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Andrew Mason
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#14
Oct5-11, 08:04 AM
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Quote Quote by bbbeard View Post
Your example does not fit the problem Confusus describes. In the problem he describes, the gas and the surroundings are at the same temperature T throughout. The difference between the reversible and irreversible process is presumably due to irreversibilities within the gas, generated during the compression process. These might be, for example, due to vortices generated by compression of a piston at a finite rate.
I disagree. There is only one way to have an isothermal compression of a gas where the temperature of the surroundings and the gas are the same throughout: it must be reversible. If you compress the gas too quickly, creating dynamic flows within the gas, when the vortices settle down the temperature of the gas will increase and this is not permitted. This is just the first law: dQ = dU + dW. If you do work on the gas, you either increase the internal energy or you increase heat flow. Neither are permitted to occur if both the gas and the surroundings are maintained at the same temperature UNLESS the work is done infinitely slowly.

Note that the conditions on irreversible heat and work with this sign convention is that dQ_irrev < T*dS and dW_irrev < P*dV.
In general that is true. The problem is that I don't see how you can have an irreversible isothermal compression of gas with the surroundings at the same temperature as the gas. Perhaps you could explain that.

AM