The ideal gas law: PV = nRT

One equation for all four variables

The combined gas law handled changing conditions for a fixed sample, but it could not answer 'how many moles?' The ideal gas law promotes amount to a full variable:

$$PV = nRT$$

Pressure times volume equals moles times the gas constant times absolute temperature. One equation now links all four measurable properties of a gas, and every simple gas law falls out of it: hold $n$ and $T$ constant and $PV$ is constant, which is Boyle's law, and so on.

The constant $R$ is the bridge between the macroscopic world of liters and atmospheres and the particulate world of moles. Its two essential forms are $R = 0.08206\ \text{L·atm/(mol·K)}$ and $R = 8.314\ \text{J/(mol·K)}$. Choose by units, never by habit.

A worked example: what volume does 2.50 mol of gas occupy at 1.20 atm and $27^\circ\text{C}$ (300 K)? $V = nRT/P = (2.50 \times 0.08206 \times 300)/1.20 = 51.3$ L. Three multiplications and a division replace an entire chain of proportional reasoning.

Density from the law: d = PM/RT

Substitute $n = m/M$, mass over molar mass, into $PV = nRT$ and rearrange:

$$d = \frac{PM}{RT}$$

where $d$ is density in mass per volume. Gas density is directly proportional to pressure and molar mass, and inversely proportional to temperature. This is why hot-air balloons float and why cold air sinks.

A worked example: what is the density of air at STP? Air averages $M \approx 29$ g/mol. $d = PM/RT = (1.00 \times 29)/(0.08206 \times 273.15) = 29/22.41 \approx 1.29$ g/L. Now heat that air to $100^\circ\text{C}$ (373 K) at the same pressure: $d = 29/(0.08206 \times 373) \approx 0.948$ g/L, about 27 percent lighter. A balloon envelope full of that hot air displaces 1.29 g/L of outside air while weighing only 0.948 g/L inside, and the difference is lift.

Weighing the invisible: molar mass from vapor

The same substitution solved for $M$ turns the gas law into an analytical instrument:

$$M = \frac{mRT}{PV}$$

Vaporize a weighed sample of an unknown volatile liquid in a flask of known volume, record $P$ and $T$, and the molar mass drops out. This is the Dumas method, nearly two centuries old and still taught because it works.

A worked example: 0.500 g of an unknown liquid vaporizes completely in a 0.250 L flask at 1.00 atm and $127^\circ\text{C}$ (400 K). $M = mRT/(PV) = (0.500 \times 0.08206 \times 400)/(1.00 \times 0.250) = 16.41/0.250 \approx 65.6$ g/mol. That molar mass narrows the identity dramatically: among common solvents it sits near 2-propanol at 60.1 g/mol or acetic acid at 60.05, close enough to guide the next test. The method needs only a balance, a flask, and a thermometer.

Stoichiometry with gases

Gas stoichiometry runs on the law too. Reactions that involve gases quote amounts in liters, but balanced equations speak in moles, and $PV = nRT$ is the translator.

A worked example: how many liters of oxygen at STP are needed to burn 10.0 g of methane? The equation $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ demands 2 mol O2 per mol CH4. Moles of methane: $10.0/16.04 = 0.623$ mol, so oxygen needed: $2 \times 0.623 = 1.247$ mol. At STP the molar volume is 22.4 L/mol, giving $1.247 \times 22.4 \approx 27.9$ L of O2. Away from STP you would use $V = nRT/P$ instead; the mole bridge is the same.

A second worked check at non-STP conditions: the same 1.247 mol at $25^\circ\text{C}$ (298 K) and 1.00 atm occupies $V = (1.247 \times 0.08206 \times 298)/1.00 \approx 30.5$ L, sensibly larger than the STP volume because the gas is warmer.

Limits and the unit-discipline habit

The law's fine print matters. It assumes point particles with no attractions, so it frays at high pressure, where particle volume matters, and near condensation, where attractions matter. It also assumes equilibrium: a gas mid-explosion is not described by one $P$, one $V$, one $T$.

Unit discipline is the practical pitfall. Temperature in kelvins, always. $R$ matched to pressure and volume units, always. And gauge pressures converted to absolute before use.

A worked example of the classic blunder: 0.100 mol at 2.00 atm and 300 K in liters. Correct: $V = (0.100 \times 0.08206 \times 300)/2.00 = 1.23$ L. With the wrong $R$: $(0.100 \times 8.314 \times 300)/2.00 = 125$ L, off by a factor of 101. The absurd answer is the tell: a tenth of a mole at 2 atm should occupy about a soda bottle, not a bathtub. When a gas-law answer looks ridiculous, check $R$ first.

Common mistakes

Real-world applications

• Computing gas density for balloon and airship design

• Finding molar masses of unknown volatile liquids

• Sizing industrial gas storage vessels

• Converting between gas mass and volume in stoichiometry