A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

We have a di-atomic ideal gas whose molecules are rigid. For such a gas the only active degrees of freedom are three translational and two rotational, giving a total of five. Using the equipartition theorem, the molar heat capacities are therefore

$$C_V = \frac{5}{2}R \qquad\text{and}\qquad C_P = C_V + R = \frac{7}{2}R.$$

The gas expands at constant pressure and does work $$W = 10\ \text{J}.$$ According to thermodynamics the work done by an ideal gas in a constant-pressure process is related to the temperature change by the expression

$$W = P\Delta V = nR\Delta T.$$

From this relation we directly obtain

$$nR\Delta T = 10\ \text{J}.$$

Next, we recall the first law of thermodynamics, which states

$$Q = \Delta U + W,$$

where $$Q$$ is the heat absorbed and $$\Delta U$$ is the change in internal energy. For an ideal gas the change in internal energy at any temperature change is

$$\Delta U = nC_V\Delta T = n\left(\frac{5}{2}R\right)\Delta T.$$

Substituting $$\Delta U$$ and $$W$$ into the first-law expression gives

$$Q = n\left(\frac{5}{2}R\right)\Delta T + nR\Delta T.$$ Combining like terms,

$$Q = n\left(\frac{5}{2}R + R\right)\Delta T = n\left(\frac{7}{2}R\right)\Delta T.$$

We already know that $$nR\Delta T = 10\ \text{J},$$ so we replace $$nR\Delta T$$ in the above equation:

$$Q = \frac{7}{2}\,(nR\Delta T) = \frac{7}{2} \times 10\ \text{J} = 35\ \text{J}.$$

Thus the heat energy absorbed by the gas in this process is $$35\ \text{J}.$$

Hence, the correct answer is Option B.