Im confused on what the significance of exact differentials are in physics, specifically in thermodynamics. In my book posted here it talks about how by requiring that $dU$ be exact it leads to equation (1.4). So my questions are, How and why do we have the requirement that $dU$ be exact? What do exact differentials tell us in physics in general and in thermodynamics and what do they allow us to do?
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2 Answers
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First, requiring $dU$ to be exact leads to (1.4) because of Clairaut's Theorem - just to clear things up.
Now, the requirement for $dU$ to be exact comes from the fact that it's a state variable. That means that if you have some process in which your system evolves into different states, eventually going back to the original one, $U$ will not have changed, no matter what said process was. Mathematically, this can be represented using line integrals as:
$$\oint dU=0 $$
The integral over a closed loop of internal energy is $0$. This can also be called path independence. Now, there is a theorem in multivariable calculus, the gradient theorem that states that for a function to hold this condition, it must be the gradient of some other function. Another common result is that $dF=\nabla F \cdot d\vec{r}$. This is also the definition of an exact differential.
In other words, $dU$ is exact because then, you have (1.3) as your book states, which can be rewritten as $dU=\nabla U \cdot d\vec{r}$ where $\vec{r}$ is the vector for the volume-pressure space. And then, you have the integral condition stated above which is required for a state variable such as $U$.
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$\begingroup$ Does this apply only for path independent conditions? like in other areas of physics like mechanics? And just to make sure essentially the byproduct of path independence means that we can write dU as $dU=\nabla U \cdot d\vec{r}$ where $\vec{r}$ $\endgroup$ Commented Sep 24, 2022 at 20:35
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1$\begingroup$ @goforhenry In mechanics, this condition applies to things such as gravitational potential energy. $\endgroup$– agaminonCommented Sep 24, 2022 at 22:13
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The other answer gave you a mathematical perspective. This will provide a physical perspective.
Physically, the difference between an exact differential and an inexact differential with respect to thermodynamic variables is that an exact differential is a differential "change" in a quantity whereas an inexact differential is a differential "amount" of a quantity.
"Changes" are associated with thermodynamic properties such as internal energy, entropy, pressure, temperature, volume, i.e., the properties of a system. "Amounts" are associated with quantities that are not thermodynamic properties, chiefly work and heat.
So the differential version of the first law is commonly written as
$$du=\delta q - \delta w$$
where $du$ is the differential change in internal energy whereas $\delta q$ and $\delta w$ are the differential amounts of energy transferred between the system and the surroundings in the form of heat and work, which are not thermodynamic properties.
Hope this helps.
