For diatomic gas, find the ratio Q : ∆U :W for isobaric process.

The correct option is 7 : 5 : 2.

Let us find the ratio of heat supplied ($Q$), change in internal energy ($\Delta U$), and work done ($W$) for a diatomic gas undergoing an isobaric process (constant pressure).

For an isobaric process, the heat supplied to the gas is given by:
$Q=n{C}_p\Delta T$
where $n$ is the number of moles, $C_p$ is the molar heat capacity at constant pressure, and $\Delta T$ is the change in temperature.

The change in internal energy is given by:
$\Delta U=n{C}_v\Delta T$
where $C_v$ is the molar heat capacity at constant volume.

According to the first law of thermodynamics, the work done during the process is:
$W=Q-\Delta U=n(C_p-C_v)\Delta T=nR\Delta T$
where $R$ is the universal gas constant.

For a diatomic gas, the degrees of freedom ($f$) at moderate temperatures is 5. Thus, the molar heat capacities are:
$C_v=\frac{5}{2}R$
and
$C_p=C_v+R=\frac{7}{2}R$

Substituting these values, we get:
$Q=n\left(\frac{7}{2}R\right)\Delta T$

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

$W=nR\Delta T$

Now, taking the ratio $Q:\Delta U:W$:
$Q:\Delta U:W=\frac{7}{2}nR\Delta T:\frac{5}{2}nR\Delta T:nR\Delta T$
Dividing each term by $nR\Delta T$ gives:
$Q:\Delta U:W=\frac{7}{2}:\frac{5}{2}:1$
Multiplying by 2 to clear the denominators yields the final ratio:
$Q:\Delta U:W=7:5:2$