What is the physical interpretation of "internal pressure"?


> 
> $$ \require{begingroup}
> \begingroup
> \newcommand{\md}[0]{\mathrm{d}} \md U = \pi\_T\,\md V + C\_V \,\md T$$
>  where $\pi\_T$ is the internal pressure and is given by $\displaystyle {\left(\partial U \over \partial V\right)\_T}$.
> 
> 
> 




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* What does the internal pressure actually physically mean? I can see it is the slope of the $U$-$V$ graph at constant temperature but that does not clear things up.
* Also why is this quantity called the internal pressure and not something else? $\endgroup$




See [orthocresol's](https://chemistry.stackexchange.com/users/16683/orthocresol) derivation of an internal pressure $\pi\_T$ relation $(1)$ [here](https://chemistry.stackexchange.com/a/59356/27806). 


$$\pi\_T = T\left(\frac{\partial P}{\partial T}\right)\_V - P\tag{1}$$




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Internal pressure $\pi\_T$ measures what you would expect from its definition:


$$\pi\_T \equiv \left(\frac{\partial U}{\partial V}\right)\_T.$$


It shows *how internal energy changes when volume is changed and temperature is constant*. For ideal gases the [change is zero](https://chemistry.stackexchange.com/a/59356/27806), something that also follows from the [equipartition theorem](https://en.wikipedia.org/wiki/Equipartition_theorem#Ideal_gas_law). 


$$\mathrm{d}U\_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)\_T = 0 \ \ \text{for an ideal gas}$$


In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles. 


As a first approximation to model real gases, the [van der Waals equation](https://en.wikipedia.org/wiki/Van_der_Waals_equation#Derivation) $(2)$ is used.


$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$


Hence,


$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$


Using relations $(1)$ and $(3)$


$$\left(\frac{\partial P}{\partial T}\right)\_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$


Again via formula $(1)$


$$\pi\_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)\_V} - P = a\frac{n^2}{V^2}\tag{4}.$$


Result $(4)$ $\pi\_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi\_T$ is called *internal pressure* will be given when $(4)$ is substituted into $(2)$:


$$\left(P+\pi\_T\right)\left(V-nb\right)=nRT,$$


or even better, $(3)$:


$$P = \frac{nRT}{V-nb} - \pi\_T.$$


* If $\pi\_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
* If $\pi\_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.


 2

Assume that the gas is enclosed in a cylinder with a movable piston but the piston is *not* moving now. If the piston - wall interface is frictionless then the *internal* pressure (the one you are asking about) is the same as the external pressure (the one that can be controlled by mechanical means, e.g., by pushing or pulling on it.) If there is no friction and the process is reversible then the internal pressure is equal to the eternal pressure (the control variable) and is a state variable one that can be described by some other equilibrium parameters, such as $U$ or $T$ or $V$. If there is friction between the piston and the wall then depending on whether one pushes the piston in or pulls it out the internal pressure maybe larger or smaller than the external one and there is no such functional relationship among these parameters.


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