Collisions of molecules against a wall - Effusion rate
Exercise 3
Consider a container filled with a gaseous substance of which a wall is pierced by a hole with surface $A$ opening onto a vacuum.
1) Evaluate the effusion rate ie the number of molecules lost per second.
The rate of effusion is the number of collisions per second against a surface $A$:
Rate of effusion=
Number of "collisions" per second against $A$ =
$\frac{d N}{d t}$=
$A\frac{P}{\sqrt{2\pi \cdot m\cdot k\cdot T}}$
where
$P$ is pressure
$N$ is the number of molecules
$t$ is time
$m$ is the mass of one molecule
$k$ is Boltzmann's constant
$T$ is Kelvin temperature
2) Deduce from there the rate of decrease of the pressure, ie the number of $Pa$ lost per second.
According to the ideal gas law:
$P$ $=$ $\frac{nRT}{V}$ $=$ $\frac{NkT}{V}$
where
$P$ is pressure
$n$ is the mole number
$N$ is the number of molecules
$R$ is the ideal gas constant
$k$ is Boltzmann's constant
$T$ is Kelvin temperature
So we have:
Rate of decrease of the pressure=
$\frac{d P}{d t}$ $=$ $\frac{\frac{d N}{d t}\cdot kT}{V} =$
$\frac{A\frac{P}{\sqrt{2\pi \cdot m\cdot k\cdot T}}\cdot kT}{V} =$
$\frac{A\cdot P}{V}\sqrt{\frac{kT}{2\pi m} }$
3) Conclude to a (theoretical) method to find the molar mass $M$ of a gaseous substance.
By transforming the above expression:
Mass of one molecule=
$m =$
$\frac{kT}{2\pi}\frac{A^2P^2}{V^2}\frac{1}{(\frac{dP}{dt})^2}$
$m$ is expressed in terms of measurable quantities!
