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I have been looking online and at previous posts for a way of calculating the internal energy of a gas or the work done on gas due to a changing volume. The problem that I keep encountering is that the temperature is always said to be isothermic whereas in my model the volume will change causing the temperature and pressure to change as well, hence is it non-isothermic.

I have tried using the fact that $W = -pdV$ or $dU = (\frac{\partial U}{\partial T})_V\,dT + (\frac{\partial U}{\partial V})_T\,dV$ but both seem to only work with isothermic conditions.

If anyone knows of any way of accurately calculating the internal energy under these conditions that would be very helpful.

Thanks.

1 Answers1

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The Van der Waals equation gives you an equation relating $p,T,V$, so using $-p = \frac{\partial F}{\partial V}_{|T}$, you can solve for $F$ up to an additive temperature dependent term (technically there is also an $n$ dependence which you can find by invoking extensivity). You can then figure out $U$ by calculating $-S = \frac{\partial F}{\partial T}_{|V}$ and using $U = F+TS$ or using the useful formula combining these two steps: $U = \frac{\partial}{\partial \beta}_{|V}(\beta F)$ with $\beta=1/T$.

Concretely, using: $$ p=\frac{nRT}{V-nb}-\frac{an^2}{V^2} $$ you get: $$ F = F_0(T)-nRT\ln(V-Nb)-\frac{an^2}{V} $$ Thus $$ U = U_0(T)-\frac{an^2}{V} $$

Note that you need additional assumptions to figure out the temperature dependence, and you can give some estimates using microscopic models (check out Mayer's expansion).

Hope this helps and tell me if something is not clear.

LPZ
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