# Need help with entropy question

Can anyone help me with this, please?

1. 1 mol of an ideal gas is compressed slowly and isothermally at 400 K in a piston-cylinder arrangement. Initial pressure = 100 kPa, final pressure 1000 kPa. The system is surrounded by a resevoir at 300 K such that heat exchange can take place between the piston-cylinder arrangement and the resevoir. System is isolated, so no heat exchange with outside world.

Calculate the entropy change of the gas, resevoir and universe if:

i. the piston is frictionless.

I'm stuck trying to do the entropy change for the gas. If I manage to do it, I should be able to do the rest.

I know delta S = INT dQ/T (haven't used tex before)

Also, from the 1st law: delta U = Qin + Won

For an isothermal change, delta U = 0 (as U depends on T only)

=> Qin = -Won

=> Qin = INT P dV

I'm not sure where to go from there, cos I can't put dQ = P dV in the integral above, can I (then substitute P = nRT/V, obviously)?

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Sure. It says in the excersize that the gas is compressed slowly and isothermally, so there always is an equilibrium. nRT/V can be put into the integral, making Q = nRT * INT (dV/V) from V1 to V2. (Which makes Q = nRT *ln(V2/V1).
Using relative volumes is enough here. Good luck!

Ok, thanks :).

I'm stuck again. How do I calculate the entropy change for the resevoir now?

1st law: delta U = Qin + Won

Won = 0, right? So delta U = Qin and I'm stuck :/.

Alright, you already showed us that delta U = 0, since it's an ideal gas. So Qin = -Won. In this case the Work on the gas is positive, the piston has to do work on the gas to compress it, so heat has to be removed from the gas, which follows fromthe formula.
Won = - INT P dV ---> so Qin = INT P dV = nRT * INT dV/V
So not Won = 0 but delta U = 0. This is because it's an isothermal process, which already suggests no change of internal energy (in the case of an ideal gas that is).
Now solve the integral and put relative values for V2 and V1 into it. There you go!