Effusion of a gas from a container

Consider a container filled with a gas at a given pressure. The container is separated from another container containing no gas (i.e. a vacuum) by a partition with a small hole of surface area ${\displaystyle A_{0}}$. The gas from the first container will travel through the hole and into the second container at a given rate. This phenomenon is called effusion and the rate at which the gas travels through the hole is called the rate of effusion.

The rate of effusion can be defined as the product of the hole's surface area, ${\displaystyle A_{0}}$, and the rate at which the particles collide with the walls of the container, ${\displaystyle Z_{w}}$.

Rate =${\displaystyle {\frac {dN}{dt}}=Z_{w}A_{0}}$

Where ${\displaystyle Z_{w}}$ is given by:

${\displaystyle Z_{w}={\frac {p}{\left(2\pi mk_{B}T\right)^{1/2}}}}$

Derivation of the time dependancy on pressure

The rate of effusion derived below is the rate with respect to the gas escaping the filled container.

Rate =${\displaystyle -{\frac {dN}{dt}}=-Z_{w}A_{0}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}$

${\displaystyle N={\frac {pV}{k_{B}T}}}$

${\displaystyle {\frac {d}{dt}}{\frac {pV}{k_{B}T}}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}$

${\displaystyle {\frac {V}{k_{B}T}}{\frac {dp}{dt}}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}$

${\displaystyle {\frac {dp}{dt}}={\frac {-A_{0}p(k_{B}T)^{1/2}}{V\left(2\pi m\right)^{1/2}}}}$

${\displaystyle {\frac {dp}{dt}}={\frac {-A_{0}p(k_{B}T)^{1/2}}{V\left(2\pi m\right)^{1/2}}}}$

Solving the differential by integration.

${\displaystyle {\frac {1}{p}}dp={\frac {-A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}dt}$

${\displaystyle \int _{p_{0}}^{p}\!{\frac {1}{p}}dp=-\int _{0}^{t}\!{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}dt}$

${\displaystyle \int _{p_{0}}^{p}\!{\frac {1}{p}}dp=-{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}\int _{0}^{t}\!dt}$

${\displaystyle \left[\ln \left(p\right)\right]_{p_{0}}^{p}=-{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}\left[t\right]_{0}^{t}}$

${\displaystyle \ln \left(p\right)-\ln \left(p_{0}\right)=-{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}\left(t-0\right)}$

${\displaystyle \ln \left({\frac {p}{p_{0}}}\right)=-{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}t}$

${\displaystyle p=p_{0}\exp \left[-{\frac {A_{0}}{V}}\left({\frac {k_{B}T}{2\pi m}}\right)^{1/2}t\right]}$