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