CHEMISTRY · LESSON 10 OF 12

Properties of gases

An ideal gas obeys PV = nRT with T in kelvin. The simple gas laws, gas densities, mixtures (Dalton) and gas stoichiometry all follow, and kinetic-molecular theory explains why.

What this lesson explains

Gases are everywhere in chemical engineering: reactor feeds and products, combustion, compressors, storage cylinders, airbags, breathing gases. The ideal gas law lets you calculate how much gas a vessel holds, what pressure it will reach when heated, or what volume of gas a reaction produces — and knowing its assumptions tells you when a real gas will deviate.

Before you begin

Pressure units

1 atm = 101.325 kPa = 760 mmHg = 1.01325 bar. 1 bar = 100 kPa. 1 Pa = 1 N/m².

Choosing R

R = 0.08206 L·atm/(mol·K) with P in atm and V in L; R = 8.314 J/(mol·K) = 8.314 L·kPa/(mol·K) with SI units. The units of R decide the units of P and V.

The idea, made visible

A gas is described by four variables: pressure P, volume V, temperature T and amount n. For an ideal gas they are linked by PV = nRT. Temperature must be absolute (kelvin): T(K) = T(°C) + 273.15. Using °C gives nonsense, such as zero volume at 0 °C.

Holding two variables fixed gives the classic laws. Boyle: at constant n and T, P ∝ 1/V (halve the volume, double the pressure). Charles: at constant n and P, V ∝ T. Gay-Lussac: at constant n and V, P ∝ T. Avogadro: at constant P and T, V ∝ n — equal volumes of gases contain equal numbers of molecules. For one sample changing conditions, use P₁V₁ / T₁ = P₂V₂ / T₂.

At constant temperature, pressure and volume are inversely proportional: every point on the curve has PV = 6 bar·L. The graph is a hyperbola, not a straight line.

Kinetic-molecular theory explains these laws. Gas molecules are in constant random motion; their own volume is negligible compared with the container; they do not attract or repel except during elastic collisions; and their average kinetic energy is proportional to the absolute temperature. Pressure is the result of molecules colliding with the walls. Heating makes them move faster and hit harder and more often.

In a mixture, each gas behaves as if it were alone: its partial pressure is P_i = n_iRT/V, and the total pressure is the sum (Dalton’s law), so P_i = χ_iP_total.

Real gases deviate from ideal behaviour at high pressure (molecules’ own volume matters) and low temperature (attractions matter). At ordinary conditions the ideal gas law is usually accurate to a few percent.

Molar volume of an ideal gas
conditions	T	P	V_m = RT/P
0 °C, 1 atm (older “STP”)	273.15 K	1 atm	22.41 L/mol
0 °C, 1 bar (IUPAC STP)	273.15 K	1 bar	22.71 L/mol
25 °C, 1 atm	298.15 K	1 atm	24.47 L/mol

Key terms

Pressure: Force per unit area, P = F/A. Gas pressure comes from molecular collisions with the walls.
Absolute temperature: Kelvin scale: T(K) = T(°C) + 273.15. 0 K is absolute zero.
Ideal gas: A model gas of point particles with no intermolecular forces; obeys PV = nRT exactly.
STP: Standard temperature and pressure. The IUPAC definition is 273.15 K and 1 bar (molar volume 22.71 L); many textbooks still use 0 °C and 1 atm (22.41 L). Check which your course uses.
Partial pressure: The pressure a gas in a mixture would exert if it alone occupied the whole volume.
Effusion: Escape of gas through a tiny hole; lighter gases effuse faster (Graham’s law).

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
P	pressure	atm, kPa, bar
V	volume	L or m³
n	amount of gas	mol
T	absolute temperature	K
R	gas constant	0.08206 L·atm/(mol·K) or 8.314 J/(mol·K)
M	molar mass	g/mol (kg/mol in u_rms)
χ_i	mole fraction of gas i	—

Ideal gas law

PV = nRT

Conditions and limits: Ideal behaviour (low to moderate pressure, not too cold); T in kelvin; units consistent with R.

