A mass of diatomic gas $$\left( {\gamma = 1.4} \right)$$   at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $${27^ \circ }C$$  to $${927^ \circ }C.$$  The pressure of the gas in final state is

Solution :
$$\eqalign{ & {T_1} = 273 + 27 = 300\,K \cr & {T_2} = 273 + 927 = 1200\,K \cr} $$
For adiabatic process,
$$\eqalign{ & {P^{1 - \gamma }}{T^\gamma } = {\text{constant}} \Rightarrow {P_1}^{1 - \gamma }{T_1}^\gamma = {P_1}^{1 - \gamma }{T_2}^\gamma \cr & \Rightarrow {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{1 - \gamma }} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^\gamma } \Rightarrow {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{1 - \gamma }} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^\gamma } \cr & {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{1 - 1.4}} = {\left( {\frac{{1200}}{{300}}} \right)^{1.4}} \Rightarrow {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{ - 0.4}} = {\left( 4 \right)^{1.4}} \cr & {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{0.4}} = {4^{1.4}}\,{\text{or,}}\,{P_2} = {P_1}{4^{\left( {\frac{{1.4}}{{0.4}}} \right)}} = {P_1}{4^{\left( {\frac{7}{2}} \right)}} \cr & = {P_1}\left( {{2^7}} \right) = 2 \times 128 = 256\,atm \cr} $$