Question

The average degree of freedom per molecule for a gas is 6. The gas performs 30 J of work when it expands at constant pressure, the heat absorbed by the gas is

• A. 180 J
• B. 90 J
• C. 120 J
• D. 140 J

Answer 1

Answer: (C)

120 J

Answer 2

Solution Explanation:

1. For a gas with 6 degrees of freedom:

   $$C_V = \frac{fR}{2} = \frac{6R}{2} = 3R$$

2. The specific heat at constant pressure is:

   $$C_P = C_V + R = 3R + R = 4R$$

3. Work done at constant pressure is given by:

   $$W = nR\Delta T$$

4. Heat absorbed by the gas:

   $$Q = nC_P\Delta T = 4nR\Delta T = 4W$$

5. Given $W = 30\,\text{J}$, so:

   $$Q = 4 \times 30\,\text{J} = 120\,\text{J}$$