Question

Two flasks A and B have an equal volume at \[{\text{100K}}\] and \[{\text{200K}}\] have pressure \[{\text{4 atm}}\] and \[{\text{1 atm}}\] respectively. The flask A contains \[{{\text{H}}_{\text{2}}}\] gas and B contains \[{\text{C}}{{\text{H}}_{\text{4}}}\] ​ gas. The collision diameter \[{\text{C}}{{\text{H}}_{\text{4}}}\]​ is twice that of \[{{\text{H}}_{\text{2}}}\]​. Calculate the ratio of the mean free path of \[{\text{C}}{{\text{H}}_{\text{4}}}\]​ to \[{{\text{H}}_{\text{2}}}\]​

Answer 1

Hint:To answer this question, you should recall the concept kinetic theory of gases and mean free path. The mean free path of a gas molecule is its average path length between collisions.
Formula used: \[\lambda = \dfrac{{RT}}{{\sqrt 2 \pi {d^2}{N_A}P}}\]
where \[R = \] gas constant, \[T = \] temperature, \[{N_A} = \] Avogadro's constant, \[P = \] Pressure and \[d = \] collision diameter

Complete step by step solution:
Let’s look at the motion of a gas molecule inside an ideal gas, a typical molecule inside an ideal gas will abruptly change its direction and speed as it collides elastically with other molecules of the same gas. Though between the collisions the molecule will move in a straight line at some constant speed, this is applicable for all the molecules in the gas. It is difficult to measure or describe this random motion of gas molecules thus we attempt to measure its mean free path\[\lambda \].
Let the collision diameter of \[{H_2} = d\].
Then the collision diameter of \[C{H_4} = 2d\].
In the question, it is given \[A \Rightarrow {H_2} \Rightarrow 100{\text{K}} \to 4{\text{atm}} \to d\]
and \[B \Rightarrow C{H_4} \Rightarrow 200{\text{K}} \to 1{\text{atm}} \to 2d\]
Substituting these values to calculate the ratio:
\[\dfrac{{\lambda (C{H_4})}}{{\lambda ({H_2})}} = \dfrac{{\left( {\dfrac{{P \times 200}}{{\sqrt 2 \times \pi \times {{\left( {2d} \right)}^2} \times {N_A} \times 1}}} \right)}}{{\left( {\dfrac{{P \times 100}}{{\sqrt 2 \times \pi \times {{\left( d \right)}^2} \times {N_A} \times 4}}} \right)}}\]

After solving this equation we arrive, the ratio is \[ = 2:1\]

Note:
You should know about the Factors affecting the mean free path
Density: As the density increases, the molecules come closer to each other, thus increasing the number of collisions, and decreasing the mean free path.
The number of molecules: As the number of molecules increases the probability of collision increases and thus the mean free path decreases.
The radius of the molecule: As the radius of the molecule increases the space between the molecules decreases causing the number of collisions to increase, thus decreasing the mean free path.
Pressure, temperature, and other physical factors also affect the density of the gas and thus affect the mean free path.