An ideal gas has a specific heat at constant pressure $C_p=5R/2$. The gas is kept in a closed vessel of volume $0.0083 m^3$ at a temperature of $300 K$ and a pressure of $1.6\times {10}^6{\mathrm{Nm}}^{-2}$. An amount of $2.49\times {10}^4\mathrm{J}$ of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.

$n=\frac{PV}{RT}=\frac{1.6\times {10}^6\times 0.0083}{8.3\times 300}=\frac{16}{3}\text{ mole }$

Now $C_p-C_v=R,\text{ therefore }{C}_v=C_p-R=\frac{5R}{2}-R$

$=\frac{3R}{2}$

When heat energy $Q$ is supplied to the gas, the increase $∆T$ in its temperature is obtained form the relation 

      $△T=\frac{Q}{n{C}_v}=\frac{2.49\times {10}^4}{\left(16/3\right)\left(3\times 8.3/2\right)}K =375 K$

$\therefore$ Final temperature $T=300+375=675\mathrm{K}$. Since the gas is kept in a closed vessel, its volume remains constant. Hence the final pressure $P\text{'}$ is obtained form the relation  

$\begin{array}{r}\frac{P^{\mathrm{\prime }}}{T^{\mathrm{\prime }}}=\frac{P}{T}\\ P=P\times \frac{P^{\mathrm{\prime }}}{T^{\mathrm{\prime }}}\\ =\frac{1.6\times {10}^6\times 675}{300}\\ =3.6\times {10}^6{\mathrm{Nm}}^{-2}\end{array}$