For the first law of thermodynamics, we have that

$Q-W=\Delta U$

Since there is a variation of the values of the thermodynamic variables, because of Q and W, there is a variation of internal energy. However, we can't say there is a variation of $Q$ and $W$ since they aren't state functions, and that's why we write just $Q-W$.

Despite that, when we consider the differential form we have

$\delta Q - \delta W=dU$

and here is my doubt. That $\delta$ doesn't indicate a variation? And in this case, how is it possible that we have a variation of $Q$ and $W$ since they aren't state functions? How should I read this equation expect as "to a variation of Q minus a variation of W corresponds a variation of U"?

PS: My doubt isn't about the fact that $\delta Q $ and $\delta W$ aren't exact differentials, instead it is about the fact they indicate variations of Q and W.

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    $\begingroup$ Do you regard Q and W as well-defined quantities? $\endgroup$ – Chet Miller Jun 12 '19 at 17:19
  • $\begingroup$ What do you mean as well-defined quantities? I regard them as the quantity of energy which allows the variation of internal energy. However, since they aren't functions I can't understand how can I consider variations of them ( since they don't have a value in A and in B, I can't consider their variation AB ) . That's why I expected the differential form to be $Q-W=dU$. Where am I getting wrong? $\endgroup$ – Stefan Jun 12 '19 at 17:24
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    $\begingroup$ They are functions. They just are not state functions. $\int \partial W$ and $\int \partial W$ depends on the integration path. $\endgroup$ – Poutnik Jun 12 '19 at 17:38
  • $\begingroup$ @Poutnik but if the thermodynamic variables aren't their variables, then they're functions of which variables? $\endgroup$ – Stefan Jun 12 '19 at 17:41
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    $\begingroup$ @Stefan Yes $Q_A$ and $Q_B$ and $W_A$ and $W_B$ are not state functions. They can't be because $Q$ and $W$ are not system properties but energy transfers that depend on the paths. But there is only COMBINATION of $Q$ and $W$ that gives a unique $\Delta U$. $\endgroup$ – Bob D Jun 12 '19 at 18:46

You doubts are well founded. Unfortunately, even after more than one century since the very clear point of view of Planck expressed in his "Treatise on Thermodynamics", people continue to be confused by an ill-chosen formalism.

Although the notation $\delta Q$ (sometimes one can find a d with a bar insted of $\delta$)is intended to convey the meaning that the quantity is not a differential, this is a poor choice since the use of $\delta$ for differences is a well established use.

So, unless one wouldn't like to use Q, as Planck did, for whatever amount of exchanged heat, one should intend $\delta Q$ and $\delta W$ in the formula $$ dU = \delta Q + \delta W $$ as heat and work small enough to justify the substitution of the difference of interna energy $\Delta U$ with its differential and nothing more. (I prefer to use the convention which uses the $+$ sign in front of both work and heat).


and here is my doubt. That 𝛿 doesn't indicate a variation?

$\delta$ does indicate a variation, but its a variation in the amount of heat and work transfer. They are inexact differentials because heat and work depend on the path. This is because heat and work are not state properties but are energy transfers.

Let's take work transfer involving a gas in a piston cylinder. The $\delta$ means we need to specify the path for work transfer.

For example, if we specify a constant pressure process we can change the $\delta$ to a $d$ and then express the differential amount of work as an exact differential, or $dW=PdV$. Integrating between the initial and final state gives $W=P(V_{f}-V_{i})$

For another example, if we specify a constant temperature path (isothermal process) where PV=constant, we can again change the $\delta$ to a $d$. For an ideal gas, where $PV=RT$, $dW=PdV=\frac {RT}{V}dV$. Integrating between initial and final states gives us $W=(RT)ln\frac{V_f}{V_{i}}$

If we don't specify the path, there are an infinite number of paths for work and heat, but the combination gives a unique change in internal energy so that always $\Delta U=Q-W$ where $Q$ and $W$ are positive when $Q$ is the total heat transferred to the system and $W$ is the total work done by the system, each determined by integration for the given paths.

Bottom line: Once we specify the path, the differential work and heat become exact differentials.

If 𝛿 indicates an infinitesimal variation, why don't we write Ξ”π‘„βˆ’Ξ”π‘Š=Ξ”π‘ˆ when we consider finite variations of U?

It is because there is no "change" in heat or work because these are not properties of the system, like internal energy, entropy, temperature, pressure, and volume, which are state properties having a unique value for each equilibrium state and therefore a unique difference in values between equilibrium states. The $\delta$ for heat and work denoted a differential amount of heat and work transferred between the system and surroundings, not a "change" in heat or work. The system and surroundings do not "possess" heat or work.

Hope this helps.

