We consider a gas in thermal equilibrium with a reservoir. If a piston does a certain amount of work on the gas W very slowly (isothermal compression), how does S change for the gas? S should decrease right? Using S = Q/T, The system loses Q and therefore S should decrease. But doesn't a system always increase in Entropy?

Question

Karlie Finley

Physics

13 September, 03:13

We consider a gas in thermal equilibrium with a reservoir. If a piston does a certain amount of work on the gas W very slowly (isothermal compression), how does S change for the gas? S should decrease right? Using S = Q/T, The system loses Q and therefore S should decrease. But doesn't a system always increase in Entropy?

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Madyson Cabrera · Answer 1

The entropy of the gas increases

Explanation:

The entropy of an isolated system tends to increase. but this gas gas is not isolated. It has a piston perfoming work on it and it can exchange heat with it's surroundings.

While the system's entropy decreases, the surroundings' entropy increases, and the entropy change of the universe, which is the sum of the entropy changes of the system and the surroundings, will be equal or greater than zero.