How do you know if something is a state function? I know Internal Energy (U) is a state function since it is the sum of energies, which is constant at a given time.

However I don't know for sure if $\Delta$U is a state function. I know that $\Delta$U$= U_{final}-U_{initial}$, and so any path from final to initial will yield the same $\Delta$U. So by this definition any $\Delta$ function should be a state function.

As a result I would expect Heat ($q$) to also be a state function since $q=\Delta$TE, but this is not the case.

Where am i going wrong in my understanding? I understand the "altitude" analogy, however i need a more technical definition with chemistry terms. Altitiude analogy
(source: cnx.org)


1 Answer 1


Where am i going wrong in my understanding?

In short: everywhere. I mean all possible ways. Trying to cheat by randomly throwing symbols and words out there without deep thinking is a way to nowhere. Thermodynamics is a tough subject because it requires a strong mental discipline, so to start from, you should know and understand the definitions. Funky pictures won't help unless you understand the actual meaning of the analogies used there.

State function

A state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.

Using this definition we could answer the question

How do you know if something is a state function?

trivially: if a quantity is uniquely determined by the equilibrium state then it is a state function, otherwise it is not. $\Delta U$ is clearly not a state function, since by its very definition ($\Delta U = U_\mathrm{final} - U_\mathrm{initial}$) it is not determined uniquely by the state of the system: neither by the final, nor by the initial one. $\Delta U$ is not even a characteristic of the state (since it clearly refers to two of them, not just one), rather it is the characteristic of a thermodynamic process of transition from one state into another.

Internal energy

I know Internal Energy ($U$) is a state function since it is the sum of energies, which is constant at a given time.

Sorry, but this sounds like a nonsense to me. What do you mean by constancy of internal energy at a given time due to it being sum of some energies? From a microscopic point of view, internal energy of a system can indeed be subdivided into kinetic energy of motion of all the system's particles and potential energy due to all kinds of interaction between particles in the system, \begin{equation} U = E_{\mathrm{micro,k}} + E_{\mathrm{micro,p}} \, . \end{equation} But what kind of constancy follows from this subdivision? Do you want to say that $\Delta U$ is always zero or what?

Note also that such definition of the total energy belongs already to the realm of statistical thermodynamics, not the classical one. In classical thermodynamics the total energy is postulated to be a state function. Essentially, the first law of thermodynamics tells us that there exist a state function called internal energy, $U$, the increase in which for a closed system during a thermodynamic process is equal to the heat supplied to the system plus the work done on the system, \begin{equation} \Delta U = q + w \, , \end{equation} or in infinitesimals for a quasistatic process, \begin{equation} \newcommand{\dif}{\mathrm{d}} \newcommand{\difbar}{\text{đ}} \dif U = \difbar q + \difbar w \, . \end{equation} Note that infinitesimal heat and work transfers in a quasistatic process are denoted by $\difbar$ rather than $\dif$, used to denote infinitesimal change in internal energy, to emphasize the fact that heat and work in contrast to internal energy are not state functions. Mathematically speaking, $\dif U$ is an exact differential, while $\difbar q$ and $\difbar w$ are inexact differentials. For more on that story, check this answer of mine. What is really important for our current purposes is the very final sentence there: essentially it was experimentally found that for a closed system $\oint \difbar q \neq 0$ and $\oint \difbar w \neq 0$, but $\oint (\difbar q + \difbar w) = 0$. This finding implies that there exist a state function $U$ such that $\dif U = \difbar q + \difbar w$. So, the law that postulates that $U$ defined by $\Delta U = q + w$ is a state function is based on experimental evidences.

  • $\begingroup$ Thanks. Does $q=\Delta$TE. Also what does this mean about Hess's law and $\Delta$H. Since $\Delta$H is the same for any chemical process from A to B. $\endgroup$
    – user22506
    Nov 2, 2015 at 19:57
  • $\begingroup$ Cavalier use of equilibrium. A state need not be at equilibrium. $\endgroup$
    – jimchmst
    Jan 3, 2023 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.