# Solving an inexact differential

An inexact differential describes a path dependent quantity, meaning I can't just subtract the value at an initial and final state to get an answer. Take for example work which is $$W=\int F\cdot dx$$ obviously it is path dependent as if I go from $$x=1$$ to $$2$$ directly versus in a circle a different amount of work is done. So assuming $$F$$ is constant

$$W=\int F\cdot dx \neq \int Fdx$$

We need $$\cos(\theta)$$ as a function of $$x$$.

But what about something even more simple with no vector calc dot product

$$d(PV)=VdP+PdV$$ $$dU=-PdV+TdS+\mu dN$$ The variables $$V$$, $$S$$ and $$N$$ are not independent of each other and we can't simply integrate both sides. But $$U$$ (the system internal energy) is an exact differential and is not path dependent on S,V or N. This is even more clear with dP.

$$dW=PdV= F\cdot dx$$ Where dW is an inexact differential

How do you integrate the right hand side? For example taking the partial $$\frac{\partial(PV)}{\partial V}=P$$ Or simply ignore dV/dP terms of full derivative

$$\int\frac{{\partial(PV)}}{\partial V} =PV +f(V)$$ $$\int\frac{{\partial(PV)}}{\partial P} =PV +f(P)$$

So we can see that it is simply probably PV but how can we do this without partials?

• shouldn't there be a minus sign in front of the PdV: dU=TdS-PdV Commented Jun 26, 2022 at 16:53
• Do you want to find the change in U or do you want to find the amount of work done? Commented Jun 26, 2022 at 16:58
• @ChetMiller so I believe -PdV is defined as the work done by the surroundings on the system while dU is the change in internal energy of the system. dV is exact as isn't dU but the integral of PdV is inexact. How is that possible? Commented Jun 26, 2022 at 17:08
• Exact means that there is only one possible value between the initial and final equilibrium states of the system. Inexact means that there are multiple possible values for W between the initial and final equilibrium states of the system, depending also on the other inexact parameter Q. Commented Jun 27, 2022 at 9:58
• @ChemEng the work done $\delta W$ is generally inexact (path-dependent), but for a thermodynamic reversible process $\delta W$ can be written as $p\mathrm{d}V$, which is exact (path-independent). That work becomes path-independent in a reversible process is important, since then $\delta Q = \mathrm{d}U + p\mathrm{d}V$ becomes exact as well. Commented Nov 15, 2023 at 10:51

Thermodynamics needs differentials. However, it is also the field where we can witness the maximum departure from the usual mathematical definitions. It is probably a good idea to try a systematic summary of mathematical definitions before discussing folklore and mathematical facts about differentials in thermodynamics.

Mathematical differentials in a nutshell

Given a real function of $$N$$ real variables ($$f(x_1,x_2,\dots,x_N$$), we can define the differential $$df$$, at a point $$\bar x=(\bar x_1,\bar x_2,\dots,\bar x_N)$$ as $$df= \sum_{i=1}^N \left[\frac{\partial{f}}{\partial{x_i}}(\bar x)\right] (x_i - \bar x_i).$$ From the definition is immediate to see that we can write $$dx_i = x_i- \bar x_i$$, and then $$df= \sum_{i=1}^N \left[\frac{\partial{f}}{\partial{x_i}}(\bar x)\right] dx_i.$$ It turns out that, at the point $$\bar x$$, $$df$$ is the best linear approximation of the variation $$\Delta f = f(x)-f(\bar x)$$, in the sense that $$\Delta f - df = \mathcal{O}(dx_1^2,dx_2^2,\dots,dx_N^2).$$ Moreover, from the definition, it is clear that $$df$$ is a linear function of the variables $$(x_1,x_2,\dots,x_N)$$, and in general a non-linear function of the variables $$(\bar x_1,\bar x_2,\dots,\bar x_N)$$.

After introducing the differential of a function, we can introduce a differential form $$dy$$, i.e. a function of $$(x_1,x_2,\dots,x_N,\bar x_1,\bar x_2,\dots,\bar x_N)$$ defined as $$dy = \sum_{i=1}^N A_i(\bar x) dx_i,$$ where, in general the functions $$A_i(\bar x)$$ are not the partial derivatives of a function of $$N$$ variables.

A differential form $$dy$$ is called exact if there is a function $$y(\bar x)$$ such that $$A_i(\bar x) = \left[\frac{\partial{y}}{\partial{x_i}}(\bar x)\right]. \tag{1}$$ Finally, we can define a line integral of the differential form $$dy$$ over the curve $$\gamma$$, whose parametric expression is $$\bar x(t)= (\bar x_1(t),\bar x_2(t),\dots,\bar x_N(t))~~~~~t \in [a,b],$$ as $$\int_{\gamma}dy =\int_a^b dt \sum_{i=1}^N A_i(\bar x(t)) \frac{d \bar x_i}{dt}. \tag{2}$$ It turns out that if the differential $$dy$$ is exact, $$\int_{\gamma}dy = y(\bar x(b)) - y(\bar x(a))$$.

