If `Q` amount of heat is given to a diatomic gas at constant volume to raise its temperature by `DeltaT`. Then for change of temperature how much amount of heat should be supplied at constant pressure ?

Options

A. `(2)/(3)Q`
B. `(5)/(3)Q`
C. `(7)/(5)Q`
D. `(3)/(2)Q`

Answer

To solve the problem, we need to determine the amount of heat that should be supplied to a diatomic gas at constant pressure to achieve the same change in temperature (\(\Delta T\)) as when heat \(Q\) is supplied at constant volume.

### Step-by-step Solution:

1. **Understanding the Heat at Constant Volume**:
- When heat \(Q\) is supplied to a diatomic gas at constant volume, the change in internal energy (\(\Delta U\)) is given by:
\[ Q = n C_v \Delta T \]
- Here, \(C_v\) is the specific heat at constant volume, and for a diatomic gas, \(C_v = \frac{5R}{2}\).

2. **Expressing Heat at Constant Volume**:
- Therefore, we can express \(Q\) as:
\[ Q = n \left(\frac{5R}{2}\right) \Delta T \]

3. **Understanding the Heat at Constant Pressure**:
- At constant pressure, the heat supplied (\(Q_p\)) is given by:
\[ Q_p = n C_p \Delta T \]
- For a diatomic gas, the specific heat at constant pressure \(C_p = \frac{7R}{2}\).

4. **Expressing Heat at Constant Pressure**:
- Thus, we can express \(Q_p\) as:
\[ Q_p = n \left(\frac{7R}{2}\right) \Delta T \]

5. **Relating \(Q_p\) to \(Q\)**:
- Now we need to relate \(Q_p\) to \(Q\). From our expressions, we have:
\[ Q = n \left(\frac{5R}{2}\right) \Delta T \]
- Rearranging this gives:
\[ n \Delta T = \frac{2Q}{5R} \]
- Substituting \(n \Delta T\) into the equation for \(Q_p\):
\[ Q_p = n \left(\frac{7R}{2}\right) \Delta T = \left(\frac{7R}{2}\right) \left(\frac{2Q}{5R}\right) \]
- Simplifying this:
\[ Q_p = \frac{7Q}{5} \]

6. **Final Result**:
- Therefore, the amount of heat that should be supplied at constant pressure to raise the temperature by \(\Delta T\) is:
\[ Q_p = \frac{7}{5} Q \]

### Conclusion:
The amount of heat that should be supplied at constant pressure to achieve the same temperature change is \(\frac{7}{5} Q\).