# Is PV energy part of internal energy, U?

The definition of enthalpy is $H = U + pV$. Doesn't the pressure and volume of a system get counted once in summing the internal energies, $U$? If so, why count it again to define $H$?

$pV$ should not be included as part of the internal energy of the system. Here is one way to think about the enthalpy $H$. It is the energy required to make a system. Of course we need to supply the energy $U$, which involves things like the motion of the particles or any potential energies.

But we also need to make room for this system. In order to do this we must push the atmosphere or whatever surrounding gas out of the way. This is where the $pV$ term comes in. It is not involved in the internal energy $U$.

Note that $p$ here is not the pressure within the system in question. Since we are talking about pushing whatever gas is present out of the way, $p$ is the pressure of the surrounding atmosphere that the system is in. This is why it is not included as part of the internal energy. This might be where your confusion lies.

# You are right, sort of.

This is actually a really good question!

First I want to remind you that generally energy is a relative rather than absolute scale: what matters are energy differences rather than absolute values.

Now the first law of thermodynamics says, $$dU = Q - W,$$ where $Q$ is the net heat (flow) into the system and $W$ is the net work done by the system. Work is always $W = F~dx$ for some force $F$ and some change in some length $dx$. But since this is all happening perpendicular to a surface, we can multiply this by $1=A/A$ without changing it to find $W = (F/A)~A~dx$ and then we can recognize that the force per unit area $F/A = p$, and a unit area times a displacement perpendicular to it is precisely a change in volume, $A~dx = dV.$ So the work done is precisely $W = p~dV.$

In this sense yes, you sort of "already have $p~V$ in there." But let's be more precise: what we're really saying is:

If there is no heat flow into the system, then changes in volume $dV$ get reflected directly in the internal energy by $dU = -p~dV.$

## Switching volume for pressure

A natural question is then: what about changes in pressure? Why aren't those being reflected directly in the internal energy? And the answer is: if you have a substance in a thermally isolated box, it is very hard to change the pressure of that substance without changing the volume of the box or adding thermal energy to it. In fact the most obvious way to do it is to stuff more particles into the box, or possibly to change one sort of chemical into another within the box (thus changing particles of one sort into another). In other words the internal energy is a state function of a couple variables: first, some notion of how much heat has been added, which we call the entropy $S$, whose changes are defined by $\Delta S = Q/T$; second, the volume of the box; third, the numbers of molecules $N = (N_1, N_2,\dots N_k)$ of the $k$ chemical reactants in the box, or so -- the internal sources of energy. There is some function $U(S, V, N)$ that characterizes the system completely.

On this account the pressure is not an independently controlled variable, it is dependent; you solve the system for it. In fact, we just haphazardly wrote an expression for it; once you know this state function $U$ you can take a partial derivative of it, $p(S, V, N) = -\left({\partial U\over \partial V}\right)_{S, N}.$ If you want to try to change $p$, well, you'll have to add/remove heat to/from the system to change $S$, or add/remove particles $N$, or change the volume $V$ -- you have to be evaluating this partial derivative at a different point of the space it's defined on!

Now what happens when you want pressure to be an independently controlled variable? Well we just said that we don't have enough independent variables -- so you have to sacrifice one of these existing variables to "make room" for it! The obvious one is to trade $V$ for $p$, this corresponds to taking our box of fixed volume and using instead a piston with a constant force on it, which allows the actual volume to fluctuate but the pressure must remain constant. This is also indirectly what happens in an open flask in your chemistry lab, as long as the molecules happen to be not leaving the flask (say a liquid or solid): the atmosphere provides the constant force.

## Internal energy becomes much less useful

So you use a piston instead of a box. What's happened? Indirectly, what's happened is that internal energy is not a great number to use to understand the system.

The problem is that, now the volume can change. So you put some heat flow $Q$ into the system, is that energy part of the internal energy $U$ now? Absolutely not. That added heat caused the system to expand, and possibly dramatically if it caused, say, near-boiling water to boil. This increased volume then did work on the outside world, which reduced $U$.

