n moles of an ideal gas with constant volume heat capacity $$C_V$$ undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is:

We have an isobaric (constant pressure) expansion of $$n$$ moles of an ideal gas. Let the constant-volume heat capacity of these $$n$$ moles be $$C_V$$. For an ideal gas the molar relation $$C_P=C_V+R$$ holds; therefore for the whole sample the constant-pressure heat capacity is

$$C_P^{(\text{total})}=C_V+nR.$$

During any infinitesimal change the first law of thermodynamics gives

$$\delta Q=\delta U+\delta W.$$

For a finite isobaric process we shall determine separately the work $$W$$ and the heat supplied $$Q$$, and finally take their ratio.

Work done:  At constant pressure the work is

$$W=P\Delta V.$$

The ideal-gas equation is $$PV=nRT$$. Differentiating it while keeping $$P$$ constant we get

$$P\,\mathrm dV=nR\,\mathrm dT \;\;\Longrightarrow\;\; P\Delta V=nR\Delta T.$$

Substituting this into the expression for work,

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

Heat supplied:  For a constant-pressure process the heat absorbed is

$$Q=C_P^{(\text{total})}\,\Delta T.$$

Using the total constant-pressure heat capacity written above, we have

$$Q=(C_V+nR)\,\Delta T.$$

Ratio of work done to heat supplied:

$$\frac{W}{Q}=\frac{nR\Delta T}{(C_V+nR)\Delta T}=\frac{nR}{\,C_V+nR\,}.$$

The factor $$\Delta T$$ cancels out algebraically, leaving us with the desired dimensionless ratio.

Hence, the correct answer is Option 3.