The discussion revolves around the entropy change for a Van der Waals gas, specifically examining different methods of calculation that yield varying results. Participants are exploring the implications of the isothermal condition on the internal energy and the equations used to derive entropy changes.
Some participants assert the correctness of the first method while questioning the reasoning behind the second method. There is an ongoing examination of the starting equations used in both approaches, with references to the first law of thermodynamics and the definition of entropy change.
Participants note that the problem states an isothermal process, which leads to discussions about the implications for internal energy in the context of a Van der Waals gas. There is a recognition of potential misunderstandings regarding the assumptions of the methods used.
I don't see any flaws in logic. I simply substitute for P and integrate.Chestermiller said:
Your equation for the effect of pressure on entropy in the first cast is correct. Where did you get the starting equation for the effect of pressure on entropy in the 2nd case from?laser1 said:
The question states "isothermal", which implies ##U=0##. So by the first law, this must mean that ##dQ=PdV##. Following this, the definition of the change in entropy is $$\Delta S = \int_i^f{\frac{dQ}{T}}$$ and I substitute ##dQ=PdV## to get the starting equation for #2.Chestermiller said:
Isothermal does not imply that du is equal to 0 for a Vdw gas.laser1 said: