Two vessels, of equal volume are connected with each other through a gas tube containing a valve. The first vessel is filled with an ideal gas and has temperature $T$ while the other one is maintained at vacuum. Now suddenly the valve is opened leading to transfer of gas. Now since the pressure in the first chamber is bound to decrease, hence its temperature must also decrease and as a result of this and the temperature in the vacuum valve must increase. But one more thing that crossed my mind while solving this is that the work done in expansion of gas against vacuum is zero and hence there must be no change is temperature since no new heat has been provided and there is vacuum in the second vessel. Can anybody please tell me, which line of thought is correct and why?
2 Answers
When there is a valve that offers significant resistance to gas flow between the two tanks, the gas will flow relatively slowly from the high pressure tank to the lower pressure tank. So the gas pressure on the high pressure side is changing slowly, and the gas remaining on the high pressure side is doing work on the gas ahead of it to force the gas ahead of it through the valve. It is doing this work by expanding adiabatically and reversibly against the back pressure of the gas ahead of it. So the gas in the high pressure tank cools adiabatically and nearly reversibly. We see this all the time in practice.
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$\begingroup$ "gas remaining on the high pressure side is doing work on the gas ahead of it to force the gas ahead of it through the valve", so part of the gas is doing work on another part. If you consider all the gas as a whole system, there is no external work done on or by the gas. So no matter what the intermediate stages may be, the final temperature must be the same as the initial temperature theoretically. For a real gas, it will cool down but that's because of the increase in potential energy, not because of the reason you described. $\endgroup$ Commented Dec 22, 2016 at 14:36
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$\begingroup$ @velut luna The temperature must be the same throughout the system only if the two tanks can communicate thermally with one another (so that, in the final steady state) the temperature in both tanks is the same. But if the two tanks are isolated thermally from one another (as implied in this problem), the temperature in the tank that had pressure initially is lower than its initial temperature in the end, while the temperature in the other tank (the one that had vacuum initially) is higher in the end. In this way, the total change in internal energy is zero (as required). $\endgroup$ Commented Dec 22, 2016 at 14:44
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$\begingroup$ I think what the OP have in mind is the case that the two tanks are not thermally insulated. $\endgroup$ Commented Dec 22, 2016 at 14:47
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$\begingroup$ Let's ask him. @Harsh Sharma, which is it? Are you assuming that the two tanks (a) can communicate with one another thermally or (b) can not communicate with one another thermally? $\endgroup$ Commented Dec 22, 2016 at 14:53
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$\begingroup$ If the OP meant they are thermally isolated,I admit that my answer is wrong and yours is correct. $\endgroup$ Commented Dec 22, 2016 at 14:56
"Now since the pressure in the first chamber is bound to decrease, hence its temperature must also decrease and as a result of this"
The above is wrong.
For ideal gas, $U=\frac{3}{2}PV=\frac{3}{2}nRT$ and so this is true when when, say, volume is constant. But you can see that if you double the volume and halved the pressure, the temperature remains the same. In fact for free expansion, $U$ is bound to be constant because there is no work done and no heat flow.
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$\begingroup$ Why is it so? Can you please explain? $\endgroup$– user118752Commented Dec 22, 2016 at 7:48
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$\begingroup$ Why pressure decreasing must lead to temperature decrease? $\endgroup$ Commented Dec 22, 2016 at 7:50
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$\begingroup$ So the temperature is equal in both the chambers becauses the molecules which are being transferred are the same molecules which were present in the first chamber $\endgroup$– user118752Commented Dec 22, 2016 at 7:54
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$\begingroup$ I don't understand your explanation. Anyway, there is no law saying that smaller pressure must lead to lower temperature. $\endgroup$ Commented Dec 22, 2016 at 7:57
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$\begingroup$ So you mean to say that the final temperature of both the chambers must be equal to initial temperature of the first chamber? If initial temperature of the chamber having gas was T then the final temperature of both the chmabers must be equal to T $\endgroup$– user118752Commented Dec 22, 2016 at 7:59