Is the difference between the pressures of real gas and ideal gas at the same conditions equal to the latent heat of volume change?

The "real gases" term in the title refers to gases with intermolecular attractive forces. Such gases necessarily have lower pressures than ideal gases at the same temperature $T$ and molar volume $v$ ($P_{real}(T,v)<P_{ideal}(T,v)$), and that because the attractive forces lower down the speeds of the gas molecules which collide at the wall boundary (in a kind of "surface tension" effect). I was wondering if the difference $P_{ideal}(T,v)-P_{real}(T,v)$ is exactly equal to the latent heat of volume change $l_v(T,v)$, which measures how much heat has to be supplied to a real gas to increase it’s volume at constant temperature.

I know that, for example, for Van der Waals gases, which have the following equation of state:

$$P_{vdw} = frac{RT}{v-b} – frac{a}{v^2}$$,

and the latent heat of volume change is really equal to the $frac{a}{v^2}$ term (this term acounts for the intermolecular attractive forces) , which is exactly the difference between ideal pressure and real pressure. So my conjecture is correct at this case, and i think its reasonable to assume it’s correctness for all gases, since the attractive forces actually make negative work on the gas molecules when it expends, so the molecules slow down, the gas temperature decreases, and the volumetric latent heat compensates for this negative work. Therefore i think that $l_v(v,T)$, which is an empirical size, needs to be equal to $P_{ideal}(T,v)-P_{real}(T,v)$.

Despite much googling and searching in the literature, i didn’t succeed in verifying this simple statement. Therefore i’d like to know if there is such a known relation.