Pressure and kinetic energy In the kinetic theory of gases, the
pressure is assumed to be equal to the force (per unit area) exerted by the individual gas atoms or molecules hitting and rebounding from the gas container's surface. Consider a gas particle traveling at velocity, v_i, along the \hat{i}-direction in an enclosed volume with
characteristic length, L_i, cross-sectional area, A_i, and volume, V = A_i L_i. The gas particle encounters a boundary after characteristic time t = L_i / v_i. The
momentum of the gas particle can then be described as p_i = m v_i = m L_i / t . We combine the above with
Newton's second law, which states that the force experienced by a particle is related to the time rate of change of its momentum, such that F_i = \frac{\mathrm{d}p_i}{\mathrm{d}t} = \frac{m L_i}{t^2}=\frac{m v_i^2}{L_i}. Now consider a large number, N, of gas particles with random orientation in a three-dimensional volume. Because the orientation is random, the average particle speed, v , in every direction is identical v_x^2 = v_y^2 = v_z^2. Further, assume that the volume is symmetrical about its three dimensions, \hat{i}, \hat{j}, \hat{k}, such that \begin{align} V ={}& V_i = V_j = V_k, \\ F ={}& F_i = F_j = F_k, \\ & A_i=A_j=A_k. \end{align} The total surface area on which the gas particles act is therefore A = 3 A_i. The pressure exerted by the collisions of the N gas particles with the surface can then be found by adding the force contribution of every particle and dividing by the interior surface area of the volume, P = \frac{N \overline{F}}{A}=\frac{NLF}{V} \Rightarrow PV = NLF = \frac{N}{3} m v^2. The total translational
kinetic energy K_\text{t} of the gas is defined as K_\text{t} = \frac{N}{2} m v^2 , providing the result PV = \frac{2}{3} K_\text{t} . 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. The mass density of a gas \rho is expressed through the total mass of gas particles and through volume of this gas: \rho = \frac {N m}{V}. Taking this into account, the pressure is equal to P = \frac{\rho v^2}{3} . Relativistic expression for this formula is P = \frac {2 \rho c^2 }{3} \left({\left(1 - \overline{v^2} / c^2\right)}^{-1/2} - 1 \right) , where c is
speed of light. In the limit of small speeds, the expression becomes P \approx \rho \overline{v^2}/3.
Temperature and kinetic energy Rewriting the above result for the pressure as PV = \frac{1}{3}Nmv^2 , we may combine it with the
ideal gas law {{NumBlk|| PV = N k_\mathrm{B} T ,|}} where k_\mathrm{B} is the
Boltzmann constant and T is the
absolute temperature defined by the ideal gas law, to obtain k_\mathrm{B} T = \frac{1}{3} m v^2, which leads to a simplified expression of the average translational kinetic energy per molecule, \frac{1}{2} m v^2 = \frac{3}{2} k_\mathrm{B} T. The translational kinetic energy of the system is N times that of a molecule, namely K_\text{t} = \frac{1}{2} N m v^2 . The temperature, T is related to the translational kinetic energy by the description above, resulting in {{NumBlk|| T = \frac{1}{3} \frac{m v^2}{k_\mathrm{B} } |}} which becomes {{NumBlk|| T = \frac{2}{3} \frac{K_\text{t}}{N k_\mathrm{B} }. |}} Equation () 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 () and (), we have {{NumBlk|| PV = \frac{2}{3} K_\text{t}.|}} Thus, the product of pressure and volume per
mole is proportional to the average translational molecular kinetic energy. Equations () and () are called the "classical results", which could also be derived from
statistical mechanics; for more details, see: The
equipartition theorem requires that kinetic energy is partitioned equally between all kinetic
degrees of freedom,
D. A monatomic gas is axially symmetric about each spatial axis, so that
D = 3 comprising translational motion along each axis. A diatomic gas is axially symmetric about only one axis, so that
D = 5, comprising translational motion along three axes and rotational motion along two axes. A polyatomic gas, like
water, is not radially symmetric about any axis, resulting in
D = 6, comprising 3 translational and 3 rotational degrees of freedom. Because the
equipartition theorem requires that kinetic energy is partitioned equally, the total kinetic energy is K =D K_\text{t} = \frac{D}{2} N m v^2. Thus, the energy added to the system per gas particle kinetic degree of freedom is \frac{K}{ND} = \frac{1}{2} k_\text{B} T . Therefore, the kinetic energy per kelvin of one mole of monatomic
ideal gas (
D = 3) is K = \frac{D}{2} k_\text{B} N_\text{A} = \frac{3}{2} R, where N_\text{A} is the
Avogadro constant, and
R is the
ideal gas constant. Thus, the ratio of the kinetic energy to the absolute temperature of an ideal monatomic gas 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. At higher temperatures (typically thousands of kelvins), vibrational modes become active to provide additional degrees of freedom, creating a temperature-dependence on
D and the total molecular energy. Quantum
statistical mechanics is needed to accurately compute these contributions.
