Can we say $\Delta U$ is a state function? According to me, $\Delta U$ is not a state function. $dU$ is an exact differential and $U$ is a state function.

Statement 1:

Basic meaning of state function (property) is something which is defined at state (equilibrium condition) like $P, V , U, S$ i.e. if we change the state then the value of that property for the final state does not depend on how we reached that final state. i.e state function defined for the state

$\Delta U = Q +W$, its value is same between 2 states (initial and final) when we select any of the path joining the 2 states. The quantity $\Delta U$ is independent of path followed. but it is not something that is associated with state. This is why $\Delta U$ is not a state function.


We are getting same value of $\Delta U$ from 2 different paths between same initial and final state. By saying this we can comment on $U$ that it is a state function. I agree with this no issues.

Statement: 3

If we are still saying $\Delta U$ is a state function.

1. Then we should say this by saying $\Delta\Delta U$ is independent of path between 2 states. And $\Delta U$ must be defined for equilibrium state. Like $P$ is a state function. Then $\Delta P$ is independent of path between 2 states. And $P$ has its meaning for equilibrium state But actually we don't say anything like this for $\Delta U$.

2. Does it mean $\Delta P$ , $\Delta V$ , $\Delta T$ all are state functions? Because we will get same value on every path between two states.

If yes then that bothers me that they do not have any meaning for a state (equilibrium condition).

If no, then what is different with $\Delta U$.

  • 1
    $\begingroup$ Is $\Delta U$ a function of the state and nothing but the state ? $\endgroup$
    – Poutnik
    Mar 1, 2022 at 13:00
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    $\begingroup$ $\Delta U$ depends on two states, not one. So it is not a state function. $\endgroup$ Mar 1, 2022 at 13:10
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    $\begingroup$ @ChetMiller It may be more grey zone that it seems. E.g. for a perfect gas we say, U = n.Cv.T, but in fact it is U(T)=U(T=0) + Delta U(T) = m(T=0)c^2 + n.Cv.T $\endgroup$
    – Poutnik
    Mar 1, 2022 at 14:15
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    $\begingroup$ Definitionally, I think it's cleaner & more precise not to call $\Delta \pu{X}$ a state function (SF) since, as has already been mentioned, $\Delta \pu{X}$ is a function of the difference between states. Also: A comment on the converse of what you wrote in statement 3, #1: If X is a SF, $\Delta \Delta \pu{X}$ is independent of path. A familiar example would be the change in $\Delta \pu{H}^{\circ}$ for a chemical reaction when we change T at constant p. But that doesn't mean that $\Delta \pu{X}$ is itself a SF. Rather, $\Delta \Delta \pu{X}$ is path-independent simply because X is a SF. $\endgroup$
    – theorist
    Mar 1, 2022 at 20:06
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    $\begingroup$ Also, I wouldn't refer to p, V, and T as SF's. That's typically reserved for thermodynamic potentials like U, G, A, H, S, and $\Omega$. p, V, and T are, by contrast, considered state variables or state parameters. I think to consider a state variable as a SF you would need to be treating it as a thermodynamic potential, i.e., you would need to be working in the "p representation", etc. And a system couldn't be described using a p or T representation, because these are intensive, so you'd have a potential that is independent of system size. Not sure if a "V" representation is possible. $\endgroup$
    – theorist
    Mar 1, 2022 at 20:26

2 Answers 2


$U$, like other state functions, is defined mathematically as a non-injective non-surjective function where the argument is the state of a system. In layman's terms non-injective non-surjective means that there can be many states that share a value of a thermodynamic property such as the energy, and values of the property that do not correspond to any known state, but each state has only one value of the property.

In classical thermodynamics a state function (such as $U$) is defined as a property of a static (or in practice quasi-static) system, not of a transition or difference between such systems (such as $\Delta U$). $\Delta U$ is also a non-injective non-surjective function, but it is a function of two states. Although you have found a feature that satisfies a mathematical property of a state function, the feature is not sufficient to render $\Delta U$ a state function. You might say the duck-type test includes that you limit yourself to features of a single system rather than of multiple systems.

Note that it is possible to define a reference state and the properties of all other states relative to the reference (as alluded to in the comments), but that does not change the basic argument.

Also, you can create systems in which a transition between two energy levels is an important property of the system (for instance in NMR or optical resonance). However those are time-dependent sytems. Classical thermodynamics focuses on static (time-independent) systems.


The reason why we use ΔU instead of U is due to the fact we cannot measure the internal energy U of a gas but only the change of U.

For any process ΔU in a closed system ΔU can be calculated by the change of temperature of the gas(ΔT) thats why ΔU is a state function.

  • 1
    $\begingroup$ I'm reviewing this answer as "looks okay". That doesn't mean I think it is correct or good. In fact, I have chosen to down-vote it. But from a technical perspective, it is an attempt to answer the question. $\endgroup$ Mar 1, 2022 at 18:49
  • $\begingroup$ When you say you can calculate $\Delta U$ for a change in T of a gas, what you mean is that you can calculate that for an ideal gas. You can't typically calculate $\Delta U$ for real gases, since you would need an equation of state. That's only available for some gases, and they're always, to some extent, an approximation. $\endgroup$
    – theorist
    Mar 1, 2022 at 20:37

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