# How much pressure would I lose on one side?

The equation for the gas flow rate (in pressure*volume/time units) $$Q = {kA\over d} (p_1-p_2)$$ where $A$ is the surface area, $d$ the wall thickness, $p_{1,2}$ the pressures on either side of the wall, and $k$ is the permeation conductivity.

How much pressure would I lose on one side of the wall? Because I get number (in pressure*volume/time units) and I need to know how much pressure I will lose over period of time.

thank you very much

• if somebody could write an small example , i would be very thankful . – user118676 Jul 7 '16 at 15:58
• what are 'pressure*volume/time' units? Usually gas flow rate is measured in 'volume/time' or perhaps 'mass/time' units. In any case if outside the container you have constant pressure (say atmospheric) and then you simply need a differential equation which relates the derivative of mass in the container with respect to time to the outflux of material $Q$ through the difference in pressure. – nluigi Jul 7 '16 at 18:29

A closed porous container consists of a gas at partial pressure $p_1$ which is less than partial pressure $p_2$ of that gas in the ambient. We shall assume that $p_2$ is constant and to further simplify matters also assume that temperature of gas inside and outside the container is the same and constant with time, say $T$. Now if $p_1(t)$ is the pressure inside the container at $t\geq 0$ then its density is, $\rho_1(t)=f(T,p_1(t))$, where $f$ is some function that characterizes the gas (for e.g. you may assume ideal gas model, in which case $f(T,p_1(t))=\frac{p_1}{R_{gas} T}$). If container volume is $V$, then mass of gas inside that container at any particular time is, $m_1(t)=\rho_1(t)V$. Since pressure difference $p_1(t)-p_2$ is reducing with time so is the flow rate, $Q(t)$.
At a particular time instant, rate at which mass of gas is being lost from the container is, $\frac{dm_1}{dt}=-\rho_1(t)Q(t)=-f(T,p_1(t))\frac{kA}{d}(p_1-p_2)$. But since we also have $m_1(t)=\rho_1(t)V$ we get by differentiating, $\frac{dm_1}{dt}=\frac{d\rho_1}{dt}V=\frac{\partial f}{\partial p_1}\frac{dp_1}{dt}V$. Equating the two we get a differential equation for $p_1(t)$ which may be solved with the initial condition, $p_1(0)=p_0$.