- #1
Philip Koeck
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- 184
For the reversible expansion of an ideal gas the heat flowing out of the surroundings and into the system is equal to the work done by the system. Since both system and surroundings have the same constant temperature the entropy increase of the system is equal to the entropy decrease of the surroundings giving a total entropy change of zero.
What would have to be different in an irreversible expansion and could this process still be isothermal?
An example I can come up with is: If the temperature of the surroundings is higher than that of the system then the entropy decrease of the surroundings is smaller than the increase in the system so the process is irreversible. However, I don't see how it still can be isothermal.
What would have to be different in an irreversible expansion and could this process still be isothermal?
An example I can come up with is: If the temperature of the surroundings is higher than that of the system then the entropy decrease of the surroundings is smaller than the increase in the system so the process is irreversible. However, I don't see how it still can be isothermal.