Lect11 - Le cture11 Applying Boltzmann Statistics Law of...

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Lecture Law of atmospheres Heat capacities C V of molecules – for real !! When equipartition fails Planck Distribution of Electromagnetic Radiation Global Warming Lecture 11 Applying Boltzmann Statistics Reading for this Lecture: Elements Ch 9 Reading for Lecture 13: Elements Ch 4D-F End of midterm material.
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Lecture Last time: Boltzmann Distribution If we have a system that is coupled to a heat reservoir at temperature T: The entropy of the reservoir decreases when the small system extracts energy E n from it. Therefore, this will be less likely (fewer microstates). The probability for the small system to be in a particular state with energy E n is given by the Boltzmann factor: where, to make P tot = 1. / n E kT n e P Z - = / Z n n E kT e - =
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Lecture The Law of Atmospheres How does atmospheric pressure vary with height? 0 Pressure: p(h) p(0) h Earth’s surface Quick Act: In equilibrium, how would T vary with height? a ) increase b ) decrease c ) constant / ( ) (0) mgh kT P h e P - = For every state of motion of a molecule at sea level, there’s one at height h that’s identical except for position. Their energies are the same except for mgh. Therefore, the ratio of probabilities for those two states is just the Boltzmann factor. The ideal gas law, pV = NkT, tells us that this is also the ratio of pressures. This is called the law of atmospheres. / ( ) (0) mgh kT p h e p - =
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Lecture Atmosphere (2) T = 227K 0 10 20 30 40 50 h (km) 1.0 0.1 0.01 0.001 p(h)/p(0) Actual data (from Kittel, Thermal Physics ) Define a characteristic height, h c : where, h c = kT/mg. Note: m is the mass of one molecule . / / ( ) (0) c h h mgh kT p h e e p - - = From this semilog plot, h c 7 km is the height at which the atmospheric pressure drops by a factor of e. 0 1 2 3 4 5 0.0 0.2 0.4 0.6 0.8 1.0 <h>=kT/mg mgh/kT p(h)/p(0)
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Lecture Act 1 What is the ratio of atmospheric pressure in Denver ( elevation 1 mi = 1609 m ) to that at sea level? (Assume the atmosphere is N 2 .) a ) 1.00 b ) 1.22 c ) 0.82 23 26 23 molecular weight 28 g/mol 6.022 10 molecules/mol 4.7 10 kg/molecule 1.38 10 J/K 273 K A m N x x k x T - - = = = = =
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Lecture Solution What is the ratio of atmospheric pressure in Denver ( elevation 1 mi = 1609 m ) to that at sea level? (Assume the atmosphere is N 2 .) a ) 1.00 b ) 1.22 c ) 0.82 23 26 23 28 g/mo 6.022 10 molecules/mol 4.7 10 kg/molecule 1.38 10 J/K 273 K A molecular weight m N l x x k x T - - = = = = = ( 29 ( 29 26 2 23 1 4.7 10 kg 9.8 m/s 1600 m exp 0.822 1.38 10 J/K 273 K p mile p sea level - - × = - = ×
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Lecture Law of Atmospheres - Discussion We have now quantitativeliy answered one of the questions that arose earlier in the course: Which of these will “fly off into the air” and how far? O
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Lect11 - Le cture11 Applying Boltzmann Statistics Law of...

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