For a gas containing `10^(23)` molecules (each having mass `10^(-22)g`) in a volume of `1 dm^(3)`, calculate the total kinetic energy of molecules if their root mean square speed is `10^(5)cm s^(-1)`. What will be its temperature?

Explanation

Total `KE=N((1)/(2)moverlineu^(2))`
`=10^(23)((1)/(2)xx10^(-25)kgxx(10^(3)ms^(-1))^(2))`
`=0.5xx10^(4)kgm^(2)s^(-2)=0.5xx10^(4)J`
Total kinetic energy is also equal to `N((3)/(2)xxkT)`. Thus
`N((3)/(2)xxkT)=0.5xx10^(4)J`
where `k` is Boltzmann constant
Hence, `T=(2)/(3)xx((0.5xx10^(4)J))/(Nk)`
`=(2)/(3)xx(0.5xx10^(4)J)/(10^(23)xx1.38066xx10^(-23)JK^(-1))=2414 K`
Hence, temperature `=2414 K`