Kinetic theory of gases

In the kinetic theory of gases, the pressure is assumed to be equal to the force (per unit area) exerted by the atoms hitting and rebounding from the gas container's surface. Consider a gas of a large number N of molecules, each of mass m, enclosed in a cube of volume V = L³. When a gas molecule collides with the wall of the container perpendicular to the x axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by:

$${\displaystyle \Delta p=p_{i,x}-p_{f,x}=p_{i,x}-(-p_{i,x})=2p_{i,x}=2mv_{x},}$$

where p is the momentum, i and f indicate initial and final momentum (before and after collision), x indicates that only the x direction is being considered, and ${\displaystyle v_{x}}$ is the speed of the particle in the x direction (which is the same before and after the collision).

The particle impacts one specific side wall once during the time interval ${\displaystyle \Delta t}$

$${\displaystyle \Delta t={\frac {2L}{v_{x}}},}$$

where L is the distance between opposite walls.

The force of this particle's collision with the wall is

$${\displaystyle F={\frac {\Delta p}{\Delta t}}={\frac {mv_{x}^{2}}{L}}.}$$

The total force on the wall due to collisions by molecules impacting the walls with a range of possible values of ${\displaystyle v_{x}}$ is

$${\displaystyle F={\frac {Nm{\overline {v_{x}^{2}}}}{L}},}$$

where the bar denotes an average over the possible velocities of the N particles.

Since the motion of the particles is random and there is no bias applied in any direction, the average squared speed in each direction is identical:

$${\displaystyle {\overline {v_{x}^{2}}}={\overline {v_{y}^{2}}}={\overline {v_{z}^{2}}}.}$$

By the Pythagorean theorem, in three dimensions the average squared speed ${\displaystyle v^{2}}$ is given by

$${\displaystyle {\overline {v^{2}}}={\overline {v_{x}^{2}}}+{\overline {v_{y}^{2}}}+{\overline {v_{z}^{2}}},}$$

Therefore

$${\displaystyle {\overline {v^{2}}}=3{\overline {v_{x}^{2}}}.}$$

and

$${\displaystyle {\overline {v_{x}^{2}}}={\frac {\overline {v^{2}}}{3}},}$$

and so the force can be written as

$${\displaystyle F={\frac {Nm{\overline {v^{2}}}}{3L}}.}$$

This force is exerted uniformly on an area L². Therefore, the pressure of the gas is

$${\displaystyle P={\frac {F}{L^{2}}}={\frac {Nm{\overline {v^{2}}}}{3V}},}$$

where V = L³ is the volume of the box.

In terms of the translational kinetic energy K of the gas, since

$${\displaystyle K=N{\frac {1}{2}}m{\overline {v^{2}}}}$$

we have

$${\displaystyle PV={\frac {2}{3}}K.}$$

This is an important, non-trivial result of the kinetic theory because it relates pressure, a macroscopic property, to the translational kinetic energy of the molecules, which is a microscopic property.

Rewriting the above result for the pressure as ${\textstyle PV={\frac {Nm{\overline {v^{2}}}}{3}}}$, we may combine it with the ideal gas law

${\displaystyle PV=Nk_{\mathrm {B} }T,}$

(1)

where ${\displaystyle k_{\mathrm {B} }}$ is the Boltzmann constant and ${\displaystyle T}$ the absolute temperature defined by the ideal gas law, to obtain

$${\displaystyle k_{\mathrm {B} }T={m{\overline {v^{2}}} \over 3},}$$

which leads to a simplified expression of the average kinetic energy per molecule,[17]

$${\displaystyle {\frac {1}{2}}m{\overline {v^{2}}}={\frac {3}{2}}k_{\mathrm {B} }T.}$$

The kinetic energy of the system is ${\displaystyle N}$ times that of a molecule, namely ${\textstyle K={\frac {1}{2}}Nm{\overline {v^{2}}}}$. Then the temperature ${\displaystyle T}$ takes the form

${\displaystyle T={m{\overline {v^{2}}} \over 3k_{\mathrm {B} }}}$

(2)

which becomes

${\displaystyle T={\frac {2}{3}}{\frac {K}{Nk_{\mathrm {B} }}}.}$

(3)

Equation (3) is one important result of the kinetic theory: The average molecular kinetic energy is proportional to the ideal gas law's absolute temperature. From equations (1) and (3), we have

${\displaystyle PV={\frac {2}{3}}K.}$

(4)

Thus, the product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.