Combined gas law

P₁V₁ / T₁ = P₂V₂ / T₂

Conditions and limits: Fixed amount of gas (n constant). Drop any variable that is also constant.

Density and molar mass

d = PM / RT; M = dRT / P

Conditions and limits: From PV = nRT with n = m/M. Units: d in g/L with R in L·atm/(mol·K) and P in atm.

Dalton’s law

P_total = P₁ + P₂ + …; P_i = χ_iP_total

Conditions and limits: Ideal gases that do not react with each other.

Molecular speed and effusion

u_rms = √3RT / M; rate₁ / rate₂ = √M₂ / M₁

Conditions and limits: u_rms with R = 8.314 J/(mol·K) and M in kg/mol. Graham’s law at the same T and P.

Why it works: Where Boyle’s law comes from

Start from PV = nRT and hold n and T constant.

The right side nRT is then a constant, call it k.
Nothing on the right changes.
PV = k, so P = k/V: pressure is inversely proportional to volume.
Rearrange.
For two states of the same sample: P₁V₁ = P₂V₂.
Both equal k.

Molecular picture: in half the volume the molecules hit each square metre of wall twice as often, so the pressure doubles.

A first worked example

Gas-law problems

Convert temperature to kelvin and choose pressure/volume units that match your value of R.
If one sample changes conditions, use the combined gas law, cancelling constant variables.
If you need n (or mass), use PV = nRT; for mass, n = m/M.
For reactions producing gases: moles from stoichiometry, then V = nRT/P.
For mixtures, find each partial pressure; for gas collected over water subtract the water vapour pressure.
Check: does heating raise P (or V)? Does compressing raise P?

Boyle’s law

Problem. An ideal gas at 2.0 bar occupies 3.0 L. At constant temperature it is compressed to 1.5 L. Find the new pressure.

P₂ = P₁V₁ / V₂ = 2.0 × 3.0 / 1.5 = 4.0 bar.
n and T constant.

Result: 4.0 bar.

What it means: Halving the volume doubled the pressure.

A different case

Heating a sealed container

Problem. A rigid container holds gas at 100 kPa and 300 K. It is heated to 450 K. Find the pressure.

Constant V and n: P₁ / T₁ = P₂ / T₂.
Gay-Lussac’s law.
P₂ = 100 × 450 / 300 = 150 kPa.
Temperatures in kelvin.

Result: 150 kPa.

What it means: Using °C (e.g. 27 °C → 177 °C) directly would wrongly predict a 6.6-fold increase. This is why aerosol cans must not be heated.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Moles in a cylinder

Problem. A 50.0 L cylinder contains nitrogen at 15.0 atm and 25 °C. How many moles and what mass of N₂ does it hold?

n = PV / RT = 15.0 × 50.0 / 0.08206 × 298.15 = 30.7 mol.
T = 25 + 273.15 = 298.15 K.
m = 30.7 × 28.02 = 859 g.
M(N₂) = 28.02 g/mol.

Result: ≈ 30.7 mol, about 0.86 kg of N₂.

What it means: At 15 atm the ideal-gas estimate is good to a few percent for nitrogen.

Molar mass from gas density

Problem. An unknown gas has density 1.25 g/L at 0 °C and 1.00 atm. Find its molar mass.

M = dRT / P = 1.25 × 0.08206 × 273.15 / 1.00 = 28.0 g/mol.
Rearranged ideal gas law.

Result: M ≈ 28.0 g/mol (consistent with N₂ or CO).

What it means: Gas density measurements are a classic way to identify gases.

