The average degrees of freedom per molecule for a gas are 6. The gas performs `25 J` of work when it expands at constant pressure. The heat absorbed by gas is

Options

A. 75 J
B. 100 J
C. 150 J
D. 125 J

Answer

To solve the problem, we need to find the heat absorbed by the gas when it expands at constant pressure. We can use the relationship between work done, heat absorbed, and the specific heat capacity of the gas.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Degrees of freedom (F) = 6
- Work done (W) = 25 J

2. **Relate Degrees of Freedom to Specific Heat Capacity:**
The specific heat capacity at constant pressure (C_p) can be calculated using the formula:
\[ C_p = \frac{F}{2} + 1 \cdot R \]
Substituting the given degrees of freedom:
\[ C_p = \frac{6}{2} + 1 \cdot R = 3R + R = 4R \]

3. **Use the Work Done Equation:**
The work done during the expansion of the gas can be expressed as:
\[ W = nR\Delta T \]
Where \(n\) is the number of moles and \(\Delta T\) is the change in temperature.

4. **Express Heat Absorbed at Constant Pressure:**
The heat absorbed by the gas at constant pressure is given by:
\[ Q_p = nC_p\Delta T \]
Substituting \(C_p\) from step 2:
\[ Q_p = n(4R)\Delta T = 4nR\Delta T \]

5. **Relate Work Done to Heat Absorbed:**
From the work done equation, we have:
\[ nR\Delta T = W = 25 \, \text{J} \]
Therefore, we can substitute \(nR\Delta T\) in the heat absorbed equation:
\[ Q_p = 4(nR\Delta T) = 4 \times 25 = 100 \, \text{J} \]

6. **Conclusion:**
The heat absorbed by the gas is:
\[ Q_p = 100 \, \text{J} \]

### Final Answer:
The heat absorbed by the gas is **100 J**.