The reaction $2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)\to 4{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$ follows first order kinetics. The pressure of a vessel containing only ${\mathrm{N}}_2{\mathrm{O}}_5$ was found to increase from 50 mmHg to 87.5 mmHg in 30 min. The pressure exerted by the gases after 60 min will be (Assume temperature remains constant) 

We have 

       $\begin{array}{rcc}2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)& ⟶4{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)& \\ p_0-2p& 4p& p\end{array}$

Total pressure, $p_{\text{total }}=\left(p_0-2p\right)+4p+p=p_0+3p$

After 30 min, $p_{\text{total }}=87.5\mathrm{mmHg}$. Hence

$p=\frac{p_{\text{total }}-p_0}{3}=\frac{(87.5-50)\mathrm{mmHg}}{3}=12.5\mathrm{mmHg}$

Partial pressure of ${\mathrm{N}}_2{\mathrm{O}}_5$ at 30 min will be

$p_{{\mathrm{N}}_2{\mathrm{O}}_5}=p_0-2p=(50-2\times 12.5)\mathrm{mmHg}=25.0\mathrm{mmHg}.$

Since the initial pressure of $50\mathrm{mmHg}$ of ${\mathrm{N}}_2{\mathrm{O}}_5$ is reduced to 25 mmHg, the half-life of the reaction will be 30 min.

After 60 min (which is equal to two half-lives), the partial pressure of ${\mathrm{N}}_2{\mathrm{O}}_5$ will be

$p_{{\mathrm{N}}_2{\mathrm{O}}_5}=12.5\mathrm{mmHg}$

For this value, the value of $p$ will be 

$p=\frac{p_0-p_{{\mathrm{N}}_2{\mathrm{O}}_5}}{2}=\frac{(50-12.5)\mathrm{mmHg}}{2}=18.75\mathrm{mmHg}$

Finally, the pressure of the gas will be

$p=p_0+3p(50+3\times 18.75)\mathrm{mmHg}=106.25\mathrm{mmHg}$