If one mole of a gas A (mol. wt-40) occupies a volume of 20 litres, under the same conditions of temperature and pressure the volume occupied by 2 moles of gas B(mol. wt=80) is

Given data:

${\mathrm{n}}_1$=40

${\mathrm{n}}_2$=80

${\mathrm{V}}_1$=20

According to ideal gas equation

PV = nRT

At constant pressure and temperature

V is directly proportional to n

Therefore

$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$

$\frac{20}{{\mathrm{V}}_2}=\frac{40}{80}$

$\frac{20}{{\mathrm{V}}_2}=\frac{1}{2}$

${\mathrm{V}}_2$=20×2=40 L

Hence, the required answer is D.