Question

R.M.S velocity of a gas molecule is equal to:
A. $\sqrt{\dfrac{{2RT}}{M}}$
B. $\sqrt{\dfrac{{3RT}}{M}}$
C. $\sqrt{\dfrac{{8RT}}{M}}$
D. $\sqrt{\dfrac{{RT}}{M}}$

Answer 1

Hint: R.M.S velocity stands for root mean square velocity. It is defined as the square root of the average of the squares of different velocities of the gas molecules.

Complete step by step answer: For calculating RMS velocity, we will start with the kinetic gas equation. The kinetic gas equation is written as:
$PV = \dfrac{1}{3}MN{u^2}$ (1)
Where, P is the pressure of a gas, V is the volume of the gas, m is the mass of one molecule of gas, N is the number of the gas molecules and u is the root mean square velocity of the molecules.
And the kinetic energy is given by the formula below:
$K.E = \dfrac{1}{2}M{u^2}$ (2)
Where m is the mass of a gas molecule and u is the velocity of a gas molecule.
For one mole of a gas the kinetic gas equation will be:
$PV = \dfrac{1}{3}M{u^2}$ (3)
As per Ideal gas equation, for one mole of a gas $PV = RT$. Now putting the above value in equation (3), we get
$RT = \dfrac{1}{3}M{u^2}$
$
   \Rightarrow 3RT = M{u^2} \\
   \Rightarrow \dfrac{{3RT}}{M} = {u^2} \\
   \Rightarrow \sqrt {\dfrac{{3RT}}{M}} = u \\
$
Hence the correct answer is option B.

Note: Alternatively, we can also give the R.M.S velocity by the below relation:
$u = \sqrt {\dfrac{{{N_1}v_1^2 + {N_2}v_2^2 + ..........}}{{{N_1} + {N_2} + .......}}} $
Some other expressions for R.M.S velocity are:
$u = \sqrt {\dfrac{{3PV}}{M}} $ , where P is pressure of gas, V is volume of gas and M is mass of one molecule of gas.

$u = \sqrt {\dfrac{{3P}}{\rho }} $ Where ρ density of gas.