The Ideal Gas Law

Fill a bicycle pump, push the handle, and feel the resistance grow. That resistance is pressure obeying a single compact law. Chemists compress the behavior of any dilute gas into one equation: $pV = nRT$ . It matters because it lets you predict pressure, volume, temperature, or amount from any three of the other variables. This article builds that law from its empirical roots. It shows the derivation, works two full numerical examples, and flags where the law breaks down.

Key takeaways

The ideal gas law unifies Boyle's, Charles's, Gay-Lussac's, and Avogadro's laws into $pV = nRT$ .
It is an equation of state: it links $p$ , $V$ , $n$ , $T$ for a hypothetical gas with no intermolecular forces and point-like particles.
At standard conditions (273.15 K, 1 bar) one mole of ideal gas occupies 22.71 L; at 1 atm it is 22.41 L.
The law is exact only in the zero-pressure, zero-density limit; real gases deviate as pressure rises or temperature falls.
Combining two states of the same fixed amount of gas gives the practical combined gas law.
Key idea: pressure, volume, temperature, and amount of an ideal gas are locked together by one linear relation, $pV = nRT$ , valid in the low-density limit.

The core insight

The ideal gas law is not a new physical discovery but a unification: it packages Boyle's inverse pressure–volume relation, Charles's direct volume–temperature relation, Gay-Lussac's direct pressure–temperature relation, and Avogadro's amount–volume relation into a single equation of state, $pV = nRT$ , that becomes exact for any real gas in the limit of vanishing pressure.

The physical picture

Picture a mole of gas molecules as tiny, structureless points darting around a container, colliding with the walls and each other but never attracting or sticking together. Each collision with a wall delivers a tiny push; billions of pushes per second, averaged over the wall's area, is what a manometer reads as pressure. If you shrink the container at constant temperature, the same number of collisions happens in less space and less time between hits, so the pressure rises — that is Boyle's law. If you heat the gas at constant volume, the molecules move faster, hit the walls harder and more often, so pressure rises — that is Gay-Lussac's law. If you let the container expand freely at constant pressure while heating, the volume must grow to keep the collision frequency per unit area constant — that is Charles's law. Add more molecules at fixed temperature and pressure, and the container must simply get bigger to keep the crowding, hence the pressure, the same — that is Avogadro's principle. The ideal gas law is the single equation that reproduces all four limiting behaviors simultaneously. It describes a mathematical idealization: a gas of point particles with no intermolecular forces, valid for any real gas at sufficiently low pressure and high temperature, where molecules are far apart and rarely interact.

Figure

Four empirical laws folding into one equation of state

The equation of state is best visualized as a surface, but a simpler and very common view is a family of isotherms on a pressure–volume plot: at fixed $T$ and $n$ , $p$ traces a hyperbola in $V$ , and hotter isotherms sit farther from the origin.

Figure

Isotherms of an ideal gas on a p–V plot, n = 1 mol

The core relations

pV = nRT

where: $p$ is pressure (Pa), $V$ is volume (m³), $n$ is amount of substance (mol), $R = 8.31446\ \text{J K}^{-1}\text{mol}^{-1}$ is the gas constant, and $T$ is absolute temperature (K).

p = \frac{nRT}{V} = \rho \frac{RT}{M}

where: $\rho$ is mass density and $M$ is molar mass, useful for gas-density and molar-mass determinations.

\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}

where: the combined gas law, valid for a fixed amount $n$ of gas moving between two states 1 and 2.

V_m = \frac{RT}{p}

where: $V_m = V/n$ is the molar volume; at $T = 273.15$ K and $p = 1$ bar, $V_m = 22.71$ L mol⁻¹ (STP); at $p = 1$ atm, $V_m = 22.41$ L mol⁻¹.

Derivation ABuilding pV = nRT from the four empirical laws

Goal. Show that Boyle's law, Charles's law, Gay-Lussac's law, and Avogadro's principle are all special cases of one equation, $pV = nRT$ , and identify what $R$ must be.

Step 1: hold $n$ and $T$ fixed. Boyle's law states that for a fixed amount of gas at constant temperature,

pV = \text{constant}(n, T).

This tells us $p$ and $V$ multiply to something that can depend only on $n$ and $T$ , not on $p$ or $V$ themselves.

Step 2: hold $n$ and $p$ fixed. Charles's law states that at constant pressure,

\frac{V}{T} = \text{constant}(n, p).

