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

`W_("adiabatic") = (nR(T_(1)-T_(2)))/(gamma-1) = 6R`
`implies (1xxR(T_(1)-T_(2)))/(5//3-1) = 6R implies T_(2) = (T-4)K`