5.3: The Frequency of Collisions with a Wall

Collision frequency per unit area

In the derivation of an expression for the pressure of a gas, it is useful to consider the frequency with which gas molecules collide with the walls of the container. To derive this expression, consider the expression for the "collision volume".

\[V_{col} = v_x \Delta t\ \cdot A\nonumber \]

in which the product of the velocity \(v_x\) and a time interval \(\Delta t \) is multiplied by \(A\), the area of the wall with which the molecules collide.

All of the molecules within this volume, and with a velocity such that the x-component exceeds \(v_x\) (and is positive) will collide with the wall. That fraction of molecules is given by

\[ N_{col} = \dfrac{N}{V} \dfrac{\langle v_x \rangle \Delta t \cdot A}{2},\nonumber \]

where the factor of (1/2) accounts for the fact that half the molecules are moving away from the wall. Dividing by \(\Delta t\) yields the frequency of collisions with the wall per unit area per unit time:

\[z_w = \dfrac{N}{V} \dfrac{\langle v_x \rangle}{2}.\nonumber \]

This can be rewritten in terms of the average speed in 3 dimensions, \(\langle v \rangle\), by making use of results from the Maxwell-Boltzmann speed distributions in one dimension and 3 dimensions. Namely:

\[ \langle |v_x| \rangle = \left( \dfrac{2 k_BT}{ \pi m} \right)^{1/2} \label{v1d} \]

and

\[\langle v \rangle = \sqrt {\dfrac{8k_B T}{\pi m}}. \label{v3D} \]

Taking the ratio of these two expression shows that

\[ \langle v \rangle = 2 \langle v_x \rangle\nonumber \]

or

\[ \langle v_x \rangle = \dfrac{1}{2} \langle v \rangle.\nonumber \]

Thus,

\[z_w = \dfrac{1}{4} \dfrac{N}{V} \langle v \rangle =\dfrac{\langle v \rangle}{4} \rho = \rho \sqrt {\dfrac{k_B T}{2 \pi m}},\label{z_w}\]

where \(\rho\) is the number density (N/V).

Alternative derivation

A different approach to determining \(z_w\) is to consider a collision cylinder that will enclose all of the molecules that will strike an area of the wall at an angle \(\theta\) and with a speed \(v\) in the time interval \(dt\). The volume of this collision cylinder is the product of its base area (\(A\)) times its vertical height (\(v\text{cos}\theta dt\)), as shown in figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): The collision cylinder for determining the number of collisions of gas molecules with a wall. (CC BY-NC; Ümit Kaya)

The number of molecules in this cylinder is \(\rho·A·v·\text{cos}\theta dt\), where \(\rho\) is the number density \(\dfrac{N}{V}\). The fraction of molecules that are traveling at a speed between \(v\) and \(v + dv\) is \(F(v)dv\). The fraction of molecules traveling within the solid angle bounded by \(\theta\) and \(\theta + d\theta\) and between \(\phi\) and \(\phi + d\phi\) is \(\dfrac{\text{sin}\theta d\theta d\phi}{4\pi}\). Multiplying these three terms together results in the number of molecules colliding with the area \(A\) from the specified direction during the time interval \(dt\)

\[dN_w = \rho·A·v·\text{cos}\theta \, dt \, · \, F(v)dv \, · \, \dfrac{\text{sin}\theta d\theta d\phi}{4\pi}\nonumber \]

This equation can be rearranged to obtain

\[\dfrac{1}{A}\dfrac{dN_w}{dt} = \dfrac{\rho}{4\pi} vF(v)dv · \text{cos}\theta \, \text{sin}\theta \, d\theta d\phi = dz_w \nonumber \]

Integrating this equation over all possible speeds and directions (on the front side of the wall only), we get

\[z_w = \dfrac{\rho}{4\pi} \int_0^{\infty} vF(v)dv · \int_0^{\pi/2}\text{cos}\theta \, \text{sin}\theta \, d\theta \int_0^{2\pi} d\phi \nonumber \]

The result is that

\[z_w = \dfrac{1}{A}\dfrac{dN_w}{dt} = \dfrac{1}{4} \dfrac{N}{V} \langle v \rangle = \rho\dfrac{\langle v \rangle}{4}\label{27.4.1} \]

