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Internal Energy of a Van der Waals Gas Watch

1. So I have this question. Im able to derive the first part, however integrating the equation is for some reason stumping me.

So I took dV over to the RHS, giving du=a/v^2 dv. Which I promptly integrated and got U=(-a/v) + c where c would just be an integration constant correct? But how am i supposed to show that c=3RT/2. I get that this is the U for an ideal gas, but Im not sure if that's relevant or not.
2. (Original post by Eremor)

So I have this question. Im able to derive the first part, however integrating the equation is for some reason stumping me.

So I took dV over to the RHS, giving du=a/v^2 dv. Which I promptly integrated and got U=(-a/v) + c where c would just be an integration constant correct?
Incorrect, since you are integrating a partial derivative where T is held constant. So the most general term with T that can differentiate back to 0 in this case gives you:

To check that, find from that expression.

But how am i supposed to show that c=3RT/2. I get that this is the U for an ideal gas, but Im not sure if that's relevant or not.
I'm not sure what they want here, but:

1. this isn't an ideal gas since in an ideal gas, particles have no volume
2. consider what would happen if - this would bring you closer and closer to the ideal gas state, since the particle volume eventually becomes negligible w.r.t V. And what would the energy tend towards?
3. Ah, i see what you mean.

Thankyou for the help!
4. (Original post by Eremor)
Ah, i see what you mean.

Thankyou for the help!
You thanked me too soon. I've realised that I made a mistake above - when you integrate, you'll end up with:

where is some arbitrary function of T. So you need to do some more work to justify the linearity in T. I'll have to think about this later, when I get the time.

[edit: there's no more to justify - just apply the reasoning that I mentioned earlier - that gives you the whole term]

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