Question

One mole of $N_2{O}_4\left(g\right)$ at $300$K is kept in a close container under one atmosphere. It is heated $600$K when $20\%$ by mass of $N_2{O}_4\left(g\right)$ decomposes to $N{O}_2\left(g\right)$. The resultant pressure is:

Answer 1

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2$

Initial : $1$ mole $0$ mole

At equilibrium moles of $N_2{O}_4=1-20$% of $1$ mole $=0.8$ mole

At equilibrium moles of ${NO}_2=0.2\times 2=0.4$ mole

$\therefore$ Total moles at equilibrium $=0.8+0.4=1.2$ mole

$P_1{V}_1=n_1{RT}_1$

$\Rightarrow 1\times V_1=1\times R\times 300\Rightarrow V_1=300R⟶\left(1\right)$

$P_2{V}_2=n_{2}R{T}_2$

$\Rightarrow P_2\times V_2=1.2\times R\times 600⟶\left(2\right)$

dividing (2) by (1),

$\frac{P_2\times V_2}{V_1}=\frac{1.2\times R\times 600}{300\times R}[V_1=V_2]$

$\Rightarrow P_2=2.4$ atm

$\therefore$ Resultant pressure of mixture is $2.4$ atm.