The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is

We are dealing with an isobaric process, which means the pressure $$P$$ of the gas remains constant throughout the transformation. For an ideal gas undergoing such a process, two thermodynamic quantities interest us: the work done $$W$$ and the heat supplied $$Q$$. Our goal is to find the ratio $$\dfrac{W}{Q}$$ for an ideal mono-atomic gas.

First, recall the expression for the work done at constant pressure. By definition,

$$W = P \, \Delta V,$$

where $$\Delta V$$ is the change in volume. Because the gas is ideal, we can use the ideal-gas equation $$PV = nRT$$. At constant pressure we have

$$P (V_2 - V_1) = P \Delta V = nR (T_2 - T_1) = nR \, \Delta T.$$

Hence,

$$W = nR \, \Delta T.$$

Now we need the heat supplied. According to the first law of thermodynamics,

$$Q = \Delta U + W,$$

where $$\Delta U$$ is the change in internal energy. For a mono-atomic ideal gas, the molar heat capacity at constant volume is

$$C_V = \frac{3}{2} R.$$

Therefore, the internal-energy change for $$n$$ moles is

$$\Delta U = n C_V \, \Delta T = n \left( \frac{3}{2} R \right) \Delta T = \frac{3}{2} n R \, \Delta T.$$

Substituting both $$\Delta U$$ and $$W$$ into the first law, we obtain

$$Q = \frac{3}{2} n R \, \Delta T + n R \, \Delta T.$$ $$Q = \left( \frac{3}{2} + 1 \right) n R \, \Delta T.$$ $$Q = \frac{5}{2} n R \, \Delta T.$$

Now we form the required ratio:

$$\frac{W}{Q} = \frac{n R \, \Delta T}{\dfrac{5}{2} n R \, \Delta T}.$$

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

$$\frac{W}{Q} = \frac{1}{\dfrac{5}{2}} = \frac{2}{5}.$$

Hence, the correct answer is Option A.