1
$\begingroup$

What is the difference between $$\Delta Q$$ and $$\mathrm dQ$$ in thermodynamics? I think one is used when the change is very small and the other with bigger changes, but I am not sure.

$\endgroup$
  • 4
    $\begingroup$ Difference between Δ, d and δ. The question has little to do with chemistry, it's rather about notations and principles of calculus. $\endgroup$ – andselisk Jul 7 at 15:09
  • 3
    $\begingroup$ @andselisk, come on my friend. Chemistry cannot exist without mathematics. Thermodynamics is a part and parcel of chemistry curricula. $\endgroup$ – M. Farooq Jul 7 at 15:52
  • $\begingroup$ Also related: Symbols of derivatives and d or Delta?. $\endgroup$ – Nilay Ghosh Jul 7 at 16:23
5
$\begingroup$

This is not about mere arbitrary notation. It is about fundamental concepts in classical thermodynamics, and the notation flows from these concepts.

So: q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact rather than exact. The dyet, $\text{đ}$, is used to indicate that (though typically it is an angled slash rather than a horizontal bar; I don't know how to make that with MathJax).

Since the differential forms of heat and work are inexact, any integration with respect to them must be a path integral. While not all physical chemistry/thermodynamics texts use this specific notation for inexact differentials, I've never seen a physical chemistry textbook that doesn't carefully use some sort of notation to distinguish exact from inexact differentials, because of the conceptual and pedagogical importance of distinguishing between state functions and path functions.

[After seeing the comment about Atkins, I went back and looked at my old Atkins pchem book and, sure enough, he doesn't use distinguishing notation for inexact differentials. I didn't know this because Atkins' thermo treatment was considered poor, so we only used him for kinetics. We used Castellan for thermo.]

Hence we have, for a closed system:

$d\text{U} = \text{đ}q + \text{đ}w$

And:

$\Delta \text{U} = q + w$

Or, alternately:

$\Delta \text{U} = \text{Q} + \text{W}$

Heat and work are path functions that reveal themselves only during a change in state—during a process. Hence, for a given process, there will be a certain amount of heat flow, q, and work flow, w. One should not use "$\Delta q$" or "$\Delta \text{Q}$", as this is conceptually incorrect: It implies that heat is a state function, and thus you could have a change in heat. No. There was (or could have been) a change in internal energy, along with some heat flow; there was no "change in heat".

More explicitly:

$$\int_{state 1}^{state2} dU = U_2 - U_2 = \Delta U$$

But:

$$\int_{state 1}^{state2} \text{đ}q \ne q_2 - q_1 = \Delta q$$

because there is no q for state 1 or state 2—q is associated with the path only, not with the state of the system.

Instead:

$$\int_\limits {\text{path from state 1 to state 2}} \text{đ}q = q$$

The only context in which it might make sense to use $\Delta q$ (or $\Delta \text{Q}$) would be if you were comparing two different processes, and calculating the difference in heat flow between those two processes. This is acceptable, but perilous, since such usage risks people falling into the common misconception of thinking that it is acceptable to use $\Delta q$ to characterize individual processes, which it is not.

Adding something from one of my comments: A key property of exact differentials that both connects them with state functions, and distinguishes them from inexact differentials, is that their cyclic integrals must always be zero. Hence: $$\oint dU = 0= \Delta U_{cyclic}$$

Conversely, it is not generally the case that:

$$\oint \text{đ}q = 0$$

Interestingly, if we carefully define the path to be reversible, and apply an integrating factor $(\frac{1}{T}) $, we can convert the inexact differential $\text{đ}q_{rev}$ into one of the most recognizable exact differentials in thermodynamics, as follows:

$$\oint \frac{\text{đ}q_{rev}}{T} = \oint dS =0$$

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Very relevant and useful answer. I am just wondering about the name of (d-bar) symbol, which is used in many phys. chem texts. I would read it as d-bar. You called it dyet. Can't find this name name by Google. $\endgroup$ – M. Farooq Jul 8 at 2:27
  • $\begingroup$ Okay found it in the book "The Elements of Typographic Style", $dyet$: A basic letter of the alphabet in Serbo-Croatian and Vietnamese. $\endgroup$ – M. Farooq Jul 8 at 2:35
  • 1
    $\begingroup$ @M.Farooq For more on the rich and varied history of the dyet, see: en.wikipedia.org/wiki/D_with_stroke Some textbooks instead use the "eth" -- see: en.wikipedia.org/wiki/Eth $\endgroup$ – theorist Jul 8 at 2:53
  • 1
    $\begingroup$ "..._ Atkins' thermo treatment was considered poor, so we only used him for kinetics_..." . A bit off-topic, but shouldn't it be "we used it"? Since a book is not having gender? $\endgroup$ – Micelle Jul 8 at 4:05
  • 1
    $\begingroup$ There seem to be caveats: (1) $dq_{rev}$ seems to be an exact differential; (2) we tend to say that state functions are not really defined at single points either, only differences or values wrt a reference (such as absolute zero T and zero entropy) even though differences between their values are path independent. $\endgroup$ – Buck Thorn Jul 8 at 5:52
4
$\begingroup$

It is rather mathematical than chemical question, as it applies to any continuous and differentiable quantity, not limited to chemistry.

$\Delta Q$ would be macroscopically measurable exchange of thermal energy. But the convention is we use $Q$ is sense of the exchange of heat, not $\Delta Q$.

$\mathrm{d}Q$ is "infinitesimal ( infinitely small) difference aka differential. It is not true aka total differential, as $\int{\mathrm{d}Q}$ depends on path and therefore should not be using "d" - but "đ". I cannot remember the proper MathJax/LaTeX symbol for it, perhaps not available.

So it is $\mathrm{đ}Q$ instead ( copied from the comment ).

So $\mathrm{d}U = \mathrm{đ}Q + \mathrm{đ}W$, where the change of internal energy between 2 states depends only on these states, but exchanged heat and work depends also on the path ( a way how the final state was reached. )

It is used oftenin context of infinitely small changes from equilibrium, that are so small the system can still be considered in equilibrium. An infinite set of such changes makes a reversible process, where the system is at any time in equilibrium.

| improve this answer | |
$\endgroup$
  • $\begingroup$ đQ. Not sure there is a latex symbol for it. Using Unicode might be easier. $\endgroup$ – Zhe Jul 7 at 15:32
  • $\begingroup$ How is it inserted on english keyboard ? $\endgroup$ – Poutnik Jul 7 at 15:47
  • 2
    $\begingroup$ Isn't $\Delta Q$ incorrect notation, as $Q$ represents a quantity of energy that is transferred, and since it is not a state function, there is no pre-existing $Q$ that is being changed by the transfer. Thus, we write $\Delta E = q + w$ rather than $\Delta E = \Delta q + \Delta w$. $\endgroup$ – Andrew Jul 7 at 15:55
  • $\begingroup$ You are probably right, it is my frequent mistake. But OTOH, it is kind of inconsistent notation, confusing me often, as there can be d, resp. đ, but not $\Delta$, in contrary to other quantities. $\endgroup$ – Poutnik Jul 7 at 15:58
  • $\begingroup$ In fact, in some sense, $\ce{Q}$ could be considered as the system thermal energy $\int_0^T{C_\mathrm{v}(T) \cdot dT}$, as the part of $\ce{U}$ $\endgroup$ – Poutnik Jul 7 at 16:02

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