At 127^oC and 1 atm. pressure, a mixture of gases contains 0.3 mole of ${\mathrm{N}}_2$, 0.2 mole of ${\mathrm{O}}_2$. The volume of the mixture is

Given data: T=137+273=400 K

0.3 moles of ${\mathrm{N}}_2$, 0.2 moles of $O_2$

PV=nRT

So the mixture of 0.3 and 0.2 is 0.5 moles

$\begin{array}{c}1\left(\mathrm{V}\right)=(0.3+02)(0.8321)\left(400\right)\end{array}$

V=16.4 lit

The required option is D