When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by $$\Delta T$$. The heat required to produce the same change in temperature, at a constant pressure is:

First, recall the general expression for the heat absorbed by an ideal gas when its temperature changes by $$\Delta T$$ in any quasi-static process:

$$Q = n\,C\,\Delta T,$$

where $$n$$ is the number of moles and $$C$$ is the molar specific heat for the process in question.

For an ideal gas, there are two important molar specific heats:

• At constant volume: $$C_V$$ (no work done, only internal energy changes).
• At constant pressure: $$C_P$$ (gas also does $$P\Delta V$$ work in addition to changing internal energy).

For a diatomic gas whose molecules are rigid (so it possesses three translational and two rotational degrees of freedom, but no vibrational energy), the equipartition theorem gives:

$$C_V = \dfrac{f}{2}\,R,$$

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

$$C_V = \dfrac{5}{2}R.$$

The well-known thermodynamic relation between $$C_P$$ and $$C_V$$ for any ideal gas is

$$C_P = C_V + R.$$

Substituting $$C_V = \dfrac{5}{2}R$$ into this relation, we obtain

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

Now, when the gas is heated at constant volume, the problem statement tells us that the supplied heat is $$Q$$ and the temperature rise is $$\Delta T$$. Using the formula for constant-volume heating, we write

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

Substituting $$C_V = \dfrac{5}{2}R$$ gives

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

Next, we want the heat $$Q_P$$ needed to produce the same temperature rise $$\Delta T$$ but at constant pressure. For that process we use

$$Q_P = n\,C_P\,\Delta T.$$

Substituting $$C_P = \dfrac{7}{2}R$$ gives

$$Q_P = n\left(\dfrac{7}{2}R\right)\Delta T.$$

To find the desired ratio, divide $$Q_P$$ by $$Q$$:

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

The factors $$n, R,$$ and $$\Delta T$$ cancel out, leaving

$$\dfrac{Q_P}{Q} = \dfrac{7}{5}.$$

Therefore,

$$Q_P = \dfrac{7}{5}\,Q.$$

Hence, the correct answer is Option B.