統計熱力學與速率理論/滲透

氣體從容器中滲透

考慮一個充滿一定壓力的氣體的容器。容器與另一個不含氣體的容器（即真空）之間用一個帶有小孔的隔板隔開，小孔的表面積為$A_{0}$。第一個容器中的氣體將以一定速率透過小孔進入第二個容器。這種現象稱為滲透，氣體透過小孔的速度稱為滲透速率。

滲透速率可以定義為小孔表面積$A_{0}$與粒子碰撞容器壁的速度$Z_{w}$的乘積。

速率 =${\frac {dN}{dt}}=Z_{w}A_{0}$

其中$Z_{w}$由下式給出

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

壓力對時間依賴性的推導

下面推導的滲透速率是相對於氣體從充滿的容器中逸出的速率。

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

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

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

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

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

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

透過積分求解微分方程。

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

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

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

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

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

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

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