Recent content by pilotpanda

Well, for an isothermal process \DeltaS = nRln\frac{V2}{V1} and since work for a reversible process is -nRTln\frac{V2}{V1}... After some struggle, I think I know the answer now. Thanks for your help!

Hi Mapes Thanks! I guess I'm just really confused about the adiabatic expansion. If the process is isothermal, I know how to approach the problem since qsys=qsur=-w, I can use \DeltaS = -w/T to calculate entropy of the surroundings and compare that with the system (either given or calculated)...

How do I calculate the entropy of the surroundings for an adiabatic expansion? I know \DeltaSsur = qrev/T for reversible process and qirrev/T for irreversible process. But q=0 for adiabatic process, so are there any other formulas that I could use to calculate \DeltaSsurr (I know there is a...