Collisions with container wall 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 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 n = N/V and that the particles obey
Maxwell's velocity distribution: f_\text{Maxwell}(v_x,v_y,v_z) \, dv_x \, dv_y \, dv_z = \left(\frac{m}{2 \pi k_\text{B} T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \, dv_x \, dv_y \, dv_z Then for a small area dA on the container wall, a particle with speed v at angle \theta from the normal of the area dA, will collide with the area within time interval dt, if it is within the distance v\,dt from the area dA. Therefore, all the particles with speed v at angle \theta from the normal that can reach area dA within time interval dt are contained in the tilted pipe with a height of v\cos (\theta) dt and a volume of v\cos (\theta) \,dA\,dt. The total number of particles that reach area dA within time interval dt also depends on the velocity distribution; All in all, it calculates to be:n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right). Integrating this over all appropriate velocities within the constraint v > 0, 0 , 0 yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time: J_\text{collision} = \frac{\displaystyle\int_0^{\pi/2} \cos(\theta) \sin(\theta) \, d\theta}{\displaystyle\int_0^\pi \sin(\theta) \, d\theta}\times n \bar v = \frac{1}{4} n \bar v = \frac{n}{4} \sqrt{\frac{8 k_\mathrm{B} T}{\pi m}}. This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed \bar{v} of the Maxwell's velocity distribution, one has to integrate over v > 0 , 0 , 0 . The momentum transfer to the container wall from particles hitting the area dA with speed v at angle \theta from the normal, in time interval dt is: [2mv \cos(\theta)]\times n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right). Integrating this over all appropriate velocities within the constraint v > 0, 0 , 0 yields the
pressure (consistent with
Ideal gas law): P = \frac{\displaystyle 2\int_0^{\pi/2} \cos^2(\theta) \sin(\theta) \, d\theta}{\displaystyle \int_0^\pi \sin(\theta) \, d\theta}\times n mv_\text{rms}^2 = \frac{1}{3} n mv_\text{rms}^2 = \frac{2}{3} n\langle E_\text{kin}\rangle = n k_\mathrm{B} T If this small area A is punched to become a small hole, the
effusive flow rate will be: \Phi_\text{effusion} = J_\text{collision} A= n A \sqrt{\frac{k_\mathrm{B} T}{2 \pi m}}. Combined with the
ideal gas law, this yields \Phi_\text{effusion} = \frac{P A}{\sqrt{2 \pi m k_\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 (v,\theta,\phi) that hit the area dA within the time interval dt are contained in the tilted pipe with a height of v\cos (\theta) \, dt and a volume of v\cos (\theta) \, dA \, dt; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of v\cos \theta: \begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &= \lambda v\cos{\theta} \left(\frac{m}{2 \pi k T}\right)^{3/2} e^{- \frac{mv^2}{2k_\mathrm{B} T}}(v^2\sin{\theta} \, dv \, d\theta \, d\phi) \end{align} with the constraint v > 0, 0 , 0 . The constant \lambda can be determined by the normalization condition \int f(v,\theta,\phi) \, dv \, d\theta \, d\phi=1 to be 4/\bar{v} , and overall: \begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &= \frac{1}{2\pi} \left(\frac{m}{k_\mathrm{B} T}\right)^2e^{- \frac{mv^2}{2k_\mathrm{B} T}} (v^3\sin{\theta}\cos{\theta} \, dv \, d\theta \, d\phi) \\ \end{align};\quad v>0,\, 0
Speed of molecules From the kinetic energy formula it can be shown that v_\text{p} = \sqrt{2 \cdot \frac{k_\mathrm{B} T}{m}}, \bar{v} = \frac {2}{\sqrt{\pi}} v_p = \sqrt{\frac {8}{\pi} \cdot \frac{k_\mathrm{B} T}{m}}, 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 kelvin, and
m is the mass of one molecule of gas in kg. The most probable (or mode) speed v_\text{p} is 81.6% of the root-mean-square speed v_\text{rms}, and the mean (arithmetic mean, or average) speed \bar{v} is 92.1% of the rms speed (
isotropic distribution of speeds). See: •
Average, •
Root-mean-square speed •
Arithmetic mean •
Mean •
Mode (statistics) Mean free path In kinetic theory of gases, 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 \sigma be the collision
cross section of one molecule colliding with another. As in the previous section, the number density n is defined as the number of molecules per (extensive) volume, or n = N/V . The collision cross section per volume or collision cross section density is n \sigma , and it is related to the mean free path \ell by\ell = \frac {1} {n \sigma \sqrt{2}} Notice that the unit of the collision cross section per volume n \sigma is reciprocal of length. == Transport properties ==