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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.

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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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  • Hi @Ipz, I am a little confused by a couple of things, the first being what the term additive temperature means, and the second being what is the Uo term, is this the initial temperature before any work or energy is applied? –  May 03 '22 at 12:51
  • I'm solving the partial differential equation $\partial_VF(V,T)=-p(V,T)$ so without appropriate boundary solutions, I can only find solutions up to an additive function of $T$, written as $F_0(T)$, since you have: $\partial_VF_0(T)=0$ (this is the multivariable analogue of the $+C$ term when you take the antiderivative). The $U_0=\frac{d}{d\beta}(\beta F_0)$ is a term depending only on temperature, which is unknown without further information. – LPZ May 03 '22 at 12:59
  • @Ipz, I have looked at the Mayers expansion and it seems I need to calculate the partition function Zo to find Fo, though I am slightly confused about what N means, I have tried using the number of particles in the gas but I cannot compute due to such large number, any idea? –  May 04 '22 at 07:56
  • I'd recommend D. Tong's lectures on statistical mechanics (you can find them in Cambridge's DAMTP website) for the Mayers expansion, he walks you through it pedagogically. $N$ is typically the number of particles. In the case of non-interacting gases, it's no problem as it is simply the exponent of the partition function (plus a factorial factor for Gibb's paradox). For the interacting gas, you need to actually perform the Mayer expansion and things get tricky. – LPZ May 04 '22 at 08:03
  • Hi @Ipz, just to clarify that U0=d/dβ(βF0) is equal to F0? –  May 13 '22 at 20:33
  • Yes, in this case but it isn’t really relevant. $U_0$ just captures the unknown remaining $T$ of $U$. – LPZ May 15 '22 at 09:47