
Two gas-argon (atomic radius = 0.07 nm, atomic weight = 40) and xenon (atomic radius = 0.1 nm, atomic weight = 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to:
A. 1.83
B. 2.04
C. 3.67
D. 4.67
Answer
161.7k+ views
Hint: The kinetic theory says that there is a collision between the gaseous particles with each other. The distance covered by molecules between the collisions is termed the mean free path. The time period the collision is termed mean free time.
Formula used: The mean free path can be expressed mathematically as,
\[\lambda = \dfrac{1}{{\sqrt 2 \pi N{d^2}}}\]
Here, M stands for molecular mass, N stands for Avogadro's number and d stands for the atomic radius.
Complete Step by Step Solution:
The mean free time is the division of the mean free path by the average velocity of an electron.
\[\tau = \dfrac{\lambda }{{{v_r}}}\]
Here, \[\tau \] stands for mean free time and \[{v_r}\] is for average velocity.
So, the mean free time equation becomes,
\[\tau = \dfrac{1}{{\sqrt 2 \pi N{d^2}{v_r}}}\] …… (1)
Now, we put \[{v_r} = \sqrt {\dfrac{{3RT}}{M}} \] in the equation (1).
\[\tau = \dfrac{1}{{\sqrt 2 \pi N{d^2}}} \times \sqrt {\dfrac{M}{{3RT}}} \]
The above expression indicates that mean free time depends on the molecular mass of radius. All the other factors are constant. So, we can say,
\[\tau = \dfrac{{\sqrt M }}{{{d^2}}}\]
So, the ratio of the two gases considering their mean free time is,
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{{M_{Argon}}}}{{{d^2}_{Argon}}} \times \dfrac{{{d^2}_{Xenon}}}{{{M_{Xenon}}}}\]
Given, \[{M_{Argon}} = 40,\,{M_{xenon}} = 140,\,{d_{Argon}} = 0.07\,nm,\,{d_{xenon}} = 0.1\,nm\]
So,
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{\sqrt {40} }}{{{{0.07}^2}}} \times \dfrac{{{{0.1}^2}}}{{\sqrt {140} }}\]
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{0.01 \times 6.32}}{{0.0049 \times 11.83}}\]
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{0.0632}}{{0.058}} = 1.09\]
So, the option closest to 1.09 is 1.83.
Therefore, option A is right.
Note: For gases, the kinetic theory says that the gaseous particles are always in random motion and their combined volumes are negligible when combined. Between the particles of gases, there is no presence of any attractive or repulsive force.
Formula used: The mean free path can be expressed mathematically as,
\[\lambda = \dfrac{1}{{\sqrt 2 \pi N{d^2}}}\]
Here, M stands for molecular mass, N stands for Avogadro's number and d stands for the atomic radius.
Complete Step by Step Solution:
The mean free time is the division of the mean free path by the average velocity of an electron.
\[\tau = \dfrac{\lambda }{{{v_r}}}\]
Here, \[\tau \] stands for mean free time and \[{v_r}\] is for average velocity.
So, the mean free time equation becomes,
\[\tau = \dfrac{1}{{\sqrt 2 \pi N{d^2}{v_r}}}\] …… (1)
Now, we put \[{v_r} = \sqrt {\dfrac{{3RT}}{M}} \] in the equation (1).
\[\tau = \dfrac{1}{{\sqrt 2 \pi N{d^2}}} \times \sqrt {\dfrac{M}{{3RT}}} \]
The above expression indicates that mean free time depends on the molecular mass of radius. All the other factors are constant. So, we can say,
\[\tau = \dfrac{{\sqrt M }}{{{d^2}}}\]
So, the ratio of the two gases considering their mean free time is,
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{{M_{Argon}}}}{{{d^2}_{Argon}}} \times \dfrac{{{d^2}_{Xenon}}}{{{M_{Xenon}}}}\]
Given, \[{M_{Argon}} = 40,\,{M_{xenon}} = 140,\,{d_{Argon}} = 0.07\,nm,\,{d_{xenon}} = 0.1\,nm\]
So,
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{\sqrt {40} }}{{{{0.07}^2}}} \times \dfrac{{{{0.1}^2}}}{{\sqrt {140} }}\]
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{0.01 \times 6.32}}{{0.0049 \times 11.83}}\]
\[\dfrac{{{\tau _{Argon}}}}{{{\tau _{Xenon}}}} = \dfrac{{0.0632}}{{0.058}} = 1.09\]
So, the option closest to 1.09 is 1.83.
Therefore, option A is right.
Note: For gases, the kinetic theory says that the gaseous particles are always in random motion and their combined volumes are negligible when combined. Between the particles of gases, there is no presence of any attractive or repulsive force.
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