# Why is the change in enthalpy, $\Delta H$, defined as $\Delta H = \Delta U+P\Delta V$?

If we define work as $w=-P\Delta V$ (so that during expansion work done by the system is negative), wouldn't it make more sense to say that $\Delta H = \Delta U-P\Delta V$ since $\Delta H = \Delta U+w$? Here, $U$ is the internal energy of the system under consideration.

Why is it represented as $\Delta H = \Delta U+P\Delta V$ in my textbook even though it follows the convention that $w=-P\Delta V$?

You're missing the point, which is that we're trying to correct for the loss of energy due to expansion.

# Boiling water

Let's think of a boiling pot of water at constant pressure, and how we want to think of the "specific heat of vaporization", which we know we'll want to write in units of $\text{J/kg.}$ We're obviously going to measure a heat $Q$ as we pour it into the system with a burner, and hopefully we'll insulate the sides of the pot to keep other heat losses down; and this boils some mass $\delta m$ of water that we can measure by putting this whole apparatus on a scale: then $Q/\delta m$ is what we want to report to anybody who is boiling water. That is the most relevant physical quantity to tell people about.

But what has really happened? Part of that energy went into an internal energy change $\delta U,$ no doubt. But not all of it did. Some of it went into the atmosphere: because the new volume for the water was much larger than the old volume of the water by some amount $\delta V$, and we know that at constant pressure $P$ the energy that the atmosphere taxes us when we make such an expansion is $P~\delta V$.

So when you're reporting $Q$ at constant pressure, you're secretly reporting $\delta U + P~\delta V,$ emphasis on the "plus". The atmosphere stole that energy from the system, and we're adding it back when we want to talk about this change in the "free energy" of the system, and we talk about free energies because the internal energy is no longer relevant to our physical situation.

# What is enthalpy, really?

Enthalpy is more precisely defined as the free energy $H = U + P~V,$ and for extremely small ("infinitesimal") changes in pressure $dP$ and volume $dV$ the change will be approximately $dU + P~dV + V~dP.$ When this expansion $dV$ happens, then in general (as you've noted!) $dU$ will have a term $-P~\delta V,$ with the "infinitesimal thermodynamic relation" being $dU = T~dS - P~dV + \sum_i \mu_i~dN_i$ for internal energy $U$, temperature $T$, entropy $S$, pressure $P$, volume $V$, number of particles of species $i$ $N_i$, and chemical potential of species $i$ $\mu_i.$

So another way to look at it is that, by this "Legendre transformation" we are changing our perspective from a $U$ which is constant when (other things are constant and...) volume is constant, to an $H$ which is constant when (other things are constant and...) pressure is constant, because that $H$ is more relevant to us in situations of constant pressure.

• So is it safe to assume that in an expansion, $\Delta H$ will always be greater than $\Delta U$ because we are trying to account for the energy lost as a result of expansion? $\Delta U$ is only gives us the internal energy for a fixed volume? Oct 6, 2015 at 19:30
• Nitpicks: $\Delta U$ is a change in internal energy not only for a fixed volume, but in general. Now, $U$ can change when the volume increases (for fluids, generally by decreasing) because particles pushing the moving wall of their expanding confines lose some internal energy to that wall. You might instead say "we only care about $\Delta U$ at fixed volume," that's fine: but it "gives us the internal energy" all the time. Other than that, yes: for most fluids, $\Delta H$ will be greater than $\Delta U$. Solids with internal stresses may resist expansion, and thus have negative pressure. Oct 6, 2015 at 19:59
• Okay, one last thing: What would $\Delta H$ represent if volume is constant? According to the derivation, you would get the equation $ΔH=ΔU+VΔP$. What does $VΔP$ represent? What energy is this accounting for? Oct 6, 2015 at 20:52
• Okay, suppose we design an engine where we have a piston with some weights on it: total pressure is atmospheric pressure plus $mg/A$ where $m$ is the mass on the piston, etc. Inside the piston, we have water in liquid form. We want to cycle the system as follows: we add heat to boil the water, raising the weight; we take the weight off the machine, changing the pressure, we absorb heat to condense the water, we put weight on the machine. Then it makes sense to calculate the total enthalpy change, $(-mg/A)~V_1 + (mg/A)V_0,$ which is just $-mg\Delta h,$ the potential energy of the mass moved. Oct 6, 2015 at 21:10
• The interpretation of this is probably just that it's the balance of the heat you needed to inject/withdraw when you were changing pressure, in order to keep it at constant volume. Oct 6, 2015 at 21:12

Note that, in this convention (taking $w=-P\Delta V$), the work is being done on the system, not by the system. This is easily seen by noting that it is positive when the system is compressed. This is the key to your confusion, and once you understand it the rest should be quite clear. I will nonetheless make the answer a bit more explicit to try and give you some intuition.

The enthalpy $H$ of a system is defined as the internal energy plus the energy it took (or would have taken) to create the system (under constant pressure). That is to say, $H=U+PV$. Therefore, it is clear that $\Delta H=\Delta U +P\Delta V$ when the pressure is constant. In terms of work done by the system $w^\mathrm{s}$, this reads $\Delta H=\Delta U +w^\mathrm{s}$. Note that $w^\mathrm{s}=-w$ where $w\equiv -P\Delta V$.

• So are you saying that a negative value for work does not indicate that work is being done by the system on the surroundings? I still don't understand where my error lies. Oct 6, 2015 at 19:21
• Think about someone pushing in the walls of a box: Will it cost the person work, or not? Now, what will $-P\Delta V$ do, increase or decrease? So what does your $w$ measure?
– Danu
Oct 6, 2015 at 20:34