Question

The average speed of nitrogen molecules in a gas is $v$. If the temperature is doubled and the ${N_2}$ molecule dissociates into nitrogen atoms, then find the average speed.
A) $v$
B) $v\sqrt 2 $
C) $2 \times v$
D) $4 \times v$

Answer 1

Hint:The rms speed of a gas molecule depends on the temperature of the gas. However here it is mentioned that as the temperature is made to increase, the molecule dissociates into atoms. This suggests that the molar mass becomes half. So the rms speed in this scenario will be proportional to the square root of the temperature and inversely proportional to the square root of the molar mass.

Formula used:
-The rms speed of a gas molecule is given by, ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where $R$ is the universal gas constant, $T$ is the temperature and $M$ is the molar mass of the molecule.

Complete step by step answer.
Step 1: List the parameters involved in the given problem.
The average speed of a nitrogen molecule i.e. ${N_2}$ molecule is given to be ${v_{rms}} = v$ .
The molar mass of the given ${N_2}$ molecule will be $M = 2 \times {\text{atomic mass}} = 2 \times 14 = 28{\text{gmo}}{{\text{l}}^{ - 1}}$ .
Let $T$ be the temperature of the gas.
When the temperature doubles i.e. $T' = 2T$ the given ${N_2}$ molecule will dissociate into nitrogen atoms. Then the molar mass of one nitrogen atom will be $M' = 14{\text{gmo}}{{\text{l}}^{ - 1}}$ .
Step 2: Express the rms speed of the nitrogen molecule and the nitrogen atom.
The rms speed of the given ${N_2}$ molecule can be expressed as ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ ------- (1)
Substituting for $M = 28{\text{gmo}}{{\text{l}}^{ - 1}}$ in equation (1) we get, ${v_{rms}} = \sqrt {\dfrac{{3RT}}{{28}}} $
Now the rms speed of the nitrogen atom will be ${v_{rms}}^\prime = \sqrt {\dfrac{{3RT'}}{{M'}}} $ -------- (2)
Substituting for $M' = 14{\text{gmo}}{{\text{l}}^{ - 1}}$ and $T' = 2T$ in equation (2) we get, ${v_{rms}}^\prime = \sqrt {\dfrac{{3R2T}}{{14}}} = \sqrt {\dfrac{{3RT}}{{28}} \times \dfrac{2}{{\left( {\dfrac{1}{2}} \right)}}} $
$ \Rightarrow {v_{rms}}^\prime = \sqrt {4 \times \dfrac{{3RT}}{{28}}} = 2v$
$\therefore $ the average speed of the nitrogen atom will be ${v_{rms}} = 2v$ .

So the correct option is C.

Note:The atomic mass of nitrogen is known to be $14{\text{u}}$. So the molar mass of nitrogen atom will be $14{\text{gmo}}{{\text{l}}^{ - 1}}$. Since the given nitrogen molecule i.e. ${N_2}$ molecule contains two such nitrogen atoms; its molar mass will be two times the molar mass of one nitrogen atom. This is obtained as $M = 2 \times 14 = 28{\text{gmo}}{{\text{l}}^{ - 1}}$. The rms speed of the gas molecule is a measure of the average speed of the particles in a gas.