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Ideal Gas Model

Level 1 - Physics topic page in Matter at Thermal Scale.

Principle

The kinetic model explains ideal-gas pressure as momentum transfer from many molecular collisions with the container walls.

Temperature measures the average translational kinetic energy of the gas molecules.

Notation

\(N\)
number of molecules
1
\(m_0\)
mass of one molecule
\(\mathrm{kg}\)
\(v_{\mathrm{rms}}\)
root-mean-square molecular speed
\(\mathrm{m\,s^{-1}}\)
\(\langle K_{\mathrm{tr}}\rangle\)
average translational kinetic energy per molecule
\(\mathrm{J}\)
\(k_B\)
Boltzmann constant
\(\mathrm{J\,K^{-1}}\)
\(R\)
molar gas constant
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(T\)
absolute temperature
\(\mathrm{K}\)

Method

Derivation 1: Relate pressure to molecular speed

In a cubical container, a molecule rebounding elastically from a wall reverses one velocity component. Averaging many molecules gives equal sharing among the three directions.

Momentum change
\[\Delta p_x=2m_0v_x\]
Direction average
\[\langle v_x^2\rangle=\frac{1}{3}\langle v^2\rangle\]
Pressure-speed relation
\[pV=\frac{1}{3}Nm_0\langle v^2\rangle\]
RMS definition
\[v_{\mathrm{rms}}=\sqrt{\langle v^2\rangle}\]

Derivation 2: Compare with the ideal-gas equation

The ideal-gas equation in molecular form is \(pV=Nk_BT\). Equating the two expressions for \(pV\) connects temperature to kinetic energy.

Molecular gas equation
\[pV=Nk_BT\]
Compare pressure forms
\[\frac{1}{3}Nm_0v_{\mathrm{rms}}^2=Nk_BT\]
RMS speed
\[v_{\mathrm{rms}}=\sqrt{\frac{3k_BT}{m_0}}=\sqrt{\frac{3RT}{M}}\]
Average kinetic energy
\[\langle K_{\mathrm{tr}}\rangle=\frac{1}{2}m_0v_{\mathrm{rms}}^2=\frac{3}{2}k_BT\]

The model assumes many molecules, random motion, negligible molecular size compared with the container volume, no long-range intermolecular forces, and elastic collisions.

Rules

These are the compact kinetic-model relations for an ideal gas.

Molecular pressure
\[pV=\frac{1}{3}Nm_0v_{\mathrm{rms}}^2\]
Molecular ideal gas
\[pV=Nk_BT\]
Average kinetic energy
\[\langle K_{\mathrm{tr}}\rangle=\frac{3}{2}k_BT\]
RMS speed
\[v_{\mathrm{rms}}=\sqrt{\frac{3k_BT}{m_0}}=\sqrt{\frac{3RT}{M}}\]

Examples

Question
Find the average translational kinetic energy of one molecule in an ideal gas at
\[300\,\mathrm{K}\]
Answer
\[\langle K_{\mathrm{tr}}\rangle=\frac{3}{2}k_BT=\frac{3}{2}(1.38\times10^{-23})(300)=6.21\times10^{-21}\,\mathrm{J}\]

Checks

  • Use molecular mass \(m_0\) with \(k_B\), and molar mass \(M\) with \(R\).
  • Temperature must be in kelvins.
  • Pressure comes from momentum transfer, not from molecules having a preferred direction.
  • Lighter molecules have larger \(v_\{\\mathrm\{rms\}}\) at the same temperature.

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Molecular Matter

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Heat Capacity