A certain amount of gas of volume $$V$$ at $$27°C$$ temperature and pressure $$2 \times 10^7 \text{ N m}^{-2}$$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $$\gamma = 1.5$$)

A gas at volume $$V$$, temperature $$27°C$$, and pressure $$P_1 = 2 \times 10^7 \text{ N/m}^2$$ first expands isothermally to volume $$2V$$, then adiabatically to volume $$4V$$. We need to find the final pressure. Given $$\gamma = 1.5$$.

In the isothermal expansion from $$V$$ to $$2V$$, we use the relation $$P_1 V_1 = P_2 V_2$$, so that $$2 \times 10^7 \times V = P_2 \times 2V$$ and thus $$P_2 = 1 \times 10^7 \text{ Pa}$$.

During the subsequent adiabatic expansion from $$2V$$ to $$4V$$, the condition $$P_2 V_2^\gamma = P_3 V_3^\gamma$$ leads to $$P_3 = P_2 \left(\frac{V_2}{V_3}\right)^\gamma = 10^7 \times \left(\frac{2V}{4V}\right)^{1.5}$$, and so $$P_3 = 10^7 \times \left(\frac{1}{2}\right)^{1.5}$$, $$P_3 = 10^7 \times \frac{1}{2\sqrt{2}}$$, $$P_3 = \frac{10^7}{2.828}$$, giving $$P_3 \approx 3.536 \times 10^6 \text{ Pa}$$.

Hence, the correct answer is Option B.