Calculate the work done when $$1\,mole$$  of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are $${10^5}\,N/{m^2}$$   and 6 litre respectively. The final volume of the gas is 2 litres. Molar specific heat of the gas at constant volume is $$\frac{{3R}}{2}.$$
[Given $${\left( 3 \right)^{\frac{5}{3}}} = 6.19$$   ]

Solution :
For an adiabatic change $$P{V^\gamma } = {\text{constant}}$$
$${P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma $$
As molar specific heat of gas at constant volume
$$\eqalign{ & {C_V} = \frac{3}{2}R \cr & {C_P} = {C_V} + R = \frac{3}{2}R + R = \frac{5}{2}R; \cr & \gamma = \frac{{{C_P}}}{{{C_V}}} = \frac{{\left( {\frac{5}{2}} \right)R}}{{\left( {\frac{3}{2}} \right)R}} = \frac{5}{3} \cr} $$
∴ From eqn. (1)
$$\eqalign{ & {P_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^\gamma }{P_1} = {\left( {\frac{6}{2}} \right)^{\frac{5}{3}}} \times {10^5}\,N/{m^2} \cr & = {\left( 3 \right)^{\frac{5}{3}}} \times {10^5} = 6.19 \times {10^5}\,N/{m^2} \cr} $$
Work done
$$\eqalign{ & = \frac{1}{{1 - \left( {\frac{5}{3}} \right)}}\,\,\left[ {6.19 \times {{10}^5} \times 2 \times {{10}^{ - 3}} - {{10}^{ - 5}} \times 6 \times {{10}^{ - 3}}} \right] \cr & = - \left[ {\frac{{2 \times {{10}^2} \times 3}}{2}\left( {6.19 - 3} \right)} \right] \cr & = - 3 \times {10^2} \times 3.19 \cr & = - 957\,joules\,\,\left[ { - ve\,{\text{sign}}\,{\text{shows}}\,{\text{external}}\,{\text{work}}\,{\text{done}}\,{\text{on}}\,{\text{the}}\,{\text{gas}}} \right] \cr} $$