Editing Kinetic theory of gases (section)

=== Temperature and kinetic energy ===
Rewriting the above result for the pressure as <math display="inline">PV = \frac{1}{3}Nmv^2 </math>, we may combine it with the [[ideal gas law]]
{{NumBlk||<math display="block"> PV = N k_\mathrm{B} T ,</math>|{{EquationRef|1}}}}
where <math> k_\mathrm{B}</math> is the [[Boltzmann constant]] and <math> T</math> is the [[Thermodynamic temperature|absolute]] [[temperature]] defined by the ideal gas law, to obtain
<math display="block">k_\mathrm{B} T = \frac{1}{3} m v^2, </math>
which leads to a simplified expression of the average translational kinetic energy per molecule,<ref>The average kinetic energy of a fluid is proportional to the [[Root-mean-square speed|root mean-square velocity]], which always exceeds the mean velocity - {{usurped|1=[https://web.archive.org/web/20071011213546/http://www.mikeblaber.org/oldwine/chm1045/notes/Gases/Kinetic/Gases08.htm Kinetic Molecular Theory]}}</ref>
<math display="block"> \frac{1}{2} m v^2 = \frac{3}{2} k_\mathrm{B} T.</math>
The translational kinetic energy of the system is <math>N</math> times that of a molecule, namely <math display="inline"> K_\text{t} = \frac{1}{2} N m v^2 </math>. The temperature, <math> T</math> is related to the translational kinetic energy by the description above, resulting in
{{NumBlk||<math display="block"> T = \frac{1}{3} \frac{m v^2}{k_\mathrm{B} } </math>|{{EquationRef|2}}}}
which becomes
{{NumBlk||<math display="block"> T = \frac{2}{3} \frac{K_\text{t}}{N k_\mathrm{B} }. </math>|{{EquationRef|3}}}}
Equation ({{EquationNote|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 ({{EquationNote|1}}) and ({{EquationNote|3}}), we have
{{NumBlk||<math display="block"> PV = \frac{2}{3} K_\text{t}.</math>|{{EquationRef|4}}}}
Thus, the product of pressure and volume per [[Mole (unit)|mole]] is proportional to the average
translational molecular kinetic energy.

Equations ({{EquationNote|1}}) and ({{EquationNote|4}}) are called the "classical results", which could also be derived from [[statistical mechanics]];
for more details, see:<ref>
[http://clesm.mae.ufl.edu/wiki.pub/index.php/Configuration_integral_%28statistical_mechanics%29 Configuration integral (statistical mechanics)] {{webarchive|url=https://web.archive.org/web/20120428193950/http://clesm.mae.ufl.edu/wiki.pub/index.php/Configuration_integral_%28statistical_mechanics%29 |date=2012-04-28 }}
</ref>

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
<math display="block"> K =D K_\text{t} = \frac{D}{2} N m v^2. </math>

Thus, the energy added to the system per gas particle kinetic degree of freedom is
<math display="block"> \frac{K}{ND} = \frac{1}{2} k_\text{B} T . </math>

Therefore, the kinetic energy per kelvin of one mole of monatomic [[ideal gas]] (''D'' = 3) is
<math display="block"> K = \frac{D}{2} k_\text{B} N_\text{A} = \frac{3}{2} R, </math>
where <math>N_\text{A}</math> 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&nbsp;J/K
* per molecule: 20.7&nbsp;[[yoctojoule|yJ]]/K = 129&nbsp;μeV/K

At [[Standard temperature and pressure|standard temperature]] (273.15&nbsp;K), the kinetic energy can also be obtained:
* per mole: 3406&nbsp;J
* per molecule: 5.65&nbsp;[[zeptojoule|zJ]] = 35.2&nbsp;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.<ref>{{cite book |last1=Chang |first1=Raymond | last2=Thoman | first2=John W. Jr. |title=Physical Chemistry for the Chemical Sciences |date=2014 |publisher=University Science Books |location=New York |pages=56–61}}</ref>