1
$\begingroup$

Imagine that I have 100 atoms of oxygen and 100 atoms of lead put together in a container. By kinetic theory of gases, they should be considered as point masses thus the partial pressure of oxygen should be equal to lead in the container, is that true ?

Is there something beyond this simple assumption which governs the distribution with some potentials taking into account the mass of the atom and the electron cloud around it?

$\endgroup$

1 Answer 1

0
$\begingroup$

Assuming the oxygen and lead vapour can be treated as ideal gases then yes the partial pressures of the two will be the same. This happens because the pressure is proportional to the momentum of the atoms multipled by their velocity (shout if you want me to explain why this is):

$$\begin{align} P &\propto pv \\ &\propto mv \times v \\ &\propto mv^2 \\ &\propto KE \end{align}$$

where $KE$ is the kinetic energy of the gas atoms. For an ideal gas the average kinetic energy of the gas atoms is $\tfrac{3}{2}kT$, where $k$ is Boltzmann's constant and $T$ is the temperature. So we conclude that the pressure is just proportional to the temperature (which is Charles' law) and does not depend on the mass of the atom.

If the interaction between atoms is too strong to be ignored our assumption that $KE = \tfrac{3}{2}kT$ is no longer valid because energy goes into the potential energy between atoms. In this case there are no simple exact equations to give the partial pressures so we use empirical equations like the Van der Waals equation of state. These tend to rely on empirically measured constants e.g. in the VdW equation the constants $a$ and $b$ have to e measured.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.