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Equations of State

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

Principle

An equation of state relates the macroscopic state variables of matter after the system has reached equilibrium.

For a gas, the most useful state variables are pressure, volume, amount of substance, and absolute temperature.

Notation

\(p\)
absolute pressure
\(\mathrm{Pa}\)
\(V\)
volume
\(\mathrm{m^{3}}\)
\(T\)
absolute temperature
\(\mathrm{K}\)
\(n\)
amount of substance
\(\mathrm{mol}\)
\(R\)
molar gas constant
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(N\)
number of molecules
1
\(k_B\)
Boltzmann constant
\(\mathrm{J\,K^{-1}}\)
\(a,b\)
van der Waals constants
varies

Method

Derivation 1: Build the ideal-gas state relation

Experiments on dilute gases show that pressure increases with amount and temperature, and decreases when the same gas occupies a larger volume.

Amount dependence
\[p\propto n\]
Temperature dependence
\[p\propto T\]
Volume dependence
\[p\propto \frac{1}{V}\]
Ideal-gas equation
\[pV=nRT\]

For a fixed amount of ideal gas, \(nR\) is constant, so state comparisons can avoid explicitly finding \(n\).

Fixed amount
\[\frac{pV}{T}=nR\]
Two states
\[\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\]

Derivation 2: Write the molecular form

Since \(N=nN_A\) and \(R=N_Ak_B\), the same equation can be written per molecule.

Molecule count
\[N=nN_A\]
Gas constants
\[R=N_Ak_B\]
Molecular equation
\[pV=Nk_BT\]

Real gases depart from the ideal model when molecular size and intermolecular attractions matter. The van der Waals equation adds a volume correction and a pressure correction.

Finite molecular size
\[V\to V-nb\]
Attraction correction
\[p\to p+a\frac{n^2}{V^2}\]
Real-gas model
\[\left(p+a\frac{n^2}{V^2}\right)(V-nb)=nRT\]

Rules

These are the compact state equations used in this section.

Ideal gas
\[pV=nRT\]
Fixed amount
\[\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\]
Molecular form
\[pV=Nk_BT\]
Molecules and moles
\[N=nN_A\]
van der Waals
\[\left(p+a\frac{n^2}{V^2}\right)(V-nb)=nRT\]

Examples

Question
Find the volume of
\[0.50\,\mathrm{mol}\]
of ideal gas at
\[300\,\mathrm{K}\]
and
\[1.20\times10^5\,\mathrm{Pa}\]
Answer
\[V=\frac{nRT}{p}=\frac{(0.50)(8.31)(300)}{1.20\times10^5}=1.04\times10^{-2}\,\mathrm{m^3}\]

Checks

  • Use absolute pressure and absolute temperature in gas equations.
  • Convert litres to cubic metres when using SI units.
  • The ideal-gas equation is a model, not a universal equation for all matter.
  • Real-gas corrections matter most at high density and low temperature.

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Heat Transfer Mechanisms

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Molecular Matter