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### ef152-lec-6-4

Course: EF 152, Spring 2008
School: Tennessee
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Word Count: 488

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6, Module Lecture 4 Molecular Basis of Thermal Physics Module 6, Lecture 4: Average Values in a Gas We will work on a __________ basis, or use ______ values. Average will be denoted by: Internal energy consists mainly of: For N molecules: 1 U = N KE = N m v 2 2 Use the average of the square of the velocity, not the square of the average velocity, or even the square of the average speed. v= v1 = -2 v2 = 4...

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6, Module Lecture 4 Molecular Basis of Thermal Physics Module 6, Lecture 4: Average Values in a Gas We will work on a __________ basis, or use ______ values. Average will be denoted by: Internal energy consists mainly of: For N molecules: 1 U = N KE = N m v 2 2 Use the average of the square of the velocity, not the square of the average velocity, or even the square of the average speed. v= v1 = -2 v2 = 4 EF 152 Spring, 2009 Lecture 6-4 1 EF 152 Spring, 2009 Lecture 6-4 v= v2 = v 2 = A. B. C. D. 1 3 9 10 2 Average Values in a Gas 2 2 2 2 v 2 = v x + v y + v z2 = v x + v y + v z2 Pressure (from a microscopic viewpoint) Pressure: arises from collisions of molecules of gas with walls of container Orient wall parallel to y-z plane. Collision changes x velocity. Since there is no preferred direction of motion: vi = v x i + v y + v z k j vf = i+ + k j P = Combining with: 1 U = N m v2 2 or U = N Consider a thin slice by the wall, with volume Adx Write dx in terms of vx: Number of molecules: Number of collisions: 3 2 m vx 2 3 N coll = EF 152 Spring, 2009 Lecture 6-4 EF 152 Spring, 2009 Lecture 6-4 1N v x dtA 2V 4 Pressure (from a microscopic viewpoint) Combine everything: 1 N dPx v dtA (2mv x ) Fx N coll (2mvx ) 1 2 V x 1N 2 p= = mv x = dt = = A A dt A dt AV Recall we are working on a statistical basis, so use averages. Combine with: Example: Kinetic Energy of 1 mole of Gas Given: Recall that 1 mole of gas at STP (0C, 1.00 atm) occupies 22.4 L (0.79 ft3). Required: Kinetic energy of a mole gas 2 of pV = Nm v x = nRT = NkT 3 2 U = N m vx 2 U= EF 152 Spring, 2009 Lecture 6-4 3 nRT 2 5 EF 152 Spring, 2009 Lecture 6-4 6 Molecular Speeds U= Solve for v: m = mass of single molecule M = molecular mass (grams/mole) Example: Average Speed of an Oxygen Molecule Given: O2 molecule at STP conditions. Required: Root-mean-square speed 3 1 1 nRT = Nm v 2 = nM v 2 2 2 2 vrms = 3RT 3kT = M m What does the subscript rms mean? Calculating rms values: 1. _______ each speed 2. Add em up 3. Divide by number of molecules 4. Take __________ EF 152 Spring, 2009 Lecture 6-4 7 EF 152 Spring, 2009 Lecture 6-4 8 Heat Capacity of Gases Isochoric: Constant Volume (isovolumetric), W=0 Additional Forms of Energy Q = U = ncv T Combine with: U= 3 nRT 2 Type of Gas Monatomic Diatomic Gas He Ar H2 N2 O2 CO CO2 SO2 H2S cv 12.47 12.47 20.42 20.76 21.10 20.85 28.46 31.39 25.95 9 3 cv = R = 12.47 J /(mol K ) 2 Diatomic will have two additional rotational energies. Vibrational motion only occurs above about 600K. Polyatomic EF 152 Spring, 2009 Lecture 6-4 EF 152 Spring, 2009 Lecture 6-4 10 Equipartition of Energy Each velocity component (either translational or angular) has, on average, an associated energy per molecule of kT. Degrees of Freedom: number of velocity components to describe the motion. Diatomic gas: ___ degrees of freedom cv = Gas Monotomic Diatomic EF 152 Spring, 2009 Lecture 6-4 5 R = 20.79 J /( mol K ) 2 cp = R + cv (5 2)R (7 2)R cv (3 2)R (5 2)R (5 3) = 1.67 (7 5) = 1.4 11 = cp cv
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American International - CHE - 311
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