# Formula of enthalpy in vacuum?

Enthalpy can be expressed as $$H = U+PV$$.

From what I have read on the internet, $$PV$$ is the energy that was required to create space for the system in the environment. Like one famous example I have come across is that if a magician has to make a rabbit suddenly appear out of nowhere, then he not only has to provide the rabbit's internal energy, $$U$$, but also an additional energy $$PV$$ so that the rabbit can displace the air around it and make space for itself.

Suppose there is a system that exists in vacuum. Then it did not need to displace any air for it to make space for itself. So can we say that in this case $$H = U$$?

• If it's a true vacuum (zero mass), then wouldn't P=0 and U=0 and thus H=0? Nov 25, 2022 at 18:09
• @BobD I meant that the system is in vacuum Nov 26, 2022 at 7:08

In vacuum, the pressure vanishes ($$P=0$$), and hence $$H = U + P V = U$$.

• This helps a lot! Nov 28, 2022 at 11:19
• But I still have one doubt. Suppose I have a small balloon which has less air so that it does not burst in vacuum. This balloon was blown under normal circumstances,i.e, under atmospheric pressure, like in the playground or at my school. So its enthalpy is PV +U. Now Suppose I shift the balloon to vacuum. The balloon will expand, but will the energy or work done by the balloon be equal to PV? Or will it be more/ less. Is it possible for the internal energy to not change. But internal energy will change because of work being done right? basically what happens to the PV term in this case Nov 28, 2022 at 11:27
• Also for pressure to be 0, my balloon will have infinite volume which is impossible. So in real life scenario, pressure will not be 0 right? Nov 28, 2022 at 14:12
• @Obinna I believe the comment by Chemomechanics answers your follow-up questions Nov 28, 2022 at 18:53
• @Chemomechanics I understand that the internal energy will be the same, but why is enthalpy also unchanged? Aren't we getting rid of the term PV here? Nov 28, 2022 at 22:54

(The answer page isn't really conducive to back-and-forth discussion, so I'll move my comments to this answer. It seems from the discussion that the intended question may not be what the system enthalpy $$H$$ is in a vacuum but what happens to enthalpy—$$\Delta H$$—when a system is moved to a vacuum.)

Let's get some enthalpy expressions that may help in examining different scenarios. By definition,

$$H\equiv U+PV$$

for internal energy $$U$$, pressure $$P$$, and volume $$V$$. Then, by expanding $$dH$$ as

$$dH=\left(\frac{\partial H}{\partial X}\right)_YdX+\left(\frac{\partial H}{\partial Y}\right)_XdY$$

for a closed system that can undergo pressure–volume work, we can derive

$$dH=T\,dS+V\,dP;$$

$$dH=C_P\,dT+V(1-\alpha T)\,dP;$$

$$dH=(C_V+VK\alpha)\,dT+K(\alpha T-1)\,dV.$$

with temperature $$T$$, entropy $$S$$, constant-pressure and constant-volume heat capacities $$C_P$$ and $$C_V$$, respectively, thermal expansion coefficient $$\alpha$$, and bulk modulus $$K$$.

Then, we can examine several possible scenarios.

1. If we create a system with energy $$U$$ in a vacuum, its enthalpy $$H=U$$ because $$P=0$$. If we move a system into a vacuum (from another), the flow work is zero, and the enthalpy transfer equals the energy transfer.

2. If we move a system into a vacuum from an atmosphere, the system will expand, and any potential energy stored in the compressed state will become immediately available to heat the material. This process is spontaneous and thus irreversible, so $$S$$ increases. (The reversible path would be quasistatic expansion to extract work, which is then dumped irreversibly into the material as heat. This offers a way to estimate the temperature change, as from $$dU=T\,dS-P\,dV$$ we can write the energy loss of the expansion as $$dU=-P\,dV$$, perhaps applying the isentropic bulk modulus relation $$K_S\equiv-V\left(\frac{\partial P}{\partial V}\right)_S$$, and then relate the energy to temperature through the constant-pressure heat capacity $$C_P$$.)

3. For a gas, we could imagine expansion to an arbitrarily large volume enclosed by a very stiff bag that holds negligible strain energy. (It's problematic to take the infinite limit of volume, as this is indistinguishable from the absence of gas, so we suddenly switch from a finite to a zero energy, enthalpy, entropy, and mass number.) For a real gas, enthalpy is generally not conserved upon free expansion except at the inversion temperature.

4. For the special case of the ideal gas, since $$PV=nRT$$ and $$dU=C_V\,dT$$, the temperature remains constant during free expansion, and the enthalpy therefore does as well. Decreases in $$P$$ exactly offset increases in $$V$$, and $$\Delta(PV)$$ therefore remains constant.

5. If instead we imagine an ideal gas enclosed in a real material (i.e., a rubber balloon) that stores strain energy upon stretching, we can expect some cooling of the gas and balloon upon expansion, in order to provide this strain energy. The predicted amount depends on the geometry, relevant pressures, and balloon material.

• The effect on the enthalpy can be estimated by integrating $dH=C_P\,dT+V(1-\alpha T)\,dP$ for the change in temperature and pressure; this was the reason for providing a variety of enthalpy-related equations. No, the strain energy stored in the material when it's compressed is not the same as the work done to push the atmosphere aside by a volume $V$. Nov 30, 2022 at 6:13
• I'm sorry it isn't clear. The $PV$ term is zero in a vacuum. Enthalpy isn't generally conserved, so any initial nonzero value of $PV$ doesn't necessarily "go" anywhere. Consider pulling on a rope with a force $F$ and then letting go—what happens to the term $F$? Nov 30, 2022 at 6:44