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Kinetic Energy of Gas Atoms and Boltzmann's Constant

This text explains the kinetic energy of one mole of gas atoms, Boltzmann's constant, and their relationship to pressure, volume, and temperature. It also discusses the experimental evidence for the kinetic theory and heat capacities of ideal gases.

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Kinetic Energy of Gas Atoms and Boltzmann's Constant

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  1. is the kinetic energy of one mole of gas atoms R / N0 is the (gas constant) / molecule and is called Boltzmann’s constant k = (R / N0) Kool result!! Since N0 m = M (molecular weight) (1/2)Mc2 = (3/2) RT

  2. Units: is a kinetic energy, therefore must also have units of energy PV ~ [pressure] [volume] = [(force / Area)] [(Area) (Length)] PV ~ force  length Energy = Force  Distance

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  5. Understand that Typical Molecular Speeds [Root Mean Square Speed] [Later we will define crms more fully.]  c = (3kT/m)1/2 (1/2)mc2 = (3/2)kT Notice: c = (3N0kT/N0m)1/2 [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 Joules/mole-deg, T = 300 deg c2 = 3RT/M = 7.47  106 Joules/Kg = 7.47  106 (m/sec)2

  6. c ~ c = 2.73  103 m/sec (Fast Moving Particle) c = 1.37  103 m/sec for He = 0.004 Kg/mole c1200 = 2 c300, Note c ~ T1/2  Note c ~ 1/m1/2  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 Lighter molecules collide with wall more frequently than heavy molecules!

  7. mc ~ ~ BUT momentum change per collision ~ mc, with 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 particles! pV = nRT

  8. 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.

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

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

  11. 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: If N1 = N2 (equal initial concentrations) (R1/R2)=(m2/m1)1/2 

  12. For m1 being H2 and m2 being O2 Find experimentally that light gases escape more quickly than heavy ones! Experimental Evidence for Kinetic Theory:Heat Capacities Definition: Heat in calories needed to raise temperature 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:

  13. KE [1 mole gas] = energy to increase T form T1 to T2 (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: Increase T from T1 to T2 : KE1=(3/2)RT1 & KE2=(3/2)RT2 By definition when T2 -T1 = 1 KE2 - KE1 = Cv Cv = (3/2) R (Independent of T for an ideal gas) (Cv 3 cal / mole - deg)

  14. T1 T2 Cp: Pressure constant so volume increases with increase in T. As we see below, work is done in this case: L A p = F / A Gas Movable Piston 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  For n=1 mole, T=1, w = R w = p∆V = nRT To raise temperature of one mole of gas by one degree must do R units of work.

  15. 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

  16. Heat Capacity Summary for Ideal Gases: Note, Cv independent of T. Cv = (3/2) R, KE change only. 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

  17. 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)perdegree of freedom.

  18. as for a monatomic gas Thus (x,y,z) 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). 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

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