1.

Calculate the average velocity of oxygen molecule at 27^@C.

Answer»

Solution :The average velocity `bar(u)` is given by
`bar(u) =sqrt((8RT)/(pi M))`
In the present case,
`T = 27^@C = 27 + 273 = 300 K, M = 32`
and `R= 8.31 xx 10^7 " ERGS " K^(-1) mol^(-1)`
`:. "" bar(u)=sqrt((8xx8.31xx10^7 xx 300)/2)=1.845xx10^5 " CM " s^(-1)`
HENCE, the average velocity of oxygen molecule at `27^@C " is " 4.46 xx 10^4 " cm " s^(-1)`


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