  • $\begingroup$ If $\delta$ indicates an infinitesimal variation, why don't we write $\Delta Q - \Delta W=\Delta U$ when we consider finite variations of U? $\endgroup$ – Stefan Jun 12 '19 at 19:35
  • $\begingroup$ @Stefan Excellent question. When I was first learning thermodynamics I wondered about the same thing. I have edited my answer to specifically address your question. Hope it helps. $\endgroup$ – Bob D Jun 12 '19 at 19:54
  • $\begingroup$ @Stefan I also revised the first paragraph in my answer to state that what I meant as "variation" was variation in the amount of heat or work transfer $\endgroup$ – Bob D Jun 12 '19 at 20:13

$\let\d=\delta \let\l=\lambda \let\D=\Delta \let\pd=\partial \def\rA{{\rm A}} \def\rB{{\rm B}} \def\PD#1#2{{\pd#1 \over \pd#2}}$ First I would stress an important difference between the "integral" form of first principle $$\D U = Q + W \tag1$$ and the so-called "differential" form $$dU = \d Q + \d W.\tag2$$ This has nothing to do with the hated "$\d$" or "$\pd$". The difference is that (1) is much less demanding as to its applicability conditions. All what is required is that initial and final states, say A and B, are well defined equilibrium states of your system. Moreover, the external conditions must be such to allow you to compute total heat and work, denoted by $Q$ and $W$. This may be possible even if along transformation the system's state isn't defined - the system may have no definite temperature, pressure, or other thermodynamic variables.

On the contrary the "differential" form makes sense only if the transformation is reversible, i.e. passes through equilibrium states. Then you are allowed to think of heat and work as differential forms (@PhilipWood: why "so-called"? it's a perfectly well defined mathematical term). For instance you'll write work as $-P\,dV$, which requires the system has a definite pressure, i.e. is in hydrostatic equilibrium. If a finite transformation is given as a curve in phase space (e.g. in $(V,P)$ plane) then that differential form can be integrated: $$W = -\!\int_\rA^\rB\!\!P\,dV.\tag3$$

If you're not familiar with differential forms don't panic: an equation like (3) is to be meant as a usual integral. Once the curve in $(V,P)$ plane is known $P$ becomes a function of $V$: $P=f(V)$ and you may write $$W = -\!\int_{V_\rA}^{V_\rB}\!\!f(V)\,dV.$$ If states A and B are close to each other then it's approximately true that $$W = P_\rA\,\D V.$$ More exactly $$W = P_\rA\,\D V + o(\D V)$$ where $o(\D V)$ means an infinitesimal of order higher than $\D V$.

We say a differential form $\l$ is an exact differential if a function $F$ of state variables exists such that $$\l = dF.$$ In plane $(V,P)$ we would have $$\l = \PD FV\,dV + \PD FP\,dP.$$ Consider the case of $\l=-P\,dV$: could it be an exact differential? If so, then we would have $$W = \int_\rA^\rB\!\!dF = F(V_\rB,P_\rB) - F(V_\rA,P_\rA)$$ i.e. $W$ would'n't depend on the transformation but only on initial and final states. We experimentally know this isn't true. Moreover a mathematical proof exists that $-P\,dV$ isn't an exact differential.

Note that only in that case you'd be allowed to write $$W = \D F$$ i.e. work would be a variation of something (a state function). But this isn't generally true, so you can't write $\D W$. This also explains why it's much much better never to use d's or $\d$'s or $\pd$'s or any other camouflage of a supposed "variation" (that doesn't exist) or "differential" (that it isn't).

Don't forget: whatever may be the bad usage frequently made in physics, a differential isn't something small or worse infinitesimal. That term has a precise mathematical meaning and we ought to learn to use it the right way.


Here it is. A system's entropy increases by at least the contribution from incoming heat, plus whatever else there might be: $$ T dS = \delta Q + \epsilon $$ To do work on a $pV$ system, you must provide enough to change the volume, and also a bit more to overcome friction: $$ \delta W = -p dV + \epsilon $$ We have $$ \delta U = \delta Q + \delta W \\ = (T dS - \epsilon) + (-p dV + \epsilon) \\ = T dS - p dV . $$ Thus it all ties up nicely. All you need to do is realise that the contribution from friction must go to increase the entropy by the corresponding amount, so the quantity $\epsilon$ is the same in the two places and cancels in the final formula. I think that many books in the past have failed to be clear about how this friction term comes in, so it is presented less clearly in many books.

For the conceptual point, you can note that if you make a change in a function of state such as internal energy, then it does indeed represent a variation in a function, no matter how the change was brought about.

  • $\begingroup$ Why to overcome friction we need to provide work? When we pushing a box against friction the work done from us to the system (box) is $$dw=F\cdot dx$$ Do you mean that we must exert a greater force when friction is there so the work is greater? Thanks in advance. $\endgroup$ – Antonios Sarikas Apr 3 '20 at 20:12

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