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# Calculate the average kinetic energy of translation of an oxygen molecule at 27∘C,the total kinetic energy of an oxygen molecule at 27∘C,the total kinetic energy in joule of one mole of oxygen at 27∘C.Given Avogadro's number=6.02×1023 and Boltzmann's constant=1.38×10−23J/(mol−K).

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Solution

## 1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at 27∘C is given asET=32kT=32×1.38×10−23×300=6.21×10−21J/mol2. An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given asET=52kT=52×1.38×10−23×300=10.35×10−21J/mol3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given asU=f2μRT=52×1×8.314×300=6235.5J/mol

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