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entropy variation during free expansion |
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| Feb13-09, 10:49 PM | #1 |
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entropy variation during free expansion
1. The problem statement, all variables and given/known data
A system that consists of [tex]n[/tex] moles of ideal gas does a free expansion (to the vacuum) from a volume [tex]V[/tex] to a volume [tex]2V[/tex]. a) What is the variation of entropy of the gas?; b) of the universe?; c) if the expansion was reversible and isothermal, what would be the variation of the entropy of the gas?; d) of the universe? 2. Relevant equations Ideal Gas Law: [tex]PV=nRT[/tex] Conservation of Energy: [tex]\texttt{d}U=T\texttt{d}S-p\texttt{d}V[/tex] 3. The attempt at a solution a) not sure how to do this. I think it should be zero. b) I know that the entropy is additive, so that the change of entropy in the universe is just the change of entropy in the system, plus the change outside the system. Supposedly there is no change of entropy outside the system, so that the answer to this question should be the same as a). c) Since the temperature doesn't change and this is an ideal gas, [tex]\texttt{d}U=0[/tex]. Therefore [tex]\texttt{d}S=\frac{p\texttt{d}V}{T}[/tex] Using the ideal gas law: [tex]\texttt{d}S=\frac{nR\texttt{d}V}{V}[/tex]. Integrating gives the result: [tex]\Delta S=nR\texttt{ln}2[/tex] d) The same as in c). Please, help me with a) and b). I believe the answer should be zero, but then I think that I can travel from the initial state to the final state through the reversible and isothermal line, just as in c) and d), in which case the answer to a) and b) should be the same as in c) and d) (since entropy is a function of state). Can someone clear this up for me? |
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| entropy, free expansion, ideal gas, thermodynamics |
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