problem18_78

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 18.78: a) The two degrees of freedom associated with the rotation for a diatomic molecule account for two-fifths of the total kinetic energy, so K rot = nRT = (1.00) (8.3145 J mol K)(300 K ) = 2.49 103 J. b) 16.0 10-3 kg mol -11 2 I = 2m(L 2) 2 = 2 6.023 10 23 molecules mol (6.05 10 m) = 1.94 10 -46 kg m 2 . c) Using the results of parts (a) and (b), 2 K rot N A 2(2.49 103 J) s = = I (1.94 10- 46 kg m 2 )(6.023 1023 molecules mol) = 6.52 1012 rad s, much larger than that of machinery. ...
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This document was uploaded on 02/05/2008.

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