Explanation

(b) In isothermal process, temperature of the gas remains constant, so the gas obeys Boyle’s law.
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \frac{p_{2}}{p_{1}} = \frac{V}{V_{1}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \frac{2p}{p} = \frac{V}{V_{1}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \frac{V}{V_{1}} = 2$ …(i)
Now, the gas is expanded adiabatically, so
$pV^{\gamma} = constant$
$\frac{p_{1}}{p_{2}} = \left( \frac{V_{2}}{V_{1}} \right)^{\gamma}$
$\Rightarrow \ \ \frac{2p}{0.75\ p} = \left( \frac{2}{1} \right)^{\gamma}$ (since volume is restored)
$\Rightarrow \log{\left( \frac{8}{3} \right) = \gamma\log 2}$
$\Rightarrow \ \ \ \log{8 - \log{3 = \gamma\log 2}}$
$\therefore\ \ \ \ \ \ \ \ \gamma = 1.41$