At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal energy, and (iii) the work done by the gas during this process.

Answer

(i) `300((1)/(2))^(2//3) K`
(ii) `7500 (2^(-2//3)-1)J`
(iii) `-7500 (2^(-2//3)-1)J`

Explanation

Given `T_(1) = 27^(@)C = 300K`
`V_(1) = V`
`V_(2)= 2V`
(i) Final temperature
In adiabatic process `TV^(gamma-1) = constant`
`:. T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1)`
or `T_(2) = T_(1) ((V_(1))/(V_(2)))^(gamma-1)`
`= 300 ((V)/(2V))^(5//3-1) gamma = (5)/(3)` for monoatomic gas
`= 300 ((1)/(2))^(2//3) K`
(ii) Change in internal energy-
`DeltaU = nC_(v)DeltaT n = 2`( Given)
`= (2) ((3)/(2)R) (T_(2)-T_(1)) C_(v) = (3)/(2)R for`
monoatomic gas
`2 xx (3)/(2)R xx [300((1)/(2))^((2)/(3)) -300] = 7500 (2^(-2//3)-1)J`
(iii) Work done
Process is adiabatic, therefore `DeltaQ = 0`
and form first law of thermodynamics,
`DeltaQ = DeltaW +DeltaU`
`DeltaW =- DeltaU`
or `=- 7500 (2^(-2//3) -1)J`