Final answer:
For the reversible isothermal expansion of an ideal gas, where the temperature remains constant, the change in internal energy is zero. Since enthalpy change for an ideal gas under constant temperature is defined as the sum of this change in internal energy and the work done (which equals the heat added for isothermal processes), the enthalpy change (ΔH) is also zero. The correct answer is option D.
Step-by-step explanation:
The question regards the thermodynamic concept of enthalpy change (ΔH) during the isothermal expansion of an ideal gas. For a reversible, isothermal expansion of an ideal gas, the change in internal energy (ΔU) is zero because the temperature is constant, and for an ideal gas, internal energy is a function of temperature only.
According to the first law of thermodynamics,
ΔU = Q - W,
where Q is the heat added to the system and W is the work done by the system. In an isothermal process for an ideal gas, ΔU = 0, so Q = W. Hence, enthalpy change ΔH, which is defined as ΔH = ΔU + PΔV for constant pressure, becomes simply ΔH = ΔU + W since PΔV is equal to the work done in the case of reversible expansion. Given that ΔU is zero, ΔH also results in zero for an isothermal process of an ideal gas.
Therefore, the enthalpy change for the scenario described in the question is D. Zero.