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OpenStudy (anonymous):
Calculate the average translational kinetic energy of a nitrogen molecule at 27 degree celcius
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OpenStudy (anonymous):
using equation T=2/3k*K.E*......K.E=3kT/2.....as T=27+273=300K......k=1.38*10power_23.......K.E=3*1.38*10power_23*300...=6.21*10power_21J{is the translational K.E of nitrogen at 27 degree
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