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

Given data is as follows below.

Pressure P=1 atm

Temperature  in Kelvin ${\mathrm{T}}_{\mathrm{K}}=127+273$

= 400 K

${\mathrm{n}}_{\mathrm{total}}={\mathrm{n}}_{{\mathrm{N}}_2}+{\mathrm{n}}_{{\mathrm{O}}_2}$

${\mathrm{n}}_{\mathrm{total}}=0.3+0.2=0.5 \mathrm{mol}$

From ideal gas equation,

PV = nRT

$\mathrm{V}=\frac{0.5\times 0.0821\times 400}{1}$

=16.4L

 The volume of the mixture is 16.4 L