$H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$.

If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.

Replacing that in $dH$ you get:

$$dH=0$$ Thus you cannot have a process for an ideal gas in which the internal energy does not change but enthalpy does. But for equations of state other tha an ideal gas it must certainly be possible.

$H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$.

If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.

Replacing that in $dH$ you get:

$$dH=0$$ Thus you cannot have a process for an ideal gas in which the internal energy does not change but enthalpy does. But for equations of state other tha an ideal gas it must certainly be possible.

For instance, if you have $P(V-b)=nRT$, you get

$$dH=b dP$$ and $dU=0$

$H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$.

If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.

Replacing that in $dH$ you get:

$$dH=-2PdV$$ Thus an isothermal expansion results in a change in enthalpy but keeps $U$ unchanged.

$H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$.

If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.

Replacing that in $dH$ you get:

$$dH=0$$ Thus you cannot have a process for an ideal gas in which the internal energy does not change but enthalpy does. But for equations of state other tha an ideal gas it must certainly be possible.