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

Given, T = `0^(@)C` = 273 K
Oxygen is diatomic molecule, therefore it has 5 degrees of freedom, 3 translationl 2 rotational.
`therefore" "KE=5/2(RT)/N_(A)=5/2xx(8.314)/(6.023xx10^(23))xx273=9.4xx10^(-21)J`

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

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