Question

Root mean square velocity for a certain di-atomic gas at room temperature $27{}^\circ C$ is found to be $1930m{{s}^{-1}}$. The gas is
$\begin{aligned} & A){{H}_{2}} \\\ & B){{O}_{2}} \\\ & C){{F}_{2}} \\\ & D)C{{l}_{2}} \\\ \end{aligned}$

Answer 1

Root mean square velocity of an atomic gas refers to the speed with which the molecules in the gas can travel at a given temperature. At a given temperature, the root means the square value of the velocity of a gas is directly proportional to the absolute temperature of the gas. At the same time, it is also inversely proportional to the mass of gas.
Formula used:
${{v}_{rms}}=\sqrt{\dfrac{3RT}{m}}$

Complete step-by-step solution:
The root means square velocity of a di-atomic gas gives an idea about how fast the molecules in the gas can travel at a given temperature. At a given temperature, it is directly proportional to the absolute temperature of the gas and inversely proportional to the mass of gas. If ${{v}_{rms}}$ represents the root mean square velocity of a diatomic gas, then, ${{v}_{rms}}$ is given by
${{v}_{rms}}=\sqrt{\dfrac{3RT}{m}}$
where
${{v}_{rms}}$ is the root mean square velocity of a diatomic gas at an absolute temperature $T$
$R$ is the ideal gas constant
$m$ is the molar mass of diatomic gas
Let this be equation 1.
Coming to our question, we are provided with a di-atomic gas, whose root mean square velocity at room temperature $27{}^\circ C$ is equal to $1930m{{s}^{-1}}$. We are required to predict di-atomic gas.
Clearly, for the given di-atomic gas, we have
${{v}_{rms}}=1930m{{s}^{-1}}$ at an absolute temperature $T=27{}^\circ C=300K$
Substituting these values in equation 1, we have
${{v}_{rms}}=\sqrt{\dfrac{3RT}{m}}\Rightarrow 1930=\sqrt{\dfrac{3\times 8.314\times 300}{m}}\Rightarrow \sqrt{m}=\dfrac{\sqrt{3\times 8.314\times 300}}{1930}=\dfrac{86.502}{1930}=0.044$
Clearly,
$m={{\left( 0.044 \right)}^{2}}\approx 0.002kgmo{{l}^{-1}}=2gmo{{l}^{-1}}$
Therefore, the mass of the given di-atomic gas is $2gmo{{l}^{-1}}$. Since we know that atomic mass of ${{H}_{2}}$ is equal to $2gmo{{l}^{-1}}$, we can conclude that the given di-atomic gas is nothing but ${{H}_{2}}$.
Hence, the correct answer is option $A$.

Note: Students need to be thorough with conversion formulas. Conversion formulas used in the above solution are as follows.
$0{}^\circ C=273K$
$1kg=1000g$
Students need to be aware of atomic masses of other given options too, to predict the correct answer. They are:

& C{{l}_{2}}-70.906gmo{{l}^{-1}} \\\ & {{O}_{2}}-32gmo{{l}^{-1}} \\\ & {{F}_{2}}-37.996gmo{{l}^{-1}} \\\ \end{aligned}$$ Also, ideal gas constant is given by $R=8.314J{{K}^{-1}}mo{{l}^{-1}}$