Notice that all the previous definitions and formulae could be generalized to the case of differential forms on a differential manifold. However, such a generalization is a kind of overshooting in Thermodynamics, where the space of the states is more naturally described as a cone in $$\mathbb R^N$$.

How differentials are used in Thermodynamics?

All previous definitions and formulas have a direct application in Thermodynamics. In particular, the line integrals of differential forms (exact or not) are defined as in formula $$(2)$$, and this should answer part of your question.

There is a unique point where some mathematical confusion is introduced in many textbooks on Thermodynamics. It is the introduction of the so-called inexact differentials.

Strictly speaking, a non-exact differential form, i.e., a differential form with no function $$y$$ satisfying condition $$(1)$$, could be called inexact. However, this is not how some people introduce the inexact differentials.

An example is the case of the internal energy as summarized by the first principle of Thermodynamics: $$\Delta U = U(B)-U(A)=Q + W$$ where $$W$$ and $$Q$$ are the total work, and the total heat exchanged in a thermodynamic transformation from an equilibrium state $$A$$ to another equilibrium state $$B$$.

The internal energy is a function of the state, and for a quasi-static reversible transformation from a state to a near state, we can approximate a small variation $$\Delta U$$ by the differential $$dU$$.

It is attempting at this point to write for a completely general process between such neighbor states. $$dU = dQ + dW.$$ However, $$Q$$ and $$W$$ are not functions of the state. Therefore the introduction of a special symbol $$\delta$$ or $$-\hspace{-1ex}d$$ to make explicit that they are not differentials of a function of state: $$dU = -\hspace{-1ex}d Q + -\hspace{-1ex}dW.$$ Unfortunately, such a notation hides the fact that in the case of a non-quasistatic process, not only $$Q$$ and $$W$$ are not differentials of a function, but they are neither functions of the state variables.

Therefore, I would leave the term inexact differential only to indicate a real differential form of the state variables whose integral is path-dependent. For instance $$- \int p(T,V,N)dV$$ (defined only for a quasi-static process).

• This is great. I always find the math in statistical physics a bit confusing and this has helped clarify a bit Commented Mar 26 at 14:41

In thermodynamics, we use $$-\hspace{-1ex}d W=PdV$$ to evaluate the work done by a process described by a path on the PV-diagram (i.e. a set of equilibrium states). The result is a line-integral, which path-dependent since $$-\hspace{-1ex}d W$$ is inexact. (One does not get something like an anti-derivative to be evaluated at the endpoints.)

Note that, for a given path on the PV-diagram, the result $$\int_{given\ path}-\hspace{-1ex}d W$$ is independent of the nature of the system. (For example, it doesn't matter if we have an ideal gas with $$PV=nRT$$ is the equation of state... or a van der Waals gas with its equation of state.) Geometrically, the work done is equal [up to definitions and sign-conventions] to the signed-area under the process path in the $$PV$$-plane.

Similarly $$dU=PdV+TdS+\mu dN$$ could be used to evaluate [as a line-integral] the change in internal energy for a process described as a path in the $$(V,S,N)$$-space of states of the system. Here, we would need $$P$$, $$T$$, and $$\mu$$ as functions of $$V$$, $$S$$, and $$N$$ [provided by the equations of state] as we move along the path for the process. Since $$dU$$ is exact, we expect that the result depends only on the endpoints.

• So let us ignore dN term for a second. If dU is a exact differential and doesn't depend on the path but I'm still taking a line integral what does it mean? How do all paths add up to the same sum? Commented Jun 26, 2022 at 16:38
• @ChemEng Work done is a line integral that depends on the special path chosen. Different paths between the same endpoints lead to different numerical results, in general (unless the differential form is exact). In certain situations (e.g. special type of paths), the "numerical result" could be expressed in terms of a convenient variable (parameter). Commented Jun 26, 2022 at 16:45
• @ChemEng (How do all paths add up to the same sum?) By analogy, consider the work done by the field in moving a point charge from point A to point B in the field of a fixed point charge. Regardless of the path taken from A to B, the work done is the same. Instead of proving this by exhaustion, one can find an argument using the paths with equipotential portions so that the nonzero contributions come from moving from one equipotential to another. Similar things can be done on the PV diagram. Commented Jun 26, 2022 at 18:14
• thanks. I see that for conservative vector fields the line integral are path independent. So if a field is a gradient of a function(aka a conservative vector field) we can see that line integrals would be path independent. Yet for dU=dw+dq we know dw and dq are inexact but their sum is exact Commented Jun 26, 2022 at 18:37
• @ChemEng Of course, these are special inexact forms (not two randomly selected ones). The exactness of this sum of inexact ones could be interpreted as inexact+exact=inexact. (Similar example: two irrational numbers whose sum is rational: $\pi+(1-\pi)=1$.) Commented Jun 26, 2022 at 19:41