The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $$\frac{x}{10}$$. Then the value of $$x$$ (round off to the nearest integer) is _________.
(Given R = 8.31 J $$mol^{-1}$$ $$K^{-1}$$)

We have an ideal diatomic gas that can translate in three directions and rotate about two perpendicular axes, but it cannot oscillate. Hence the total number of active degrees of freedom is five. For an ideal gas the molar heat capacities are related to the degrees of freedom by the formula

$$C_V = \frac{f}{2}\,R \qquad\text{and}\qquad C_P = C_V + R,$$

where $$f$$ is the number of degrees of freedom and $$R$$ is the universal gas constant. Substituting $$f = 5$$ we obtain

$$C_V = \frac{5}{2}R,$$

and therefore

$$C_P = \frac{5}{2}R + R = \frac{7}{2}R.$$

The number of moles present is $$n = 3.00\,\text{mol}$$ and the rise in temperature is $$\Delta T = 40.0^{\circ}\text{C} = 40.0\,\text{K}.$$

The process occurs at constant pressure. First, let us evaluate the change in internal energy. For an ideal gas

$$\Delta U = n\,C_V\,\Delta T.$$

Substituting the known values gives

$$\Delta U = 3.00 \left(\frac{5}{2}R\right)(40.0) = 3.00 \times \frac{5}{2} \times 40.0\,R = 3.00 \times 100\,R = 300\,R.$$

Next, we calculate the work done by the gas at constant pressure. The formula for the work in an isobaric (constant-pressure) process is

$$W = n\,R\,\Delta T.$$

Substituting once more we get

$$W = 3.00\,R\,(40.0) = 120\,R.$$

Now we form the ratio of the change in internal energy to the work done:

$$\frac{\Delta U}{W} = \frac{300\,R}{120\,R} = \frac{300}{120} = 2.5 = \frac{25}{10}.$$

The problem states that this ratio equals $$\dfrac{x}{10},$$ so by direct comparison we have $$x = 25.$$

So, the answer is $$25$$.