Question

If the critical constants of a hypothetical gas are $V_c=150c{m}^{3}mo{l}^{-1},P_C=50$ atm and $T_C$ = 300 K. Then the radius of the molecule is: [Take atm $mo{l}^{-1}{K}^{-1}$]

• A. ${\left(\frac{75}{2\pi N_A}\right)}^{\frac{1}{3}}$
• B. ${\left(\frac{75}{8\pi N_A}\right)}^{\frac{1}{3}}$
• C. ${\left(\frac{3}{\pi N_A}\right)}^{\frac{1}{3}}$
• D. ${\left(\frac{3}{256\pi N_A}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is D ${\left(\frac{3}{256\pi N_A}\right)}^{\frac{1}{3}}$

$P_c=\frac{a}{27b^2},T_c=\frac{8a}{27Rb}$
$\frac{P_c}{T_c}=\frac{R}{8b}\Rightarrow b=\frac{300\times \left(\frac{1}{12}\right)}{8\times 50}=\frac{1}{16}$
$b=4\times \frac{4}{3}\pi r^3.N_A$
$\frac{1}{16}=\frac{16}{3}\pi r^3.N_A$
$r={\left(\frac{3}{256\pi N_A}\right)}^{\frac{1}{3}}$