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dU=0 for expansion and compression of an ideal gas

Posted: Fri Jan 12, 2018 10:38 pm
by Lizzie Zhang 2C
In the book, it is explained that the expansion or the compression of an ideal gas results in deltaU=0 because the kinetic energies and the potential energies of the ideal gas is unchanged. I'm just wondering how to explain this by knowing dU=dQ+dW? Because either compression or expansion would cost work... And I remember that the work done by the system as it expands is equal to the area beneath the ideal gases isotherm graph (P against V) somewhere...maybe from my TA?

I just found it a bit confusing... Thanks for anyone who's willing to explain!

Re: dU=0 for expansion and compression of an ideal gas

Posted: Sat Jan 13, 2018 10:22 am
by Curtis Tam 1J
I believe it's delta zero given that you are referring to isothermal expansion/compression of an ideal gas. This means that any work done to move the piston inwards or outwards is restored by heat flow. This makes sense because if deltaU= q + w = 0, then you have q = - w, which basically means the previous sentence.