Asked by lisa
determine the kinetic energy of falling by an average of one degree of freedom of the nitrogen molecule at temperature T = 1000 K, as well as the kinetic energy of rotational motion.
Answers
Answered by
drwls
What do you mean by "the kinetic energy of falling" ?
"Falling by one degree of freedom" makes no sense. Make sure you copied the problem correctly.
There are three degrees of freedom of translational motion because space has three dimensions. Any atom or molecule has kT/2 energy per degree of translational degree of freedom. k is the Boltzmann constant. T must be in Kelvin
A diatomic molecule like N2 has an addtional two degrees of rotational freedom, with kT/2 energy for each, except at extremely low temperatures, where only the lowest rotational states are populated
"Falling by one degree of freedom" makes no sense. Make sure you copied the problem correctly.
There are three degrees of freedom of translational motion because space has three dimensions. Any atom or molecule has kT/2 energy per degree of translational degree of freedom. k is the Boltzmann constant. T must be in Kelvin
A diatomic molecule like N2 has an addtional two degrees of rotational freedom, with kT/2 energy for each, except at extremely low temperatures, where only the lowest rotational states are populated
Answered by
lisa
определить кинетическую энергию, приходящуюся в среднем на одну степень свободы молекулы азота при температуре Т=1000 К, а также кинетическую энергию вращательного движения.
Answered by
Neeraj
Good Question
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