When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by `Delta T`. The heat required to produce the same change in temperature, at a constant pressure is , where xQ is ___________ .

Explanation

To solve the problem, we need to find the heat required to produce the same change in temperature (ΔT) at constant pressure (Qp) when we know the heat supplied at constant volume (Q).

### Step-by-Step Solution:

1. **Understanding Heat at Constant Volume (Q)**:
- When heat \( Q \) is supplied to a diatomic gas at constant volume, the change in internal energy \( \Delta U \) is equal to \( Q \).
- The formula for the change in internal energy is given by:
\[ Q = \Delta U = n C_v \Delta T \]
- Here, \( n \) is the number of moles, \( C_v \) is the specific heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

2. **Heat at Constant Pressure (Qp)**:
- The heat required at constant pressure can be expressed as:
\[ Q_p = n C_p \Delta T \]
- Here, \( C_p \) is the specific heat capacity at constant pressure.

3. **Relating \( C_p \) and \( C_v \)**:
- For diatomic gases, the specific heat capacities are related as follows:
\[ C_v = \frac{5R}{2} \quad \text{and} \quad C_p = \frac{7R}{2} \]
- We can express \( C_p \) in terms of \( C_v \):
\[ C_p = C_v + R \]
- Substituting the values, we get:
\[ C_p = \frac{5R}{2} + R = \frac{7R}{2} \]

4. **Substituting \( C_p \) into \( Q_p \)**:
- Now substituting \( C_p \) into the equation for \( Q_p \):
\[ Q_p = n C_p \Delta T = n \left(\frac{7R}{2}\right) \Delta T \]
- We can express \( Q_p \) in terms of \( Q \):
\[ Q_p = n \left(\frac{7}{5} C_v\right) \Delta T \]
- Since we have \( Q = n C_v \Delta T \), we can substitute \( Q \) into the equation:
\[ Q_p = \frac{7}{5} Q \]

5. **Final Answer**:
- Thus, the heat required to produce the same change in temperature at constant pressure is:
\[ Q_p = \frac{7}{5} Q \]
- Therefore, \( xQ = \frac{7}{5} \).