- #1
kidsmoker
- 88
- 0
Homework Statement
http://img30.imageshack.us/img30/9954/72990922.jpg
The Attempt at a Solution
I would say the answer to (a) was no but that would be a guess. Does it have to do with the fact that it's done instantaneously, whereas to achieve anything like an irreversible process in reality you have to perform it very slowly?
For (b), Q=0 so the change in internal energy is equal to the work done. This is
[tex]W=\frac{C_{V}}{R}(p_{1}V_{1}-p_{2}V_{2})[/tex]
But I only know the volumes, not the pressures. Is there some way to use the fact that it expands into a vacuum?
I think the entropy change will be zero because no heat is going in or out.
Not sure about the temperature change bit. I would have expected the temperature to change in both. For an ideal gas, if there is some work done then the temperature definitely must change, but how does this differ for a real gas?
Thanks.
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