The pressure inside a tyre is 4 times that of atmosphere. If the tyre bursts suddenly at temperature $$300\,K,$$  what will be the new temperature?

Solution :
Under adiabatic change
$$\eqalign{ & \frac{{{T_2}}}{{\;{T_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}}\,\,{\text{or}}\,\,{T_2} = {T_1}{\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}} \cr & \therefore {T_2} = 300{\left( {\frac{4}{1}} \right)^{\frac{{1 - \left( {\frac{7}{5}} \right)}}{{\left( {\frac{7}{5}} \right)}}}}; \cr & \gamma = 1.4 = \frac{7}{5}\,{\text{for}}\,{\text{air}} \cr & {\text{or}}\,\,{T_2} = 300{\left( 4 \right)^{ - \frac{2}{7}}} \cr} $$