At a certain temperature and total pressure of ${10}^5$ Pa, Iodine vapour contains 40% by volume of I-atoms.

${\mathrm{I}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{I}\left(\mathrm{g}\right)$

$\text{ Calculate }{\mathrm{K}}_{\mathrm{p}}\text{ for the equilibrium, }$

Given, 

I-atoms in iodine vapour  =  40% by volume

Mole fraction of I atoms = $\frac{40}{100}$

${\mathrm{I}}_2 \mathrm{Molecules} \mathrm{in}$Iodine vapours = 60% by volume

Mole fraction of ${\mathrm{I}}_2= \frac{60}{100}$

Partial pressure of I-atoms

${\mathrm{p}}_1=\frac{40}{100}\times {10}^5=4\times {10}^4 \mathrm{Pa}$

Similarly, partial pressure of iodine molecules $\left({\mathrm{I}}_2\right)$,

${\mathrm{p}}_2=\frac{60}{100}\times {10}^5=6\times {10}^4 \mathrm{Pa}$

${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{\mathrm{I}}^2}{{\mathrm{p}}_{{\mathrm{I}}_2}}=\frac{{[4\times {10}^4]}^{ 2}}{[6\times {10}^4]}$

$=2.67\times {10}^4 \mathrm{Pa}$