What would be the enthalpy change for a isothermal expansion?

Question

I was doing a question on thermodynamics and came across this one stating

During isothermal expansion of gas of an ideal gas :

a) Internal energy decreases

b) Enthalpy increases

c) Enthalpy reduces to zero

d) Enthalpy remains unchanged

I cannot agree to any of these as $\Delta H = \Delta U + W$, $\Delta U = 0$ as $\Delta T = 0$ but $W \neq 0$ so $\Delta H = W$ which is $\neq 0$ and as $W \lt 0$ the enthalpy should decrease but how much how do I know and to add to the confusion the answer given is enthalpy remains unchanged in the light of the equation $\Delta H = \Delta U +\Delta n_\mathrm gRT$ and $\Delta U$ and $\Delta n = 0$ so $\Delta H = 0$ is it right.

Answer 1

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).

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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

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

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.