Equations (1) and (4) are called the "classical results", which could also be derived from statistical mechanics; for more details, see:[18]

Since there are ${\displaystyle 3N}$ degrees of freedom in a monatomic-gas system with ${\displaystyle N}$ particles, the kinetic energy per degree of freedom per molecule is

${\displaystyle {\frac {K}{3N}}={\frac {k_{\mathrm {B} }T}{2}}}$

(5)

In the kinetic energy per degree of freedom, the constant of proportionality of temperature is 1/2 times Boltzmann constant or R/2 per mole. This result is related to the equipartition theorem.

Thus the kinetic energy per Kelvin of one mole of (monatomic ideal gas) is 3 [R/2] = 3R/2. Thus the kinetic energy per Kelvin can be calculated easily:

• per mole: 12.47 J / K
• per molecule: 20.7 yJ / K = 129 μeV / K

At standard temperature (273.15 K), the kinetic energy can also be obtained:

• per mole: 3406 J
• per molecule: 5.65 zJ = 35.2 meV.

Although monatomic gases have 3 (translational) degrees of freedom per atom, diatomic gases should have 6 degrees of freedom per molecule (3 translations, two rotations, and one vibration). However, the lighter diatomic gases (such as diatomic oxygen) may act as if they have only 5 due to the strongly quantum-mechanical nature of their vibrations and the large gaps between successive vibrational energy levels. Quantum statistical mechanics is needed to accurately compute these contributions. [19]

For an ideal gas in equilibrium, the rate of collisions with the container wall and velocity distribution of particles hitting the container wall can be calculated[20] based on naive kinetic theory, and the results can be used for analyzing effusive flow rates, which is useful in applications such as the gaseous diffusion method for isotope separation.

Assume that in the container, the number density (number per unit volume) is ${\displaystyle n=N/V}$ and that the particles obey Maxwell's velocity distribution:

$${\displaystyle f_{\text{Maxwell}}(v_{x},v_{y},v_{z})\,dv_{x}\,dv_{y}\,dv_{z}=\left({\frac {m}{2\pi kT}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{B}T}}}\,dv_{x}\,dv_{y}\,dv_{z}}$$

Then for a small area ${\displaystyle dA}$ on the container wall, a particle with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal of the area ${\displaystyle dA}$, will collide with the area within time interval ${\displaystyle dt}$, if it is within the distance ${\displaystyle vdt}$ from the area ${\displaystyle dA}$. Therefore, all the particles with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal that can reach area ${\displaystyle dA}$ within time interval ${\displaystyle dt}$ are contained in the tilted pipe with a height of ${\displaystyle v\cos(\theta )dt}$ and a volume of ${\displaystyle v\cos(\theta )dAdt}$.

The total number of particles that reach area ${\displaystyle dA}$ within time interval ${\displaystyle dt}$ also depends on the velocity distribution; All in all, it calculates to be:

$${\displaystyle nv\cos(\theta )\,dA\,dt\times \left({\frac {m}{2\pi k_{B}T}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{B}T}}}\left(v^{2}\sin(\theta )\,dv\,d\theta \,d\phi \right).}$$

Integrating this over all appropriate velocities within the constraint ${\displaystyle v>0,0<\theta <\pi /2,0<\phi <2\pi }$ yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time:

$${\displaystyle J_{\text{collision}}={\frac {\int _{0}^{\pi /2}\cos \theta \sin \theta d\theta }{\int _{0}^{\pi }\sin \theta d\theta }}\times n{\bar {v}}={\frac {1}{4}}n{\bar {v}}={\frac {n}{4}}{\sqrt {\frac {8k_{\mathrm {B} }T}{\pi m}}}.}$$

This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed ${\displaystyle {\bar {v}}}$ of the Maxwell's velocity distribution, one has to integrate over${\displaystyle v>0,0<\theta <\pi ,0<\phi <2\pi }$.

