A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming `gamma` to be 1.5, the new pressure of the gas is

Options

A. 2P
B. P
C. 4P
D. P/2

Explanation

To solve the problem step by step, we will follow the processes of isothermal expansion and adiabatic compression of the ideal gas.

### Step 1: Isothermal Expansion
We start with an ideal gas at pressure \( P \) and volume \( V \). The gas is expanded isothermally to four times its original volume.

Using the ideal gas law for isothermal processes, we know that:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_1 = P \) (initial pressure)
- \( V_1 = V \) (initial volume)
- \( V_2 = 4V \) (final volume after expansion)

Substituting the known values into the equation:
\[ P \cdot V = P_2 \cdot (4V) \]

### Step 2: Solve for \( P_2 \)
Now, we can solve for \( P_2 \):
\[ P_2 = \frac{P \cdot V}{4V} = \frac{P}{4} \]

### Step 3: Adiabatic Compression
Next, the gas is compressed adiabatically back to its original volume \( V \). For an adiabatic process, we use the relation:
\[ P_2 V_2^\gamma = P_f V_f^\gamma \]
Where:
- \( P_2 = \frac{P}{4} \) (pressure after isothermal expansion)
- \( V_2 = 4V \) (volume after isothermal expansion)
- \( V_f = V \) (final volume after adiabatic compression)
- \( \gamma = 1.5 \)

Substituting these values into the equation:
\[ \frac{P}{4} \cdot (4V)^{\gamma} = P_f \cdot V^{\gamma} \]

### Step 4: Simplify the Equation
Now we simplify the equation:
\[ \frac{P}{4} \cdot (4^\gamma \cdot V^\gamma) = P_f \cdot V^\gamma \]

Cancelling \( V^\gamma \) from both sides:
\[ \frac{P}{4} \cdot 4^\gamma = P_f \]

### Step 5: Substitute \( \gamma \)
Now substituting \( \gamma = 1.5 \):
\[ 4^{1.5} = (2^2)^{1.5} = 2^{3} = 8 \]
Thus, we have:
\[ P_f = \frac{P}{4} \cdot 8 = 2P \]

### Final Answer
The new pressure of the gas after the adiabatic compression is:
\[ \boxed{2P} \]