Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `10^5 N//m^2` and 6 litres respectively. The final volume of the gas are 2 litre. Molar specific heat of the gas at constant volume is `3R//2`.

Explanation

For an adiabatic change `PV^(gamma)` = constant `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma)`.
As molar specific heat of gas at constant volume `C_(v) = (3)/(2)R`
`C_(P) = C_(V) + R = (3)/(2)R + R = (5)/(2)R implies gamma = (C_(P))/(C_(V)) = (5/2R)/(3/2R) = (5)/(3)`
`therefore P_(2) = [(V_(1))/(V_(2))]^(gamma) P_(1) = [(6)/(2)]^(5//3) xx 10^(5) = (3)^(5//3) xx 10^(5) = 6.19 xx 10^(5) N//m^(2)`
Work done ` W= (1)/(1-gamma) [P_(2)V_(2)- P_(1)V_(1)] = (1)/(1-(5//3)) [6.19 xx 10^(5) xx 2 xx 10^(-3) - 10^(-5) xx 6 xx 10^(-3)]`
`= -[(2xx10^(2)xx3)/(2) (6.19-3)] = -3xx10^(2) xx 3.19 = -957` joules
-ive sign shows external of work done on the gas