The momentum transfer to the container wall from particles hitting the area ${\displaystyle dA}$ with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal, in time interval ${\displaystyle dt}$ is:

$${\displaystyle [2mv\cos(\theta )]\times nv\cos(\theta )\,dA\,dt\times \left({\frac {m}{2\pi k_{B}T}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{B}T}}}\left(v^{2}\sin(\theta )\,dv\,d\theta \,d\phi \right).}$$

Integrating this over all appropriate velocities within the constraint ${\displaystyle v>0,0<\theta <\pi /2,0<\phi <2\pi }$ yields the pressure (consistent with Ideal gas law):

$${\displaystyle P={\frac {2\int _{0}^{\pi /2}\cos ^{2}\theta \sin \theta d\theta }{\int _{0}^{\pi }\sin \theta d\theta }}\times nmv_{\text{rms}}^{2}={\frac {1}{3}}nmv_{\text{rms}}^{2}={\frac {2}{3}}n\langle E_{kin}\rangle =nk_{\mathrm {B} }T}$$

If this small area ${\displaystyle A}$ is punched to become a small hole, the effusive flow rate will be:

$${\displaystyle \Phi _{\text{effusion}}=J_{\text{collision}}A=nA{\sqrt {\frac {k_{\mathrm {B} }T}{2\pi m}}}.}$$

Combined with the ideal gas law, this yields

$${\displaystyle \Phi _{\text{effusion}}={\frac {PA}{\sqrt {2\pi mk_{\mathrm {B} }T}}}.}$$

The above expression is consistent with Graham's law.

To calculate the velocity distribution of particles hitting this small area, we must take into account that all the particles with ${\displaystyle (v,\theta ,\phi )}$ that hit the area ${\displaystyle dA}$ within the time interval ${\displaystyle dt}$ are contained in the tilted pipe with a height of ${\displaystyle v\cos(\theta )dt}$ and a volume of ${\displaystyle v\cos(\theta )dAdt}$; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of ${\displaystyle v\cos \theta }$:

$${\displaystyle {\begin{aligned}f(v,\theta ,\phi )\,dv\,d\theta \,d\phi &=\lambda v\cos {\theta }{\times }\left({\frac {m}{2\pi kT}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}}(v^{2}\sin {\theta }\,dv\,d\theta \,d\phi )\\\end{aligned}}}$$

with the constraint ${\textstyle v>0,\,0<\theta <{\frac {\pi }{2}},\,0<\phi <2\pi }$. The constant ${\displaystyle \lambda }$ can be determined by the normalization condition ${\displaystyle \int f(v,\theta ,\phi )\,dv\,d\theta \,d\phi =1}$ to be ${\textstyle {4}/{\bar {v}}}$, and overall:

$${\displaystyle {\begin{aligned}f(v,\theta ,\phi )\,dv\,d\theta \,d\phi &={\frac {1}{2\pi }}\left({\frac {m}{k_{\mathrm {B} }T}}\right)^{2}e^{-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}}(v^{3}\sin {\theta }\cos {\theta }\,dv\,d\theta \,d\phi )\\\end{aligned}};\quad v>0,\,0<\theta <{\frac {\pi }{2}},\,0<\phi <2\pi }$$

