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A fresh air is composed of ntirogen N_(2)(78%) and oxygen O_(2)(21%). Find the rms speed of N_(2)andO_(2) at 20^(@)C. |
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Answer» Solution :Absolute temperature `T=20^(@)C+273=293K` Gas CONSTANT `R=8.32J mol^(-1)K^(-1)` For, Nitrogen `(N_(2)`, Molar mass (M) `=28 g PER mol =28xx10^(-3)kg/mol` `therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(28xx10^(-3)))=sqrt((7313.28)/(28xx10^(-3)))` `(nu_(rms))_(N_2)=511ms^(-1)` For, Oxygen `(O_(2))`, Molar mass (M) =32 g per mol `=32xx10^(-3) kg//mol` `therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(32xx10^(-3)))=sqrt((7313.28)/(32xx10^(-3)))` `(nu_(rms))_(O_2)=478ms^(-1)` |
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