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Chemistry 18 Online
OpenStudy (anonymous):

What is the average kinetic energy of a gas at 285 Kelvin? (R = 8.314 J/K-mol) 2.37 x 103 J/mol 7.11 x 103 J/mol 3.55 x 103 J/mol 1.77 x 103 J/mol

OpenStudy (anonymous):

\[T= 2/3k <1/2mv ^{2}>\] using above equation \[T=2/3k<K.E> \] \[<K.E> = 3kT/2\] T = 285 Kelvin k = 1.38 x 10^-23 JK^-1

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