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mv x

Wall Collisions.  Vessel wall.  Vessel wall. Approach. mv. mv y. - mv x. mv x. mv y. Recoil. mv. Before Collision with wall. After Collision with wall. Y. X. Z. ct. wall. A. This cylinder contains all the atoms which will strike A in

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mv x

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  1. Wall Collisions  Vessel wall Vessel wall Approach mv mvy -mvx mvx mvy Recoil mv Before Collision with wall After Collision with wall

  2. Y X Z ct wall A This cylinder contains all the atoms which will strike A in a time t (It also contains quite a few atoms that will not collide with the wall during t).

  3. Fatom/atom = (-2mc)/∆t This is the force exerted ON an atom due to a single collision. Since the momentum change for the wall is the negative of that for the atom:

  4. One Particle Momentum Change for Elastic Wall Collision  Wall m m m∆v = m(-c - (+ c)) = -2mc v = - c v = + c

  5. Then: [(Momentum change) / sec] = (∆ Pwall / impact)  ( impacts / sec) = (∆ Pwall / sec)     F = (2mc)  I = (∆ Pwall / sec) Our problem now is to determine ∆t. There is no easy way to do this so we resort to a trick:

  6. A ct  Fwall = [(2mc)][(1 / 6)(N / V)(Ac)] = (1 / 3)(N / V)mc2A Calculating I Total atoms in collision cylinder = (N / V) (Act) (1/3)(1/2)(N / V) (Act) = (1/6) (N / V) (Act) Directions/axis # of axes

  7. P = (1/3) (N / V)mc2 or  PV = (2/3) N [(1/2) mc2] Let N0 = Avogadro’s #; n = # moles in V = N / N0 PV = N (RT / N0) = (2/3) N [(1/2) mc2] or 

  8. is the kinetic energy of one mole of gas atoms Kool result!!

  9. Units: PV ~ [pressure] [volume] PV ~ force  length

  10. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

  11. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

  12. Understand that Typical Molecular Speeds [Root Mean Square Speed]  c = (3kT/m)1/2 (1/2)mc2 = (3/2)kT c = (3RT/M)1/2  c2 = 3RT/M c2 = 3RT/M = 7.47  106 Joules/Kg = 7.47  106 (m/sec)2

  13. c = 2.73  103 m/sec (Fast Moving Particle) Why do Light and Heavy Gases Exert Same Pressure at Constant V,T, n (# moles)? (p = nRT/V) wall collision frequency/unit area = (1/6) (N/V) (Ac t)/(At) = (1/6) (N/V) c However, since

  14. BUT momentum change per collision ~ mc, with Two effects cancel since (1/m1/2) x (m1/2) is independent of m

  15. Experimental Evidence for Kinetic Theory: Effusion Put very small hole in box and measure # of molecules coming through. If hole is really small , molecules won’t know it’s there and will collide with hole at same rate as they collide with the wall.

  16. Effusion of Gases: The Movie Note:  Hole Must be very small!

  17. Effusion of a Gas through a Small Hole Vacuum Gas

  18. If hole area = A, rate at which molecules leave = (1/6) (N / V) Ac = R 

  19. Find experimentally that light gases escape more quickly than heavy ones! Experimental Evidence for Kinetic Theory: Heat Capacities Two kinds: Cp(add heat at constant pressure) Cv (add heat at constant volume)

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