The ratio of the kinetic energy of a diatomic | vedclass

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(D) The average kinetic energy of a gas molecule is given by $K = \frac{f}{2} k_B T$,where $f$ is the number of degrees of freedom,$k_B$ is the Boltzm

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$7/5$

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(D) The average kinetic energy of a gas molecule is given by $K = \frac{f}{2} k_B T$,where $f$ is the number of degrees of freedom,$k_B$ is the Boltzmann constant,and $T$ is the absolute temperature.For a diatomic gas molecule at $NTP$ (low temperature),the degrees of freedom $f_1 = 5$ ($3$ translational + $2$ rotational).At high temperatures,the vibrational mode also becomes active,so the degrees of freedom $f_2 = 7$ ($3$ translational + $2$ rotational + $2$ vibrational).Assuming the temperature $T$ is the same for the comparison of energy states,the ratio of kinetic energy at high temperature to that at $NTP$ is $\frac{K_{high}}{K_{NTP}} = \frac{f_2}{f_1} = \frac{7}{5}$.

The ratio of the kinetic energy of a diatomic gas molecule at a high temperature to that at $NTP$ is:

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