Confusion about entropy of mixing of the same ideal gas in two different containers with different temperatures

Question

Two identical thermally insulated vessels interconnected by a tube with a valve contain one mole of the same ideal gas each. The gas temperature in one vessel is equal to T1 and in the other, T2. The molar heat capacity of the gas of constant volume equals Cv. The valve having been opened, the gas comes to a new equilibrium state. Find the entropy increment $\Delta$S of the gas. Demonstrate that $\Delta$S > 0.

I came across a solution here. But I am unable to understand some of the underlying concept.

How can we consider this as isochoric? Each gas now occupies double the volume it had occupied earlier. We can't neglect the change in volume while calculating entropy change.

According to me, entropy change for each gas should be $\Delta$S = nR $ln$ $\frac{V_{2}}{V_{1}}$ +nC$_{v}$ $ln$ $\frac{T_{3}}{T_{1}}$

Here. T$_{3}$ =(T$_{1}+$T$_{2}$)/2, as energy remains conserved during mixing.

So, how have they omitted the change in volume factor in the solution?

Also, is there any other approach to be followed in case the gases initially have different temperatures?

Any help would be appreciated.

Edit: I think that the answer given here only partly answers my question. It is still not clear to me , that how can I take into account the fact that the gases are indistinguishable in this case and why neglecting the Volume terms gives the right answer.

Not completely . But, How should I take into account the fact that the gases are indistinguishable? — Rishi Shekher, Dec 22 '22 at 13:26
If the gases are indistinguishable, then each expanding into the space of the other is also indistinguishable and doesn’t affect the entropy. Note that the link above in turn contains links to five closely related discussions. Please consider reviewing them. — Chemomechanics, Dec 22 '22 at 14:04

score 2 · Accepted Answer · edited Dec 22 '22 at 14:18

2

If V is the initial volume, then the initial pressures of the gases are $\frac{RT_1}{V}$ and $\frac{RT_2}{V}$, and, relative of a reference state of $T_0$ and 1 bar, the initial entropy of the system is $$S_1=C_p\ln{(T_1/T_0)}-R\ln{(RT_1/V)}+C_p\ln{(T_2/T_0)}-R\ln{(RT_2/V)}$$In the final state, the temperature is $\frac{(T_1+T_2)}{2}$and the final pressure is $\frac{2R(T_1+T_2)}{(2)(2V)}$, so the final entropy relative to the same reference state is $$S_2=2C_P\ln{(T_1+T_2)/(2T_0)}-2R\ln{(R(T_1+T_2)/(2V)}$$So the change in entropy is $$\Delta S=S_2-S_1=C_v\ln{\frac{(T_1+T_2)^2}{4T_1T_2}}$$

edited Dec 22 '22 at 14:18

hyportnex

18,748

answered Dec 22 '22 at 13:31

Chet Miller

33,586
3
21
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Thanks a lot for your help. But why can't I say that the gas in each part has expanded to twice it's initial volume? – Rishi Shekher Dec 22 '22 at 15:00
It is the same gas, and volume per mole is the same in each chamber before and after. – Chet Miller Dec 22 '22 at 16:56

Confusion about entropy of mixing of the same ideal gas in two different containers with different temperatures

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