# The Van der Waals law

## The law

The → ideal gas law considers the molecules or atoms of a gas as point particles that exert a pressure by striking the wall of the container that contains them.
In fact, Van der Waals found it necessary to introduce corrections due to the fact
- that the particles of a gas do not only hit the walls, but also suffer collisions between them.
- that these particles are not punctual but have a specific volume
These two corrections are translated by both → Van der Waals parameters (constants) $a$ and $b$ which differ from one gas to another:

$(P\; +\;\frac{n^2\cdot a}{V^2})(V- n\cdot b)$$ = $$n\;R\;T $

with:
$a$ Parameter (takes into account interactions)
$b$ Parameter (takes into account the reduction of the total volume because of the particulate volumes of the particles)
$P$ Pressure
$V$ Volume
$n$ Number of moles
$T$ Kelvin temperature
$R$ Ideal gas constant = $0,08205 \frac{L\;atm}{mole\;K} $

## Validity

Let's see what happens when the volume becomes very large for a fixed number of moles of gas:
$\lim_{V\to \infty}\frac{n^2\cdot a}{V^2}$ $ = $$0$
The first factor of the first member of the law is thus reduced to $P$
$\lim_{V\to \infty}(V-n\cdot b)$ $ = $ $\lim_{V\to \infty}V(1-\frac{n\cdot b}{V})$ $ = $ $\lim_{V\to \infty}\;V$
The second factor of the first member of the law is thus reduced to $V$
and then we find the ideal gas law of perfect gases:
$P\;V$$ = $$n\;R\;T $

In practice we use the Van der Waals law for concentrated gases !

## Examples

1) Let's calculate the pressure exerted by one mole of $ O_2 $ confined in a volume of $ 1 \; L $
a) by the ideal gas law:
$P$ $=$ $\frac{n\cdot R\;\cdot T}{V}$ $=$ $\frac{1\cdot 0,08205\cdot 273,15}{1}$ $=$ $22,41\;atm$
b) by the law of Van der Waals:
$P$ $ = $$ \frac{n\cdot R\cdot T}{V- n\cdot b}-\frac{n^2\cdot a}{V^2}$$ = $$\frac{1\cdot0,08205\cdot273,15}{1-0,0318\cdot 1} - \frac{1^2\cdot 1,36}{1^2} $$ = $$21,79\;atm$
2,8% deviation !
Now let's calculate the pressure exerted by one mole of $ O_2 $ confined in a volume of $ 10 \; L $( less dense gas !)
a) by the ideal gas law:
$P$ $=$ $\frac{n\cdot R\;\cdot T}{V}$ $=$ $\frac{1\cdot 0,08205\cdot 273,15}{10}$ $=$ $2,241\;atm$
b) by the law of Van der Waals:
$P$ $ = $$ \frac{n\cdot R\cdot T}{V- n\cdot b}-\frac{n^2\cdot a}{V^2}$$ = $$\frac{1\cdot0,08205\cdot273,15}{10-0,0318\cdot 1} - \frac{1^2\cdot 1,36}{10^2} $$ = $$2,235\;atm$
0,3% deviation !