A vessel of volume V = 5.0 litre contains 1.4 gm of nitrogen at temperature $\mathrm{T}=1800\mathrm{K}$. Find the pressure of the gas if 30% of its molecules are dissociated into atom at this temperature.

$V=5.0\text{ litre }=5\times {10}^{-3}{\mathrm{m}}^3;m=1.4\mathrm{gm}M=28\mathrm{gm};n=\frac{m}{M}=\frac{1}{20};P=?$

30% dissociated into atoms (nitrogen is diatomic).

 Let the number of moles $n_0=\frac{m}{M}=\frac{1}{20}$

$$\begin{gathered} 30\mathrm{\%}\text{ diatomic }\to \text{ monoatomic }\left(\frac{3{\mathrm{N}}_0}{10}\right)\to \frac{6{\mathrm{N}}_0}{10}={\mathrm{n}}_1 \\ \mathrm{and} 70\mathrm{\%}\text{ remain diatomic }\Rightarrow {\mathrm{n}}_2=\frac{7{\mathrm{N}}_0}{10} \end{gathered}$$

$n_{eq}\text{ of mixture }\left(n_1+n_2\right)=\frac{13N_0}{10}\therefore \text{ From ideal gas equation }PV=n_{\mathrm{mix}}RT$

$$\begin{gathered} P\left(5\times {10}^{-3}\right)=\left(\frac{13n_0}{10}\right)\left(\frac{25}{3}\right)\left(1800\right) \\ \text{Here, }{n}_0=\frac{1}{20} \end{gathered}$$

$\Rightarrow \mathrm{P}=1.94\times {10}^5{\mathrm{Nm}}^{-2}$