## dU=0 for expansion and compression of an ideal gas

Lizzie Zhang 2C
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### dU=0 for expansion and compression of an ideal gas

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!

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

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

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.