Explanation

(d) The change in entropy of an ideal gas
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}S = \frac{\mathrm{\Delta}Q}{T}$ …(i)
In isothermal process, there is no change in internal energy of gas ie, $\mathrm{\Delta}U = 0$
$\therefore\ \ \ \ \ \ \mathrm{\Delta}U = \mathrm{\Delta}Q - W$
$\Rightarrow \ \ \ \ \ 0 = \mathrm{\Delta}Q - W$
$\Rightarrow \ \ \ \ \ \ \mathrm{\Delta}Q = W$
$\therefore\ \mathrm{\Delta}Q =$work done by gas in isothermal process which went through from $\left( p_{1},V_{1},T \right)\ to\ (p_{2},V_{2},T)$
or $\mathrm{\Delta}Q = nRTlog_{e}\left( \frac{V_{2}}{V_{1}} \right)$ …(ii)
For 1 mole of an ideal gas, n=1
So, from Eqs.(i) and (ii), we get
$\mathrm{\Delta}S = Rlog_{e}\left( \frac{V_{2}}{V_{1}} \right)$
$= R\ In\ \left( \frac{V_{2}}{V_{1}} \right)$