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Thermodynamics Rate of Effusion
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[QUOTE="SolsticeFire, post: 2889005, member: 272774"] [h2]Homework Statement [/h2] Effusion is the slow leakage of atoms through a small hole in a container. We'll consider a box at pressure P, with volume V and temperature T. The hole will have the area A, and is in one of the six walls of the container. a) At what rate do atoms (assume mass M) pass through the hole? Use avg (v[SUB]x[/SUB][SUP]2[/SUP])[SUP]1/2[/SUP] as an approximation for avg (v[SUB]x[/SUB]). Equipartition can be used to evaluate avg (v[SUB]x[/SUB][SUP]2[/SUP]). [h2]Homework Equations[/h2] Equipartition: mv[SUB]x[/SUB][SUP]2[/SUP] = kT (k = Boltzman's constant) Ideal Gas Law: PV = NkT [h2]The Attempt at a Solution[/h2] It asks for rate at which atoms pass through the hole and so far I did: --> I need to get number of atoms per second units for answer so I assumed the answer should be in form of N atoms/s. ---> Assuming ideal gas situation the frequency at which the atoms hit the wall with the hole is v[SUB]x[/SUB]/2L. (I'm not sure whether this assumption is so accurate, but I figured at in ideal gas conditions an atom to hit the wall the difference in time has to be 2L/v[SUB]x[/SUB]). ---> The factor of area covered by the hole on the side wall can be written as a ratio (Ahole/Aside). So now we have frequency = [N (Ahole/Aside)v[SUB]x[/SUB]]/2L Then using equipartition and ideal gas: PV = NkT; N = PV/kT and Mvx^2 = kT; vx = sqrt (kT/M) Rewriting the formula I got: [PV/kT (Ahole/Aside) sqrt(kT/M)]/2L simplifying we get: (Ahole/Aside) [PV/{sqrt(kTM)x 2L}] However this implies that as Volume increases rate of atoms going through increases since volume is in the numerator; also as T decreases, Rate goes up. It doesn't add up. Is there a mistake in my line of reasoning here? Or have made a bad assumption with f = v[SUB]x[/SUB]/2L? :S. I have been working on this problem for three days now, and I can't seem to find a mistake in my algebra, but the solution makes no sense. I need this solution to do part b) and c) of the same question. Any guidance will be greatly appreciated. Thanks! [/QUOTE]
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