For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is

Options

A. `u = ((2E)/(m))^(1//2)`
B. `u = ((E)/(2m))^(1//2)`
C. `u = ((3E)/(2m))^(1//2)`
D. `u = ((E)/(3m))^(1//2)`

Answer

A

Explanation

According to the kinetic gas eqauation ,
`pV = (1)/(3) mnc^(2)`
For a single molecule of mass `m , n = 1` .
Thus ,
`u_(rms) = sqrt((3 pV)/(m))` (1)
According to gas equation , for one mole of gas , we have
`pV = RT`
For a single molecule , we have
`pV = (R )/(N_(0))T = kT` (2) lt brgt where `k` is called the Boltzmann constant.
Average kinetic energy per molecule `(E)` is given as
`E = (3)/(2) kT`
or `kT = (2)/(3) E` (3)
Combining Eqs. (1),(2), and (3) , we have
` u_(rms) = sqrt((3 xx (2)/(3)E)/(m)) = sqrt((2E)/(m))`