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Two containers of gases at different temperature are isolated from the surrounding and separated from other by a partition that allows heat exchange. What would have to happen if the entropy were to decrease? To increase? What is likely to happen?

Two containers of gases at different temperature are isolated from the surrounding and separated from other by a partition that allows heat exchange. What would have to happen if the entropy were to decrease? To increase? What is likely to happen?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
8 years ago
Let us consider two containers of gases at different temperatures T1 and T2 (T1> T2) and Q1 is heat absorbed by T1 and Q2 is the heat delivered at T2.
If the process is reversible, the total entropy of the system will not be changed, because the heat Q1 that absorbed by T1 is equal to the heat Q2 that delivered at T2 and opposite changes in entropy. Thus the net change in entropy of the system will be zero. So for a reversible cycle there is no change in the entropy of anything, including the reservoirs.
In the case of irreversible process, if we put together two objects that are at different temperatures T1 and T2, a certain amount of heat will flow from one to the other by itself. The entropy of the gas at temperature T1 decreases by ΔQ/ T1 and the entropy of the gas at temperature T2 increases by ΔQ/ T2. The heat will, of course, flow only from the higher temperature T1 to the lower temperature T2, so that ΔQ is positive if T1 is greater than T2. So the change in entropy (ΔS) of the whole world is positive, and it is the difference of two fractions.
ΔS = ΔQ/ T1 - ΔQ/ T2
Therefore in any process that is irreversible, the entropy of the whole world is increased. Only in reversible process does the entropy remain constant. Since no process is absolutely reversible, there is always at least a small gain in the entropy; a reversible process is an idealization in which we have made the gain of entropy minimal

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