Is there a thermodynamic process with constant internal energy but variable enthalpy?

Question

Lately I've been trying to wrap my head around those two concepts and the difference between them.

In an ideal gas, $PV$ is said to be an energy related to the kinetic energy of molecules boucing against the boundaries of the system. However, the internal energy $U$ being defined as "the sum of all potential and kinetic energies of each molecule" seems to indicate $PV$ is already part of $U$.

As of my current understanding, I get that $PV$ is counted twice in $H$ since $H=U+PV$ (once in the $U$ as explained above and the remaining $PV$ term). I don't think this makes a lot of sense and is most likely wrong, but I cannot think of a different understanding for $PV$.

If there is a process in whcich $H$ changes but not $U$, I might finally be able to grasp those concepts once and for all.

Answer 1

The molar internal energy of an ideal gas is only a function of temperature. By kinetic theory, ideal gas particles have no volume, no interaction potentials, and no loss of energy on collisions. So the internal energy of an ideal gas is distributed solely in the translational kinetic energy of the particles. Strictly speaking, we should also make the case for a system of an ideal gas in zero external field (e.g. gravity). In any case, to be clear, the gas particles in an ideal gas have no potential energy terms. Finally also, the kinetic energy does not need to be defined by the particles "bouncing against the boundaries of the system". That part defines the pressure that the gas exerts on the walls due to them having kinetic energy and reversing their momentum vector when they bounce.

The definition of enthalpy adds the capacity of the system as a whole to do mechanical work. We need not know whether the system contains an ideal gas, a real gas, a liquid, or a solid. We also need not know whether the system contains "particles" or whether it is indivisible all the way through.

Now, consider two identical rooms filled with the same ideal gas and at the same temperature. The molar internal energies are equal $\bar{U}_1 = \bar{U}_2$. For an ideal gas, $\bar{H} \equiv \bar{U} + p\bar{V} = \bar{U} + RT$. Therefore the molar enthalpies of the gases are also the same.

Finally, we can write the difference form for the state function as $\Delta \bar{H} = \Delta \bar{U} + \Delta (p\bar{V})$. For an ideal gas, this also becomes $\Delta \bar{H} = \Delta \bar{U} + R\Delta T$.

We can now state for an ideal gas that

• An isothermal process will change neither the internal energy nor the enthalpy.

• A process that changes only the temperature will change both internal energy and enthalpy.

• An isobaric or isochoric process will change temperature (see above).

When we move to real gases, liquids, and solids, we cannot use the statement $\Delta (p\bar{V}) = R\Delta T$. What we find is that we may in this case have processes that change the internal energy without changing the enthalpy or that change the enthalpy without changing the internal energy. Why? In a nutshell, using real gases as the example case, the particles now have other modes (rotation, vibration) to store internal energy and they also have (attractive) interactions. Any energy that is put into the system to change $p, T$ now also must be distributed not only to translational kinetic energy of particles but also to other modes, and it must offset any interactive potential energies between the molecules themselves as they are pushed closer or further apart from each other. So, consider when we have a hypothetical process that would distribute energy in a real substance to the same internal degree yet reduce the capacity of the system to do external mechanical work. In this case, $\Delta U = 0$ even as $\Delta H$ decreases.

The details of real systems and the processes that could do something akin to the above are a treatise in their own right.

Answer 2

$H=U+pV$, 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 $dU=0$