A closed vessel contains one mol of ${\mathrm{N}}_2{\mathrm{O}}_4\left(g\right)$ at 27°C and one atmospheric pressure. On heating to 327°C, 20% by mass of this gas decomposed to brown coloured ${\mathrm{N}}_2{\mathrm{O}}_4\left(g\right)$. The resultant pressure will be:

mol. mass(= 1 mol) of ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)=(2\times 14)+(4\times 16)=28+64=92g$

g. mol. mass of $2{\mathrm{NO}}_2=2[14+(2\times 16\left)\right]=92\mathrm{g}$

At equilibrium, mass of  $N_2{O}_4\left(g\right)=92-\left(92\times \frac{20}{100}\right)=92-18.4=73.6\mathrm{g}$

$\therefore$ no. of mol of ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$

$\begin{array}{l}=\frac{\text{ mass of }{\mathrm{N}}_2{\mathrm{O}}_4}{\text{ g.mol.mass of }{\mathrm{N}}_2{\mathrm{O}}_4}\\ =\frac{73.6\mathrm{g}}{92\mathrm{g}}=0.8\mathrm{mol}\end{array}$

In the reaction,

${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{NO}}_2\mathrm{g}$ no. of mol of ${\mathrm{NO}}_2=\frac{1}{2}\times 1$ mol of ${\mathrm{N}}_2{\mathrm{O}}_4=\frac{1}{2}\times 0.8\mathrm{mol}$ = 0.4 mol

$\begin{array}{l}{T}_1=27+273=300\mathrm{K}\\ T_2=327+273=600\mathrm{K}\end{array}$

Total no. of mol $=0.8+0.4=1.2\mathrm{mol}$

Since the volume (V) is constant, thus

$$\begin{gathered} \left(1\right) P_1=0.8\mathrm{mol},V_1=V;T_1=300\mathrm{K},n_1=1\mathrm{mol} \\ \left(2\right) {\mathrm{P}}_2=0.4\mathrm{mol},{\mathrm{V}}_2=\mathrm{V};{\mathrm{T}}_2=600\mathrm{K},{\mathrm{n}}_2=1.2\mathrm{mol} \end{gathered}$$

But $\frac{P_1{V}_1}{n_{1}R{T}_1}=\frac{P_2{V}_2}{n_{2}R{T}_2}$

Or $\frac{P_1}{n_1{T}_1}=\frac{P_2}{n_2{T}_2};\frac{1}{1\times 300}=\frac{P_2}{1.2\times 600}$

$\therefore P_2=\frac{1.2\times 600}{300}=2.4\mathrm{atm}.$