Calculate the temperature at which the root mean square velocity, the average velocity, and the most proable velocity of oxygen gas are all equal to `1500 m s^(-1)`.

Explanation

(`a`) Root mean square velocity is
`sqrt((3RT_(1))/(M))=1.5xx10^(3)m s^(-1)`
`:. T_(1)=((1.5xx10^(3) m s^(-1))^(2)(32xx10^(3) kg mol^(-1)))/(3xx8.314 J K^(-1) mol^(-1))`
(Molar mass of oxygen `= 32 g mol^(-1)`)
`2886 K`
(`b`) Average velocity `=((8RT_(2))/(pi M))^(1//2)=1.5xx10^(3) m s^(-1)`
`:. T_(2)=((1.5xx10^(3) m s^(-1))^(2)(32xx10^(-3) kg mol^(-1))(3.1416))/(8xx8.314 J K^(-1) mol^(-1))`
`=3399 K`
(`c`) Most probable velocity `= sqrt((2RT_(3))/(M))=1.5xx10^(-3)ms^(-1)`
`:. T_(3)=((1.5xx10^(-3) m s^(-1))^(2)(32xx10^(-3) kg mol^(-1)))/(2xx8.314 J K^(-1) mol^(-1))`
`=4330 K`