From the kinetic energy formula it can be shown that

$${\displaystyle v_{\text{p}}={\sqrt {2\cdot {\frac {k_{\mathrm {B} }T}{m}}}},}$$

$${\displaystyle {\bar {v}}={\frac {2}{\sqrt {\pi }}}v_{p}={\sqrt {{\frac {8}{\pi }}\cdot {\frac {k_{\mathrm {B} }T}{m}}}},}$$

$${\displaystyle v_{\text{rms}}={\sqrt {\frac {3}{2}}}v_{p}={\sqrt {{3}\cdot {\frac {k_{\mathrm {B} }T}{m}}}},}$$

where v is in m/s, T is in kelvins, and m is the mass of one molecule of gas. The most probable (or mode) speed ${\displaystyle v_{\text{p}}}$ is 81.6% of the root-mean-square speed ${\displaystyle v_{\text{rms}}}$, and the mean (arithmetic mean, or average) speed ${\displaystyle {\bar {v}}}$ is 92.1% of the rms speed (isotropic distribution of speeds).

See:

• Average,
• Root-mean-square speed
• Arithmetic mean
• Mean
• Mode (statistics)

In books on elementary kinetic theory[21] one can find results for dilute gas modeling that are used in many fields. Derivation of the kinetic model for shear viscosity usually starts by considering a Couette flow where two parallel plates are separated by a gas layer. The upper plate is moving at a constant velocity to the right due to a force F. The lower plate is stationary, and an equal and opposite force must therefore be acting on it to keep it at rest. The molecules in the gas layer have a forward velocity component ${\displaystyle u}$ which increase uniformly with distance ${\displaystyle y}$ above the lower plate. The non-equilibrium flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.

Let ${\displaystyle \sigma }$ be the collision cross section of one molecule colliding with another. As in the previous section, the number density ${\displaystyle n}$ is defined as the number of molecules per (extensive) volume, or ${\displaystyle n=N/V}$. The collision cross section per volume or collision cross section density is ${\displaystyle n\sigma }$, and it is related to the mean free path ${\displaystyle l}$ by

$${\displaystyle l={\frac {1}{{\sqrt {2}}n\sigma }}}$$

Notice that the unit of the collision cross section per volume ${\displaystyle n\sigma }$ is reciprocal of length. The mean free path is the average distance traveled by a molecule, or a number of molecules per volume, before they make their first collision.

Let ${\displaystyle u_{0}}$ be the forward velocity of the gas at an imaginary horizontal surface inside the gas layer. The number of molecules arriving at an area ${\displaystyle dA}$ on one side of the gas layer, with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal, in time interval ${\displaystyle dt}$ is

$${\displaystyle nv\cos({\theta })\,dA\,dt\times \left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}\,e^{-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}}(v^{2}\sin {\theta }\,dv\,d\theta \,d\phi )}$$

These molecules made their last collision at a distance ${\displaystyle l\cos \theta }$ above and below the gas layer, and each will contribute a forward momentum of

$${\displaystyle p_{x}^{\pm }=m\left(u_{0}\pm l\cos \theta \,{du \over dy}\right),}$$

where plus sign applies to molecules from above, and minus sign below. Note that the forward velocity gradient ${\displaystyle du/dy}$ can be considered to be constant over a distance of mean free path.

Integrating over all appropriate velocities within the constraint

$${\displaystyle {\begin{cases}v>0\\0<\theta <\pi /2\\0<\phi <2\pi \end{cases}}}$$

yields the forward momentum transfer per unit time per unit area (also known as shear stress):

$${\displaystyle \tau ^{\pm }={\frac {1}{4}}{\bar {v}}n\cdot m\left(u_{0}\pm {\frac {2}{3}}l\,{du \over dy}\right)}$$

The net rate of momentum per unit area that is transported across the imaginary surface is thus

$${\displaystyle \tau =\tau ^{+}-\tau ^{-}={\frac {1}{3}}{\bar {v}}nm\cdot l\,{du \over dy}}$$

