1. Identify the process

The gas is suddenly compressed, so there is no time for heat exchange with surroundings. Hence, the process is adiabatic.

For an adiabatic process of an ideal gas,

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

where $\gamma = \dfrac{C_p}{C_v}$.

2. Write initial and final states

Initial state:

$P_1 = P_0, \quad V_1 = V_0$

Final state:

$P_2 = ?, \quad V_2 = \frac{V_0}{4}$

Using the adiabatic relation,

$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$

Substitute the values:

$P_0 V_0^{\gamma} = P_2 \left(\frac{V_0}{4}\right)^{\gamma}$

3. Solve for final pressure

$P_2 = P_0 \frac{V_0^{\gamma}}{\left(\frac{V_0}{4}\right)^{\gamma}}$

$P_2 = P_0 \frac{V_0^{\gamma}}{\frac{V_0^{\gamma}}{4^{\gamma}}}$

$P_2 = P_0 \cdot 4^{\gamma}$

So,

$P_2 = P_0 4^{\gamma}$

4. Check options

• A: $P_0 4^{1/\gamma}$ ❌
• B: $P_0$ ❌
• C: $4P_0$ ❌
• D: $P_0 4^{\gamma}$ ✅

5. Final answer

The final pressure is

$\boxed{P_0 4^{\gamma}}$

So the correct option is D.