Question

The time taken for a given volume of gas ${{E}}$ to effuse through a hole is \[75{{ }}s\] . Under identical conditions, the same volume of a mix of ${{CO}}$ and \[{{{N}}_2}\] (containing \[40\% \] of \[{{{N}}_2}\] by volume) effused in \[70{{ }}s\] . Calculate,
The relative molar mass of ${{E}}$ and,
The RMS velocity (in \[ms\% {{ }} - {{ }}1\] units) of ${{E}}$ at \[0{{{ }}^ \circ }{{C}}\] .

Answer 1

Hint:Effusion is the movement of molecules of a gas through the opening in a container. The opening of the container should be less than the mean free path of the particles, which means that it should be less than the average distance covered by the molecules between collisions.

Complete answer:
In the question, a mixture of two gases, carbon monoxide and nitrogen are taken.
Molecular mass of \[{{{N}}_2}\] \[ = \] \[14 + 14\] \[ = \] \[28\]
Molecular mass of ${{CO}}$ \[ = \] \[12 + 16\] \[ = \] \[28\]
Therefore, molar mass of the mixture \[ = \] \[40\% \] of \[{{{N}}_2}\]\[ + \] \[60\% \] of ${{CO}}$
\[ = \] \[0.4{{ }} \times {{ }}28\]+ \[0.6{{ }} \times {{ }}28\]
  \[ = \] \[28\]
From Graham’s law of effusion, we know that, rate of effusion of a gas is inversely proportional to the square root of molar mass of the gas or mixture of gases.
Rate \[\alpha \] $\dfrac{1}{{{t}}}$ \[\alpha \] $\dfrac{1}{{\sqrt {{M}} }}$
Time taken for a given volume of gas ${{E}}$ to effuse through a hole $ = $ \[75{{ }}s\]
Time taken for a given volume of mixture of gases ${{CO}}$ and \[{{{N}}_2}\] to effuse through a hole $ = $ \[70{{ }}s\]
From the relation of rate, time and molar mass, we can derive a relation between time and molar mass, that is, time is proportional to molar mass.
${{t }}\alpha {{ }}\sqrt {{M}} $
Now, if any one of the quantities is unknown, we can calculate it by dividing the time and square root of molar masses.
$\dfrac{{{{{t}}_1}}}{{{{{t}}_2}}}{{ = }}\sqrt {\dfrac{{{{{M}}_1}}}{{{{{M}}_2}}}} $
We can substitute the values of time taken for effusion of gas ${{E}}$ and mixture of nitrogen and carbon monoxide gases and the molar mass of the mixture to find the unknown molar mass.
$\dfrac{{75}}{{70}}{{ = }}\sqrt {\dfrac{{{{{M}}_1}}}{{28}}} $
Hence, the equation can be rearranged as follows,
$\dfrac{{75}}{{70}}{{ }} \times {{ }}\sqrt {28} {{ = }}\sqrt {{{{M}}_1}} $
Squaring both the sides, we get,
${\left( {\dfrac{{75}}{{70}}} \right)^2}{{ }} \times {{ }}28{{ = }}{{{M}}_1}$
Therefore, on solving, we get,
${{{M}}_1}$ $ = $ $32.14{{ g mo}}{{{l}}^{ - 1}}$

Hence, the relative molar mass of ${{E}}$$ = $ $32.14{{ g mo}}{{{l}}^{ - 1}}$ .

Root mean square velocity is used to predict the speed of molecules at a given temperature.
The formula for finding the root mean square velocity of moles is given by,
${{{v}}_{{{rms }}}} = {{ }}\sqrt {\dfrac{{3{{RT}}}}{{{M}}}} $
Given, ${{T}}$ $ = $ $273{{ K}}$
${{R}}$ $ = $ $8.314$ ${{J }}{{{K}}^{ - 1}}{{mo}}{{{l}}^{ - 1}}$
${{M}}$ $ = $ $32.14{{ g mo}}{{{l}}^{ - 1}}$
Substituting the values of ${{R}}$ , ${{T}}$ and ${{M}}$ in the equation, we get,
So, ${{{v}}_{{{rms }}}} = {{ }}\sqrt {\dfrac{{3{{ }} \times {{ }}8.314{{ J }}{{{K}}^{ - 1}}{{ mo}}{{{l}}^{ - 1}}{{ }} \times {{ 273 K}}}}{{32.14{{ g mo}}{{{l}}^{ - 1}}{{ }} \times {{ 1}}{{{0}}^{ - 3}}}}} $
On calculation, we get,
${{{v}}_{{{rms}}}}{{ = 460}}{{.28 m }}{{{s}}^{ - 1}}$

Hence, The RMS velocity of ${{E}}$ at \[0{{{ }}^ \circ }{{C}}\]is ${{460}}{{.28 m }}{{{s}}^{ - 1}}$ .

Note:
The time taken for effusion or diffusion of a gas is directly proportional to the square root of its molar mass. Root mean square velocity of a gas is the same as the mean of squares of velocities of individual gas molecules. Remember the formula for calculating RMS velocity.