Example 27.4.1

Calculate the collision frequency per unit area (\(Z_w\)) for oxygen at 25.0°C and 1.00 bar using equation \(\ref{27.4.1}\):

\[z_w = \dfrac{1}{4} \dfrac{N}{V} \langle v \rangle \nonumber \]

Solution

N molecules = \(N_A\) x \(n\), so that

\[ \dfrac{N}{V} = \dfrac{(N_A) \cdot n}{V} = \dfrac{(N_A) \cdot P}{R \cdot T} \nonumber \]

\[ \dfrac{(6.022 x 10^{23} \, mole^{-1})(1.00 \, bar)}{(0.08319 \, L \cdot bar \cdot mole^{-1} \cdot K^{-1})(298 \, K)} = 2.43 \times 10^{22} \, L^{-1} = 2.43 \times 10^{25} \, m^{-3} \nonumber \]

and

\[ \langle v \rangle = \left({\dfrac{8RT}{\pi M}} \right)^{\dfrac {1}{2}} = \left({\dfrac{8(8.314 J \cdot K^{-1} \cdot mol^{-1})(298K)}{\pi \cdot (0.031999 \, kg)}} \right)^{\dfrac {1}{2}} = 444 \, m\cdot s^{-1} \nonumber \]

Thus

\[z_w = \dfrac{1}{4} (2.43 \times 10^{25} m^{-3})(444 \, m\cdot s^{-1}) \left({\dfrac{1 \, m}{100 \, cm}} \right)^2 = 2.70\times 10^{23} s^{-1} \cdot cm^{-2} \nonumber \]

The factor of N/V is often referred to as the “number density” as it gives the number of molecules per unit volume. At 1 atm pressure and 298 K, the number density for an ideal gas is approximately 2.43 x 10¹⁹ molecule/cm³. (This value is easily calculated using the ideal gas law.) By comparison, the average number density for the universe is approximately 1 molecule/cm³.

Exercise 27.4.1

Calculate the collision frequency per unit area (\(Z_w\)) for hydrogen at 25.0°C and 1.00 bar using equation \(\ref{27.4.1}\):

\[z_w = \dfrac{1}{4} \dfrac{N}{V} \langle v \rangle \nonumber \]

Answer

\[ \langle v \rangle = 1770 \, m\cdot s^{-1} \nonumber \] and \[Z_w = 1.08\times 10^{24} s^{-1} \cdot cm^{-2} \nonumber \]

Effusion

Effusion is defined as a loss of material across a boundary. A common example of effusion is the loss of gas inside of a balloon over time.

 

The rate at which gases will effuse from a balloon is affected by a number of factors. But one of the most important is the frequency with which molecules collide with the interior surface of the balloon. As shown in equation \(\ref{z_w}\) this is inversely proportional to the mass (m) or molar mass (M) depending on whether you are working on a per-particle or molar basis.

This can be used to compare the relative rates of effusion for gases of different molar masses or measure vapor low vapor pressures.

The Knudsen Cell Experiment

A Knudsen cell is a chamber in which a thermalized sample of gas is kept, but allowed to effuse through a small orifice in the wall. The gas sample can be modeled using the Kinetic Molecular Theory model as a collection of particles traveling throughout the cell, colliding with one another and also with the wall. If a small orifice is present, any molecules that would collide with that portion of the wall will be lost through the orifice.

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Figure \(\PageIndex{2}\): Effusion of gas particles through an orifice. (CC BY-SA 3.0; Astrang13).

This makes a convenient arrangement to measure the vapor pressure of the material inside the cell, as the total mass lost by effusion through the orifice will be proportional to the vapor pressure of the substance. The vapor pressure can be related to the mass lost by the expression

\[ P = \dfrac{g}{A \Delta t} \sqrt{\dfrac{2 \pi RT}{M}} \nonumber \]

where \(g\) is the mass lost, \(A\) is the area of the orifice, \(\Delta t\) is the time the effusion is allowed to proceed, \(T\) is the temperature and \(M\) is the molar mass of the compound in the vapor phase. The pressure is then given by \(P\). A schematic of what a Knudsen cell might look like is given below.

Contributors and Attributions

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

• Tom Neils, Grand Rapids Community College
• Jonathan Gutow (UW Oshkosh)