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 the work done if the compression is isothermal. `(gamma=5//3)`

Explanation

During an adiabatic process `T_(1)V_(1)=T_(2)V_(2)`
Here, `T_(1)=273K, V_(2)=(V_(1))/(2), gamma=(5)/(3)`
`:.273 xx V_(1)^((5)/(3)-1)=T_(2)((V_(1))/(2))^((5)/(3)-1)`
`T_(2)=(2)^(2//3) xx 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)`
`=8.34 xx 3)/(22.4xx2) (-158.6)= -89J`
(ii) Work done during isothermal compression is
`W=2.3026nRT log_(10) ((V_(2))/(V_(1)))`
n = number of moles `=(1)/(22.4)`,
`T=273K, R=8.314J mol^(-1)K^(-1) rArr (V_(2))/(V_(1)) =(1)/(2)=0.5`
`:.W=(2.306xx8.314xx273 log_(10) (0.5))/(22.4) rArr W= -70J`