Question

For a real gas (mol. mass $=30$) if density at critical point is $0.40g.c{m}^3$ and its $T_c=\frac{2\times {10}^5}{821}K$, then calcium Vander Waal's constant $a$ (in atm $L^{2}mo{l}^{-2})$.

Answer 1

Given: $d=0.40$ g$/c{m}^3$, $T_C=\frac{2\times {10}^5}{821}$K

At critical point,

$T_C=\frac{8a}{27Rb}=\frac{2\times {10}^5}{821}$

$\frac{8\times a}{27\times 0.0821\times b}=\frac{2\times {10}^5}{821}$

$\therefore b=\frac{2a}{135}$

Critical pressure $\left(P_C\right)=\frac{a}{27b^2}=\frac{dRT}{M}$ (using ideal gas equation: $Pv=nRT$, $Pv=\frac{m}{M}RT$, $\therefore P=\frac{dRT}{M}$)

$\therefore a=\frac{\left(27b^2\right)\times dRT}{M}$

$=\frac{27\times 4a^2\times 0.40\times {10}^3\times 0.0821\times 2\times {10}^5}{(135{)}^2\times 821\times 30}$

$a=0.633$ atm $l^2/mo{l}^2$.