Ratio of Heat Supplied Internal Energy Change and Work Done for Diatomic Gas

To find the ratio of the heat supplied ($\Delta Q$), the change in internal energy ($\Delta U$), and the work done ($\Delta W$) when heat is supplied to a diatomic gas at constant pressure, we use the principles of thermodynamics.

1. Heat Supplied ($\Delta Q$) at Constant Pressure: For $n$ moles of a gas undergo a temperature change $\Delta T$ at constant pressure, the heat supplied is given by: $\Delta Q = n C_p \Delta T$

For a diatomic gas, the degrees of freedom are $f = 5$. The molar specific heat at constant pressure ($C_p$) is: $C_p = \left(\frac{f}{2} + 1\right) R = \left(\frac{5}{2} + 1\right) R = \frac{7}{2} R$

Thus, $\Delta Q = n \left(\frac{7}{2} R\right) \Delta T = \frac{7}{2} n R \Delta T$

2. Change in Internal Energy ($\Delta U$): The change in internal energy depends only on the change in temperature and is given by: $\Delta U = n C_v \Delta T$

The molar specific heat at constant volume ($C_v$) for a diatomic gas is: $C_v = \frac{f}{2} R = \frac{5}{2} R$

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

3. Work Done ($\Delta W$): Using the First Law of Thermodynamics ($\Delta Q = \Delta U + \Delta W$): $\Delta W = \Delta Q - \Delta U$ $\Delta W = n C_p \Delta T - n C_v \Delta T = n (C_p - C_v) \Delta T = n R \Delta T$

4. Ratio $\Delta Q : \Delta U : \Delta W$: $\Delta Q : \Delta U : \Delta W = \left(\frac{7}{2} n R \Delta T\right) : \left(\frac{5}{2} n R \Delta T\right) : \left(n R \Delta T\right)$

Dividing all terms by $n R \Delta T$: $\Delta Q : \Delta U : \Delta W = \frac{7}{2} : \frac{5}{2} : 1 = 7 : 5 : 2$

Therefore, the ratio of $\Delta Q : \Delta U : \Delta W$ is $7 : 5 : 2$.

Correct Option: D