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  4. To double the translational kinetic energy of the molecules of a gas one has to

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Q. To double the translational kinetic energy of the molecules of a gas one has to

AMUAMU 1997

Solution:

: The mean kinetic energy of translation $ =\frac{3R}{2}T $ $ {{E}_{K}}\propto T $ Hence to double $ {{E}_{K}}, $ absolute temperature T be double.