Question

Initially, the root mean square (rms) velocity of ${N_2}$ molecules at a certain temperature is u. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be:
A.2u
B.14u
C.4u
D.$\dfrac{u}{2}$

Answer 1

The root mean square velocity (rms value) is the square root of the mean of squares of the velocity of individual gas molecules which is represented by: ${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$. Since the initial root mean square velocity of ${N_2}$ molecules is given to be ‘u’ and the molar mass of ${N_2}$ molecules is 28 g/mol. After dissociation of nitrogen molecules dissociates into nitrogen atoms, and the molar mass becomes 14 g/mol. Now substituting the values, we can find the new rms velocity.

Complete step by step answer:
Given in the question is,
Initial root mean square velocity of N2 molecules, ${V_{rms}} = u$
Temperature = T
We know, the root mean square velocity (rms value) is equal to
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
Where
${V_{rms}} =$ Root-mean-square velocity
M = Molar mass of the gas (Kg/mole)
R= Molar gas constant
T = Temperature in Kelvin.
Initially,
The molecular mass of ${{\mathbf{N}}_{\mathbf{2}}}$ is $= M = 28\dfrac{g}{{mol}}$
${V_{rms}} = ({N_2}) = \sqrt {\dfrac{{3RT}}{{M({N_2})}}} = \sqrt {\dfrac{{3RT}}{M}} = u$
Now, after the temperature is doubled, the new temperature becomes “2T” becomes equal to “2 and ${N_2}$ molecules dissociate to become N atoms.
The molecular mass of ${N_2}$ is $= M = 14\dfrac{g}{{mol}}$
New rms velocity $= {V_{rms}}(N) = \sqrt {\dfrac{{3RT}}{{M(N)}}} = \sqrt {\dfrac{{3R(2T)}}{{14}}} = 2u$

Therefore, the correct answer is option (A).

Note: The reason we use the root mean square velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions. This is a key formula as the velocity of the particles is what determines both the diffusion and effusion rates.