Combining the above kinetic equation with Newton's law of viscosity

$${\displaystyle \tau =\eta \,{du \over dy}}$$

gives the equation for shear viscosity, which is usually denoted ${\displaystyle \eta _{0}}$ when it is a dilute gas:

$${\displaystyle \eta _{0}={\frac {1}{3}}{\bar {v}}nml}$$

Combining this equation with the equation for mean free path gives

$${\displaystyle \eta _{0}={\frac {1}{3{\sqrt {2}}}}{\frac {m\cdot {\bar {v}}}{\sigma }}}$$

Maxwell-Boltzmann distribution gives the average (equilibrium) molecular speed as

$${\displaystyle {\bar {v}}={\frac {2}{\sqrt {\pi }}}v_{p}=2{\sqrt {{\frac {2}{\pi }}\cdot {\frac {k_{\mathrm {B} }T}{m}}}}}$$

where ${\displaystyle v_{p}}$ is the most probable speed. We note that

$${\displaystyle k_{B}\cdot N_{A}=R\quad {\text{and}}\quad M=m\cdot N_{A}}$$

and insert the velocity in the viscosity equation above. This gives the well known equation for shear viscosity for dilute gases:

$${\displaystyle \eta _{0}={\frac {2}{3{\sqrt {\pi }}}}\cdot {\frac {\sqrt {mk_{\mathrm {B} }T}}{\sigma }}={\frac {2}{3{\sqrt {\pi }}}}\cdot {\frac {\sqrt {MRT}}{\sigma \cdot N_{A}}}}$$

and ${\displaystyle M}$ is the molar mass. The equation above presupposes that the gas density is low (i.e. the pressure is low). This implies that the kinetic translational energy dominates over rotational and vibrational molecule energies. The viscosity equation further presupposes that there is only one type of gas molecules, and that the gas molecules are perfect elastic and hard core particles of spherical shape. This assumption of elastic, hard core spherical molecules, like billiard balls, implies that the collision cross section of one molecule can be estimated by

$${\displaystyle \sigma =\pi \left(2r\right)^{2}=\pi d^{2}}$$

The radius ${\displaystyle r}$ is called collision cross section radius or kinetic radius, and the diameter ${\displaystyle d}$ is called collision cross section diameter or kinetic diameter of a molecule in a monomolecular gas. There are no simple general relation between the collision cross section and the hard core size of the (fairly spherical) molecule. The relation depends on shape of the potential energy of the molecule. For a real spherical molecule (i.e. a noble gas atom or a reasonably spherical molecule) the interaction potential is more like the Lennard-Jones potential or Morse potential which have a negative part that attracts the other molecule from distances longer than the hard core radius. The radius for zero Lennard-Jones potential is then appropriate to use as an estimate for the kinetic radius.

Following a similar logic as above, one can derive the kinetic model for thermal conductivity[21] of a dilute gas:

Consider two parallel plates separated by a gas layer. Both plates have uniform temperatures, and are so massive compared to the gas layer that they can be treated as thermal reservoirs. The upper plate has a higher temperature than the lower plate. The molecules in the gas layer have a molecular kinetic energy ${\displaystyle \varepsilon }$ which increases uniformly with distance ${\displaystyle y}$ above the lower plate. The non-equilibrium energy flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.

Let ${\displaystyle \varepsilon _{0}}$ be the molecular kinetic energy of the gas at an imaginary horizontal surface inside the gas layer. The number of molecules arriving at an area ${\displaystyle dA}$ on one side of the gas layer, with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal, in time interval ${\displaystyle dt}$ is

$${\displaystyle nv\cos(\theta )\,dA\,dt\times \left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{B}T}}}(v^{2}\sin(\theta )\,dv\,d\theta \,d\phi )}$$

These molecules made their last collision at a distance ${\displaystyle l\cos \theta }$ above and below the gas layer, and each will contribute a molecular kinetic energy of

$${\displaystyle \varepsilon ^{\pm }=\left(\varepsilon _{0}\pm mc_{v}l\cos \theta \,{dT \over dy}\right),}$$

where ${\displaystyle c_{v}}$ is the specific heat capacity. Again, plus sign applies to molecules from above, and minus sign below. Note that the temperature gradient ${\displaystyle dT/dy}$ can be considered to be constant over a distance of mean free path.

