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

The work done in an adiabatic process is given by
`W=(nR(T_(i)-T_(f)))/(gamma-1)`
In this case, `W=6R,n=1,gamma=5//3,T_(i)=T`
`therefore6R=(1xxR)/(5/3-1)(T-T_(f))`
`therefore6R=(3R)/2(T-T_(f))`
`thereforeT-T_(f)=4" "thereforeT_(f)=(T-4)K`