At `27^(@)C` two moles of an ideal monoatomic gas occupy a volume `V`. The gas expands adiabatically to a volume `2V`. Calculate : ltbr. (i) the final temperature of the gas,
(ii) change in its internal energy and
(iii) the work done by the gas during the process.
take `R = (25)/(3)J//mol-K`

Answer

(i) `189K (ii) -2767J (iii) 2767J`

Explanation

(i) For adiabatic process `T_(1)V_(1)^(gamma -1) = T_(2)V_(2)^(gamma-1)`
`implies 300V^(5/3-1) = T_(2).(2V)^(5//3-1) implies T_(2) = 189K`
(ii) Change in internal energy `DeltaU = nC_(V)DeltaT = n(3)/(2)R(T_(2)-T_(1)) = 2 xx (3)/(2)xx 8.314 xx (189 - 300) = -2767J`
(iii) Work done by gas = `(P_(1)V_(1)-P_(2)V_(2))/(gamma-1) = (nR(T_(1) - T_(2)))/(gamma-1) = 2767J`