Integrating over all appropriate velocities within the constraint

$${\displaystyle {\begin{cases}v>0\\0<\theta <\pi /2\\0<\phi <2\pi \end{cases}}}$$

yields the energy transfer per unit time per unit area (also known as heat flux):

$${\displaystyle q_{y}^{\pm }=-{\frac {1}{4}}{\bar {v}}n\cdot \left(\varepsilon _{0}\pm {\frac {2}{3}}mc_{v}l\,{dT \over dy}\right)}$$

Note that the energy transfer from above is in the ${\displaystyle -y}$ direction, and therefore the overall minus sign in the equation. The net heat flux across the imaginary surface is thus

$${\displaystyle q=q_{y}^{+}-q_{y}^{-}=-{\frac {1}{3}}{\bar {v}}nmc_{v}l\,{dT \over dy}}$$

Combining the above kinetic equation with Fourier's law

$${\displaystyle q=-\kappa \,{dT \over dy}}$$

gives the equation for thermal conductivity, which is usually denoted ${\displaystyle \kappa _{0}}$ when it is a dilute gas:

$${\displaystyle \kappa _{0}={\frac {1}{3}}{\bar {v}}nmc_{v}l}$$

Following a similar logic as above, one can derive the kinetic model for mass diffusivity[21] of a dilute gas:

Consider a steady diffusion between two regions of the same gas with perfectly flat and parallel boundaries separated by a layer of the same gas. Both regions have uniform number densities, but the upper region has a higher number density than the lower region. In the steady state, the number density at any point is constant (that is, independent of time). However, the number density ${\displaystyle n}$ in the layer increases uniformly with distance ${\displaystyle y}$ above the lower plate. The non-equilibrium molecular flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.

Let ${\displaystyle n_{0}}$ be the number density of the gas at an imaginary horizontal surface inside the layer. The number of molecules arriving at an area ${\displaystyle dA}$ on one side of the gas layer, with speed ${\displaystyle v}$ at angle ${\displaystyle \theta }$ from the normal, in time interval ${\displaystyle dt}$ is

$${\displaystyle nv\cos(\theta )\,dA\,dt\times \left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}e^{-{\frac {mv^{2}}{2k_{B}T}}}(v^{2}\sin(\theta )\,dv\,d\theta \,d\phi )}$$

These molecules made their last collision at a distance ${\displaystyle l\cos \theta }$ above and below the gas layer, where the local number density is

$${\displaystyle n^{\pm }=\left(n_{0}\pm l\cos \theta \,{dn \over dy}\right)}$$

Again, plus sign applies to molecules from above, and minus sign below. Note that the number density gradient ${\displaystyle dn/dy}$ can be considered to be constant over a distance of mean free path.

Integrating over all appropriate velocities within the constraint

$${\displaystyle {\begin{cases}v>0\\0<\theta <\pi /2\\0<\phi <2\pi \end{cases}}}$$

yields the molecular transfer per unit time per unit area (also known as diffusion flux):

$${\displaystyle J_{y}^{\pm }=-{\frac {1}{4}}{\bar {v}}\cdot \left(n_{0}\pm {\frac {2}{3}}l\,{dn \over dy}\right)}$$

Note that the molecular transfer from above is in the ${\displaystyle -y}$ direction, and therefore the overall minus sign in the equation. The net diffusion flux across the imaginary surface is thus

$${\displaystyle J=J_{y}^{+}-J_{y}^{-}=-{\frac {1}{3}}{\bar {v}}l{dn \over dy}}$$

Combining the above kinetic equation with Fick's first law of diffusion

$${\displaystyle J=-D{dn \over dy}}$$

gives the equation for mass diffusivity, which is usually denoted ${\displaystyle D_{0}}$ when it is a dilute gas:

$${\displaystyle D_{0}={\frac {1}{3}}{\bar {v}}l}$$