### Ideal Gas Mixtures

If you have $n_i$ moles of each species (for example $n_A$ moles
of A), you can try to calculate the pressure or volume that that
gas alone (i.e., ignoring the other gases that are around)
exerts/occupies.

##### DEFINITION

**Partial pressure** refers to the pressure that
would be exerted by a species (in a mixture) if there were no other
species present.

##### DEFINITION

The pure component volume, $V_A$, refers to the volume that
would be occupied by a species (in a mixture) if there were no
other species present.

So, in an ideal gas mixture EACH COMPONENT satisfies the ideal
gas law provided the partial pressure or pure component volumes are
used!

$p_AV = n_ART$

or

$PV_A = n_ART$

In this way, the sum of the component pressures (partial
pressures) or volumes (pure component volumes) should sum to the
total pressure or volume:

$p_A + p_B + ... = P$

$V_A + V_A + ... = V$

This is easy to see if you divide either the partial pressure
equation or the pure component volume equation by the ideal gas law
for the total mixture:

$\frac{p_AV = n_ART}{PV = nRT}$

$\frac{PV_A = n_ART}{PV = nRT}$

Note that RT cancels in both equations and that V cancels in the
first and P cancels in the second, also that $n_A/n = y_A$. We can
then rearrange the result to get:

$p_A = y_AP$

$V_A = y_AV$

So The volume fraction (or pressure fraction) of an ideal gas is
equal to the mol fraction! ($V_A/V = n_A/n$)

##### Note:

Since ideal gas molecules do not interact, isobaric mixing of
ideal gases is effectively the same as reducing the pressure or
specific volume of each component (since the partial pressure will
be lowered). Not surprisingly, you simply use the same ideal gas
expressions that we derived earlier.

##### Test Yourself:

1 mol of nitrogen and 2 mol oxygen, both initially at 1 bar and
298 K are mixed at constant P and T in an insulated container. Find
the entropy change of the system.