Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston of A is free to move, while that B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is

Options

A. `30 K`
B. `18 K`
C. `50 K`
D. `42 K`

Explanation

To solve the problem, we will analyze the two cylinders A and B containing an ideal diatomic gas and apply the principles of thermodynamics.

### Step-by-Step Solution:

1. **Identify the Processes**:
- Cylinder A has a free-moving piston, which means it undergoes an **isobaric process** (constant pressure).
- Cylinder B has a fixed piston, which means it undergoes an **isochoric process** (constant volume).

2. **Heat Transfer in Each Process**:
- For the isobaric process (Cylinder A), the heat added is given by:
\[ Q_A = n C_p \Delta T_A \]
where \( C_p \) is the specific heat at constant pressure, \( \Delta T_A \) is the temperature change in cylinder A, and \( n \) is the number of moles of gas.
- For the isochoric process (Cylinder B), the heat added is given by:
\[ Q_B = n C_v \Delta T_B \]
where \( C_v \) is the specific heat at constant volume, and \( \Delta T_B \) is the temperature change in cylinder B.

3. **Equate the Heat Transfers**:
- Since the same amount of heat is given to both cylinders, we have:
\[ Q_A = Q_B \]
Substituting the expressions from above:
\[ n C_p \Delta T_A = n C_v \Delta T_B \]
- The number of moles \( n \) cancels out from both sides:
\[ C_p \Delta T_A = C_v \Delta T_B \]

4. **Use the Specific Heats for Diatomic Gas**:
- For a diatomic gas, the specific heats are:
\[ C_p = \frac{7R}{2}, \quad C_v = \frac{5R}{2} \]
- Substitute these values into the equation:
\[ \frac{7R}{2} \Delta T_A = \frac{5R}{2} \Delta T_B \]

5. **Solve for \( \Delta T_B \)**:
- Rearranging the equation gives:
\[ \Delta T_B = \frac{C_p}{C_v} \Delta T_A \]
- Substituting \( C_p \) and \( C_v \):
\[ \Delta T_B = \frac{\frac{7R}{2}}{\frac{5R}{2}} \Delta T_A = \frac{7}{5} \Delta T_A \]
- Given that \( \Delta T_A = 30 \, K \):
\[ \Delta T_B = \frac{7}{5} \times 30 = 42 \, K \]

### Final Answer:
The rise in temperature of the gas in cylinder B is **42 K**.

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