One mole of an ideal gas at an initial temperature true of `TK` does `6R` joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is `5//3`, the final temperature of the gas will be

Options

A. `(T+2.4)K`
B. `(T-2.4)K`
C. `(T+4)K`
D. `(T-4)K`

Answer

D

Explanation

`n=1,T_1=TK,DeltaW=6R,gamma=(5)/(3)`
`DeltaW=(nR(T_1-T_2))/(gamma-1)`
`6R=(R(T_1-T_2))/((5)/(3)-1)=(2)/(3)`
`6Rxx(2)/(3)=R(T_1-T_2)implies4=T-T_2`
`T_2=(T-4)K`