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

Options

A. `25J`
B. `50J`
C. `75J`
D. `100J`

Answer

To find the heat absorbed by the gas during its expansion at constant pressure, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat absorbed by the system,
- \(W\) is the work done by the system.

Since the gas is expanding at constant pressure, we can also relate the change in internal energy to the specific heat capacities \(C_V\) and \(C_P\).

### Step 1: Calculate the degrees of freedom and specific heat capacities

Given that the average degree of freedom \(F\) is 6, we can find \(C_V\) and \(C_P\):

\[ C_V = \frac{F}{2}R = \frac{6}{2}R = 3R \]

\[ C_P = C_V + R = 3R + R = 4R \]

### Step 2: Calculate the change in internal energy (\(\Delta U\))

At constant pressure, the change in internal energy can be expressed as:

\[ \Delta U = n C_V \Delta T \]

However, we need to relate this to the work done. Since we are not given the number of moles \(n\) or the change in temperature \(\Delta T\), we will use the relationship involving work done.

### Step 3: Use the first law of thermodynamics

From the first law of thermodynamics:

\[ \Delta U = Q - W \]

Rearranging gives:

\[ Q = \Delta U + W \]

### Step 4: Relate \(\Delta U\) to \(W\)

For an ideal gas, at constant pressure, the work done \(W\) is given as:

\[ W = P \Delta V \]

Given that \(W = 25J\), we can substitute this into the equation:

\[ Q = \Delta U + 25J \]

### Step 5: Calculate the change in internal energy (\(\Delta U\))

Since we are at constant pressure, we can also relate the change in internal energy to the specific heat capacity at constant volume:

\[ \Delta U = n C_V \Delta T \]

However, we need to find \(Q\) directly. We can use the relationship between \(C_P\) and \(C_V\):

\[ \Delta U = n C_V \Delta T = n (C_P - R) \Delta T \]

But we still need \(n\) and \(\Delta T\).

### Step 6: Use the heat capacity relation

Since we are not given \(n\) or \(\Delta T\), we can assume that the heat absorbed \(Q\) can be expressed in terms of \(W\) and the specific heat capacities:

\[ Q = W + n C_V \Delta T \]

### Step 7: Substitute known values

Since we don't have enough information to find \(n\) and \(\Delta T\), we can assume that the heat absorbed \(Q\) can be approximated as:

\[ Q = W + (W/R) \cdot (C_P - C_V) \]

Given \(W = 25J\), we can simplify this further.

### Final Calculation

However, without specific values for \(R\) or \(n\), we can conclude that the heat absorbed \(Q\) will be greater than \(25J\) due to the work done.

### Conclusion

The heat absorbed by the gas is:

\[ Q = 25J + \Delta U \]

Where \(\Delta U\) will depend on the specific conditions of the gas.