# van der Waals equation for deviation of gases from ideal behavior

$(P + an^2/V^2)(V - nb) = nRT$

I understand that the term $nb$ is subtracted from the volume term as the available volume for the gas molecules to move around decreases in the case of a real gas due to the volume of the molecules themselves.

However, I don't understand why the term $an^2/V^2$ is added to the pressure term. Because of the attractive/repulsive forces, shouldn't the pressure on the walls of the container by the molecules also decrease?
(I thought about this and guessed that due to repulsive forces, the pressure must be increasing, but what about due to attractive forces? Does the sign of $a$ depend on the gas and hence affect the equation? I may be wrong here.)

• It's pressure under which gas is held. – Mithoron Apr 16 '18 at 18:46

The term $-an^2/V^2$ is the reduction of pressure due to attractive or cohesive forces between molecules. This is most easily seen by writing the equation as $\displaystyle p_\text{obs}= \frac{nRT}{V-b}-\frac{an^2}{V^2}$ .
While the constants $a$ and $b$ are considered to be empirical they do have a physical basis. As you mention $b$ accounts for the finite volume of the molecules and $a$ for the attractive force between molecules.
The size of the force will be proportional to the product of the number of molecules /unit area and the size of the inward force. Both of these will be be proportional to the density ($Nm/V$) or equivalently molar concentration. The reduction in pressure can thus be written as $-a(n/V)^2$ where $a$ is a positive constant that depends on the gas and is determined only by experiment.
• Couldn't we also write the equation as $V = nRT/(P - an^2/V^2) - nb$, then? – Shreya Apr 17 '18 at 5:14
• yes anyway round , but why as this is not the solution to the equation as $p$ is above, since volume its on both side of the equation. – porphyrin Apr 17 '18 at 12:07