Question

For oxygen at 25$^{o}C$, the collision diameter is 0.361 nm. What is the mean free path for oxygen molecule at:
(a) 1 atm pressure and (b) 0.1 Pa pressure

Answer 1

Hint: Collision diameter ($\sigma $) is the closest distance between the centers of the two molecules undergoing collision.
The average distance travelled by a molecule between two successive collisions is the mean free path of the molecules and is calculated as
     \[\lambda =\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}}\times \dfrac{RT}{P{{N}_{o}}}\]

Complete step by step answer:
The kinetic theory of gases states that the molecules of gas are randomly moving with different velocities in different directions and thus, are constantly colliding with one another.
The distance a molecule travels before colliding with another molecule is free path.
It is very trivial to say the molecule travels the exact same distance before collision every time as it is fairly possible that the molecule will cover a small distance sometimes and large distances at other times.
So we define the free path of a molecule as the mean free path, which is the average of the different values of the distances travelled by the molecule.

The mean free path is given as
     \[\lambda =\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}n}\]
where, $\sigma $ is the collision diameter (in m)
$n$ is the number of molecules per unit volume of the gas
We have been given the collision diameter of oxygen, i.e. $\sigma =0.361$nm.
Given the temperature of the gas, T = 25$^{o}C$.
We have to find the mean free path, $\lambda $ at two different pressures, i.e. 1 atm and 0.1 Pa
So, for our calculations we can modify the above expression for $\lambda $ in terms of pressure (P) and temperature (T) as
From the ideal gas equation we have, PV = NRT.
On rearranging, $\dfrac{P}{RT}=\dfrac{N}{V}$. Let $\dfrac{N}{V}={n}'$ such that $\dfrac{P}{RT}={n}'$, which represents the number of moles of gas per unit volume of the gas.
But we need the number of molecules per unit volume. So we have to multiply $\dfrac{P}{RT}$ by Avogadro’s number to get the number of molecules per unit volume of the gas. Thus,
     \[n=\dfrac{P{{N}_{o}}}{RT}\]
Replacing the value of $n$ by \[\dfrac{P{{N}_{o}}}{RT}\] in the above expression for mean free path, we get
     \[\lambda =\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}n}=\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}}\times \dfrac{RT}{P{{N}_{o}}}\]
Here, P is pressure in $N{{m}^{-2}}$ or Pa
R is gas constant and has the value = $8.314J{{K}^{-1}}mo{{l}^{-1}}$
${{N}_{o}}=6.022\times {{10}^{23}}mo{{l}^{-1}}$
(a) Let us find the value of $\lambda $ at P = 1 atm.
- 1 atm = 101325 Pa = 101325 $N{{m}^{-2}}$
We have, $\sigma =0.361$nm = $0.361\times {{10}^{-9}}$ m = $3.61\times {{10}^{-10}}$ m
T = 25$^{o}C$
To convert temperature in Celsius ($^{o}C$) to Kelvin (K),
 T (K) =T ($^{o}C$ ) + 273.15
  T (K) = 25 + 273.15
           = 298.15 K
Take T = 298 K for calculations.

Substituting $\sigma =3.61\times {{10}^{-10}}$ m, T = 298 K, P = 101325 $N{{m}^{-2}}$R = $8.314J{{K}^{-1}}mo{{l}^{-1}}$ in the mean free path expression, we get
     \[\begin{align}
  & \lambda =\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}}\times \dfrac{RT}{P{{N}_{o}}} \\
 & \,\,\,\,=\dfrac{1}{\sqrt{2}\times 3.14\times {{(3.61\times {{10}^{-10}})}^{2}}{{m}^{2}}}\times \dfrac{8.314J{{K}^{-1}}mo{{l}^{-1}}\times 298K}{101325N{{m}^{-2}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}} \\
\end{align}\]

We know that 1 J = 1$Nm$, on simplifying and cancelling the units, we get
     \[\begin{align}
  & \lambda =\dfrac{1}{\sqrt{2}\times 3.14\times 13.03\times {{10}^{-20}}{{m}^{2}}}\times \dfrac{2477.572J}{610179.15N{{m}^{-2}}\times {{10}^{23}}} \\
 & =\dfrac{1}{\sqrt{2}\times 3.14\times 13.03\times {{10}^{-20}}{{m}^{2}}}\times \dfrac{2477.572Nm}{610179.15N{{m}^{-2}}\times {{10}^{23}}} \\
 & =\dfrac{1}{57.87\times {{10}^{-20}}}\times \dfrac{2477.572}{610179.15\times {{10}^{23}}}m \\
\end{align}\]
Multiplying the numerator and denominator by ${{10}^{9}}$, we can further simplify as
     \[=\dfrac{2477.572\times {{10}^{-3}}m}{35311520.1}\times \dfrac{{{10}^{9}}}{{{10}^{9}}}=70.1\times {{10}^{-9}}m\]
Therefore, the mean free path, $\lambda $= 70.1 nm at 1 atm pressure.

(b) Here, the given pressure, P = 0.1 Pa = 0.1$N{{m}^{-2}}$
Taking the given values of $\sigma =3.61\times {{10}^{-10}}$ m, T = 298 K and R = $8.314J{{K}^{-1}}mo{{l}^{-1}}$in the mean free expression, we obtain
     \[\begin{align}
  & \lambda =\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}}\times \frac{RT}{P{{N}_{o}}} \\
 & \,\,\,\,=\frac{1}{\sqrt{2}\times 3.14\times {{(3.61\times {{10}^{-10}})}^{2}}{{m}^{2}}}\times \frac{8.314Nm{{K}^{-1}}mo{{l}^{-1}}\times 298K}{0.1N{{m}^{-2}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}} \\
\end{align}\]
Cancelling the units and calculating, we get $\lambda $ as
     \[\begin{align}
  & \lambda =\frac{1}{57.87}\times \dfrac{2477.572\times {{10}^{-3}}}{0.1\times 6.022}m \\
 & =\dfrac{2477.572\times {{10}^{-3}}}{34.85}m=71.1\times {{10}^{-3}}m \\
\end{align}\]
Therefore, the mean free path of oxygen molecules is $7.11\times {{10}^{-3}}$ m at 0.1 Pa pressure.

Note: Points to be kept in mind while converting units are as follows:
1 atm = 101325 Pa
1 Pa = 1$N{{m}^{-2}}$
1 J = 1$Nm$
The given problem involves long calculations, so carefully solve the question step by step to avoid any confusion or calculation error.