At `27^(@)C`, two moles of an ideal mono-atomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
(c ) the work done by the gas during the process. (R = 8.31 J/mol K)

Explanation

(a) In case of adiabatic change,
`PV^(r ) =` const. with `PV=nRT`
So that `T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)` [with `gamma=(5//3)`]
i.e., `300xx v^(2//3) = T(2V)^(2//3) or T=300//(2)^(2//3) = 189K`
(b) As `DeltaU=nC_(v)DeltaT = (nR DeltaT)/(gamma-1) [ "as" C_(v)=(R )/((gamma-1))]`
So, `DeltaU=(2xx8.31 xx (189-300))/(((5)/(3)-1)) = -2767.23J`
Negative sign means internal energy decreases.
(c ) According to first law of thermodynamics,
`DeltaQ=DeltaU+DeltaW` and as for adiabatic change
`DeltaQ= 0 rArr DeltaW = - DeltaW = 2767.23J`