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    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.
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    (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:

    U = cT -\frac{a}{V}

    To check that, find (\frac{\partial U}{\partial V})_T 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 V \to \infty - 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?
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    Ah, i see what you mean.

    Thankyou for the help!
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    (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:

    U =cf(T) - \frac{a}{V}

    where f(T) 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 V \to \infty reasoning that I mentioned earlier - that gives you the whole term]
 
 
 
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