A certain amount of gas of volume V at 27 ° C temperature and...

Given here, ${P}_{1} = 2 \times {10}^{7}N{m}^{- 2} = 2 \times {10}^{7}Pa$

Using Boyle's law,

${P}_{1}{V}_{1} = {P}_{2}{V}_{2}$

Since ${V}_{2} = 2{V}_{1}$.

Hence, ${P}_{2} = \dfrac{{P}_{1}}{2}$ (isothermal expansion)

${P}_{2} = 1 \times {10}^{7}Pa$

Now, during adiabatic expansion,

${P}_{2}{\left({V}_{2}\right)}^{\gamma} = {P}_{3}{\left(2{V}_{2}\right)}^{\gamma}$

${P}_{3} = \dfrac{1 \times {10}^{7}}{{2}^{1 . 5}} = 3 . 536 \times {10}^{6}Pa$