1.

Calculate the ratio of KE of molecule of oxygen and neonj gas at `27^(@)C.`

Answer» For oxygen gas number of degree of freedonis `5`
`therefore" "(KE)_("oxygen")=5/2kT`
None is monatomic, thrrefore it has `3` degrees of freedom
`" "(KE)_("neon")=3/2kT`
Hence, the ratio of their `KE=((KE)_("oxygen"))/((KE)_("neon"))=(5//2kT)/(3//2kT)=5/3`


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