Question

The compression factor (compressibility factor) for one mole of a van der Waals gas at $0^{0}C$ and 100-atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant ${}^{\prime }{a}^{\prime }$ in $\left(atm{L}^{2}mo{l}^{-2}\right)$.

• A. 1.123
• B. 1.253
• C. 2.411
• D. 0.622

Answer 1

The correct option is B 1.253

Compressibility factor = $\frac{PV}{RT}$

0.5 = 100 $\frac{V}{0.082}$ 273

= 0.1117 L

When the volume of a gas molecule is negligible, van der Waal's equation becomes,

$P+\frac{a}{V^2}(V-0)=RT$

Substituting the values

A = $(0.082\times 0.1119\times 273)$ - $(100\times 0.1119\times 0.1119)$

$=1.253atm{L}^{2}mo{l}^{-}$${}^2$