Question

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

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

Answer 1

Answer: (B)

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,
$\frac{d{P}_c}{d{V}_c}=0$
$\therefore 0=\frac{-R{T}_c}{{\left(V_c-b\right)}^2}+\frac{2a}{V_c^3}$
or $\therefore 0=\frac{R{T}_c}{{\left(V_c-b\right)}^2}+\frac{2a}{V_c^3}$ ......(ii)
Also at critical point, $\frac{d^2{P}_c}{d{V}_c^2}=0$
$\therefore 0=\frac{2R{T}_c}{{\left(V_c-b\right)}^3}-\frac{6a}{V_c^4}$
or $\frac{2R{T}_c}{{\left(V_c-b\right)}^3}=\frac{6a}{V_c^4}$ .....(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
$\frac{R{T}_c}{4b^2}=\frac{2a}{27b^3}or{T}_c=\frac{8a}{27bR}$ ....(v)
Putting the values of $V_c$ and $T_c$ in (i), we get $P_c=\frac{R}{2b}\left(\frac{8a}{27bR}\right)-\frac{a}{9b^2}=\frac{a}{27b^2}$