Gas volume from a reaction

Problem. What volume of CO₂ at 25 °C and 1.00 atm is produced when 10.0 g of CaCO₃ decomposes? CaCO₃(s) → CaO(s) + CO₂(g).

n(CaCO₃) = 10.0 / 100.09 = 0.0999 mol.
M(CaCO₃) = 100.09 g/mol.
n(CO₂) = 0.0999 mol (1 : 1).
Mole ratio.
V = nRT / P = 0.0999 × 0.08206 × 298.15 / 1.00 = 2.44 L.
Ideal gas law.

Result: ≈ 2.44 L of CO₂.

What it means: Coefficients give mole ratios; the gas law converts moles to volume.

Partial pressures

Problem. A mixture contains 0.60 mol N₂ and 0.20 mol O₂ at a total pressure of 2.0 atm. Find each partial pressure.

χ(N₂) = 0.60 / 0.80 = 0.75; χ(O₂) = 0.25.
Mole fractions.
P(N₂) = 0.75 × 2.0 = 1.5 atm; P(O₂) = 0.25 × 2.0 = 0.50 atm.
P_i = χ_i P_total.

Result: N₂ 1.5 atm, O₂ 0.50 atm (sum 2.0 atm).

What it means: Oxygen’s partial pressure, not the total pressure, controls how much oxygen dissolves in blood or water.

Graham’s law

Problem. How many times faster does H₂ (M = 2.0 g/mol) effuse than O₂ (M = 32.0 g/mol)?

rate(H₂) / rate(O₂) = √32.0 / 2.0 = √16 = 4.0.
Lighter molecules move faster at the same temperature.

Result: H₂ effuses 4.0 times as fast.

What it means: Rates depend on the square root of the mass ratio, not the ratio itself.

A gas collected over water

Problem. Hydrogen is collected over water at 25 °C. The total pressure is 755 mmHg and the volume is 250 mL. The vapour pressure of water at 25 °C is 23.8 mmHg. How many moles of H₂ were collected?

The collected gas is a mixture of H₂ and water vapour: P(H₂) = 755 − 23.8 = 731.2 mmHg.
Dalton’s law: subtract the water vapour’s partial pressure.
Convert: 731.2/760 = 0.9621 atm; V = 0.250 L; T = 298.15 K.
Units to match R = 0.08206 L·atm/(mol·K).
n = PV / RT = 0.9621 × 0.250 / 0.08206 × 298.15 = 9.83 × 10⁻³ mol.
Ideal gas law for the dry hydrogen.

Result: About 9.83 × 10⁻³ mol H₂.

What it means: Using the total pressure would overestimate the hydrogen by about 3%.

Common misunderstandings

Misunderstanding: Using °C in gas laws.
Correct idea: Always convert to kelvin.
Misunderstanding: Mismatched units with R.
Correct idea: With 0.08206 use atm and L; with 8.314 use Pa and m³ (or kPa and L).
Misunderstanding: Using 22.4 L/mol at any conditions.
Correct idea: That value applies only at 0 °C and 1 atm.
Misunderstanding: Adding pressures of gases in different containers.
Correct idea: Dalton’s law applies to gases sharing the same volume.
Misunderstanding: Forgetting to take the square root in Graham’s law.
Correct idea: Rate ratio = √(M₂/M₁).

Keep in mind

PV = nRTIdeal gas law
P₁V₁ / T₁ = P₂V₂ / T₂Combined gas law
d = PM / RT; M = dRT / PDensity and molar mass
P_total = P₁ + P₂ + …; P_i = χ_iP_totalDalton’s law
u_rms = √3RT / M; rate₁ / rate₂ = √M₂ / M₁Molecular speed and effusion

Scope of this lesson

Real-gas equations (van der Waals) are described qualitatively only.
Speed distributions (Maxwell–Boltzmann) are introduced by name; only the rms speed is calculated.

Next: Thermochemistry. Reactions release or absorb heat. Measure heat with q = mcΔT (calorimetry); express reaction heats as ΔH; combine them with Hess’s law, enthalpies of formation or bond enthalpies.

Original study text. Sources and credits.

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