2321avg33(1.3810J/K)[(32.0+273.15)K] = 6.3210J.22KkT−−==××The ratio ε/Kavgis (6.76 ×10–20J)/(6.32 ×10–21J) = 10.7. 27. (a) We use ε= LV/N, where LVis the heat of vaporization and Nis the number of molecules per gram. The molar mass of atomic hydrogen is 1 g/mol and the molar mass of atomic oxygen is 16 g/mol so the molar mass of H2O is (1.0 + 1.0 + 16) = 18 g/mol.
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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.