Two sample `A` and `B` of a gas initially at the same pressure and temperature are compressed from volume `V` to `V//2` (A isothermally and `B` adiabatically). The final pressure of `A` is

Options

A. `P_A gt P_B`
B. `P_A = P_B`
C. `P_A gt P_B`
D. `P_A = 2P_B`

Answer

To solve the problem step by step, we will analyze the two processes: isothermal compression for sample A and adiabatic compression for sample B.

### 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. Therefore, we can write:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 \) is the initial pressure,
- \( V_1 \) is the initial volume,
- \( P_2 \) is the final pressure,
- \( V_2 \) is the final volume.

### Step 2: Assign Initial Conditions
Let’s assume:
- Initial pressure \( P_1 = P \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = \frac{V}{2} \)

### Step 3: Calculate Final Pressure for Sample A
Using the equation from Step 1:

\[ P_1 V_1 = P_2 V_2 \]

Substituting the known values:

\[ P \cdot V = P_2 \cdot \frac{V}{2} \]

Now, we can solve for \( P_2 \):

\[ P_2 = \frac{P \cdot V}{\frac{V}{2}} \]

This simplifies to:

\[ P_2 = 2P \]

Thus, the final pressure for sample A (isothermal process) is:

\[ P_A = 2P \]

### Step 4: Understand the Adiabatic Process for Sample B
In an adiabatic process, no heat is exchanged with the surroundings. The relationship between pressure and volume for an adiabatic process is given by:

\[ PV^\gamma = \text{constant} \]

Where \( \gamma \) is the adiabatic index (ratio of specific heats).

### Step 5: Assign Initial Conditions for Sample B
Let’s assume:
- Initial pressure \( P_1 = P \)
- Initial volume \( V_1 = V \)
- Final volume \( V_2 = \frac{V}{2} \)

### Step 6: Calculate Final Pressure for Sample B
Using the equation for the adiabatic process:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Substituting the known values:

\[ P \cdot V^\gamma = P_2 \cdot \left(\frac{V}{2}\right)^\gamma \]

Now, we can solve for \( P_2 \):

\[ P_2 = P \cdot \frac{V^\gamma}{\left(\frac{V}{2}\right)^\gamma} \]

This simplifies to:

\[ P_2 = P \cdot \frac{V^\gamma}{\frac{V^\gamma}{2^\gamma}} \]

Which further simplifies to:

\[ P_2 = P \cdot 2^\gamma \]

Thus, the final pressure for sample B (adiabatic process) is:

\[ P_B = P \cdot 2^\gamma \]

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

Since \( \gamma > 1 \), it follows that \( 2^\gamma > 2 \). Therefore:

\[ P_B > P_A \]

### Conclusion
The final pressure of sample A (isothermal) is \( 2P \), and the final pressure of sample B (adiabatic) is \( P \cdot 2^\gamma \), where \( P_B > P_A \).

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