Find experimentally that light gases escape more quickly than heavy ones! Experimental Evidence for Kinetic Theory: Heat Capacities Two kinds: Cp (add.

Presentation on theme: "Find experimentally that light gases escape more quickly than heavy ones! Experimental Evidence for Kinetic Theory: Heat Capacities Two kinds: Cp (add."— Presentation transcript:

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)

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

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 L A T1T1T1T1 T2T2T2T2 Gas p = F / A   Movable Piston

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

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

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

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, ,  (x,y,z)  Z X Y

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

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.)

L = c  1s V/sec = {π 2}[c] = {A}  [ L / t ]  Hit Miss   Hit Hit   Miss  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 ! Note: A=   2 Gas Kinetic Collision Cylinder     

 =2  2    =2  Red Molecule R sweeps out a Cylinder of volume π2c per second (c = speed). Gas Kinetic Collision Cylinder

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

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