The actual change of $U$ was of course $Q - p~dV,$ but that uses some change in volume $dV$ that you do not know yet. If you want to know how this internal energy changes with heat flows (hence changes in $S$), you would have to write $u(S, p, N) = U(S, V(S, p, N), N)$ and then you could come up with $$\frac{\partial u}{\partial S} = \frac{\partial U}{\partial S} + \frac{\partial U}{\partial V} \cdot \frac{\partial V}{\partial S} = T - p ~\left(\frac{\partial V}{\partial S}\right)_{p,N}.$$ Even then, about the only interesting thing I can tell you about this last expression is that you can rewrite it by a Maxwell relation into another form that doesn't look so useful; we can write $({\partial V}/{\partial S})_{p,N} = ({\partial T}/{\partial p})_{S,N}.$ So if you knew how temperature changes with pressure when there is no heat flow, you could work out how volume changes in response to heat flow. Which is cool but it's not really all that helpful.

Step back a second, what is our real problem? The heat didn't flow into the "bucket" of internal energy -- not completely -- but rather some of it flowed into the "bucket" of the outside world.

Can you see the solution? If not, do not worry -- I did not understand it at first either, in my chemistry classes the enthalpy seemed very arbitrary at first.

## The rise of enthalpy

But the answer I am trying to lead you to is this: the solution is simply that you need a bigger bucket. You need a bucket that comprises both the energy that flows into the internal energy, plus the energy that flows into the outside world. Then you can say "hey, all of that heat energy definitely went into this bigger bucket!" and everything is theoretically simple again.

The astonishing thing is that the atmosphere is so big that you can actually do this with absolutely no direct reference to its internal composition. And this goes back to the very first fact I said: the zero of energy was meaningless in the first place. What this means is, I don't need to reconstruct the exact energy of the atmosphere from first principles to set this zero exactly. Instead, I just need something which recapitulates $p~dV$ when I change the volume at constant pressure; it's the bigness of the atmosphere that makes the pressure approximately constant. But that is the easiest integral ever: it just integrates to $p~V + C,$ and I can take $C$ to be zero. So I can say that all of the heat energy went into the "enthalpy bucket" $U + p~V$, of which an indeterminate amount went into the internal energy subchamber $U$ and an indeterminate amount went into the atmosphere, which is approximately $p~V + C$ for some unknown and irrelevant $C$.

The genius here is that $p$ and $V$ are both measurable properties of my system. I can just ask the chemists to make a note not just of energies added or subtracted, but also the pressure and volume of the system before and after whatever reaction happened, and instead of saying "such-and-so-much energy change" I will ask them to report these "such-and-so-much enthalpy change." It is very practical. And I know that it has a direct correspondence both with any energies due to chemical reactions (changes in $N$) and any energies due to heat added (changes in $S$). So I ask the chemists to write down the enthalpy per unit mass to boil water, and then I know that if I put a joule of energy into some boiling-temperature water, it will vaporize exactly such-and-so mass of that water, in an unsealed box. That's not going to be the right answer for a sealed box, but that doesn't matter.

• Thanks Chris! I'll accept your invitation to "dumb this down":A simplified version of the 1st Law is: ΔU = Q – W. (Q is heat added, -W is work done by the system) 1. Start with a constant volume system whose internal energy is U. 2. Add gas to increase internal pressure, thus doing work on the system, +W. 3. The new internal energy is U + W, where W is a function of P and V. I conclude that PV potential energy is part of the sum of energies in U. That brings me back to my original confusion about the definition of H: If H = U + PV and U includes PV, doesn’t PV get counted twice? – Frank Lynch Jul 23 '18 at 5:24
• @FrankLynch In your example you said the system has a constant volume. Adding gas to the system is not doing work on the system. When using $pV$ to calculate work done on the system, you are meaning to use $p\Delta V$ which must involve a volume change. Therefore, in this example no work is being done on the system. The reason we have $V$ and not $\Delta V$ in $H$ is because the initial volume is $0$. And once again, $pV$ is not a potential energy, and $pV$ is not part of $U$, so it is not double counted. – BioPhysicist Jul 23 '18 at 9:27
• @FrankLynch A different way to look at your example. You might be adding in additional energy due to how you are adding gas into the system which is changing the internal energy. However, this is not energy addition by "$p\Delta V$" work. This is energy addition due to the work you do on the gas molecules themselves when you put them into the system. – BioPhysicist Jul 23 '18 at 9:39
• Work was done by something in the environment (a compressor?) on the system. I think this changed the internal potential energy by VΔp. – Frank Lynch Jul 23 '18 at 13:23
• @FrankLynch Why are you so set that $pV$ must be explicitly part of $U$? Just because a volume or pressure change can end up changing $U$ does not mean that the additional $pV$ term in $H$ is included explicitly as a part of $U$. the $pV$ term in $H$ is a completely different thing than an explicit part of $U$. – BioPhysicist Jul 23 '18 at 16:50

