Calculate the pressure exerted by `10^(23)` gas molecules each of mass `10^(-22) g` in a container of volume 1 litre the rms speed is `10^(5) cm s^(-1)`

Explanation

From the given data, we have
`N=10^(23)`
`m=10^(-22)g=10^(-25)kg`
`V=1 dm^(3)=10^(-3)m^(3)`
`(overlineu^(2))^(1//2)=10^(5)cm s^(-1)=10^(3)m s^(-1)`
Therefore, from the kinetic gas equation
`P=(1)/(3)(mNoverlineu^(2))/(V)`
We get
`P=(1)/(3)xx((10^(-25)kg)(10^(23))(10^(3)m s^(-1))^(2))/(10^(-3)m^(3))`
`=(1)/(3)xx10^(7)kgm^(-1)s^(-2)=(1)/(3)xx10^(7)Pa`