How do I calculate the molar volume and pressure correction terms in the van der Waals equation of state for $"CO"_2$ if the density of $"CO"_2$ at a certain temperature is $"4.4 g/L"$, while $a = "3.6 L"^2cdot"atm/mol"^2$ and $b = "0.04 L/mol"$?

Well, real gases have intermolecular forces, don't they? And thus, we use the van der Waals equation of state to account for such forces: $P = (RT)/(barV...

First off, the van der Waals equation looks like this: $\mathbf(P = (RT)/(barV - b) - a/(barV^2))$ where: $P$ is the pressure in $"bar"$s. $R$ is the universal...

Van der waals forces or what is also called London Dispersion forces (LDF) increase with the increase of surface area. For example, we know that the only possible intermolecular forces...

The attractive van der Waals forces between helium atoms are so weak that the normal boiling point of liquid helium is only about 4.2 K, but without them, helium could...

The stronger the Van der Waals forces, the higher the boiling point. To melt or boil a simple molecular substance, we only have to overcome the intermolecular forces. Looked at...

Van der Walls include any kind of inter- or intra-molecular force other than covalent and ionic bonds, including all interactions between: charges and dipoles dipoles and dipoles monopoles and...

This is the conventional order. Ionic bonds are of course non-molecular interactions.

As the name suggests, vanderwaal's force is merely a force of attraction b/w two molecules. It occurs when the protons of one molecule attract the electrons of the other....

The van der Waals equation is $color(blue)(bar(ul(|color(white)(a/a) (P + (n^2a)/V^2)(V - nb) = nRTcolor(white)(a/a)|)))" "$ $P + (n^2a)/V^2 = (nRT)/(V - nb)$ $P = (nRT)/(V - nb)- (n^2a)/V^2$...

$a = "5.493 L"^2cdot"bar/mol"^2$ $b = "0.06412 L/mol"$ $P = "3.875 bar"$ Using the result derived in full that: $a = (27R^2T_c^2)/(64P_c)$ $b = (RT_c)/(8P_c)$ and the...