An ideal gas of volume V and pressure P expands isothermally to volume 16 V and then compressed adiabatically to volume V . The final pressure of gas is [`gamma = 1.5]`

Options

A. P
B. 3P
C. `4 P `
D. `6 P`

Answer

To solve the problem step by step, we will analyze the two processes the ideal gas undergoes: isothermal expansion and adiabatic compression.

### Step 1: Isothermal Expansion
The gas expands isothermally from volume \( V \) to volume \( 16V \). In an isothermal process, the product of pressure and volume remains constant, which can be expressed as:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 = P \) (initial pressure)
- \( V_1 = V \) (initial volume)
- \( P_2 \) is the pressure after expansion
- \( V_2 = 16V \) (final volume after expansion)

Substituting the known values into the equation:

\[ P \cdot V = P_2 \cdot 16V \]

### Step 2: Solve for \( P_2 \)
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):

\[ P = P_2 \cdot 16 \]

Now, solving for \( P_2 \):

\[ P_2 = \frac{P}{16} \]

### Step 3: Adiabatic Compression
Next, the gas is compressed adiabatically back to volume \( V \). The relationship for an adiabatic process is given by:

\[ P_2 V_2^\gamma = P_3 V_3^\gamma \]

Where:
- \( P_3 \) is the final pressure after adiabatic compression
- \( V_3 = V \) (final volume after compression)
- \( \gamma = 1.5 \)

Substituting the known values into the equation:

\[ \left(\frac{P}{16}\right) \cdot (16V)^{1.5} = P_3 \cdot V^{1.5} \]

### Step 4: Simplify the Equation
Now we can simplify the left side:

\[ \frac{P}{16} \cdot (16^{1.5} \cdot V^{1.5}) = P_3 \cdot V^{1.5} \]

This simplifies to:

\[ \frac{P \cdot 16^{1.5}}{16} = P_3 \]

### Step 5: Calculate \( 16^{1.5} \)
Calculating \( 16^{1.5} \):

\[ 16^{1.5} = (16^{3/2}) = (4^2)^{3/2} = 4^3 = 64 \]

### Step 6: Substitute Back to Find \( P_3 \)
Now substituting back into the equation for \( P_3 \):

\[ P_3 = \frac{P \cdot 64}{16} \]

### Step 7: Final Calculation
Now simplifying:

\[ P_3 = 4P \]

### Final Answer
The final pressure of the gas after the adiabatic compression is:

\[ \boxed{4P} \]