Ideal Gas Law

3.4 Ideal Gas Law

Load a piston cylinder or lock its volume, change one variable and see which one PV = nRT forces to move. Switch to a three-gas mixture and read partial pressure straight off mole fraction

PV = nRTPartial PressureMole FractionP vs V, V vs T

Topic 3.4

Ideal Gas Law

Explain the relationship between the macroscopic properties of a sample of gas or mixture of gases using the ideal gas law.

The macroscopic properties of an ideal gas are linked by PV = nRT. Every classical gas law is a special case: hold n and T fixed and you get Boyle's law (P ∝ 1/V); hold n and P fixed and you get Charles's law (V ∝ T); hold n and V fixed and you get Gay-Lussac's law (P ∝ T). The combined form P₁V₁/T₁ = P₂V₂/T₂ is on the equation sheet for fixed-amount problems.

For a mixture of ideal gases, each component exerts its pressure independently of the others. So each gas's partial pressure is proportional to its mole fraction, and the total is the sum:

P_A = P_total × X_A, where X_A = (moles A)/(total moles)
P_total = P_A + P_B + P_C + …

This is why partial pressure depends only on how many particles of a gas are present, never on their identity or mass.

EK 3.4.A.3 asks you to work with graphical representations of the P, V, T, n relationships — recognizing, for example, that P vs V is a hyperbola while P vs 1/V is a straight line through the origin.

Practical notes. Temperature must always be in kelvin. Pick R to match your pressure unit: 0.08206 L·atm·mol⁻¹·K⁻¹ with atm, or 8.314 J·mol⁻¹·K⁻¹ with pascals and cubic meters. And a very common exam move: rearranging PV = nRT with n = m/M gives M = mRT/PV, the molar mass of an unknown gas from measurable quantities.

Key points

Temperature in the ideal gas law is always absolute (kelvin). This is the single most common arithmetic error in Unit 3.
Partial pressure depends on mole fraction only — identity and molar mass are irrelevant.
M = mRT/PV lets you find the molar mass of an unknown gas from mass, P, V, and T.
At STP (273.15 K, 1.0 atm) one mole of ideal gas occupies 22.4 L.

Equations

$P V = n R T$
on the exam sheetThe master equation. Everything else in this topic follows from it.
- $P$ pressure
- $V$ volume (L)
- $n$ moles
- $R$ gas constant
- $T$ absolute temperature (K)
$\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$
on the exam sheetFor a fixed amount of gas changing conditions.
$P_{A} = P_{t o t a l} \times X_{A}$
on the exam sheetX_A = moles A ÷ total moles.
- $X_{A}$ mole fraction of A
$P_{t o t a l} = P_{A} + P_{B} + P_{C} + \dots$
on the exam sheetDalton’s law of partial pressures.

Common mistakes

Celsius in PV = nRT. Always convert: K = °C + 273.15.
Mismatched R and units. Using 0.08206 with kPa, or 8.314 with litres and atm, produces answers off by orders of magnitude.
Volume in mL. R in L·atm requires litres.
Assuming 22.4 L/mol at any conditions. That value is STP-specific.
Forgetting water vapor. A gas collected over water is a mixture: P_gas = P_total − P_H₂O.

Worked example

A 3.00 L vessel at 27 °C contains 0.200 mol N₂ and 0.300 mol O₂. Find the total pressure and the partial pressure of each gas.

Convert temperature: T = 27 + 273 = 300 K. Total moles = 0.500 mol.

Total pressure:
P = nRT/V = (0.500)(0.08206)(300)/(3.00) = 4.10 atm

Mole fractions:
X(N₂) = 0.200/0.500 = 0.400
X(O₂) = 0.300/0.500 = 0.600

Partial pressures:
P(N₂) = 0.400 × 4.10 = 1.64 atm
P(O₂) = 0.600 × 4.10 = 2.46 atm

Check: 1.64 + 2.46 = 4.10 atm ✓. Note that O₂ has the larger partial pressure purely because there is more of it — its greater molar mass is irrelevant.

Full notes for topic 3.4 →