No. For the ideal gas, for example, the internal energy is just the total kinetic energy of the atoms. This depends solely on the temperature, and is independent of the volume that the gas occupies, or its pressure --- except that these three quantities are related by the gas law $PV=nRT$.

• Wikipedia says internal energy is U(S,V,N), where N is the mole number. With V as part of the definition of U and representing pressure as N/V, I thought PV potential energy might be part of U. If not, I get the idea that U is just thermal energy: kinetic energy of atoms. – Frank Lynch Jul 22 '18 at 20:40
• @FrankLynch is this $U(S,V,N)$ for ideal gases or more of a general thing? What is the link you are looking at? – BioPhysicist Jul 22 '18 at 20:54
• It is a general thing: en.wikipedia.org/wiki/internal_energy – Frank Lynch Jul 22 '18 at 21:02
• @FrankLynch $pV$ is not a potential energy, and just because $U$ can be written to have a volume dependence does not mean $pV$ is part of the internal energy. See my answer for more details. – BioPhysicist Jul 22 '18 at 21:27
• The units of pV are: p (Force/Length^2), V (Length^3) so p times V is Force x Length--an energy term. A pressurized volume has the ability to do work so it has potential energy. It may be excluded from the definition of U. That is what I am trying to understand. – Frank Lynch Jul 22 '18 at 22:01
• $U$ is the energy which the system consists of. For a magician to summon a rabbit into existence, he must add this amount of energy to create the rabbit.

• $pV$ is the energy, which must be done to make the surroundings fit this newly created system. To summon the rabbit, the magician must first make room for it by pushing the air away. This requires work $pV$. That same rabbit summoned in outer space would not require any extra energy than just $U$ spent, since there are no surroundings (no air) interfering.

• Is this from Schroder's book? – BioPhysicist Jul 22 '18 at 20:37
• If my rabbit has a rigid skin and I created him with all of his interior full of high pressure gas, will he have greater U? – Frank Lynch Jul 22 '18 at 21:44
• @FrankLynch He sure will. But such internal pressure is not covered by the $p$ in the $pV$ part of enthalpy. That $p$ covers only pressure between system and surroundings, because the $pV$ term covers work done on/by the surroundings. Internal pressures simply increase potential energies, i.e. they contribute to the internal energy, for example as the stored energy in a compressed spring strapped in a rubber band. – Steeven Jul 23 '18 at 5:10
• @AaronStevens It sure is physics.stackexchange.com/questions/356412/… – Steeven Jul 23 '18 at 5:12

The answer is, "No, PV potential energy is not included in internal energy." Using data from https://webbook.nist.gov/chemistry/fluid/ for helium at 25C and 1 bar, internal energy, U, is 3.74kJ/mol. At 25C and 2 bar, U is also 3.74kJ/mol--pressure has no significant affect on U.

NIST also lists enthalpy, H, at both of the above P-T points. In both cases, H is greater than U by the amount P x V --the same amount at 1 or 2 bar because as P doubles volume halves for a nearly ideal gas, like He.

Thanks to all who contributed via physics.stackexchange to help me understand the enthalpy function, H = U + PV

Frank Lynch