Question

In van der Waal?? equation the critical pressure $P_c$ is given by

• A. $3b$
• B. $\frac{a}{27 b^2}$
• C. $\frac{27 a}{b^2}$
• D. $\frac{b^2}{a}$

Answer 1

Answer: (B)

$\frac{a}{27 b^2}$

Answer 2

The van der waal?? 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 $P_{c} = \frac{R}{2b} \left(\frac{8a}{27bR}\right) - \frac{a}{9b^{2}} = \frac{a}{27b^{2}}$