Spherical gas chamber, change of pressure

[Jotun.uu]

Homework Statement

A sphere is filled with a gas which has a constant temperature T. The pressure in the sphere starts at p0 and the spheres radius is r. Assume that a small area (A) of the sphere is held at a very low temperature, while the rest of the surface holds the temperature T. Also assume that all the molecules that hit the surface A condense and do not return to their gas state. Calculate the time it takes for the pressure to fall from p0 to p.

Given information
r = 10 cm
A = 1 cm^2
T = 300 K (kelvin)
m = 2.98*10^-26 kg
p0 = 10 torr
p = 1*10^-4 torr

Homework Equations

(1) Amount of collisions ns = p / sqrt(2*PI*m*Kb*T)
(2) pV = NKbT, where Kb is Boltzmanns constant
(3) Collisions per area ns*A

The Attempt at a Solution

If you multiply the amount of collisions with the area being held at a low temperature i reckon you should get the amount of molecules hitting it, which then integrated should give a time.
Thus, dN/dp = (dN/dp)*(dp/dt). From (2) i know that dN/dp = V/(KbT).
and dNdp/dpdt is the amount of molecules which change the pressure, so dNdp/dpdt = -ns*A (i make it negative because they decrease go down as they condense).

So far i think i have it mostly right, from looking at the correct answer written by my teacher. But if you see a problem with the way i think please tell me, as my teachers answer is very non descriptive in its thought process.

So now i have (from (1) and(3)) dNdp/dpdt = -A*p / sqrt(2*PI*m*Kb*T).
With some math-e-magic [dN/dp = V / (Kb*T)];
V*dp/(Kb*T*dt) = -A*p / sqrt(2*PI*m*Kb*T).
I isolate dp/p = (-A*Kb*T / (V * sqrt(2*PI*m*Kb*T))) dt

And here is where things go awfully wrong for me! I must be making some fundamental mistake when integrating or something! I feel like i should integrate with regard to the time so i should get something like

ln(p0/p) = -t*V *sqrt(2*PI*m*Kb*T) / (A*Kb*T) and then isolate t but my teachers answer tells me this is wrong...

please help!

t should become 3.24 seconds. The equation my teacher gets for t is -(V/A)ln(p/p0)(2*PI*M/(R*T)) where M = m * Avogadros constant and R = Kb * Avogadros constant. I don't understand where the square root has gone for 1 thing!

 

[Jotun.uu]

(Bump). I Really do need an answer to this!

 

[kloptok]

I can't see where you go wrong, everything you say seems to be correct. I calculated it by your method and found t =3.24 sec..

Step by step (k is Boltzmann's constant):

[tex]
1. \ \ \ \frac{-n_s A k T}{V} = \frac{d p}{d t} \\
[/tex]
[tex]
2. \ \ \ -\frac{k T A}{V \sqrt{2 \pi m k T}} \int ^t _0 dt = \int ^p _{p_0} \frac{dp}{p} \\
[/tex]
[tex]
3. \ \ \ \frac{A}{V} \sqrt{\frac{kT}{2 \pi m}} (-t) = \ln{\frac{p}{p_0} } \\
[/tex]
[tex]
4. \ \ \ t=\frac{V}{A} \sqrt{\frac{2\pi m}{kT} } \ln{\frac{p_0}{p}}
[/tex]

which numerically gives t=3.24 seconds. Your teacher seems to be wrong about his/her equation, there is definitely a square root in there (and there definitely is one if you want to get t=3.24 s).

 

[Jotun.uu]

Thank you so much for your answer. Makes perfect sense. I knew i was not crazy!

I will report this discrepancy to my teacher post haste =)Edit: Are you sure you are susposed to reverse ln(p(p0)? Is it to get rid of the minus sign? Because -ln(p/p0) = ln( (p/p0)^-1 ) ??

 

[kloptok]

That's true, of course you don't have to reverse the logarithm. I just did it since it felt more comfortable to get a positive sign in the expression for t. (I know that p0>p so ln(p0/p) and thus t has a positive sign.)

It is generally true that
[tex]
\ln{\frac{a}{b}}=\ln{\bigg ( \frac{b}{a} \bigg )^{-1}}=-\ln{\frac{b}{a}}
[/tex]

I'm glad I could be of help! And sure, you can't always trust what the professor says and that's what being a physicist is really all about - always question things.

 

[Jotun.uu]

Yeah. Having a positive expression for time did always sit better with me as well ;)