Two samples, A and B of a gas at the same initial temperature and pressure are compressed from volume V to V/2, A isothermally and B adiabatically. The final pressure of A will be

Options

A. greater than that of B
B. equal to that of B
C. less than of B
D. twice that of B

Explanation

To solve the problem, we need to analyze the two processes (isothermal and adiabatic) that the gas samples A and B undergo when compressed from volume V to V/2.

### Step 1: Understand the Isothermal Process for Sample A
In an isothermal process, the temperature remains constant. According to the ideal gas law, we have:
\[ PV = nRT \]
Since the temperature (T) is constant, the product \( PV \) remains constant.

### Step 2: Apply the Isothermal Condition
Let the initial pressure of sample A be \( P \) and the initial volume be \( V \). When the volume is compressed to \( V/2 \), we can write:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_1 = P \) (initial pressure)
- \( V_1 = V \) (initial volume)
- \( P_2 \) is the final pressure we want to find
- \( V_2 = V/2 \) (final volume)

Substituting the values into the equation:
\[ P \cdot V = P_2 \cdot \frac{V}{2} \]

### Step 3: Solve for Final Pressure \( P_2 \)
Rearranging the equation gives:
\[ P_2 = \frac{P \cdot V}{\frac{V}{2}} \]
\[ P_2 = 2P \]

Thus, the final pressure of sample A after isothermal compression is:
\[ P_A = 2P \]

### Step 4: Understand the Adiabatic Process for Sample B
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure and volume for an adiabatic process is given by:
\[ PV^\gamma = \text{constant} \]
Where \( \gamma \) (gamma) is the heat capacity ratio (C_p/C_v) and is greater than 1.

### Step 5: Apply the Adiabatic Condition
Let the initial pressure of sample B also be \( P \) and the initial volume be \( V \). When the volume is compressed to \( V/2 \), we can write:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Where:
- \( P_1 = P \)
- \( V_1 = V \)
- \( P_2 \) is the final pressure we want to find
- \( V_2 = V/2 \)

Substituting the values into the equation:
\[ P \cdot V^\gamma = P_2 \cdot \left(\frac{V}{2}\right)^\gamma \]

### Step 6: Solve for Final Pressure \( P_2 \)
Rearranging the equation gives:
\[ P_2 = P \cdot \frac{V^\gamma}{\left(\frac{V}{2}\right)^\gamma} \]
\[ P_2 = P \cdot \frac{V^\gamma}{\frac{V^\gamma}{2^\gamma}} \]
\[ P_2 = P \cdot 2^\gamma \]

Thus, the final pressure of sample B after adiabatic compression is:
\[ P_B = P \cdot 2^\gamma \]

### Step 7: Compare Final Pressures
Now we have:
- Final pressure of A: \( P_A = 2P \)
- Final pressure of B: \( P_B = P \cdot 2^\gamma \)

Since \( \gamma > 1 \), it follows that:
\[ 2^\gamma > 2 \]
Thus:
\[ P_B > P_A \]

### Conclusion
The final pressure of gas sample A (isothermal process) is less than that of gas sample B (adiabatic process). Therefore, the final pressure of A will be \( 2P \).

### Final Answer
The final pressure of A will be \( 2P \).