# Does entropy really not increase here?

Two vessels separated by a partition have equal volume $V_0$ and equal temperature $T_0$. They both contain the same ideal gas, and the particles are indistinguishable. The left vessel has pressure $P_0$ and the right vessel has pressure $2P_0$. After the partition is removed and the system equilibrates, what is the net change in entropy?

At first my intuition tells me that $\Delta S=0$ since, for the whole system, $dQ=0$ and $dW=0$ throughout the entire process. Moreover, if we look at only one vessel - say, the left one - the change in entropy is given by:

\begin{align}\Delta S_1&=\frac{\Delta E_1}{T_0}\\ &=\frac{\Delta (c_V NT)}{T_0}\\ &=\frac{c_V T_0\Delta (N)}{T_0}\\ &=c_V\frac{N_0}{2} \,\,\,\,\textrm{(from ideal gas formula)} \end{align}

If we do it for the right vessel, we simply get $\Delta S_2 = -\Delta S_1$, so once again $\Delta S=0$. But the accepted answer to this Physics SE question says otherwise. Is there something I'm missing?

[EDIT] To show where I got $\Delta N$, notice that if the left vessel has $N_0$ moles of gas at first, the right vessel has $2N_0$ moles of gas.

$$N_2=\frac{P_2T_0}{V_0}=\frac{2P_0T_0}{V_0}=2N_0$$

So in the end, the total system will have $N_0+2N_0=3N_0$ moles of gas. Since the volumes are equal, in the end each will have $\frac{3}{2}N_0$ moles of gas.

• @RubenVerresen See my edit. – Arturo don Juan Sep 6 '17 at 18:28
• Yeah, but the particles aren't just mixing (if they were we'd just have Gibbs paradox, i.e. $\Delta S=0$ since they're indistinguishable). They're undergoing a thermodynamic change, and that's why I think the thermodynamic calculation is necessary, but my calculation is in contradiction with the answer to the question I linked. I think that other person is wrong, but they have a really good answering record so I'm totally confident. – Arturo don Juan Sep 6 '17 at 18:45
• correction: I'm *not totally confident. – Arturo don Juan Sep 6 '17 at 19:42

The change in entropy is certainly not zero. It is greater than zero for this spontaneous process. Just because the Q in an irreversible process is zero does not mean that the entropy change is zero. The entropy change is the integral of dQ/T only for a reversible path.

I get $\frac{3}{2}P_0$ for the final pressure. The initial number of moles in the left container is $\frac{P_0V_0}{RT_0}$ and the initial number of moles in the right container is $\frac{2P_0V_0}{RT_0}$. If the initial moles in the left container goes from $P_0$ to $1.5P_0$ (compression) at temperature $T_0$, what is its change in entropy? If the initial moles in the right container goes from $2P_0$ to $1.5P_0$ (expansion) at temperature $T_0$, what is its change in entropy? What is the total change in entropy for the process?

You can check your result against my final answer of $$\Delta S=\frac{P_0V_0}{T_0}\ln{(32/27)}$$

EDIT: If you increase the pressure on an ideal gas isothermally and reversibly, then dU=0. So, $$TdS=PdV=d(PV)-VdP=-VdP=-\frac{nRT}{P}dP$$Integrating, we get:$$\Delta S=-nR\ln(P_2/P_1)$$

• Thanks for the answer Chester. The only way I know how to calculate the change in entropy in an ideal gas is to use a formula derived under the assumption that the entropy is a function of state, e.g. $\Delta S = c_V \Delta N$ if $T$ and $V$ are constant. – Arturo don Juan Sep 6 '17 at 19:30
• Also, if after letting the system equilibriate I stick the partition back in a third of the way and slowly move it back to the center, wouldn't I have reversed the process? All the macroscopic variables would go back to their original values. Basically, isn't this process reversible? – Arturo don Juan Sep 6 '17 at 19:33
• If you put the partition back an want to get back to the original state reversibly, you will have to remove heat from the right container and add heat to the left container. Otherwise, if you do it adiabatically and reversibly, the gas in the right container will get hotter and the gas in the left container will get colder. – Chet Miller Sep 6 '17 at 19:39
• Are. you not familiar with the equation for an ideal gas $$\Delta S=n(C_v\ln{(\frac{T_2}{T_1})}+R\ln{(\frac{V_2}{V_1})})$$ or $$\Delta S=n(C_p\ln{(\frac{T_2}{T_1})}-R\ln{(\frac{P_2}{P_1})})$$ – Chet Miller Sep 6 '17 at 19:42
• @Arturo: Entropy is a state function, so the change in entropy between two thermodynamic equilibrium states is independent of the process by which the system goes between those two states. You have an irreversible process, sure, but you can still use an imaginary reversible process between the same two states in order to compute the change in entropy. – march Sep 6 '17 at 20:00