
A cubical box with porous walls containing an equal number of ${O_2}\,and\,{H_2}$ molecules is placed in a larger evacuated chamber. This entire system is maintained at temperature T that is constant. The ratio of Velocity V of ${O_2}$ molecules to that of the Velocity V of ${H_2}$ molecules, found in the chamber outside the box after a very short interval is given by?
(A) $\dfrac{1}{{2\sqrt 2 }}$
(B) $\dfrac{1}{4}$
(C) $\dfrac{1}{{\sqrt 2 }}$
(D) $\sqrt 2 $
Answer
161.1k+ views
Hint:Let start with finding the relation between the root mean square velocity and the masses of the molecules of Hydrogen and oxygen. Then on the basis of the relation between the masses of Hydrogen and Oxygen we get the relation between the root mean square velocity of the molecules given.
Formula Used:
${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, R is constant
T is temperature
And M is the molecular mass of the particle.
Complete answer:
First start with the given information in the question:
Let Velocity be ${V_{rms}}$ , the root mean square velocity.
Now, the root mean square velocity of Hydrogen be ${V_H}$ ,
and the root mean square velocity of Oxygen be ${V_O}$
Now, we know that ${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
So, from the above equation we can find that root mean square velocity is inversely proportional to the mass of the molecule.
${V_{rms}} \propto \dfrac{1}{\surd{M}}$
Now, we know that;
${M_H} = 2$ and ${M_0} = 32$
Putting the above value of the masses of Hydrogen and Oxygen in the equation, we get;
$\dfrac{{{{\left( {{V_{rms}}} \right)}_{{O_2}}}}}{{{{\left( {{V_{rms}}} \right)}_{{H_2}}}}} = \sqrt {\dfrac{{{{\left( M \right)}_{{H_2}}}}}{{{{\left( M \right)}_{{O_2}}}}}} = \sqrt {\dfrac{1}{{16}}} = \dfrac{1}{4}$
Hence the correct answer is Option(B).
Note: Here we have used the formula of root mean square velocity in terms of the universal gas constant, temperature of the gas and the mass of the gas molecule. As all the other quantities were constant hence we get the required relation for the root mean square velocity and the mass of the gas.
Formula Used:
${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, R is constant
T is temperature
And M is the molecular mass of the particle.
Complete answer:
First start with the given information in the question:
Let Velocity be ${V_{rms}}$ , the root mean square velocity.
Now, the root mean square velocity of Hydrogen be ${V_H}$ ,
and the root mean square velocity of Oxygen be ${V_O}$
Now, we know that ${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
So, from the above equation we can find that root mean square velocity is inversely proportional to the mass of the molecule.
${V_{rms}} \propto \dfrac{1}{\surd{M}}$
Now, we know that;
${M_H} = 2$ and ${M_0} = 32$
Putting the above value of the masses of Hydrogen and Oxygen in the equation, we get;
$\dfrac{{{{\left( {{V_{rms}}} \right)}_{{O_2}}}}}{{{{\left( {{V_{rms}}} \right)}_{{H_2}}}}} = \sqrt {\dfrac{{{{\left( M \right)}_{{H_2}}}}}{{{{\left( M \right)}_{{O_2}}}}}} = \sqrt {\dfrac{1}{{16}}} = \dfrac{1}{4}$
Hence the correct answer is Option(B).
Note: Here we have used the formula of root mean square velocity in terms of the universal gas constant, temperature of the gas and the mass of the gas molecule. As all the other quantities were constant hence we get the required relation for the root mean square velocity and the mass of the gas.
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