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Find experimentally that light gases escape more quickly than heavy ones!

Find experimentally that light gases escape more quickly than heavy ones!. Experimental Evidence for Kinetic Theory: Heat Capacities. Two kinds: C p (add heat at constant pressure) C v (add heat at constant volume). KE [1 mole gas] =. (1) Increases kinetic energy of molecules:.

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Find experimentally that light gases escape more quickly than heavy ones!

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  1. 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)

  2. KE [1 mole gas] = (1) Increases kinetic energy of molecules: KE = (1/2) mc2, c2 ~ T (2) Perform work. Increase T from T1 to T2 : KE1=(3/2)RT1 & KE2=(3/2)RT2 Cv = (3/2) R

  3. T1 T2 L A p = F / A Gas Movable Piston work = F L = (Ap)  L = p x (AL) work = p x (AL) = p(V2 - V1) = nR(T2-T1) = nRT  w = p∆V = nRT

  4. Cp = heat added to increase KE + heat added to do work For n = 1 mole and T2-T1 = T= 1 degree: Cp = (KE) + w = (3/2) R + R Remember that Cv = (3/2) R

  5. Heat Capacity Summary for Ideal Gases: Note, Cv independent of T. Cv = (3/2) R, KE change only. Cp/Cv = 1.67 Find for monatomic ideal gases such as He, Xe, Ar, Kr, Ne Cp/Cv = 1.67

  6. For diatomics and polyatomics find Cp/Cv < 1.67! This would make Cp/Cv < 1.67 A possible solution: KE = (1/2)kT (or 1/2 RT on a mole basis)perdegree of freedom.

  7. (x,y,z) A degree of freedom is a coordinate needed to describe position of a molecule in space. A diatomic molecule is a line (2 points connected by a chemical bond). It requires 5 coordinates to describe its position: x, y, z, ,  Z  Y  X

  8. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

  9. Collision Frequency and Mean Free Path  Focus on one molecule (say a red one) flying through a background of other molecules (say blue ones). Make the simplifying assumption that only the red one is moving. (Will fix later.)

  10. Gas Kinetic Collision Cylinder Note: A= 2 L= c1s Miss Hit  Hit    A    Hit V/sec = {π 2}[c] = {A}  [L / t ]  Miss The red molecule sweeps out a cylinder of volume 2c in one second. It will collide with any molecules whose centers lie within the cylinder. Note that the (collision) cylinder radius is the diameter  of the molecule NOT its radius !

  11. Red Molecule R sweeps out a Cylinder of volume π2c per second (c = speed). Gas Kinetic Collision Cylinder No Collision Collision 2  =2  =2 Collision

  12. z = [volume swept out per second]  [molecules per unit volume]

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