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Entropy of gas expansion, isothermal/reversible vs irreversible (Joule), ideal vs VDW

  • Thread starter Jesssa
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  • #1
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Hello,

This is a question I've been working on out of blundell and blundell,

http://imageshack.us/a/img560/3342/entwopy.jpg [Broken]

The red box is my answers to the question which I am pretty sure are ok.

I am having trouble with the very last part of the question.

By the logic of the first part, the entropy would be the same with the VDW gas in both cases right (cases a and b)? Since entropy is a state variable.

But how can you find the change in temperature?

Thank you
 
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Answers and Replies

  • #2
ehild
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You know that ΔU=Q+W. If it is free expansion into vacuum, what is the heat exchanged and the work done by the gas during the process?

ehild
 
  • #3
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Ah right,

The work done by the gas will be zero and there will be no heat exchanged (assuming the enclosure is insulated).

So

ΔU=0=(3/2)nRΔT-n2a/αV + n2a/V


(3/2)RΔT=na/αV -n2a/V =na(1/α-1)/V

(3/2)RΔT=-na(-1/α+1)/V = -na(-1/α+1)/V=-na(α-1)/αV

So ΔT<0 i.e. decreases by an amount proportional to (α-1)/α

Thanks!
 
  • #4
ehild
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Revise the change of entropy, too, as not only the volume but also the temperature changes.

ehild
 
  • #5
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Oh, did you suggest that because there is a mistake in the working I posted in the picture and the second post, or do you mean in general?
 
  • #6
ehild
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You said the change of entropy would be the same for the van der Waals gas in both processes a and b (isotherm expansion and expansion into vacuum) as the gas arrives to the same state. It is not true for the VDW gas as it cools down in process b.

ehild
 
  • #7
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Oh sorry,

Thanks again for all your help!
 

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