Final answer:
In isothermal expansion of an ideal gas, internal energy does not change (ΔEint = 0) because the temperature is constant. The heat absorbed (Q) is equal to the work done (W). Contrary to the question's statement, for an ideal gas, the enthalpy change (ΔH) is also zero during isothermal expansion.
Step-by-step explanation:
During an isothermal expansion of an ideal gas, the temperature of the system remains constant. As the internal energy (Eint) of an ideal gas is only a function of temperature, the change in internal energy is zero (ΔEint = 0) for such a process. According to the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system (Q) minus the work done by the system (W), we deduce that the heat absorbed is exactly equal to the work done by the gas during isothermal expansion; hence, Q = W. The heat capacity at constant pressure for an ideal gas is given by Cp = Cv + R, where Cv is the heat capacity at constant volume, and R is the gas constant.
Contrary to the statement in the original question, during an isothermal expansion, the enthalpy (H) of an ideal gas, which depends on both internal energy and pressure-volume work (H = U + pV), may change since pressure and volume are changing. However, in the case of an ideal gas undergoing isothermal expansion, if we assume that the number of moles and the gas constant remain constant, the enthalpy change (ΔH) can be shown to be zero as well because ΔH = Δ(U + pV) = ΔU + Δ(pV), where ΔU is zero and Δ(pV) is zero for an isothermal process (from the ideal gas law pV = nRT).