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Q.
In van der Waal’s equation the critical pressure $P_c$ is given by
Kinetic Theory
Solution:
The van der waal’s equation of state is
$\left(P + \frac{a}{V^{2}}\right)\left(V -b\right) = RT $
or $P = \frac{RT}{V -b} - \frac{a}{V^{2}} $
At the critical point,
$P = P_{c}, V = V_{c} $ and $T=T_{c}$
$ \therefore P_{c} = \frac{RT}{V_{c} - b} - \frac{a}{V_{c}^{2}}$ ....(i)
At the critical point on the isothermal,
$ \frac{dP_{c}}{dV_{c}} = 0$
$ \therefore 0 = \frac{-RT_{c}}{\left(V_{c} -b\right)^{2}} + \frac{2a}{V_{c}^{3}}$
or $ \therefore 0= \frac{RT_{c}}{\left(V_{c} -b\right)^{2}} + \frac{2a}{V_{c}^{3}} $ ......(ii)
Also at critical point, $\frac{d^{2}P_{c}}{dV^{2}_{c}} = 0$
$ \therefore 0 = \frac{2RT_{c}}{\left(V_{c} - b\right)^{3}} - \frac{6a}{V_{c}^{4}} $
or $\frac{2RT_{c}}{\left(V_{c} - b\right)^{3}} = \frac{6a}{V_{c}^{4}} $ .....(iii)
$\frac{1}{2} \left(V_{c} - b\right) = \frac{1}{3} V_{c} or V_{c} = 3b$ ....(iv)
Putting this value in (ii), we get
$ \frac{RT_{c}}{4 b^{2}} = \frac{2a}{27b^{3}} or T_{c} = \frac{8a}{27b R}$ ....(v)
Putting the values of $V_c$ and $T_c$ in (i), we get
$ P_{c} = \frac{R}{2b} \left(\frac{8a}{27bR}\right) - \frac{a}{9b^{2}} = \frac{a}{27b^{2}} $