What is the physical interpretation of "internal pressure"?
See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here. 

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



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, something that also follows from the equipartition theorem. 

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

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 $(2)$ 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 $\pi_T &gt; 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
If $\pi_T &lt; 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.