# lec 2 - Chem 107B TA Office Hours Kinetic Theory of Gases...

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Chem 107B TA Office Hours

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Kinetic Theory of Gases Model: Gas consists of large number of molecules far apart on average (~10 23 in 1-liter). Molecules have small size compared to distance from neighbor. Collisions between molecules are elastic and random (thermal energy = k B T). No interactions between molecules (no attraction or repulsion). Question: How do we describe macroscopic properties of a gas (e.g. press, vol or temp) in terms of molecular motion (i.e. speed and energy of individual molecules)? PV = nRT PV Energy (i.e. motion) of gas molecules Goal: Derive an expression for Pressure and Temp in terms of average speed of individual molecules. Work Energy
Summary of Kinetic Theory Equations ) ( Law Gas Ideal T Nk PV B = ) ( 3 2 Theory Kinetic v Nm PV = B k v m T 3 2 = M RT m T k v B 3 3 2 = = trans E N PV 3 2 = T k E B trans 2 3 = Translational kinetic energy proportional to PV and Temperature: Mean squared velocity ( ) proportional to PV and Temperature: 2 v Nm PV v 3 2 =

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Statistics of Speed (c) and Energy (E) Speed (c) Fraction (f(c)) Energy (E) Fraction (f(E)) Boltzmann Maxwell ) exp( ) ( T k E E f B α Probability c 1 < c < c 2 : 2 1 ) ( c c dc c f c 1 c 2 Probability E > E a : E a ) 2 exp( ) ( 2 2 T k mc c c f B ) exp( ) ( T k E dE E f B a E a = Mean Speed (<c>): = = 0 ) ( dc c cf c c Mean Energy (<E>): = = 0 ) ( dE E Ef E E
Maxwell Distribution of Speed T k mc B B e T k m c c f 2 2 3 2 2 2 4 ) ( = π v = velocity, has a direction (vector) c = speed, no direction (scalar) Fraction of molecules (dN/N) with speeds c to c+dc: Maxwell Distribution 2 2 2 z y x v v v c + + = dc e T k m c N dN T k mc B B 2 2 3 2 2 2 4 = = 2 1 ) ( c c dc c f N dN Mean Speed (<c>): = = 0 ) ( dc c cf c c

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Average Molecular Speed ( ) T k mc B B e T k m c c f 2 2 3 2 2 2 4 ) ( = π = 0 ) ( dc c cf c dc e c T k m c T k mc B B = 0 2 3 2 3 2 2 4 From integral table: 2 2 0 3 2 1 a dx e x ax = In our case: T k m a B 2 = and thus M RT m T k c B ππ 8 8 = = 2 2 3 2 2 1 2 4 = T k m T k m B B where R = N A k B and M = mN A c c or c =
rms ) T k mc B B e T k m c c f 2 2 3 2 2 2 4 ) ( = π From integral table: In our case, T k m a B 2 = M RT m T k c B rms 3 3 = = 5 2 0 4 8 3 a dx e x ax = [] 2 / 1 0 2 2 / 1 2 ) ( = = dc c f c c c rms 2 / 1 0 2 2 4 2 3 2 4 = dc e c T k m c T B k mc B rms 2 / 1 5 2 3 2 8 3 2 4 = T k m T k m B B where R = N A k B and M = mN A

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## This note was uploaded on 07/20/2010 for the course CHE CHE 118A taught by Professor Lievens during the Winter '08 term at UC Davis.

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lec 2 - Chem 107B TA Office Hours Kinetic Theory of Gases...

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