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1 3 mc2 = (R / N )T 2 2 0

R / N0 is the () / molecule and is called Boltzmann’s constant k = (R / N0)

1 3 mc2 = kT Kool result!! 2 2 1 3 N ( mc2) = RT 0 2 2

 1 2 N  mc  is the kinetic of one of gas atoms 0 2 

→→ 2 Since N0 m = M (molecular weight) (1/2)Mc = (3/2) RT Units: 2 Units: mc is a , therefore 2    2 mc2 PV = n  N   must also have units of energy   0  must also have units of energy 3  2 

PV ~ [] []

= [(force / Area)] [(Area) (Length)]

PV ~ force ×× length

Energy = Force ×× Distance Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Typical Molecular Speeds

= 2 = [Root Mean Square Speed] Understand that c c crms

[Later we will define crms more fully.] (1/2)mc2 = (3/2)kT →→ c = (3kT/m)1/2

1/2 Notice: c = (3N0kT/N0m) [N0 is Avagadro’s number]

But N0k = R and N0m = M (molecular weight) c = (3RT/M)1/2 →→ c2 = 3RT/M Consider Gas with At wt of 0.001 Kg/mole (H atoms!) R= 8.314 /mole-deg, T = 300 deg c2 = 3RT/M = 7.47 ×× 106 Joules/Kg = 7.47 ×× 106 (m/sec)2 Typical Molecular Speeds

= 2 = Understand that c c crms [Root Mean Square Speed]

(1/2)mc2 = (3/2)kT →→ c = (3kT/m)1/2

Consider Gas with At wt of 0.001 Kg/mole (H atoms!) m = (0.001 Kg / mole) / (6.02 ×× 1023 molecule / mole) = 1.66 ×× 10-27 Kg/molecule k = 1.38 ×× 10-23 Joules/molecule - deg, T = 300 deg c2 = 3kT/m = 7.47 ×× 106 Joules/Kg = 7.47 ×× 106 (m/sec)2 c = 2.73 ×× 103 m/sec (Fast Moving ) c = 1.37 ×× 103 m/sec for He = 0.004 Kg/mole

Note c ~ T1/2 →→ c1200 = 2 c300, = = = 1/2 c 4c (M 2,M 32) Note c ~ 1/m →→ H2 O2 H2 O2

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 1 c ~ Lighter molecules collide with wall more m frequently than heavy molecules! BUT momentum change per collision ~ mc, with

m mc ~ ~ m m

Thus, heavier molecules transfer more momentum per impact

Two effects cancel since (1/m1/2) x (m1/2) is independent of m

Result is that p scales like n, the number of moles or the number of ! pV = nRT 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. Effusion of Gases: The Movie

Note: QuickTime™ and a ← Video decompressor ← Hole are needed to see this picture. Must be very small! Effusion of a Gas through a Small Hole

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

For two different molecular speeds c1, c2 and concentrations N1/V, N2/V: R   N   N   c  1 = 1 / 2  1         R2 V V c2

If N1 = N2 (equal initial concentrations) 3kT   R1 =  c1  = m1 1/2   →→ (R(R1/R/R2)=(m)=(m2/m/m1)) R2 c2 3kT m2 R 32 →→ 1 = = 4 For m1 being H2 and m2 being O2 R2 2 Find experimentally that light gases escape more quickly than heavy ones!

Experimental Evidence for Kinetic Theory: Capacities

Definition: Heat in needed to raise of 1 mole of a substance 1° centigrade.

Two kinds: Cp (add heat at constant pressure) Cv (add heat at constant volume) Adding heat means adding energy. Energy goes two places: (1) Increases kinetic energy of molecules: KE = (1/2) mc2, c2 ~ T

(2) Perform work. As we shall see shortly, no work is done at constant volume.

Cv: volume constant. All heat goes to KE. Compute as follows: 1 3 3 KE [1 mole gas] = mc2 × N = kTN = RT 2 0 2 0 2

Increase T from T1 to T2 : KE1=(3/2)RT1 & KE2=(3/2)RT2 3 KE − KE = R(T − T ) =energy to increase T form T to T 2 1 2 2 1 1 2 ° By definition when T2 -T1 = 1 KE2 - KE1 = Cv

Cv = (3/2) R (Independent of T for an ) ≅≅ (Cv 3 cal / mole - deg) Cp: Pressure constant so volume increases with increase in T. As we see below, work is done in this case:

A L p = F / A Gas ←←Movable Piston

T1 T2 work = F ×× L = (Ap) ×× L = p x (AL) ∆∆ But, AL = V2 - V1 = V (volume change) ∆∆ work = p x (AL) = p(V2 - V1) = nR(T2-T1) = nR T ∆∆ °° w = p∆V = nR∆∆T →→ For n=1 mole, T=1 , w = R

To raise temperature of one mole of gas by one degree must do R units of work. Heat Capacity at constant pressure has two terms:

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

∆∆ For n = 1 mole and T2-T1 = T= 1 degree:

∆∆ ∆∆ KE change = (KE) = (3/2) R T = (3/2) R, (same as for Cv )

Work = w = R ∆∆T = R

∆∆ Cp = (KE) + w = (3/2) R + R

Predict Cp = (5/2) R for an ideal gas.

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 = (3/2) R + R, KE change + work. Also Independent of T

Cp/Cv = [(5/2)R]/[(3/2)R] = 5/3

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!

Since work argument above

P(V2 - V1) = RT is simple and holds for all gases,

This suggests KE > (3/2)RT for diatomics,

This would make Cp/Cv < 1.67

A possible solution:

Equipartition Theorem: This is a very general law which states that for a molecule or atom:

KE = (1/2)kT (or 1/2 RT on a mole basis) per degree of freedom. A degree of freedom is a coordinate needed to describe position of a molecule in space.

Example: A point has 3 degrees of freedom because it requires three coordinates to describe its position: (x, y, z).  1  3 Thus KE = 3 kT = kT as for a  2  2 A diatomic molecule isis aa lineline (2(2 pointspoints connectedconnected byby aa chemical bond). It requires 5 coordinates to describe its position: x, y, z, θθ, ϕϕ Z θ (x,y,z)  1  5 KE = 5 kT = kT ϕ Y 2 2

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