When `1mol` of a monoatomic ideal gas at `TK` undergoes adiabatic change under a constant external pressure of `1atm`, changes volume from `1 L to 2L`. The final temperature (in K) would be

Options

A. `(T)/(2^(2//3))`
B. `T +(2)/(3xx0.0821)`
C. `T`
D. `T -(2)/(3 xx 0.0821)`

Answer

D

Explanation

Process `DeltaH = 0`
`DeltaE = DeltaW` ltbr. `DeltaW =- P DeltaV =- 1`
`DeltaE = nC_(V)DeltaT`
`C_(V) = (R )/(gamma-1), gamma=(5)/(3)` for monatomic sas
`(gamma=(C_(P))/(C_(V)))`
`C_(V) = (R )/((5)/(3)-1) =(3R)/(2)`
`n = 1`
`DeltaE = 1 xx (3R)/(2) xx (T_(2)-T) =- 1`
`T_(2) = T -(2)/(3 xx 0.0821)`