Question

The compressibility factor for $1$ mole of a vander Waals gas at $273K,8.21$ atm pressure having volume $1.092$ litre is $0.4$. Assuming the volume of gas molecules to be negligible. The value of vander Walls constant $a$ will be ?

• A. $12.53L^{2}mo{l}^{-2}atm$
• B. $14.53L^{2}mo{l}^{-2}atm$
• C. $18.37L^{2}mo{l}^{-2}atm$
• D. $10.21L^{2}mo{l}^{-2}atm$

Answer 1

The correct option is B $14.53L^{2}mo{l}^{-2}atm$

$Z=\frac{pV}{nRT}$
$0.4=\frac{8.21\times V}{1\times 0.082\times 273}$
$V=1.09066L$
$(p+\frac{a}{V^2})(V-b)=RTfor1mol$
Neglecting b, $(p+\frac{a}{V^2})V$$=RT$ or
$pV+\frac{a}{V}=RT$ or
$\frac{pV}{RT}+\frac{a}{VRT}=1$ or
$a=(1-\frac{pV}{RT})VRT=(1-0.4)\times 1.09066\times 0.082\times 273$
$a=14.6493atm{L}^{2}mo{l}^{-2}$