Explanation

(d) In an adiabatic process
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}Q = 0\ $
So, from 1st law of thermodynamics
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W = - \mathrm{\Delta}U$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - nC_{V}\ \mathrm{\Delta}T$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - n\left( \frac{R}{\gamma - 1} \right)(T_{f} - T_{i})$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{nR}{\gamma - 1}(T_{i} - T_{f})$ ----(i)
Here W=6R J, n=1 mol
R=8.31 J/mol-K, $\gamma = \frac{5}{3},\ \ \ \ \ \ T_{i} = TK$
Substituting given values in Eq. (i), we get
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6R = \frac{R}{(5/3 - 1)}(T - T_{f})$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6R = \frac{3R}{2}(T - T_{f})$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T - T_{f} = 4\ $
$\therefore\ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T}_{f} = (T - 4)K$