Compute the internal energy of a non-ideal gas – van der waals

Given: $\displaystyle (p + \alpha\frac{n^2}{V^2})(V-nb) = nRT$
Find: internal energy U

SOLUTION:
The internal energy of a van der waals gas is given by

$\displaystyle U_{real} = U_{ideal} + \int_{V_o}^{V_1} T^2 (\frac{\partial}{\partial T}\frac{p}{T})dV$

To solve the integral, we first find p using algebra

$\displaystyle (p + \alpha\frac{n^2}{V^2})(V-nb) = nRT$

$\displaystyle pV - pnb + \frac{\alpha n^2}{V} - \frac{\alpha n^3b}{V} = nRT$

$\displaystyle p(V-nb) = nRT - \frac{\alpha n^2}{V} + \frac{\alpha n^3b}{V}$

Divide both sides by T(V-nb)

$\displaystyle \frac{p}{T} = \frac{nR}{V-nb} - \frac{\alpha n^2}{TV(V-nb)} + \frac{\alpha n^3b}{TV(V-nb)}$

Differentiate wrt T

$\displaystyle \frac{\partial}{\partial T}\frac{p}{T} = \frac{\alpha n^2}{T^2V(V-nb)} - \frac{\alpha n^3b}{T^2V(V-nb)}$

The integral becomes

$\displaystyle = \int_{V_o}^{V_1} T^2 (\frac{\alpha n^2}{T^2V(V-nb)} - \frac{\alpha n^3b}{T^2V(V-nb)}) dV$

$\displaystyle = \int_{V_o}^{V_1} (\frac{\alpha n^2}{V(V-nb)} - \frac{\alpha n^3b}{V(V-nb)}) dV$

$\displaystyle = (\alpha n^2 - \alpha n^3b) \int_{V_o}^{V_1} \frac{dV}{V(V-nb)}$

$\displaystyle = (\alpha n^2 - \alpha n^3b) \int_{V_o}^{V_1} \frac{dV}{V^2} - \int_{V_o}^{V_1} \frac{dV}{Vnb}$

$\displaystyle = (\alpha n^2 - \alpha n^3b) (\frac{1}{V_o} - \frac{1}{V_1} - \frac{1}{nb} ln\frac{V_1}{V_o})$

Thus,

$\displaystyle U_{real} = \frac{3}{2}nR + (\alpha n^2 - \alpha n^3b) (\frac{1}{V_o} - \frac{1}{V_1} - \frac{1}{nb} ln\frac{V_1}{V_o})$

Comment for corrections and further questions. Thanks!

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