If a monoatomic ideal gas of volume 1 litre at N. T.P . Is compressed
adiabatically to half of its volume , find the work done on the gas . Also find

Explanation

During an abiabatic process `T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)`
Here , `T_(1)=273K,V_(2)=(V_(1))/(2),gamma=(5)/(3).`
`rArr273V_(1)^((5)/(3)-1)=T_(2)(V_(1)/(2))^((5)/(3)-1)`
`T_(2)=(2)^(2/3)273=431.6K,`
number of moles , `n=(1" litre")/(22.4 " litre")=(1)/(22.4)`
Work done `(nR)/(gamma-1)(T_(1)-T_(2))`
`=(8.314)/(22.4((5)/(3)-1))(273-431.6)`
`=(8.34xx3)/(22.4xx2)(-158.6)=-89J`