EDIT: I was missing the entropy change due to temperature change and now everything makes sense

Suppose I have $n$ mol of an ideal gas in a chamber with a piston with volume $V$ at pressure $P_{1}$ and at a temperature $T$. I place a weight which exerts a pressure $P-P_{1}$ on the piston. The gas undergoes isothermal compression until its volume reaches $V'=\frac{nRT}{P}$. I can compute its entropy change with the formula $nR \ln(\frac{V'}{V})$.

So now I want to know the entropy change of the surroundings, for the sake of simplicity let's assume that it is an ideal gas at equilibrium with the other gas, so its pressure is $P_{1}$ and its temperature is $T$, with an initial volume of $V_{e}>>V,V'$ and a final volume $V_{e}'>>V,V'$, volume conservation allows us to say that $V_{e}'=V_{e}+V-V'$. Its temperature change is (because of energy conservation) $\Delta T=\frac{(P-P_{1})(V-V')}{Nc_{v}}$ its entropy change is (or should be)

\begin{align}Nc_v \ln(1+\frac{\Delta T}{T})+NR \ln(\frac{V_{e}'}{V_{e}})&=NR \ln(\frac{V_{e}+V-V'}{V_{e}})+Nc_v \ln(1+\frac{\Delta T}{T})\\&\approx NR\frac{V-V'}{V_{e}}+Nc_v\frac{\Delta T}{T}\\&=\frac{P_1 (V-V')}{T}+\frac{(P-P_{1})(V-V')}{T}.\end{align}

However, I know that it is in fact $\frac{P (V-V')}{T}$, because modelling the surroundings as an infinite thermal reservoir all the heat exchanged is reversible and the amount of heat exchanged is $P (V-V')$.

Why am I wrong?

  • $\begingroup$ Let's be more precise. You have a small chamber of volume V attached to a large chamber of volume Ve, and the two chambers interact with one another through a frictionless piston. Both chambers are initially at temperature T1 and pressure P1, are insulted from the rest of the surroundings, and you place an added weight on a second piston connected to the small chamber. This causes the volume in the small chamber to decrease, and this results in a change in the volume and pressure (and temperature) of the gas in the larger chamber. Is this anything like what you are describing. $\endgroup$ Jun 10, 2018 at 17:22
  • $\begingroup$ Notice that the discrepency arises because $P_1\neq P$. I.e. whatever's going wrong, the issue is how we're dealing with the massive piston. $\endgroup$
    – jacob1729
    Jun 10, 2018 at 17:41
  • $\begingroup$ @ChesterMiller Not quite. The small chamber is inside the big one. In my head the big one is the atmosphere and the small one is a cylinder with a piston where a weight is placed. $\endgroup$
    – Gamabunto
    Jun 10, 2018 at 18:54
  • $\begingroup$ If you combine the two terms in your final equation for the entropy change of the "surroundings," you get the result you expected. So, what's the problem? $\endgroup$ Jun 11, 2018 at 11:27
  • $\begingroup$ I found this a very interesting and highly thought provoking problem. What ever made you conceive of it? $\endgroup$ Jun 11, 2018 at 15:03

1 Answer 1


OK. Let's do it for a large finite outer chamber.

There is a cylinder with a massless piston between the two chambers, and the smaller chamber is vertically below the cylinder. We place a mass m on top of the piston at time zero, and then let the two chambers and the mass re-equilibrate.

Before determining the change in entropy, we must establish the final equilibrium state of the system. This is accomplished using the first law of thermodynamics in conjunction with the ideal gas law and the force balance on the mass.

Let the subscript S represent the final state of the small chamber and L represent the final state of the large outer chamber. The final force balance on the mass will be: $$P_S-P_L=\frac{mg}{A}\tag{1}$$ Applying the first law of thermodynamics to this system requires us to take into account the change in potential energy of the mass and to recognize the change in internal energy of the gases plus the change in potential energy of the mass must be zero (since no external work is done by the gases in the two chambers and there is no heat transfer through the outer shell: $$\Delta U+mg\Delta z=0$$or $$(n+n_e)C_v(T_f-T)+\frac{mg}{A}(V'-V)=0\tag{2}$$ where $$n=\frac{P_1V}{RT}\tag{3}$$and $$n_e=\frac{P_1V_e}{RT}\tag{4}$$ Eqn. 1 means that the temperatures of the gases in the two chambers rise slightly as a result of the potential energy dissipated by the falling of the mass.

From the ideal gas law, we also have that: $$P_S=\frac{nRT_f}{V'}\tag{5}$$and $$P_L=\frac{n_eRT_f}{(V_e+V-V')}\tag{6}$$ Combining Eqns. 1 and 3-6 yields:$$\frac{T_f}{T}\left(\frac{V}{V'}-\frac{V_e}{(V_e+V-V')}\right)=\frac{mg}{P_1A}\tag{7}$$ Combining Eqns. 2-4, we obtain: $$\frac{T_f}{T}=1+(\gamma-1)\frac{mg}{P_1A}\frac{(V-V')}{(V_e+V)}\tag{8}$$ The change in entropy of the "surroundings" is given by the equation: $$\Delta S_e=n_eC_v\ln{(T/T_f)}+n_eR\ln{[(V_e+V-V')/V_e}]$$ If I combine this with the previous equations and then take the limit as $V_e$ becomes infinite, I obtain: $$\Delta S_e=\frac{mg}{A}\frac{(V-V')}{T}+P_1\frac{(V-V')}{T}\tag{9}$$ Using your notation, this becomes: $$\Delta S_e=(P-P_1)\frac{(V-V_1)}{T}+P_1\frac{(V-V')}{T}$$ Now, if I combine the two terms on the right hand side, I obtain: $$\Delta S_e=P\frac{(V-V_1)}{T}$$ But this is the result you expected. So, what's the problem?


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