Volume is directly proportional to absolute temperature. This is the piece that fixes the temperature dependence of the "constant" found in Step 1.

Step 3: hold $n$ and $V$ fixed. Gay-Lussac's law states that at constant volume,

\frac{p}{T} = \text{constant}(n, V).

This is consistent with Step 2 rather than independent of it: combining Steps 1–3 for fixed $n$ gives

\frac{pV}{T} = \text{constant}(n).

Step 4: vary $n$ at fixed $p, T$ . Avogadro's principle states that equal volumes of gas at the same $p$ and $T$ contain equal numbers of moles, so $V \propto n$ at fixed $p, T$ . Substituting into Step 3's result shows the constant on the right must itself be directly proportional to $n$ :

\frac{pV}{T} = nR

where $R$ is a proportionality constant independent of $p$ , $V$ , $T$ , and $n$ — the same for every ideal gas, because Avogadro's principle guarantees equal molar volumes regardless of chemical identity.

Step 5: identify $R$ . Rearranging Step 4 and evaluating at any single known state pins down $R$ numerically. Using the measured molar volume at STP, $V_m = 22.71\times10^{-3}\ \text{m}^3\text{mol}^{-1}$ at $T = 273.15$ K and $p = 1.000\times10^{5}$ Pa,

R = \frac{pV_m}{T} = \frac{(1.000\times10^{5}\ \text{Pa})(22.71\times10^{-3}\ \text{m}^3\text{mol}^{-1})}{273.15\ \text{K}} = 8.314\ \text{J K}^{-1}\text{mol}^{-1}.

Boxed result.

\boxed{pV = nRT}

Worked examples

Example 1 — Foundational: finding pressure of a fixed amount of gas. A rigid 5.00 L steel cylinder holds 0.220 mol of nitrogen gas at 298 K. Find the pressure, assuming ideal behavior. Rearrange the ideal gas law for pressure:

p = \frac{nRT}{V}.

Convert volume to SI units:

V = 5.00\ \text{L} = 5.00\times10^{-3}\ \text{m}^3.

Substitute:

p = \frac{(0.220\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})}{5.00\times10^{-3}\ \text{m}^3}.

Compute the numerator:

(0.220)(8.314)(298) = 545.1\ \text{J} = 545.1\ \text{Pa}\,\cdot\,\text{m}^3.

Divide by volume:

p = \frac{545.1}{5.00\times10^{-3}} = 1.090\times10^{5}\ \text{Pa} = 1.09\times10^{5}\ \text{Pa}.

Convert to atmospheres for a sanity check:

p = \frac{1.090\times10^{5}\ \text{Pa}}{1.013\times10^{5}\ \text{Pa atm}^{-1}} = 1.08\ \text{atm}.

That is a modest overpressure above 1 atm, entirely reasonable for a small amount of gas in a 5 L cylinder near room temperature — the answer passes the sanity check.

Example 2 — Applied: molar mass of an unknown vapor from density. A chemist vaporizes 0.3625 g of an unknown volatile liquid completely in a 250.0 mL flask at 373.0 K, and measures the pressure of the vapor as 1.00 atm (1.013×10⁵ Pa). Find the molar mass, and suggest what the substance might be. Start from the density form of the ideal gas law:

p = \frac{m}{M}\cdot\frac{RT}{V}

so that

M = \frac{mRT}{pV}.

Convert all quantities to SI units:

m = 0.3625\ \text{g} = 3.625\times10^{-4}\ \text{kg}, \quad V = 250.0\ \text{mL} = 2.500\times10^{-4}\ \text{m}^3.

Substitute into the rearranged law:

M = \frac{(3.625\times10^{-4}\ \text{kg})(8.314\ \text{J K}^{-1}\text{mol}^{-1})(373.0\ \text{K})}{(1.013\times10^{5}\ \text{Pa})(2.500\times10^{-4}\ \text{m}^3)}.

Compute the numerator:

(3.625\times10^{-4})(8.314)(373.0) = 1.1240\ \text{kg}\,\cdot\,\text{J}\,\cdot\,\text{K}^{-1}\text{mol}^{-1}\text{K}\ \text{(numerically)} = 1.124\ \text{J}\,\cdot\,\text{kg} \cdot \text{units cancel to kg}\,\cdot\,\text{mol}^{-1}\text{ after full division}.

