I was thinking about how would one find the enthalpy in terms of $V,T$ but not on the usual ideal gas, but instead on a Van Der Waals gas with constant temperature, that is, a gas that satisfies: $$(P+\frac{an^2}{V^2})(V-nb)=nRT$$ with $a,b$ constants. We all know that the enthalpy is defined as $$H=U+PV\Rightarrow dH=dU+VdP+PdV$$ and since we are taking constant temperature (isothermic) to make our lifes easier, and $dU=TdS-PdV$ we get $$dH=dU+PdV+VdP=TdS+VdP$$ since and by definition of the entropy in terms of pressure and temperature $dS=C_p \frac{dT}{T}+\frac{nR}{P}dP=\frac{nR}{P}dP$ (since temperature is constant) we can get $$dH=T(\frac{nR}{P}dP)+VdP=(\frac{nRT}{P}+V)dP$$ but am I supposed to substitute one of the variables $n,V,T$ with the Van Der Waals? On any case I could get a very long expression and I don't think it is possible, how can I enhance this?
Van der Waals Gas – Calculating Enthalpy of a Van der Waals Gas
energyentropygasthermodynamics
Related Solutions
Solve Van der Waal's equation and approximate the $PV$ term by $RT$, and you will have your required equation. Although we are introducing an error, it will hardly affect the final result since there will be both $a$ and $b$.
For Van der Waal gas: $$\left(p+\frac{n^2q}{v^2}\right)\left(v-nb\right) = nRT$$ $$\Rightarrow pv+\frac{n^2q}{v}-pnb-\frac{n^3ab}{v^2}=nRT$$ $$\Rightarrow pv = nRT - \frac{n^2q}{v}+pnb+\frac{n^3ab}{v^L}$$ $$\Rightarrow V = \frac{nRT}{p} - \frac{n^2q}{pV}+nb+\frac{n^3ab\cdot P}{pV^2\cdot P}$$ $$\Rightarrow V = \frac{nRT}{P}-\frac{nq}{RT}+nb+\frac{nabP}{R^2T^2}$$ Note: Replacing $Pv$ by $nRT$ in the terms, the error becomes negligibly small and it hardly affects the final result. $$\because \left(\frac{\partial V}{\partial T}\right)_p = \frac{nR}{P}+\frac{nq}{RT^2}-\frac{2nabP}{R^2T^3}$$ We know, then, that $$M_{JT}=\left(\frac{\partial T}{\partial P}\right)_H$$ $$\Rightarrow M_{JT}=-\frac{1}{cp}\left(\frac{\partial H}{\partial P}\right)_T = -\frac{1}{cp}\left[V-\left(\frac{\partial V}{\partial T}\right)_P T\right]$$
Note from editor: some of the handwriting was a bit unclear, so here is the original picture posted for reference (I did, however, rotate the image so it is easier to read):
Statistical mechanics
Van der Waals equation can be derived from virial/cluster expansion in terms of the average/mean density of the liquid/gas:
$$
\rho(\mathbf{r})=\left\langle \sum_{i=1}^N\delta(\mathbf{r}-\mathbf{r}_i)\right\rangle,
$$
where $\mathbf{r}_i$ is the position of the i-th atom/molecule.
Thermodynamics
Another way to look at it is that pressure is the derivative of the free energy in respect to volume, so one can obtain the expression for the free energy by integrating the pressure expressed from van der Waals equation in respect to volume. One can then analyze liquid-gas transition in terms of Landau theory of phase transitions, with density serving as the order parameter.
See, e.g., section 8.5 in Lecture notes on Statistical Physics by Lydéric Bocquet
See also Wikipedia for the expression for the Helmholz free energy in more conventional terms.
Simple mean-field derivation of Van der Waals equation
Simple mean-field derivation of VdW equation can be performed along the lines of the derivation given in Wikipedia. I sketch it here in even more simplified/clarified form.
The energy of a gas of N interacting particles is $$ H(\mathbf{x}_1, \mathbf{p}_1;...;\mathbf{x}_N, \mathbf{p}_N)= \sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m}+\frac{1}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^NU(|\mathbf{x}_i-\mathbf{x}_j|), $$ where the potential is attractive, i.e., $U(|\mathbf{x}_i-\mathbf{x}_j|)<0$ (unless the molecules/atoms are very close to each other)
The partition function is $$ Z=\int \prod_{i=1}^Nd^3\mathbf{x}_id^3\mathbf{p}_ie^{-\beta H(\mathbf{x}_1, \mathbf{p}_1;...;\mathbf{x}_N, \mathbf{p}_N)}= \int d^3\mathbf{p}_1...d^3\mathbf{p}_Ne^{-\beta \sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m}}Z_c=\\\left(2\pi mk_B T\right)^{\frac{3N}{2}}Z_c $$ where the configuration integral is $$ Z_c=\int d^3\mathbf{x}_1...d^3\mathbf{x}_Ne^{-\frac{\beta}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^NU(|\mathbf{x}_i-\mathbf{x}_j|)}, $$ and $\beta=1/(k_BT)$.
We now perform the mean-field approximation replacing the pairwise interactions between particles by an interaction of each particle with an averaged field of all other particles: $$ \frac{1}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^NU(|\mathbf{x}_i-\mathbf{x}_j|)\approx \frac{1}{2}\sum_{i=1}^N \frac{1}{V}\int_V d^3\mathbf{x_j}U(|\mathbf{x}_i-\mathbf{x}_j|)\approx\\ \frac{1}{2}\sum_{i=1}^N \frac{N}{V}\int_V d^3\mathbf{y}U(|\mathbf{y}|)=-\frac{a'N^2}{V}, $$ where the minus sign is due to the fact that the potential is attractive, so that $a'>0$. The configuration integral is then $$ Z_c=\int d^3\mathbf{x}_1...d^3\mathbf{x}_N e^{\beta\frac{a'N^2}{V}}= (V-b'N)^N e^{\beta\frac{a'N^2}{V}}, $$ where we took account for the fact that the particles cannot be too close to each other, and hence the volume integral for each particle is reduced by the factor accounting for the volume of all other particles.
We now can write down the free energy of the gas as $$ F=-\frac{1}{\beta}\log Z=-\frac{a'N^2}{V}-\frac{N}{\beta}\log(V-b'N)-\frac{3N}{2\beta}\log\left(2\pi mk_BT\right) $$ The gas pressure is given by (since $dF=-SdT-pdV$) $$ P=-\left(\frac{\partial F}{\partial V}\right)_T= \frac{Nk_BT}{V-Nb'}-\frac{a'N^2}{V^2} $$ Converting to molar fractions $n=N/N_A$, $b=N_Ab'$, $a=a'N_A^2$, we obtain the usual form of the VdW equation $$ P=\frac{nRT}{V-nb}-\frac{an^2}{V^2}\Leftrightarrow \left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT. $$
Best Answer
Your equation for dS is for an ideal gas. For a real gas, it reads: $$dS=\frac{C_P}{T}dT-\left(\frac{\partial V}{\partial T}\right)_PdP$$