Effusion of a gas from a container
edit
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
A
0
{\displaystyle A_{0}}
The rate of effusion can be defined as the product of the hole's surface area,
A
0
{\displaystyle A_{0}}
Z
w
{\displaystyle Z_{w}}
d
N
d
t
=
Z
w
A
0
{\displaystyle {\frac {dN}{dt}}=Z_{w}A_{0}}
Z
w
{\displaystyle Z_{w}}
Z
w
=
p
(
2
π
m
k
B
T
)
1
/
2
{\displaystyle Z_{w}={\frac {p}{\left(2\pi mk_{B}T\right)^{1/2}}}}
Derivation of the time dependancy on pressure
edit
The rate of effusion derived below is the rate with respect to the gas escaping the filled container.
−
d
N
d
t
=
−
Z
w
A
0
=
−
p
A
0
(
2
π
m
k
B
T
)
1
/
2
{\displaystyle -{\frac {dN}{dt}}=-Z_{w}A_{0}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}
N
=
p
V
k
B
T
{\displaystyle N={\frac {pV}{k_{B}T}}}
d
d
t
p
V
k
B
T
=
−
p
A
0
(
2
π
m
k
B
T
)
1
/
2
{\displaystyle {\frac {d}{dt}}{\frac {pV}{k_{B}T}}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}
V
k
B
T
d
p
d
t
=
−
p
A
0
(
2
π
m
k
B
T
)
1
/
2
{\displaystyle {\frac {V}{k_{B}T}}{\frac {dp}{dt}}={\frac {-pA_{0}}{\left(2\pi mk_{B}T\right)^{1/2}}}}
d
p
d
t
=
−
A
0
p
(
k
B
T
)
1
/
2
V
(
2
π
m
)
1
/
2
{\displaystyle {\frac {dp}{dt}}={\frac {-A_{0}p(k_{B}T)^{1/2}}{V\left(2\pi m\right)^{1/2}}}}
d
p
d
t
=
−
A
0
p
(
k
B
T
)
1
/
2
V
(
2
π
m
)
1
/
2
{\displaystyle {\frac {dp}{dt}}={\frac {-A_{0}p(k_{B}T)^{1/2}}{V\left(2\pi m\right)^{1/2}}}}
1
p
d
p
=
−
A
0
V
(
k
B
T
2
π
m
)
1
/
2
d
t
{\displaystyle {\frac {1}{p}}dp={\frac {-A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}dt}
∫
p
0
p
1
p
d
p
=
−
∫
0
t
A
0
V
(
k
B
T
2
π
m
)
1
/
2
d
t
{\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}
∫
p
0
p
1
p
d
p
=
−
A
0
V
(
k
B
T
2
π
m
)
1
/
2
∫
0
t
d
t
{\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}
[
ln
(
p
)
]
p
0
p
=
−
A
0
V
(
k
B
T
2
π
m
)
1
/
2
[
t
]
0
t
{\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}}
ln
(
p
)
−
ln
(
p
0
)
=
−
A
0
V
(
k
B
T
2
π
m
)
1
/
2
(
t
−
0
)
{\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)}
ln
(
p
p
0
)
=
−
A
0
V
(
k
B
T
2
π
m
)
1
/
2
t
{\displaystyle \ln \left({\frac {p}{p_{0}}}\right)=-{\frac {A_{0}}{V}}{\left({\frac {k_{B}T}{2\pi m}}\right)}^{1/2}t}
p
=
p
0
exp
[
−
A
0
V
(
k
B
T
2
π
m
)
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]}
See also: Wikipedia:Effusion 
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aabcc51e97245026273e2fc42d56d5aa6419920e)
![{\displaystyle p=p_{0}\exp \left[-{\frac {A_{0}}{V}}\left({\frac {k_{B}T}{2\pi m}}\right)^{1/2}t\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/062769e1d070259190f54c0f227868bad826744f)