An ideal gas undergoes isothermal expansion at constant pressure. During the process:

We have an ideal gas that expands isothermally at constant pressure. The task is to examine how its enthalpy $$H$$ and entropy $$S$$ change during the process.

First, recall the relation between enthalpy and temperature for an ideal gas. For any ideal gas, enthalpy is a function of temperature only, and it is given by the expression

$$H = n\,C_p\,T,$$

where $$n$$ is the number of moles and $$C_p$$ is the molar heat capacity at constant pressure. Because the process is isothermal, we have

$$T_2 = T_1 \; \Longrightarrow \; \Delta T = T_2-T_1 = 0.$$

Hence the change in enthalpy is

$$\Delta H = n\,C_p\,(T_2-T_1)=n\,C_p\,(0)=0.$$

So, during the isothermal expansion the enthalpy remains constant.

Now we turn to the entropy change. The definition of entropy change for any reversible path is

$$\Delta S = \int_{1}^{2} \frac{\delta Q_{\text{rev}}}{T}.$$

For an isothermal process of an ideal gas, the change in internal energy is zero because $$U= n\,C_vT$$ and $$\Delta T=0$$. From the first law of thermodynamics, $$\delta Q_{\text{rev}} = \delta W_{\text{rev}}$$ under these conditions because $$\delta W_{\text{rev}} = P\,dV$$ and $$\Delta U = 0$$. Substituting $$\delta Q_{\text{rev}} = P\,dV$$ into the entropy integral gives

$$\Delta S = \int_{V_1}^{V_2} \frac{P\,dV}{T}.$$

Using the ideal-gas equation $$P\,V = n\,R\,T\; \Longrightarrow \; P = \dfrac{nRT}{V},$$ we substitute for $$P$$:

$$\Delta S = \int_{V_1}^{V_2} \frac{nRT}{V\,T}\,dV = \int_{V_1}^{V_2} \frac{nR}{V}\,dV = nR \int_{V_1}^{V_2} \frac{dV}{V} = nR \,\ln\!\left(\frac{V_2}{V_1}\right).$$

Because the gas expands, $$V_2 > V_1$$, which implies $$\dfrac{V_2}{V_1} > 1$$ and therefore

$$\ln\!\left(\frac{V_2}{V_1}\right) > 0.$$

Thus,

$$\Delta S = nR \,\ln\!\left(\frac{V_2}{V_1}\right) > 0,$$

so the entropy increases.

Combining the two results, we see that enthalpy stays unchanged while entropy rises. This description matches Option A.

Hence, the correct answer is Option A.