Lectures25-35 - dP dt = − ° k B T V ± pA(2 πmk B T 1 2...

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CHEM 350 Lectures 25-35, March 8, 10, 12, 15, 17, 19, 22, 24, 26, 29, 31 2010 Graham’s law example: A: known molar mass B: unknown molar mass At fixed T , and P , for a given hole size: rate of escape of A rate of escape of B = M B M A = m B m A Time dependence of the pressure of an e ff using gas: dP dt = k B T V dN dt where we used p = Nk B T V The rate of change of molecules is equal to the collision frequency with the hole ( Z W A 0 : rate of e ff usion) and therefore: dN dt = Z W A 0 note the negative sign (the concentration decreases). In terms of P an d T , we have: dN dt = P (2 π mk B T ) 1 / 2 A 0 Using the above in the pressure expression:
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Unformatted text preview: dP dt = − ° k B T V ± pA (2 πmk B T ) 1 / 2 dP dt = − ° k B T 2 πm ± 1 / 2 pA V integrating, ² P P dP ° P ° = − ° k B T 2 πm ± 1 / 2 A V ² t dt ° ln P P = − ° k B T 2 πm ± 1 / 2 A V t or P = P e − t/τ where τ = V A ° 2 πm k B T ± 1 / 2 exponential decay oF pressure with time. The halF-liFe is the time it takes For the pressure to decay to halF its initial value: P 2 = P e − t 1 / 2 /τ solving For t 1 / 2 : t 1 / 2 = τ ln 2 1...
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