I recently ran into the following interesting problem I am now stuck on:

$$S = \frac{n}{2}(\sigma+5R\ln\frac{U}{n} +2R \ln\frac{V}{n}) $$ The air inside a house with open windows is initially in thermal equilibrium with air outside the house at 273 K. A few hours after the heating in the house is turned on, the temperature of the air inside the house reaches an equilibrium value of 300 K. Using the entropy S given above, determine how the energy density of the air inside the house compares at the two temperatures. You may assume that the air inside and outside the house obeys the ideal gas equation pV = nRT.

I have tried to solve it in different ways, one of the difficulties I run into is determining what's constant. Because the windows are open, the pressure must be 1 atm. At the same time the house has constant volume, however the number of particles inside change with the temperature change.

Being in thermal equilibrium suggests $T_{initial}$ = 273 K, and we know that $T_{final}$ = 300 K, although it is not clear to me how this can be an "equilibrium temperature" without the outside temperature also being 300 K. I suppose that the air just outside the window has some sort of stable temp. gradient, but this makes it hard to apply an equation of (equilibrium-) state.

From the above equation I judge that we have a diatomic gas (it is possible to "read off" $C_v = 5R/2$ from the entropy expression).

We should therefore have: $U = \frac{5RnT}{2} $, hence the energy density in the house: $$ \rho = \frac{U}{V} = \frac{3p}{2} $$ but we have just established that the pressure is constant, hence the internal energy density should be unchanged? Is this correct? How can I show this using the expression for entropy above (as the question asks us to do)?

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    $\begingroup$ Ha, I just wrote a note discussing this. See the very bottom of the page here and the associated links. Does Kreuzer and Payne's paper address your question? $\endgroup$ – Chemomechanics Mar 24 '18 at 21:53
  • $\begingroup$ @Chemomechanics thank you very much for the link, I really liked the "discussion note" of yours, and also the fact that the internal energy density stays the same - surprisingly unintuitive! :) Kreuzer had some equations with entropy, but as far as I can see he didn't use the form of S I was asked to use in the question. Any advice on how to go about that? I thought S could perhaps be worked out for the remaining gas only (V=const) somehow and then one might show that it is constant, hence U = constant? I have not been able to make progress. $\endgroup$ – Jhonny Mar 26 '18 at 21:07
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    $\begingroup$ I think you've already done it. You used the entropy equation to determine the molar heat capacity, and then you used that to determine that the energy density is unchanged. I also agree with you that "equilibrium value" in the problem is a bit confusing; "stable uniform value" might be better. $\endgroup$ – Chemomechanics Mar 26 '18 at 22:04

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