Question

Case challenge! Space Accident.

Answer 1

On board a space station there is a small explosion. The explosion gives a number of small holes in the space station. As head you should decide whether evacuation is necessary. The space station measures 55 m x 12 m x 3m. There are 250 holes from the explosion and the average area of each hole is 1.0 mm2. The technicians aboard promises that they can put a plate over the area with the holes in 12 hours. The air in the space station is a mixture of O2 (40%) and N2 (60%) with a total pressure of 0.5 atm so that the partial pressure of O2 under normal conditions is 0.2 atm. In order to live, people requires a partial pressure of O2 between 0.1 atm. and 0.3 atm. We can assume that the molecular mass of O2 and N2 are the same and equal to 32 g / mol. The temperature of the space station is 295K.

It is now your mission to decide if evacuation is necessary using calculations.
Creative answers are also more than welcome :)

Requirements to solve the problem:
Be familiar with infinitesimal calculus (differential calculus and integral calculus)
Be familiar with Graham's law of effusion.
Have some imagination. :)

Answer 2

sounds like a great problem that you and some friends should talk about over coffee or something

Answer 3

Well when I had to do the problem it was my teacher that was testing my trivia in phys chem.

Answer 4

Wtf

Answer 5

so i got the rms velocity of 479 m/s at 295 K, since they weigh the same, it's the same for both N2 and O2.

I just don't know how to find the flow rate through the hole.

Answer 6

would that be the way of going about it?

ps. didn't write that, it's from MIT

Answer 7

ill just sit a drink juice

Answer 8

False

Answer 9

What are you trying to determine in this question?

Answer 10

if they should evacuate the station, more specifically, if the partial pressure of oxygen in the station will decrease below 0.1 atm in next 12 hours following.

Answer 11

In reality, I would think they should evacuate :p

Answer 12

sooo i've been doing more thinking..

volume and pressure the moles of gas (N2 and O2)=40896.07

since thats at the partial pressure of oxygen of 0.2 atm, the total loss of gas should not be of 20448.04 moles. Dividing by 250 (holes) the by 12 (hrs) and 60 (min), i arrived at a max flux that should not exceed 0.1136 moles/min*hole

I would say that the initial flux rate cannot exceed this value, because the rate will slow down as the pressure decreases, so as long as the initial rate is not higher than 0.1136 moles/min*hole, there will be P(O2)>0.1 atm for the next 12 hours ... i still don't know how to find the rate of gas flow though.

Answer 13

If I had to convert the text problem into math we are interested in following expression:
\[\LARGE rate _{pressure}=\frac{ dp }{ dt }\]

Answer 14

Lets approximate and assume O2 and N2 are ideal gasses:
\[\LARGE pV=nRT\]
\[\LARGE pV=\frac{ N }{ N _{A} }RT\]
\[\LARGE N=\frac{ pVN _{A} }{ RT }\]
\[\LARGE dN=d \left( \frac{ pVN _{A} }{ RT } \right)=\frac{ VN _{A} }{ RT } dp\]
\[\LARGE dp=\frac{ RT }{ VN _{A} } dN\]
\[\LARGE \frac{ dp }{ dt }=\frac{ RT }{ VN _{A} } \frac{ dN }{ dt }\]

Now we have an expression for the rate where dN/dt is Graham's law of effusion.

Answer 15

Use Graham's law of effusion and use separation of variables along with integration over the right lengths.

Answer 16

Forgot to write you need to be familiar with Knudsen method too b/c then you can write Graham's law of effusion as:
\[\LARGE -\frac{ dN }{ dt }=\frac{ pA _{0} }{ \sqrt{2 \pi mkT} }=\frac{ pA _{0}N _{A} }{ \sqrt{2 \pi MRT} }\]
last equal sign is true b/c:
\[\LARGE R=N_{A}k\]
\[\LARGE M=mN _{A}\]

Answer 17

ohh sweet, dude. you took the situation in terms of change in pressure not the loss of gas !

I actually saw that last formula somewhere and i figured that in order to do the question you had to derive it from Graham's law. Thanks for posting the question, i liked this.

Answer 18

It is not done yet! :D you can do the rest :)

Answer 19

But you are right @aaronq when you say evacuation is not needed. so nice :D

Answer 20

haha woo for not evacuating and abandoning the mission. i guess i'll have to try to finish this up later

Answer 21

@aaronq If you like I can try derive the effusion rate (dN/dt) as I've written it?

Answer 22

awesome dude ! thanks for writing this out :)

Answer 23

The only problem with it is that I do not know how to show it is -dN/dt... :/ can only show it is the rate using a drawing.

Answer 24

whatever, this is good enough dude lol

i'm gonna try going through the derivation, this might come in handy one day

Answer 25

It is just a try so there may be mistakes.

Answer 26

okay, if i find something that seems funny, i'll let you know. How's your summer?

Answer 27

OMG...This is truly a total challenge lol :P Interesting though! Well from what I could follow at least hahaa :P

Answer 28

\[\LARGE \frac{ dp }{ dt }=\frac{ RT }{ VN _{A} } \frac{ dN }{ dt }\]
\[\LARGE \frac{ dp }{ dt }=\frac{ RT }{ VN _{A} } \frac{ -pA _{0}N _{A} }{ \sqrt{2 \pi MRT} }\]
\[\LARGE \frac{ dp }{ dt }=\frac{ -pA _{0} }{ V }\frac{ RT }{ \sqrt{2 \pi MRT} }=\frac{ -pA _{0} }{ V }\sqrt{\frac{ RT }{ 2 \pi M }}\]
Seperation of variables:
\[\LARGE \frac{ 1 }{ p } dp=\frac{ A _{0} }{ V }\sqrt{\frac{ RT }{ 2 \pi M }} dt\]
To make this a bit more easy we define a new function (just to make the integration more simple):
\[\LARGE \tau = \frac{ A _{0} }{ V }\sqrt{\frac{ RT }{ 2 \pi M }}\]
It most then follow that:
\[\LARGE \frac{ 1 }{ p } dp=\tau dt\]
We integrate this from the time 0 to t and the pressure to the time 0 (p_0) to the pressure to the time t (p_1):
\[\LARGE \int\limits_{p _{0}}^{p _{1}} \frac{ 1 }{ p } dp=\int\limits_{0}^{t}\tau dt\]
TO BE CONTINUED.

Answer 29

Correction:
\[\LARGE \frac{ 1 }{ p } dp=-\tau dt\]
\[\LARGE \int\limits\limits_{p _{0}}^{p _{1}} \frac{ 1 }{ p } dp=\int\limits\limits_{0}^{t}-\tau dt\]

Answer 30

We solve the integrals (remember that tau is not depending on t):
\[\LARGE \ln(p _{1})-\ln(p _{2})=-\tau t\]
\[\LARGE \ln \left( \frac{ p _{1} }{ p _{0} } \right)=-\tau t\]
We take the exponential function on both sides:
\[\LARGE \frac{ p _{1} }{ p_{0} }=e ^{-\tau t}\]
We then get:
\[\LARGE p _{1}=p_{0}*e ^{-\tau t}\]
Substutite the expression for tau:
\[\LARGE p=p _{0} \exp \left(- \frac{ A _{0} }{ V }\sqrt{\frac{ RT }{ 2 \pi M }}*t \right)\]
This most be our final equation as it contain only known variables.

Answer 31

Sorry I make some mistakes :/
\[\LARGE \ln(p _{1})-\ln(p _{0})=-\tau t\]

Answer 32

great @Frostbite !