[Cooper]

Hi,

Q. Calculate the entropy change for an isothermal expansion where 1 mole of ideal gas expands from 1 dm^3 to 100 dm^3 in a vacuum of 100 dm^3 at 300K.

The answer says you can calculate it assuming it is reversible, and because entropy is a state function, that also equals the entropy change of the irreversible process.

But I am having trouble finding the answer if you treated it as irreversible when you calculated the entropy change.

[tex]\Delta S=\int\frac{dq}{T}=\int\frac{dU}{T}-\int\frac{dw}{T}=\int\frac{dU}{T}-\int\frac{p_{ext}dV}{T}[/tex]
Since the process is isothermal (making dU = 0) and the external pressure is zero (making dw = 0), it seems like the math is telling me the entropy change should be zero. Of course I get it is not actually zero, but I don't understand what mistake I am making computationally...any help?

Thanks

[mjc123]

Q. Calculate the enthalpy change

I assume you mean entropy, and haven't been answering the wrong question. (It happens!)

But I am having trouble finding the answer if you treated it as irreversible when you calculated the entropy change

You can't. dS = dq_rev/T, so your equation is wrong. That is why you have to calculate it as if it were reversible, and equate the answer because it is a state function.

[Cooper]

Q. Calculate the enthalpy change

I assume you mean entropy, and haven't been answering the wrong question. (It happens!)

Haha, sorry, yes I meant entropy! I will fix that now.

But I am having trouble finding the answer if you treated it as irreversible when you calculated the entropy change

You can't. dS = dq_rev/T, so your equation is wrong. That is why you have to calculate it as if it were reversible, and equate the answer because it is a state function.

Oh, right...that makes sense. Thank you!

[cseil]

I don't understand if you've found your solution.
Anyway I'd use Maxwell's relations to calculate it. It's very easy.

Use the equation
## dA = -sdT - pdV ##
and find an equivalence for ##dS/dV##.