Question

The mass of an oxygen molecule is about 16 times that of a hydrogen molecule. At room temperature the ‘rms’ speed of oxygen molecules is v. The ‘rms’ speed of the hydrogen molecules at the same temperature will be:
A) v/16
B) v/4
C) 4v
D) 16v

Answer 1

Hint: We know that the rms speed also known as the root-mean-square speed is defined as the measurement of the speed of the particles of any gaseous substance when we take into consideration the average velocity of the gaseous particles, squared with the molecules present in the gas.

Complete step by step answer:
We know that the mass of ${{O}_{2}}$ can be written as $M_{0}=M_{H_{2}} \times 16$
Now, let us find the rms speed of oxygen.
The expression to find the rms speed of oxygen or ${{O}_{2}}$ is given as: $\sqrt{\dfrac{3RT}{{{M}_{{{O}_{2}}}}}}$.
Here, R is the molar gas constant, T is the temperature in Kelvin, ${{M}_{{{O}_{2}}}}$ is the molar mass of oxygen.
We consider the rms speed from the above expression of oxygen to be v.
Now, let us find the rms speed of hydrogen
The expression to find the rms speed of oxygen or ${{H}_{2}}$ is given as: $\dfrac{\sqrt{3RT}}{{{M}_{{{H}_{2}}}}}$
Here, R is the molar gas constant, T is the temperature in Kelvin, ${{M}_{{{H}_{2}}}}$is the molar mass of hydrogen.
From the expression in equation 1, we can develop the equation as:
$\Rightarrow \dfrac{\sqrt{3RT}}{{{M}_{{{H}_{2}}}}}=\dfrac{\sqrt{3RT}}{{{M}_{{{O}_{2}}}}/16}$
Since the value of $\sqrt{\dfrac{3RT}{{{M}_{{{O}_{2}}}}}}$ was considered to be v, the value of
$\Rightarrow \dfrac{\sqrt{3RT}}{{{M}_{{{O}_{2}}}}/16}$ should be 4v, because of the multiplication of 16.
Hence, the rms speed of hydrogen at the same temperature will be 4v.

So, the correct option is option C.

Note: We use the rms speed because in case of a sample of gas particles the resultant velocity is always 0 as because the particles will always move in all the directions. Thus, the average velocity cannot be considered in this situation.