Question

For a van der Waals gas compressibility factor at 273.15 K is 1.5. The molar volume of this gas at 273.15 K is (assume P = 1 atm)

• A. 14.93 L
• B. 22.4 L
• C. 33.6 L
• D. 28.2 L

Answer 1

The correct option is C 33.6 L

Given: $Z=1$

$T=273.15K$

$P=1atm$

$R=0.082Lat{m}^{-1}{K}^{-1}$

$Z=\frac{PV}{RT}$

$\Rightarrow 1.5=\frac{1\times V}{0.082\times 273.15}$

$\therefore V_m=1.5\times 0.082\times 273.15$

$=33.59\simeq 33.6L$

$\therefore$ Molar volume of the gas is $33.6L$ .