One mole of an ideal gas at an initial temperature 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

In an adiabatic process `Q =0`
So, from the first law of thermodynamics
`W= -DeltaU =-nC_(V)DeltaT=-n((R)/(gamma-1))`
`implies W = (nR)/(gamma-1)(T_i-T_f)`..(i)
Here, `W = 6R J, n=1 mol`
`R = 8.31 J//mol-K gamma = 5/3 , T_(i)=TK`
Substituting given values in eq. (i) we get
`:. 6R = R/((5//3-1)) (T-T_f)`
`implies T = T_(f)=4`
`:. T_(f) = (T-4)K`.