2
$\begingroup$

From Statistical Physics, 2nd Edition by F. Mandl:

Two vessels contain the same number $N$ molecules of the same perfect gas. Initially the two vessels are isolated from each other, the gases being at the same temperature $T$ but at different pressures $P_1$ and $P_2$. The partition separating the two gases is removed. Find the change in entropy of the system when equilibrium has been re-established, in terms of the initial pressures $P_1$ and $P_2$. Show that this entropy change is non-negative.

I'm a little confused about a few things.

  1. Is there a temperature change in this process? Intuitively, I would say no because $(T+T)/2=T$. My other guess would be that the temperature must change because we now have a third pressure, $P_3$, that is different from the pressure of the other two and also because we have increased the volume.
  2. I believe this is an irreversible process, correct? Because you can't realistically separate the gasses into that which came from vessel A and that which came from vessel B.
  3. Can the change in volume simply be called $V_A+V_B$? I thought it would be that simple but thinking more about it I feel as though the change in pressure and possible change in temperature might change things.

  4. When the partition is removed, is there a heat exchange between the 2 gases? My intuition says no because heat can only flow when there is a temperature difference and in this case both vessels are at temperature $T$.

My attempt:

So all in all I need to solve $\Delta S= \int \frac{dQ}{T}$

$$\Delta S= \int \frac{dQ}{T}$$ $$=\int \frac{dE+dW_{by}}{T}$$ We know that $dE=0$ and that $dW=PdV$

$$=\frac{1}{T}\int PdV$$ $$=\frac{P\Delta V}{T}$$

This is where I'm stuck - I don't think there is a valid thing to put in for $\Delta V$ because there were 2 systems that formed into 1 bigger system. If the final system is $V_1+V_2$, then what was its previous size? $V_1$ or $V_2$? Or can I say that it was $\frac{V_1+V_2}{2}$?

$\endgroup$
2
$\begingroup$

The total volume of the two rigid containers does not change, so the combined system does no work W on the surroundings. The two containers are presumably insulated, so no heat Q is exchanged with the surroundings. So, from the first law of thermodynamics, the change in internal energy of the combined system is zero. Since, for an ideal gas, internal energy is a function only of temperature, the final temperature of the combined system is equal to the initial temperature of the separate systems.

The process is irreversible, but not for the reason you gave. Since the same gas is present in both containers, the system can be returned to its original state, but not without incurring a change in the surroundings, involving heat transfer.

Quarky Quanta's intuition was correct with regard to the final equilibrium pressure of the combined system, provided n is the total number of moles of gas in the two original containers.

COMPLETION OF PROBLEM SOLUTION: $$V_1=\frac{NkT}{P_1}$$ $$V_2=\frac{NkT}{P_2}$$ $$V_1+V_2=\frac{NkT(P_1+P_2)}{P_1P_2}$$ $$P_F=\frac{2NkT}{(V_1+V_2)}=\frac{2P_1P_2}{(P_1+P_2)}$$ $$\Delta S=Nk\ln{\frac{P_1}{P_F}}+Nk\ln{\frac{P_2}{P_F}}=2Nk\ln{\left[\frac{(P_1+P_2)/2}{\sqrt{P_1P_2}}\right]}$$ So the change in entropy is determined by the ratio of the arithmetic mean of the initial pressures to their geometric mean (a ratio which is always greater than 1).

$\endgroup$
  • $\begingroup$ It doesn't specify that the 2 containers are rigid though....also, how would we reverse this process using heat transfer? $\endgroup$ – whatwhatwhat Mar 6 '16 at 17:11
  • 1
    $\begingroup$ In my judgment, the problem statement clearly implies that the containers are rigid. To return the system to its original state using heat transfer, you could divide the final volume in half, and isothermally and reversibly compress half the molecules in a piston/cylinder arrangement to the original high pressure while holding the cylinder in contact with a constant temperature bath. You could expand the other half of the gas in an analogous way to its original low pressure. All the gas would then be in its original state. $\endgroup$ – Chet Miller Mar 6 '16 at 18:55
  • 1
    $\begingroup$ Incidentally, the reversible isothermal process I described for returning the system to its original state would involve a decrease in entropy. This means that the irreversible process that took the system from its original state to its final state involved an increase in entropy (as expected). $\endgroup$ – Chet Miller Mar 7 '16 at 0:52
  • $\begingroup$ Do the systems exchange heat with each other when the partition is removed? I ask because the change in entropy can also be defined by $\frac{d Q}{T}$. $\endgroup$ – whatwhatwhat Mar 7 '16 at 2:55
  • 1
    $\begingroup$ The gases in the two containers combine and equilibrate after the partition is removed. But you can't determine the change in entropy by determining dQ/T for the actual irreversible process. The first step in determining the change in entropy for an irreversible process is to TOTALLY FORGET ABOUT THE ACTUAL IRREVERSIBLE PROCESS PATH and focus instead only on the initial and final equilibrium states. You need to then define a reversible path between these two end states, and determine the integral of dQ/T for that path. That's what they mean when they say determine the integral of dQrev/T. $\endgroup$ – Chet Miller Mar 7 '16 at 3:38
1
$\begingroup$

Let me try to explain this in a conceptual way, if that does not make sense we can try a more rigorous approach. Assume that $P_1 > P_2$ and correspondingly $V_1 < V_2$. If you think of the entire volume of a container as being tiny cells that can be occupied by the gas particles, then entropy simply measuring the number of different configurations (permutations) that the gas can take. Now if you assumed that there were gas particles in $V_2$ but $V_1$ was empty, then removing the barrier will increase the number of cells, hence increasing the allowed micro-states and correspondingly the entropy. The same thing is true if there were particles in $V_1$ but $V_2$ was empty. In this case, both the volumes have particles and therefore it makes sense that removing the barrier increases the entropy because that increases the number of allowed micro-states. I have assumed here that the number of cells in the volume is much more than the number of gas particles present, which I think is a fair assumption.

To be more specific about your questions, my intuition is that (1) the temperature will not change, (2) the process is irreversible and (3) the new volume is $V_1 + V_2$ and the new pressure is $p = \frac{n R T}{V_1 + V_2}$.

$\endgroup$
  • $\begingroup$ But what about the fact that in the final state, there are now $2N$ molecules present? Wouldn't that mean that $p=\frac{2NRT}{V_1+V_2}$? $\endgroup$ – whatwhatwhat Mar 7 '16 at 2:44
  • 1
    $\begingroup$ My mistake, I think you are correct. The pressure should in fact have n = 2N. $\endgroup$ – Quarky Quanta Mar 7 '16 at 3:28
  • $\begingroup$ What would you say about the heat exchange between the 2 gasses? I've edited my question to add a point #4. $\endgroup$ – whatwhatwhat Mar 7 '16 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.