Collection of Solved Problems in Physics

Mean Free Path of Hydrogen

Task number: 3943

• Hint 1

  The Assignment mentions normal conditions. What does it mean?

• Normal Conditions

  T = 273.15 K

  p = 101 325 Pa

• Hint 2 – Molecule Diameter

  To determine unknown diameter d of hydrogen molecule, use a relationship for mean free path \(\bar{\lambda}\).

• Equation for Mean Free Path

  The mean free path \(\bar{\lambda}\) basically tells us the average distance travelled by the molecule between two collisions.

  The following equation applies

  \[ \bar{\lambda}=\frac{kT}{{\pi}\sqrt{2}pd^{2}}, \]

  where k is the Boltzmann constant, T is thermodynamic temperature, p is pressure and d is the diameter of the molecule.

• Hint 3 – Mean Free Path Under Lower Pressure

  To determine mean free path \(\bar{\lambda}_2\) under lower pressure p₂ use the fact that distance is inversely proportional to pressure.

• Given Values

  \(\bar{\lambda}_{1}=130\,\mathrm{nm}=1.3\cdot{10^{-7}}\,\mathrm{m}\)	mean free path of hydrogen molecule under normal conditions
  p2 = 133.3·10−4 Pa	lower pressure

  ---

  Table values:

  T1 = 273.15 K	normal temperature
  p1 = 101 325 Pa	normal pressure
  k = 1.38 ·10−23 JK−1	Boltzmann constant

• Analysis

  Mean free path indicates an average distance travelled by a molecule between two collisions. It is determined by a relation from statistical physics which states that it is directly proportional to thermodynamic temperature and inversely proportional to pressure and molecule diameter.

  We can now determine unknown molecule diameter from the above mentioned relation.

  When determinig mean free path under lower pressure we use the fact that this distance is inversely proportional to pressure.

• Solution

  Mean free path \(\bar{\lambda}_1\) is given by

  \[\bar{\lambda}_1=\frac{kT_1}{p_1 \pi d^{2} \sqrt{2}},\]

  where T₁ is thermodynamic temperature under normal conditions, p₁ normal pressure, d molecule diameter and k Boltzmann constant.

  We can determine unknown diameter d of the molecule from this formula:

  \[d^{2}=\frac{kT_{1}}{p_{1} \pi \bar{\lambda_{1}} \sqrt{2}}\] \[d=\sqrt{\frac{kT_{1}}{p_{1} \pi \bar{\lambda_{1}} \sqrt{2}}}.\]

  During the calculation of mean free path \(\bar{\lambda}_2\) under lower pressure p₂ we use the fact that this distance is inversely proportional to pressure. Provided that temperature remains constant (which is this case), the relationship must apply between the values of mean free path and related pressure

  \[\frac{\bar{\lambda}_{1}}{\bar{\lambda}_{2}}=\frac{p_{2}}{p_{1}}.\]

  After adjusting this relation we obtain a final equation for mean free path \(\bar{\lambda}_{2}\) under lower pressure p₂:

  \[\bar{\lambda}_{2}=\frac{p_{1}\bar{\lambda}_{1}}{p_{2}}.\]

• Číselné dosazení

  \[d=\sqrt{\frac{kT_{1}}{p_{1} \pi \bar{\lambda_{1}} \sqrt{2}}}=\sqrt{\frac{1.38\cdot{10^{-23}}\cdot{273,.5}}{101\,325\cdot \pi\cdot{1.3}\cdot{10^{-7}}\cdot \sqrt{2}}}\,\mathrm{m}\] \[d\dot{=}2.54\cdot{10^{-10}}\,\mathrm{m}=0.254\,\mathrm{nm}\] \[\bar{\lambda}_{2}=\frac{p_{1}\bar{\lambda}_{1}}{p_{2}}=\frac{101\,325\cdot{1.3}\cdot{10^{-7}}}{133.3\cdot{10^{-4}}}\, \mathrm{m}\dot{=}0.99\, \mathrm{m}\]

• Answer

  The molecule diameter is approximately 0.254 nm.

  The mean free path of these molecules under lower pressure is approximately 0.99 m.

• Note

  A closer justification of given relation for mean free path of a molecule is given in the task Mean Free Path of Argon Molecule.