\(P _{1} V _{1}= P _{2} V _{2}\)
Since \(V_{2}=2 V_{1}\) Hence \(P_{2}=P_{1} / 2\) (isothermal expansion)
\(P _{2}=1 \times 10^{7} Pa\)
\(P _{2}\left( V _{2}\right)^{\gamma}= P _{3}\left(2 V _{2}\right)^{\gamma}\)
\(P _{3}=\frac{1 \times 10^{7}}{2^{1.5}}=3.536 \times 10^{6}\)