Question

The compression factor (compressibility factor ) for one mole of a Vander Waals gas at $0^{\circ }C$ and atmospheric pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the Vander Waals constant 'a':

• A. $ 106.6 dm^{3} atmmol^{−2} $
• B. $ 225.1 dm^{3} atmmol^{−2} $
• C. $ 125.6 dm^{3} atmmol^{−2} $
• D. None of these

Answer 1

The correct option is C $ 125.6 dm^{3} atmmol^{−2} $

$Z=\frac{PV}{RT}$
$0.5=\frac{1\times V}{RT}$
$V=0.5RT$
$(P+\frac{a}{V^2})V=RT$
$(1+\frac{a}{0.25R^2{T}^2}0.5RT=RT$
$0.5+\frac{2a}{R^2{T}^2}=1$
$2a=0.5R^2{T}^2$
$a=0.25R^2{T}^2$
$a=0.25\left(0.081{)}^2\right(273{)}^2$
$a=125.6d{m}^{3}atmmo{l}^{-2}$