Lect02 - Wi l l i am Thomson (1824 1907) a.k.a. Lord Kelvin...

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Lecture 2, p 1 William Thomson (1824 – 1907) a.k.a. “Lord Kelvin” Said he wouldn’t shave until the Cubs won a World Series. Became Professor at University of Glasgow at age 22!
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Lecture 2, p 2 Thermal reservoir increasing T Volume Pressure Today: Equipartition First Law of Thermodynamics Internal energy Work and heat Lecture 2: Ideal Gases
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Lecture 2, p 3 Equipartition & Absolute Temperature One of the principal aims of this course is to gain a generally applicable definition of absolute temperature from statistical mechanics (counting). For now: absolute temperature T is proportional to the average translational kinetic energy of a particle in a gas: <KE trans > = constant x T In fact, each independent quadratic term in the energy of a particle in the gas (e.g., ½ mv x 2 , ½ κ x 2 , ½ Ιϖ 2 , but not mgh ) is found to have the average energy: <Energy mode > = ½ kT Equipartition Theorem k = Boltzmann constant = 1.38 x 10 -23 J/K is independent of the particle’s properties (mass, charge, etc. ) We will show later why and when this theorem applies.
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Lecture 2, p 4 Equipartition of Energy: Examples Free point particles: only translational kinetic energy (x, y, z components) mv x 2 > + <½ mv y 2 > + <½ mv z 2 > = 3(½ kT) Average energy is independent of mass! Diatomic molecules: translational and rotational kinetic energy < ½ mv x 2 > + <½ mv y 2 > + <½ mv z 2 > + <½ I ϖ 1 2 > + <½ I 2 2 > = 5(½ kT) At normal temperatures, vibrations aren’t active in molecules. We’ll see why later. Mass on a spring: Kinetic and potential energy kx 2 > + <½ mv x 2 > = kT Vibrations in solids: Kinetic + potential energy in 3 dimensions κ x >+<½ mv > + <½ κ y >+<½ mv > + <½ κ z >+<½ mv > = 3 kT v x=0 Equilibrium position ½kT per quadratic degree of freedom http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
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Lecture 2, p 5 Bowling Ball Puzzle Why does the Bowling ball slow down? Explain using the Equipartition Theorem. (Hint: Treat the ball as one big particle.) What’s the average speed of the bowling ball after it comes into equilibrium with the room? Assume that M = 10 kg and T = 300 K. Neglect the motion in the z direction.
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Lecture 2, p 6 Bowling Ball Solution Why does the Bowling ball slow down? Explain using the Equipartition Theorem. (Hint: Treat the ball as one big particle.) What’s the average speed of the bowling ball after it comes into equilibrium with the room? Assume that M = 10 kg and T = 300 K. Neglect the motion in the z direction. The center of mass velocity of this “large particle” is given by the Equipartition Theorem: 2 rms 1 v v M = 2 2 2 23 2 22 2 2 11 1 1 1 1 ( ) 2 2 2 2 2 2(1.38 10 J/K)(300 K) 8.28 10 m /s 10 kg 2.9 10 m/s x y rms m v m v v kT kT kT v m v - - - = + = + × = = = × = ×
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Lecture 2, p 7 From Last Lecture: The Ideal Gas Law 2 3 TRANS N p KE V = ( 29 2 2 2 3 1 2 2 TRANS x y z KE m v v v kT = + + = Recall that we derived this pressure-energy relation:
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Lect02 - Wi l l i am Thomson (1824 1907) a.k.a. Lord Kelvin...

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