Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: PKemNoob on February 18, 2013, 06:37:49 AM
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Hi, I am in a pchem course and I am having trouble with this question:
Starting from the expression dH = dU + d(PV) show that for an ideal gas V1/T1 = V2/T2 if the pressure is kept constant. There is supposed to be some integration involved.
I know that CpdT = CvdT + nRdT :rarrow: dH = dU + d(PV), but where do I go from here?
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You could start by going dH = dU + PdV + VdP, cancel out the last term, and replacing dH and dU with heat capacity terms
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Umm yeah, that's exactly what I did in my original post to start out, as you can see above.
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Ok, you can replace dH with CpdT, can you think how to replace dU for an ideal gas?
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like it says in my original post: dh is CpdT, dU is CvdT and d(PV) is nRdT .:. CpdT= CvdT + nRdT.
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Oh right, sorry, my brain wasn't clicking there.
I know that CpdT = CvdT + nRdT :rarrow: dH = dU + d(PV), but where do I go from here?
So then CpdT - CvdT = nRdT = PdV (since dP is 0)
Integration on both sides give you the Ideal Gas Law, with n, R and P all constants. This can be simplified to Charles' Law. Don't know if this is satisfactory proof for you :-\
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I've tried taking that approach before but it lead me only to bad places... perhaps I am doing the integration wrong??? Can you work it step by step so I can compare it to my attempts at deriving it?
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Nevermind, I got it... thanks, idk why I didn't see that.. maybe bc i dont sleep lol