1. Given data

• Initial pressure: $P_1 = 2 \times 10^7\,\text{Pa}$
• Initial temperature: $27^\circ\text{C} = 300\,\text{K}$
• Initial volume: $V_1 = V$
• First process: isothermal expansion until volume doubles
• Second process: adiabatic expansion until volume doubles again
• Ratio of specific heats: $\gamma = 1.5 = \frac{3}{2}$

We need the final pressure.

2. First expansion: Isothermal

For an isothermal process of an ideal gas,

$P_1V_1 = P_2V_2$

After first expansion,

$V_2 = 2V$

So,

$P_2 = \frac{P_1V_1}{V_2} = \frac{(2\times 10^7)(V)}{2V} = 10^7\,\text{Pa}$

Thus, after isothermal expansion:

$P_2 = 1.0\times 10^7\,\text{Pa}, \quad V_2 = 2V$

3. Second expansion: Adiabatic

For an adiabatic process,

$PV^\gamma = \text{constant}$

Here the gas expands from:

• Initial state for adiabatic step: $(P_2, V_2)$
• Final state: $(P_3, V_3)$

The volume is redoubled, so

$V_3 = 2V_2 = 4V$

Using adiabatic relation:

$P_2V_2^\gamma = P_3V_3^\gamma$

Therefore,

$P_3 = P_2\left(\frac{V_2}{V_3}\right)^\gamma$

Substitute values:

$P_3 = 10^7\left(\frac{2V}{4V}\right)^{1.5}$

$P_3 = 10^7\left(\frac{1}{2}\right)^{3/2}$

Now,

$\left(\frac{1}{2}\right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}} \approx \frac{1}{2.828} \approx 0.3536$

Hence,

$P_3 = 10^7 \times 0.3536 = 3.536 \times 10^6\,\text{Pa}$

4. Final answer

$\boxed{P_3 = 3.536 \times 10^6\,\text{Pa}}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.