What would be the enthalpy change for a isothermal expansion?
You have already mentioned that the change of internal energy is zero $(\Delta U=0)$ since, for an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$, and  in a closed system $n$ is constant $(\Delta n=0)$ and during an isothermal process also $T$ remains constant $(\Delta T=0)$.

Enthalpy $H$ is defined as

$$H=U+pV$$

and the ideal gas law states that

$$pV=nRT$$

Thus

$$H=U+nRT$$

Since $n$ (closed system) and $T$ (isothermal process) are constant, the product $nRT$ is constant, and therefore, according to the ideal gas law, also the product $pV$ is constant.

Furthermore, since $U$ is constant during the given process, the sum $H=U+nRT$ as well as  $H=U+pV$ has to remain unchanged (answer d).



Note that your assumption

$$\Delta H = \Delta U + W$$

(where $W=p\Delta V$ is the pressure-volume work) only applies to processes under constant pressure $p$, since enthalpy is defined as

$$H=U+pV$$

and thus

$$\begin{align}
\mathrm dH&amp;=\mathrm dU+\mathrm d(pV)\\
&amp;=\mathrm dU+V\mathrm dp+p\mathrm dV
\end{align}$$

which simplifies to

$$\mathrm dH=\mathrm dU+p\mathrm dV$$

at constant pressure $(\mathrm dp=0)$.

However, the pressure does not remain constant during the process given in the question.