Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `105 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`.

Answer

`-972J`

Explanation

`PV^(gamma)` = constant `implies (10^(5))(6)^(5//3) = (P_(2)) (2)^(5//3)`
`implies P_(2) = (10^(5))(3)^(5//3) Nm^(-2)`
` W =(P_(1)V_(1) - P_(2)V_(2)) /(gamma-1)`
=`((10^(5))(6 xx10^(-3)) - (10^(5) xx 3^(5//3))(2xx10^(-3)))/((5)/(3)J) = -972J`