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

Answer

A

Explanation

`T_(1) =T, W= 6R` joules, `gamma = 5/3`
`W =(P_(1)V_(1)-P_(2)V_(2))/(gamma -1) = (nRT_(1) - nRT_(2))/(gamma -1) = (nR(T_(1)-T_(2))/(gamma -1))`
n=1, `T_(1) = T rArr (R(T -T_(2))/(5//3-1)) = 6R rArr T_(2) = (T-4)K`