# Effusion changing pressure

1. Dec 21, 2017

### Physgeek64

1. The problem statement, all variables and given/known data
A thin-walled container of volume 10−3 m3 is evacuated to a pressure of 10−7 mbar. The vessel is surrounded by air at 1 bar and 290 K. If a small hole of area 10−17 m2 is made in the wall of the container, how long does it take the pressure inside to rise to 10−6 mbar?

2. Relevant equations

3. The attempt at a solution
$\frac{dN}{dt}=A\phi = \frac{A<v> N}{4V}$
$\frac{dN}{N}=\frac{A<v>}{4V} t$
$t=\frac{4V}{A<v>}\ln{\frac{P_2}{P_1}}$

However this requires us to know m and i can't see how to figure this out. Therefore i am assuming my solution is wrong....

2. Dec 21, 2017

Your original differential equation is incorrect. You are assuming the effusion only goes in one direction is this problem (that is ok), but the equation should read $\frac{dN}{dt}=\frac{A n \bar{v}}{4 }$ where $n$ is the number of particles per unit volume of the outside air. ($N$ is the number of particles of air inside the container). This $n$ is constant and does change. One number you will need is $\bar{v}$. Average speed $\bar{v}$ in a Maxwell-Boltzmann distribution is given by $\bar{v}= (\frac{8}{\pi}) (\frac{kT}{M})^{1/2}$ , editing: $\bar{v}=(\frac{8 kT}{\pi M})^{1/2}$, where $M$ is the (approximate) mass of a molecule of air. Since $N_2$ has M.W.=28, and $O_2$ has M.W.=32 , using an average M.W.=30 is an ok approximation, even though the air is 78% $N_2$ and 21% $O_2$.

Last edited: Dec 21, 2017
3. Dec 21, 2017

### Physgeek64

Oh okay, so will i end up with a simple linear relationship between t and N?

4. Dec 21, 2017

Yes. The pressure inside the container is small enough compared to the outside pressure that the rate of effusion into it is much higher than any effusion back out. And yes, the effusion rate into it will be constant. $\\$ And don't forget to include the initial condition of $N_o$ that you start with already inside the container. $\\$ Just for an additional detail: In its entirety $\frac{dN}{dt}=A (\frac{n \bar{v}}{4}-\frac{N \bar{v}}{4 V})$, and this second term on the right side can be considered small compared to the first term for this problem. (No need to solve the more detailed exact differential equation, because the second term is negligible here).

Last edited: Dec 21, 2017
5. Dec 21, 2017

@Physgeek64 See also the additions to post 4 above. $\\$ Also one minor correction to the above: $\bar{v}=( \frac{8 kT}{\pi M})^{1/2 }$.