When 20 g of CaCO3 were put into 10 litre flask and heated to 800°C, 35% of CaCO₃ remained unreacted at equilibrium. K_p for decomposition of CaCO₃ is:

$\underset{\mathrm{Mole} \mathrm{after} \mathrm{dissociation}}{\underset{\mathrm{Mole} \mathrm{before} \mathrm{dissociation}}{ }} \underset{\left[\frac{20}{100}\times \frac{35}{100}\right]}{\underset{\frac{20}{100}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}}\rightleftharpoons \underset{\left[\frac{20}{100}\times \frac{65}{100}\right]}{\underset{0}{\mathrm{CaO}\left(\mathrm{s}\right)}}+\underset{\left[\frac{20}{100}\times \frac{65}{100}\right]}{\underset{0}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}}$
$\therefore$ Mole of CO₂ formed $=\frac{20\times 65}{{10}^4}=1.3\times {10}^{-1}$
$\begin{array}{l}\because \mathrm{PV}=\mathrm{nRT}\\ {\mathrm{P}}_{{\mathrm{CO}}_2}=\frac{1.3\times {10}^{-1}}{10}\times 0.0821\times 1073=1.145\mathrm{atm}\end{array}$
Now, ${\mathrm{K}}_{\mathrm{p}}=\mathrm{p}{\mathrm{CO}}_2=1.145\mathrm{atm}$