ch19-p036 - temperature. We obtain T = 1 2 Mv p 2 / R =...

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36. (a) From the graph we see that v p = 400 m/s. Using the fact that M = 28 g/mol = 0.028 kg/mol for nitrogen (N 2 ) gas, Eq. 19-35 can then be used to determine the absolute
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Unformatted text preview: temperature. We obtain T = 1 2 Mv p 2 / R = 2.710 2 K. (b) Comparing with Eq. 19-34, we conclude v rms = 3/2 v p = 4.910 2 m/s....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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