SL: Entropy of Mixing (Ideal Gases)

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.


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


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!

$p_AV = n_ART$
$PV_A = n_ART$

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! ($V_A/V = n_A/n$)


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.