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Some typical numbers: STP: 6.02  1023 molecules / 22.4 liters = 1/[2(N/V)] Assumes volume swept out is independent of whether collisions occur (not.

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Presentation on theme: "Some typical numbers: STP: 6.02  1023 molecules / 22.4 liters = 1/[2(N/V)] Assumes volume swept out is independent of whether collisions occur (not."— Presentation transcript:

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2 Some typical numbers: STP: 6.02  1023 molecules / 22.4 liters = 1/[2(N/V)] Assumes volume swept out is independent of whether collisions occur (not a bad assumption in most cases) z = 2 c (N/V)

3 P = 1 Torrz = 5.45  106 sec -1 (P = 1/760 atm) (P = 1/760 atm)

4 Distribution of Molecular Speeds Real gases do not have a single fixed speed. Rather molecules have speeds that vary giving a speed distribution. This distribution can be measured in a laboratory (done at Columbia by Polykarp Kusch) or derived from theoretical principles.

5 Collimating slits Synchronized rotating sectors Detector  Box of Gas at temperature T Molecular Beam Apparatus for Determining Molecular Speeds Whole apparatus is evacuated to roughly 10-6 Torr!

6 0 Maxwell-Boltzmann Speed Distribution ∆N is the number of molecules in the range c to c +∆c and N = Total # of molecules. x y = e -x 1 (c 2 )∆c (  N/N) = 4  [m/( 2  kT )] 3/2 e -[(1/2)mc / kT] 2x = (1/2)mc2/kT ∆N/N is the fraction of molecules with speed in the range c to c+∆c

7 For our case x = (1/2) mc2/kT = /kT, where  = K.E. 0 x y = e -x 1 c c2c2c2c2 parabola

8 c2c2c2c2 peak c 2 orc

9 0 o C 1000 o C 2000 o C Fractional # of Molecules (  N/N) Speed, c (m/s ) Typical Boltzmann Speed Distribution and Its Temperature Dependence High Energy Tail (Responsible for Chemical reactions) (c 2 )∆c (  N/N) = 4  [ m /( 2  kT )] 3/2 e -[(1/2)mc /kT] 2[Hold ∆c constant at say c = 0.001 m/s]

10 1) The Root Mean Square Speed: crms = (3RT/M)1/2 If N is the total number of atoms, c1 is the speed of atom 1, and c2 the speed of atom 2, etc.: 2) crms = [(1/N)(c12 + c22 +c32 + ………)]1/2 3) cmp is the value of c that gives (N/N) in the Boltzmann distribution its largest value.

11 Fractional # of Molecules (  N/N) Speed, c (m/s) 1000 o C C rms (1065 m/s) C avg (980 m/s) C mp (870 m/s) Boltzmann Speed Distribution for Nitrogen = [2RT/M] 1/2 = [8RT/  M] 1/2 = [3RT/M] 1/2

12 Cleaning Up Some Details A number of simplifying assumptions that we have made in deriving the Kinetic Theory of Gases cause small errors in the formulas for wall collision frequency, collision frequency (z), mean free path (), and the meaning of c, the speed: 1) The assumption that all atoms move only perpendicular to the walls of the vessel is obviously an over simplification. ct A

13 2) For the collision frequency, z, the correct formula is The (2)1/2 error here arises from the fact that we assumed only one particle (red) was moving while the others (blue) stood still. 3) Even though the formula for wall collisions used in deriving the pressure was incorrect, the pressure formula is correct! This is because of offsetting errors made in deriving the wall collision rate, I, and the momentum change per impact, 2mc.

14 = cavge/{(2)1/2(N/V)  2 cavge} A final question that arises concerns which c, cavge, crms, or cmp is the correct one to use in the formulas for wall collision rates (I), molecule collision rates (z), mean free path () and pressure (p). For p the correct form of c is crms while for I, or z considered as independent quantities, cavge is correct.

15 Summary of correct Kinetic Theory formulas: Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

16 Chemical Kinetics The Binary Collision Model Must actually have a hydrogen molecule bump into a chlorine molecule to have chemistry occur. Reaction during such a collision might look like the following picture:

17 H2H2H2H2 Cl 2 HH ClCl HClHCl +

18 Collision Frequency Real gases consist of particles of finite size that bump into each other at some finite rate. Assume first that the red molecule has a constant speed C and the green ones are standing still.

19  AB =  A +  B 2  A BBBB  AB =  A +  B V c =  (  AB ) 2 C If a green molecule has some piece in this volumeCollision!

20 A is the radius of molecule A, B the radius of B  AB =  A +  B

21 There is one subtlety. In deriving z, we assumed the red molecule flew through a cloud of motionless green ones at a speed of C. In reality, of course, all the molecules are moving. Where =mA mB/(mA+mB)  is called the reduced mass and can be thought of as a kind of (geometric) average of the masses of A,B. <urel> is the mean speed of molecule A with respect to molecule B.


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