Compute the denominator:

(1.013\times10^{5})(2.500\times10^{-4}) = 25.33\ \text{Pa}\,\cdot\,\text{m}^3 = 25.33\ \text{J}.

Divide:

M = \frac{1.124}{25.33} = 0.04438\ \text{kg mol}^{-1} = 44.4\ \text{g mol}^{-1}.

A molar mass near 44 g mol⁻¹ is consistent with carbon dioxide (44.01 g mol⁻�1) or propane (44.10 g mol⁻¹); given a volatile liquid vaporizing at 373 K, this points toward a small organic compound like acetaldehyde derivatives or, more simply, confirms the vapor density method used industrially for unknown identification. The number is physically sensible: small, positive, and in the range of common light gases — not an absurd result like a negative or astronomically large mass.

Limiting cases & common traps

Situation	What happens	Why
$p \to 0$	Ideal gas law becomes exact for any real gas	Molecules are far apart; intermolecular forces and molecular volume become negligible
High pressure	Real gases deviate; $pV_m/RT \neq 1$	Finite molecular volume and attractive forces matter
Low temperature, near condensation	Large deviations, law fails badly	Intermolecular attractions dominate; gas approaches a phase transition
Fixed $n$ , comparing two states	Use the combined law, $p_1V_1/T_1 = p_2V_2/T_2$	$R$ and $n$ cancel between the two states
Mixing gases	Total pressure needs Dalton's law of partial pressures	Ideal gas law alone gives total moles, not individual contributions

The chemist behind it

The law's modern algebraic form was assembled in 1834 by Émile Clapeyron, who combined Boyle's and Charles's/Gay-Lussac's laws into a single relation using the newly developed calculus of Carnot's ideal heat engine analysis. Clapeyron's version used a gas-specific constant; it was later generalized with Avogadro's principle so that a single universal constant $R$ serves every gas, giving the equation the compact, chemically powerful form used today.

Origins · 1 minBenoît Paul Émile Clapeyron and how this was discovered

Benoît Paul Émile Clapeyron

1799–1864 · French engineer who first wrote the combined ideal gas law in analytic form while extending Sadi Carnot's work on heat engines

Ask the tutor: How does the ideal gas law change once we add the van der Waals correction terms for real gases?

Why is R the same number for every gas

Avogadro's principle guarantees that equal volumes of any ideal gas at the same pressure and temperature contain the same number of moles. That means the proportionality constant linking $pV/T$ to $n$ cannot depend on molecular identity, mass, or shape — it must be universal. This is precisely why $R$ , unlike gas-specific constants used in older engineering formulas, works identically for helium, nitrogen, or water vapor, as long as each behaves ideally under the conditions used.

What are the units of R and why do so many versions exist

$R$ appears in whatever pressure–volume–temperature unit system is convenient: 8.314 J K⁻¹mol⁻¹ in SI, 0.08206 L atm K⁻¹mol⁻¹ for lab-scale gas calculations, or 8.314 kPa L K⁻¹mol⁻¹. All represent the same physical constant; the value changes only because pressure and volume units differ. Always match your $R$ value's units to the units you're using for $p$ and $V$ , or convert first.

Is pV = nRT ever exactly true for a real gas

Strictly, no real gas obeys $pV = nRT$ exactly at any finite pressure, because real molecules have volume and attract each other weakly. However, the law becomes asymptotically exact as pressure approaches zero, since intermolecular separations grow large and interactions become negligible. For many gases near room temperature and 1 atm, deviations are under one percent, which is why the law is such a reliable working approximation.

How does the ideal gas law connect to Dalton's law of partial pressures

When several ideal gases share one container, each behaves as though the others were absent, because ideal molecules do not interact. The ideal gas law applied to the total moles gives total pressure, while applying it to each gas's own moles gives its partial pressure. Dalton's law states these partial pressures sum to the total, a direct consequence of the linearity of $pV = nRT$ in $n$ .

Why do STP molar volume values differ between 22.41 L and 22.71 L

The discrepancy comes from which pressure standard is used. Older STP defined standard pressure as 1 atm (101.325 kPa), giving 22.41 L mol⁻¹. The current IUPAC-recommended STP uses 1 bar (100 kPa) exactly, giving 22.71 L mol⁻¹. Both use 273.15 K. Always check which convention a problem or textbook uses before quoting molar volume.