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Calculate the average kinetic energy of oxygen molecule at `0^(@)`C. `(R=8.314 J mol^(-1)` `K^(-1),N_(A)`=`6.02xx10^(23))`

Text Solution

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Oxygen is diatomic molecule, therefore it has 5 degrees of freedom, 3 translational 2 rotational.
`therefore K.E.= (5)/(2) k_(B) T`
`K. E. = (5)/(2) (RT)/(N_A)`
`T= 0^@ C= 273 K = (5)/(2) xx (8.314)/(6.023x 10^(23))xx273= 9.4 xx 10^(-21) J`.
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