$$\begin{gathered} {\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3 \\ 1\text{ mole }{\mathrm{N}}_2\text{ and }4\text{ moles }{\mathrm{H}}_2\text{ are taken } \end{gathered}$$

in 15 L flask at 27°C. After complete conversion of N₂ into NH₃, 5 L of H₂O is added. Pressure set up in the flask is

                   ${\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3$

Taken           1      4            0

Reacts          1     3  

H₂ Ieft = 1 mol NH₃ formed = 2 mol but NH₃(g) dissolves in water forming NH₄OH. Residual gas is only H₂ which exerts pressure.

Volume = 15 - 5 = 10 L (5 L is occupied by H₂O added)

$\therefore \mathrm{p}=\frac{\mathrm{n}}{\mathrm{V}}\mathrm{RT}=\frac{1\times 0.0821\times 300}{10}\mathrm{atm}$