Question

The root mean square velocity of hydrogen molecules at 300K is \[{\text{1930 m}}{{\text{s}}^{{\text{ - 1}}}}\] . The rms velocity of oxygen molecules at \[{\text{1200 K}}\]will be:
(A) \[{\text{482}}{\text{.5 m/s}}\]
(B) \[965.0{\text{ }}m/s\]
(C) \[1930{\text{ }}m/s\]
(D) \[3860{\text{ }}m/s\]

Answer 1

Hint:In this problem,we are going to apply the concept of root mean square velocity.Root mean square velocity of any molecule is directly proportional to the temperature of a compound molecule and inversely proportional
\[{{\text{C}}_{{\text{rms}}}}{\text{ = a}}\sqrt {{\text{T/m}}} \]
Where ${{\text{C}}_{{\text{rms}}}}$= Root mean square velocity of the molecule
T= Temperature at which molecule present
m= mass of the molecule

Complete step by step answer:
Given: Root mean square velocity of hydrogen molecules at \[{\text{300 K}}\]is \[{\text{1930 m}}{{\text{s}}^{{\text{ - 1}}}}\].
Here we have to find root mean square velocity of oxygen molecule at \[{\text{1200 K}}\]
According to the formula for root mean square velocity, Root mean square velocity directly proportional to the temperature of a compound molecule and inversely proportional to the mass of that molecule.
Hence the ratio of root mean square velocity of oxygen molecule to root mean square velocity of hydrogen molecule is given by:
${{\text{C}}_{\text{o}}}{\text{/}}{{\text{C}}_{\text{H}}}{\text{ = }}\sqrt {\dfrac{{{{\text{T}}_{\text{o}}}{{\text{m}}_{\text{o}}}}}{{{{\text{T}}_{\text{H}}}{{\text{m}}_{\text{H}}}}}} $
Where ${C_o}$= Root mean square velocity of oxygen molecule
${{\text{C}}_{\text{H}}}$= Root mean square velocity of hydrogen molecule
${T_o}$= Temperature at which oxygen molecule present
${T_H}$= Temperature at which Hydrogen molecule present
${m_o}$= Mass of oxygen molecule
${m_H}$= Mass of Hydrogen molecule
$\Rightarrow{C_o}/{C_H} = \sqrt {\dfrac{{{T_o} \times {m_H}}}{{{T_H} \times {m_o}}}} $
$\Rightarrow{C_o}/{C_H} = \sqrt {\dfrac{{1200 \times 2}}{{300 \times 32}}} $
$\Rightarrow{C_o}/{C_H} = \sqrt {\dfrac{{2400}}{{12800}}} $
$\Rightarrow{C_o}{C_H} = \dfrac{1}{2}$
\[\Rightarrow{C_o} = \dfrac{1}{2} \times {C_H}\]
\[\Rightarrow{C_o} = \dfrac{1}{2} \times 1930\]
\[\therefore{{\text{C}}_o} = {\mathbf{965}}.{\mathbf{0}}{\text{ }}{\mathbf{m}}/{\mathbf{s}}\]

Hence option (B) is the correct option.

Note:While solving such types of numericals students made mistakes regarding the value of molar mass of the molecules.So, appropriate values of these are necessary in order to get the correct answer.Here,mass of the oxygen and hydrogen molecule should be taken appropriately.