# How much approximately the pressure would drop?

A container made from from aluminium ($0.5~\mathrm{mm}$ thickness $10^3~\mathrm{cm^3}$) closed system , it has 50 psi (nitrogen) pressure inside at temperature $80^\circ~\mathrm C.$

Approximately :

How much pressure will it be lost after 10 years ? 0.1 psi or 1 psi or more ...

I know that Gas permeation through solid is extremely slow and negligible in most cases. It's hard to get general idea without approximate number .

• What is inside the box? If any measurable amount of gas leaks, it will be more if it's hydrogen than if it's, say, isobutane. And what is on the outside -- vacuum? Jul 16 '16 at 21:15
• inside the box is air Jul 16 '16 at 21:17
• outside is also air Jul 16 '16 at 21:18
• Any answer that you get will be unverifiable, unless you are willing to wait 10 years. If you REALLY need an answer for this, try a radioactive alpha emitter with a relatively short half life, and approximately the same density as the gas you are interested in. ANY leakage into a larger container will be immediately detectable, and the rate of leakage will give you your answer. Jul 17 '16 at 0:21

Consider Fick's second law of diffusion, in one dimension, where $u$ is the concentration of the diffusing gas (in $\mathrm{mol/m^3}$) and $D$ the diffusion coefficient:

$$\frac{\partial u}{\partial t}=D\frac{\partial u}{\partial x}$$ If we assume the concentration of gas outside the container to be much smaller than inside (a reasonable assumption), then the right hand concentration gradient can be determined from Fick's first law to be (with $\tau$ the thickness of the container walls): $$\frac{\partial u}{\partial x}= -\frac{u-u_0}{\tau}$$ Where $u_0$ is the nitrogen concentration outside the box (assumed constant). With the Ideal Gas law: $$pV=nRT$$ $$p=\frac{n}{V}RT=uRT$$ $$u=\frac{p}{RT}$$ Also: $$u_0=\frac{\chi p_a}{RT}$$ Where $p_a$ is atmospheric pressure and $\chi=0.78$ the molar fraction of nitrogen in air. With those relations and $A$ the total surface area of the container, we get: $$-\frac{V}{RT}\frac{dp}{dt}=\frac{AD}{\tau}\Big(\frac{p}{RT}-\frac{\chi p_a}{RT}\Big)$$ $$-V\frac{dp}{dt}=\frac{AD}{\tau}(p-\chi p_a)$$ $$-\frac{dp}{p-\chi p_a}=\frac{AD}{\tau V}dt$$ $$\int_{p_0}^{p(t)}\frac{dp}{p-\chi p_a}=-\frac{AD}{\tau V}\int_0^tdt$$ $$\ln\Bigg(\frac{p(t)-\chi p_a}{p_0-\chi p_a}\Bigg)=-\frac{AD}{\tau V}t$$ We'll call: $$\alpha=-\frac{AD}{\tau V}$$ $$\implies \Large{p(t)=\chi p_a+(p_0-\chi p_a)e^{-\alpha t}}$$ Note that this expression does not contain $T$. $D$ however is temperature dependent as the following figure shows, for diffusion of gases through solids:

As specific values for aluminium/nitrogen are hard to find we'll use the above values to get a crude estimate. First we need to calculate $\alpha$ from:

$A=600 \times 10^{-4}\mathrm{m^2}$

$D=10^{-12}\mathrm{m^2s^{-1}}$

$\tau=0.5\times 10^{-3}\mathrm{m}$

$V=10^{-3}\mathrm{m^3}$

$\implies \alpha \approx 10^{-7}\mathrm{s^{-1}}$

For 10 years:

$t=10\times365\times24\times60\times60=3.15 \times 10^8\mathrm{s}$

This puts the upper estimate of $-\alpha t \approx -30$ and thus $p(\text{10 years})\approx p_0e^{-30} \approx \chi p_a=11.5\mathrm{psi}$.

10 years is of course also quite a long time and $0.5\:\mathrm{mm}$ quite thin for a container holding a gas starting at $5.5\:\mathrm{Bar}$!

• but the temperature $80^\circ~\mathrm C.$ .. should not we use 10^{-16} at least ? Jul 18 '16 at 12:41
• @user118676: You can use $10^{-16}$, which of course decreases the pressure loss enormously. But most sources I've seen give values of $10^{-12}$ for gas/steel at RT. The graph above is also at higher than 80 C temperature range. W/o a more precise value for $D_{Al,N2}$ it's not possible to give a definitive answer to the question.
– Gert
Jul 18 '16 at 13:34