Question

At room temperature a diatomic gas is found to have r.m.s. speed of $1930m{{s}^{-1}}$. The gas is:
A. ${{H}_{2}}$
B. $C{{l}_{2}}$
C. ${{O}_{2}}$
D. ${{F}_{2}}$

Answer 1

Hint: You could either recall the expression for RMS speed or derive the same from the average kinetic energy expression. All the values to be substituted in the expression are directly given in the question. You could simply substitute them and get the answer.
Formula used:
Expression for RMS speed,
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

Complete answer:
We are given a diatomic gas whose root mean square (RMS) speed at room temperature is found to be,$1930m{{s}^{-1}}$
We are asked to find which among the given options is the diatomic gas mentioned in the question.
We know that the average kinetic energy of a molecule is directly proportional to the absolute temperature of the gas. Also, it is independent of pressure, volume or the nature of the ideal gas. This relation is given by,
$\dfrac{1}{2}m\overline{{{v}^{2}}}=\dfrac{3}{2}{{k}_{B}}T$ ………………… (1)
Where,
m = molar mass of the molecule
${{k}_{B}}$ = Boltzmann constant
T = absolute temperature
$\overline{{{v}^{2}}}$ = RMS speed of the molecule
The RMS speed, that is, the root mean square speed is the measure of the speed of particles in a gas. And it can be defined as the square root of the average velocity squared of the total molecules in a gas.
We could rearrange equation (1) to get,
$\overline{{{v}^{2}}}=\dfrac{3{{k}_{B}}T}{m}$
RMS speed is also represented as,${{v}_{rms}}$
Therefore,
${{v}_{rms}}=\sqrt{\dfrac{3{{k}_{B}}T}{m}}$
Let us replace m by M, where M is the molar mass of the gas,
$\Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3\left( {{k}_{B}}\times {{N}_{A}} \right)T}{M}}$
Where,
${{N}_{A}}$ = Avogadro number
Also,
${{k}_{B}}\times {{N}_{A}}=R=8.314Jmo{{l}^{-1}}{{K}^{-1}}$
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$ …………………………. (2)
We can see that the RMS takes into account both the molecular weight of the gas and the temperature of the gas and these factors also affect the kinetic energy of a gas.
 Now we could directly substitute the given values in the question into equation (2),
RMS speed,
${{v}_{rms}}=1930m{{s}^{-1}}$
Boltzmann constant,
$R=8.314Jmo{{l}^{-1}}{{K}^{-1}}$
Room temperature,
$T=300K$
Rearranging (2),
$\Rightarrow M=\dfrac{3RT}{{{v}_{rms}}^{2}}$
$\Rightarrow M=\dfrac{3\times 8.314\times 300}{{{1930}^{2}}}$
$\Rightarrow M=0.002kg=2g$
Molar mass is found to be 2kg, which we from our prior knowledge know is the molar mass of a hydrogen molecule. Therefore, the given diatomic gas is hydrogen.

Hence, the answer to the given question is option A.

Note:
Make sure that you use the expression in which the mass used is the molar mass of the gas. Doing so, you will be able to directly compare the known values of molar masses of the gases given in the options. Hence, you will get the answer easily.