Work is defined when there is a change of volume. Mathematically, $W=-\int P\, dV$. When pressure is constant $W= P\, \Delta V$. For an ideal gas expansion we have $W= nRT\ln\frac{V_2}{V_1}$. If its constant volume then $W=0$ in both cases.

Enthalpy is defined as $H=E+P\, V$, where $E$ is the internal energy. If it is a change with constant volume, $Q=\Delta E$ and $\Delta H=Q+ V\, P$. In this case, what is $V\,\Delta P$? It cant be defined as work since no volume changes. Maybe a form of energy transfer that is not work but similar to it? Does it have particular name? Or can it be included in a kind of works?


2 Answers 2


The scenario you are describing is known as an isochoric process. Because there is no change in volume, the thermodynamic work done is zero. Some refer to this 'energy transfer that is not work but similar to it' as isochoric work.

The utility of isochoric processes is that all energy added to the system result in a one to one increase of internal energy.

$\Delta U = Q$

Here is a quick review of classical thermodynamics by Dr. Zhigilei to put it.

  • $\begingroup$ Thank you for your answer. What I wonder now is that, if I can say -VΔP is isochoric work, when in isochoric process, does Q have to be ΔU + VΔP? Why we don't consider VΔP energy when we calcualte Q? My physical chemistry textbook also says dQ=dU+PdV. (well it says a little about non-expension work but not enough for me to comprehend why no VdP) It was not a big question until yesterday because I thought W is only defined when V is changed. But when considering ΔH in isochoric process, ΔH = Qv + VΔP (which is equal to nCpΔT consequently). ΔH consider VΔP. Why Q excluded this energy? help me.. $\endgroup$
    – Zillai
    Mar 27, 2018 at 3:00
  • $\begingroup$ The only explanation I can give to myself is that.... it's because just a definition. Q is developed in physics with piston engine. in this situation, maybe VdP was not a point of consideration because expasion work was much important. so Q is just defined with expansion work. In chemistry however, non expansion work probably is important factor. that's why they define the function H which can include all the work from Δ(PV). That's why I have to consider VΔP in ΔH. This is really my imagination, maybe misconception but I just wrote it anyway. $\endgroup$
    – Zillai
    Mar 27, 2018 at 3:10
  • $\begingroup$ One last thing. I searched it on youtube channel and found some of lecture regard to the subject. Many of them explain that W=0 in isochoric system. Yes, that's what I've know so far. and I found another lecture, which explains W=-VdP. Drawing a P-V graph and a constant V line on it, others said it has no PΔV because of vertical line to V axis, and the other guy said VΔP, making an area not by V but by P axis. $\endgroup$
    – Zillai
    Mar 27, 2018 at 3:22
  • $\begingroup$ The only differences between the lecture was a definition of system. One with W=0 mentioned closed system, while the other with W=-VdP mentioned open system. but.... constant volume process with open system?? ahh..... something is missing..... but I don't think that's not my missing part. or.... is it?? enlighten me please.... $\endgroup$
    – Zillai
    Mar 27, 2018 at 3:26

Enthalpy H and internal energy U are both state functions

Irrespective of the pressure and volume changes from initial to final state, you can consider H or U to describe processes.

Enthalpy H is most useful for processes at constant pressure

If you are interested in the amount of heat a reaction gives off, you have to consider that a closed system can exchange energy with the surrounding via heat or work. In the special case of a reaction (or process) where the only work is P-V-work, and where the pressure is constant, the heat transferred is equal to the change in enthalpy. That is why reactions with negative change in enthalpy are called exothermic. If their is non-P-V work, however, an exothermic reaction can happen without giving off heat (e.g. reaction in a battery).

Internal energy is closely related to heat when the volume is constant

In the special case of a reaction (or process) where there is no work, the heat transferred is equal to the change in internal energy. This happens, for example, when the volume is constant and there is no non-P-V work (a constant volume implies that there is no P-V-work).

What about enthalpy changes when the pressure does change?

In that case, there is no easy relationship between heat and enthalpy. At constant volume, VΔP is the difference between enthalpy and internal energy. There is no P-V-work, and it makes sense to work the the internal energy U instead.

In this case, what is VΔP?

It is the difference between H and U in the case of constant volume, like stated in the question. It does not have any physical meaning. The way H is defined makes it easy to interpret for constant pressure processes. The flip-side is that it is difficult to interpret for all other processes, including those happening at constant volume.

Maybe a form of energy transfer that is not work but similar to it? Does it have particular name? Or can it be included in a kind of works?

No, there is no work associated with it. It is just what is mathematical left over from the difference between U and H when the volume is constant.


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