I have always been confused between the different kind of energies in thermodynamics so today I thought of finally getting it right. Here's my analogy(which is wrong and I need help in improving it, please read it once):

Let us suppose we take a gas at temperature $T_1<T_{room}$ in a beaker fitted with a piston and conducting walls.

Its present energy can be represented by the following diagram:
When the temperature increases both the pressure volume energy and internal enrgy increase by fixed amounts, independent of path(being state variables), according to the equation $\Delta(PV)=nR\Delta T$ and $\Delta U=3/2\ nR \Delta T$

As enthalpy is the sum of both these energies, even it is a state variable, and its change is independent of the path followed.

Now event though $\Delta (PV)$ is a state function and there is a fixed change in its value, there are 2 ways of changing (increasing in this case) it. One by increasing volume and the other by increasing pressure.
The problem is that in the process of increasing pressure-volume energy by increasing volume, the system has to do some work, which means it has to throw out some energy. But we can't let that happen. So we need to provide the system with some extra energy that it can throw out to do work. We haven't increased the energy of the system by doing that as it has come back to the surroundings in the form of work.
There is this fundamental change in energy of the system that is absorbed by the system from the surroundings and there is this extra energy $P\Delta V$ that is also absorbed but then thrown out in the form of work. The total energy absorbed is termed as heat $q$ and the energy thrown out is called work $(-w)$. They depend on the path as it depends on the way we increase the $PV$ energy.

The biggest problem with this analogy is it cannot explain $\Delta U=q+w$. According to my analogy $\Delta U+\Delta (PV)=q+w$. This means pressure-volume energy is a part of internal energy and not something different. But then what would be the significance of enthalpy $H=U+PV$?
Moreover I don't know where to put:
1. gibbs free energy as I don't have an intuitive understanding of it, I only know the mathematical formula and the definition: maximum useful work that the system can do and
2. energy lost due to entropy ($T\Delta S$).

It would be great if someone could help me improve my understanding of thermodynamics as I am tired of getting confused each time I open this chapter. An approach coinciding with this one (something like a venn diagram) would be appreciated but anything is welcome. Please help, I really need it!


2 Answers 2


This is not an attempt at a complete answer but I think its relevant to share this great explanation from "An Introduction to Thermal Physics" by Schroeder:

enter image description here

  • 1
    $\begingroup$ So basically the whole of H is required to summon up a rabbit but we need to only provide the differences as work. Then the energy required to summon up the rabbit is not just U but U+PV. Hence pressure volume energy doesn't count under internal energy. Am I right? Btw thank you this image is pretty useful. $\endgroup$
    – oshhh
    Mar 3, 2017 at 2:42
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    $\begingroup$ Yes. The internal energy of the rabbit would remain same if he were in vacuum. The energy TS flows in spontaneously because of the difference in temperature. If you were creating him in Zero Kelvin, you'll have to do the full work. $\endgroup$ Mar 3, 2017 at 4:55
  • $\begingroup$ Okay, but what about the proof of $\Delta U+\Delta(PV)=q+w$ in my question? If we consider pressure volume energy not to be a part of internal energy to be different, then the 1st law is violated $\endgroup$
    – oshhh
    Mar 3, 2017 at 5:04
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    $\begingroup$ Let us talk in terms of changes. $\Delta U$ would be the total measure of change in energy of the system. This change can be manifested either by work done or heat exchanges. Hence, $\Delta U = W + \Delta Q$. (You may be using a different sign convention for work but it doesn't matter.) $\Delta H$ is the measure of heat changes at constant pressure. It is a very useful parameter when the system is in contact with a constant pressure reservior such as the atmosphere and hence it is the measure used for heat of reactions. $\endgroup$ Mar 3, 2017 at 8:02
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    $\begingroup$ Now to create such a balloon in vacuum we only need to expand (no work done) and then supply the heat. This is heat supplied at constant volume which equals change in internal energy. But to bring the balloon to atmospheric pressure you need to do extra work ($=P\Delta V = PV$ as we are starting from zero volume). Or you could do the same thing by supplying heat to balloon at constant pressure. Both ways you get enthalpy change. $\endgroup$ Mar 3, 2017 at 8:31

I agree that many chemistry (and physics) texts present these terms in a very confusing way.

Overall, the ways these are used are:

  1. Internal energy is useful for incompressible things because they're at constant volume. Conservation of energy for liquids, ionic stuff, and solutions can be handled with internal energy. A classic example is analyzing a battery.
  2. Enthalpy is useful for compressible things because those processes are often not at constant volume. Conservation of energy for phase changes and gas phase processes usually require enthalpy. A classic example is a gas expanding through a valve or turbine.
  3. Gibbs energy is useful for determining which version of a system will exist. A classic example is determining which direction a reaction will go.

I didn't really understand them until I had to use them, and specifically I needed to do engineering problems with enthalpy. I would recommend working problems with each of these terms.

It sounds like you're a chemistry student, so your textbook can probably supply internal energy problems. Second, skip to later in the book to the chapter on spontaneous reactions and work some of the simple problems that require you to use Gibbs energy (side note: IUPAC prefers no 'free') to determine whether a process is spontaneous. Third, find an engineering thermodynamics textbook and work some of the problems involving isenthalpic processes (isenthalpic being constant enthalpy, although an engineering text will call these adiabatic which is only slightly different).

To answer your question specifically, I think the trouble is that if you let your system change volume, then you by definition no longer have a constant volume system and you need to use enthalpy, not internal energy.

  • $\begingroup$ But how are they related to each other...I don't mean the mathematical formulae but the real intuitive relations. $\endgroup$
    – oshhh
    Feb 28, 2017 at 2:20
  • $\begingroup$ @OsheenSachdev I'm not exactly sure what you mean. I can't think of an example where you analyze one system with both internal energy and enthalpy and try to learn something about their difference. Sometimes its just appropriate to use one, and sometimes the other. There are some cases where you would consider the enthalpic vs entropic contributions to Gibbs energy. Critical solution temperatures are one such application that you might be interested to read about. Can you clarify what you mean by "intuitive relations" thats different than "how one would use them"? $\endgroup$
    – ericksonla
    Mar 1, 2017 at 3:08

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