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

Experimental Evidence for Kinetic Theory:Experimental Evidence for Kinetic Theory:Heat CapacitiesHeat Capacities

Two kinds: Two kinds: CCpp (add heat at constant pressure)(add heat at constant pressure)

CCvv (add heat at constant volume) (add heat at constant volume)

KE [1 mole gas] = KE [1 mole gas] = 12

mc2 ×N0 =32kTN0 =

32RT

(1) Increases kinetic energy of molecules:(1) Increases kinetic energy of molecules:

KE = (1/2) mcKE = (1/2) mc22, c, c2 2 ~ T~ T

(2) Perform work.(2) Perform work.

CCvv = (3/2) R = (3/2) R

Increase T from TIncrease T from T11 to T to T2 2 :: KEKE11=(3/2)RT=(3/2)RT11 & KE & KE22=(3/2)RT=(3/2)RT22

work = work = FF L = ( L = (ApAp) ) L = p x (AL) L = p x (AL)

work = p x (AL) = p(Vwork = p x (AL) = p(V22 - V - V11) = nR(T) = nR(T22-T-T11) = nR) = nRTT

w = p∆V = nRw = p∆V = nRTT

LLAA

TT11 TT22

GasGasp = F / A

Movable PistonMovable Piston

CCpp = heat added to increase KE + heat added to do work = heat added to increase KE + heat added to do work

For n = 1 mole and TFor n = 1 mole and T22-T-T11 = = T= 1 degree:T= 1 degree:

CCpp = = (KE) + w = (3/2) R + R(KE) + w = (3/2) R + R

Remember that CRemember that Cvv = (3/2) R = (3/2) R

Heat Capacity Summary for Ideal Gases:Heat Capacity Summary for Ideal Gases:

CCvv = (3/2) R, KE change only. = (3/2) R, KE change only. Note, CNote, Cvv independent of T. independent of T.

CCpp/C/Cvv = 1.67 = 1.67

Find for monatomic ideal gases such as He, Xe, Ar, Kr, NeFind for monatomic ideal gases such as He, Xe, Ar, Kr, Ne CCpp/C/Cvv = 1.67 = 1.67

For diatomics and polyatomics find CFor diatomics and polyatomics find Cpp/C/Cvv < 1.67! < 1.67!

This would make CThis would make Cpp/C/Cvv < 1.67 < 1.67

KE = (1/2)kT (or 1/2 RT on a mole basis)KE = (1/2)kT (or 1/2 RT on a mole basis) perper degree of freedomdegree of freedom..

A possible solution:A possible solution:

A degree of freedom is a coordinate needed to describeA degree of freedom is a coordinate needed to describe position of a molecule in space.position of a molecule in space.

A diatomic molecule is a line (2 points connected by aA diatomic molecule is a line (2 points connected by a chemical bond). It requires 5 coordinates to describe itschemical bond). It requires 5 coordinates to describe its position: x, y, z, position: x, y, z, , ,

KE =512kT ⎛

⎝ ⎞ ⎠=

52kT

(x,y,z)

ZZ

XX

YY

Bonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus * Bonus * Bonus * Bonus

Collision Frequency and Mean Free PathCollision Frequency and Mean Free Path

Focus on one molecule (say a Focus on one molecule (say a redred one) flying through a one) flying through a background of other molecules (say blue ones). background of other molecules (say blue ones).

Make the simplifying assumption that only theMake the simplifying assumption that only the redred one is moving. (Will fix later.) one is moving. (Will fix later.)

LL = c= c1s1s

V/sec = {π V/sec = {π 22}[c]}[c]= {A} = {A} [[ LL / t / t ]]

HitHitMissMiss

HitHit

HitHit

MissMiss

The red molecule sweeps out a cylinder of volume The red molecule sweeps out a cylinder of volume 22cc in one second. It will collide with any molecules in one second. It will collide with any molecules

whose centerswhose centers lie within the cylinder. Note that lie within the cylinder. Note that the (collision) cylinder radius is the diameter the (collision) cylinder radius is the diameter of of

the molecule NOT its radius the molecule NOT its radius !!

Note: Note: A= A= 22

Gas Kinetic Collision CylinderGas Kinetic Collision Cylinder

=2=22 2

=2=2

Red Molecule R sweeps out a Red Molecule R sweeps out a Cylinder of volume πCylinder of volume π22c per c per second (c = speed).second (c = speed).

Gas KineticGas KineticCollisionCollisionCylinderCylinder

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