Answer: 2.303 R

We have, heat absorbed by the gas $q_{rev}= -W_{rev}$ = 2.303 nRT $log \frac{V_2}{V_1}$

Change in Entropy $\Delta S = \frac{q_{rev}}{T}$ = 2.303 nR $log \frac{V_2}{V_1}$

Here, n=1 and $\frac{V_2}{V_1} = 10$

$\therefore \Delta S = 2.303 \times 1 \times R \times log 10 = 2.303 \times R \times 1 =$ ** 2.303 R**