
A gas mixture contains an equal number of molecules ${N_2}$ and $S{F_6}$, some of it is passed through a gaseous effusion apparatus. Calculate how many molecules of ${N_2}$ are present in the product gas for every $100$ molecules of $S{F_6}$.
Answer
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Hint: Diffusion is the process in which the molecules move from a region of higher concentration to a region of lower concentration. The rate of diffusion depends on the nature of the interaction between the material and the medium.
Complete step by step answer:
Step I:
Let there be \[{m_1}\]molecules of ${N_2}$ gas
And ${m_2}$ molecules of $S{F_6}$ gas
The rate of diffusion of the gas is inversely proportional to the square root of the molecular mass of the gas. If $M$ is the molecular mass, then the rate of diffusion is given by
$r = \dfrac{1}{{\sqrt M }}$---(i)
Step II:
Therefore rate of diffusion of ${N_2}$ gas is given by
${r_1} = \dfrac{1}{{{M_1}}}$
Where ${M_1}$ is the molar mass of ${N_2}$
Similarly the rate of diffusion of $S{F_6}$ gas can be written as
${r_2} = \dfrac{1}{{{M_2}}}$
Where ${M_2}$ is the molar mass of $S{F_6}$
Step III:
Hence rate of diffusion for both the gases is given by
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\dfrac{1}{{\sqrt {{M_1}} }}}}{{\dfrac{1}{{\sqrt {{M_2}} }}}}\]
\[\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]---(iii)
Step IV:
Molecular mass of ${N_2} = 28$
Molecular mass of $S{F_6} = 146$
Substitute the values in equation (iii),
$\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{146}}{{28}}} $
$\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {5.214} $
$\dfrac{{{r_1}}}{{{r_2}}} = 2.28$
Step V:
Therefore the number of nitrogen molecules present in each $100$ molecule of $S{F_6}$ is $100 \times 2.28 = 228$ molecules.
Note:
It is to be noted that the rate of diffusion depends on the mass of solute and temperature. The process of diffusion continues, till the concentration becomes equal. Also if the size of the particles are small they will move faster and diffusion will also be faster. But if the size of the particles are very large they will not be able to move faster and rate of diffusion will be slow. The kinetic energy of the molecules depends on temperature. At high temperature, they will move faster and at low temperature they will move slow.
Complete step by step answer:
Step I:
Let there be \[{m_1}\]molecules of ${N_2}$ gas
And ${m_2}$ molecules of $S{F_6}$ gas
The rate of diffusion of the gas is inversely proportional to the square root of the molecular mass of the gas. If $M$ is the molecular mass, then the rate of diffusion is given by
$r = \dfrac{1}{{\sqrt M }}$---(i)
Step II:
Therefore rate of diffusion of ${N_2}$ gas is given by
${r_1} = \dfrac{1}{{{M_1}}}$
Where ${M_1}$ is the molar mass of ${N_2}$
Similarly the rate of diffusion of $S{F_6}$ gas can be written as
${r_2} = \dfrac{1}{{{M_2}}}$
Where ${M_2}$ is the molar mass of $S{F_6}$
Step III:
Hence rate of diffusion for both the gases is given by
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\dfrac{1}{{\sqrt {{M_1}} }}}}{{\dfrac{1}{{\sqrt {{M_2}} }}}}\]
\[\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]---(iii)
Step IV:
Molecular mass of ${N_2} = 28$
Molecular mass of $S{F_6} = 146$
Substitute the values in equation (iii),
$\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{146}}{{28}}} $
$\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {5.214} $
$\dfrac{{{r_1}}}{{{r_2}}} = 2.28$
Step V:
Therefore the number of nitrogen molecules present in each $100$ molecule of $S{F_6}$ is $100 \times 2.28 = 228$ molecules.
Note:
It is to be noted that the rate of diffusion depends on the mass of solute and temperature. The process of diffusion continues, till the concentration becomes equal. Also if the size of the particles are small they will move faster and diffusion will also be faster. But if the size of the particles are very large they will not be able to move faster and rate of diffusion will be slow. The kinetic energy of the molecules depends on temperature. At high temperature, they will move faster and at low temperature they will move slow.
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