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: Understand the Initial Conditions
We have an ideal gas with:
- Initial volume, \( V_1 = V \)
- Initial pressure, \( P_1 = P \)
- Initial temperature, \( T \) (constant during isothermal expansion)

### Step 2: Isothermal Expansion
The gas expands isothermally to a new volume:
- Final volume after expansion, \( V_2 = 16V \)

According to the ideal gas law, for an isothermal process, the pressure and volume are inversely related:
\[ P_1 V_1 = P_2 V_2 \]
Where \( P_2 \) is the pressure after expansion.

Substituting the known values:
\[ P \cdot V = P_2 \cdot 16V \]

We can simplify this:
\[ P_2 = \frac{P}{16} \]

### Step 3: Adiabatic Compression
Now, the gas is compressed adiabatically back to volume \( V \). For an adiabatic process, we use the relation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Where:
- \( \gamma = 1.5 \)
- \( P_1 = P_2 = \frac{P}{16} \) (pressure after isothermal expansion)
- \( V_1 = 16V \) (volume after isothermal expansion)
- \( V_2 = V \) (final volume after adiabatic compression)

Substituting the values:
\[ \left(\frac{P}{16}\right) (16V)^{1.5} = P_f V^{1.5} \]

### Step 4: Simplifying the Equation
Calculate \( (16V)^{1.5} \):
\[ (16V)^{1.5} = 16^{1.5} V^{1.5} = 64 V^{1.5} \]

Now substituting this back into the equation:
\[ \left(\frac{P}{16}\right) (64 V^{1.5}) = P_f V^{1.5} \]

### Step 5: Canceling \( V^{1.5} \)
We can cancel \( V^{1.5} \) from both sides:
\[ \frac{P \cdot 64}{16} = P_f \]

### Step 6: Final Calculation
Now simplifying the left side:
\[ \frac{64P}{16} = 4P \]
Thus, the final pressure \( P_f \) is:
\[ P_f = 4P \]

### Conclusion
The final pressure of the gas after the adiabatic compression is \( 4P \).

---