Discussion
32 kg of oxygen is mixed with 28 kg of nitrogen at the same temperature. The gases are at the same pressure of 103 kPa before and after mixing. If \(\bar{R}\) is the universal gas constant in kJ/kmol.K, the change in entropy of the mixture is
Answer: A
The entropy change is \(1.38\bar{R}\).
32 kg of O2 is 1 kmol and 28 kg of N2 is 1 kmol, so each mole fraction is 0.5. Mixing ideal gases at the same temperature and pressure gives
\[ \begin{aligned} \Delta S &= -\bar{R}\sum n_{i}\ln x_{i} = -\bar{R}\,[\ln 0.5 + \ln 0.5] \\ &= 2\bar{R}\ln 2 = 1.386\bar{R} \end{aligned} \]
The option \(0.69\bar{R}\) counts only one gas's contribution, \(\ln 2\). \(\bar{R}\) and \(0.34\bar{R}\) do not follow from the formula. The pressure value, 103 kPa, does not matter, since it is the same before and after.
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