:0601 Á 8:31447 log e 3 þ 0:722 Á 8:31447 log e 1:5
¼ 0:7926 kJ=K
We now consider the thought experiment of a reversible change of the oxygen
and nitrogen systems from their initial states to the final mixture state as defined in
Fig. 5.7. The reversible change takes place as follows: Assuming availability of a
vacuum of total 1.5 m
3 , one part of which 1 m
3 is to be used for the first step of the
oxygen system and the second part of which 0:5 m
3 is to be used for the first step of
the nitrogen system.
124
5 Entropy and the Entropy Principle