Gas Laws — Boyle, Charles, Avogadro, and the Ideal Gas

Understanding Gases: The Starting Point

Gases behave very differently from solids and liquids. They fill any container completely, compress easily, and exert pressure in all directions. Understanding gas behaviour requires understanding three macroscopic variables: pressure (P), volume (V), and temperature (T) — and how they relate to each other.

The gas laws are the experimental relationships discovered by systematically varying two of these variables while keeping the third constant. They culminate in the ideal gas law, one of the most useful equations in all of science and engineering.

For CBSE Class 11, gas laws form the foundation of thermodynamics. For JEE Main, numerical problems on gas laws appear in almost every paper. For NEET, conceptual questions on which law applies in which situation are standard.

Key Terms & Definitions

Pressure (P): Force per unit area exerted by gas molecules colliding with container walls. SI unit: Pascal (Pa = N/m²). Also measured in atm (1 atm = 101325 Pa), bar (1 bar = 10⁵ Pa), mmHg (760 mmHg = 1 atm).

Volume (V): The space occupied by the gas. SI unit: m³. Common unit in chemistry: litre (L) or mL.

Temperature (T): ALWAYS use absolute temperature (Kelvin) in gas law calculations. $T(\text{K}) = T(\text{°C}) + 273$ .

Amount of gas (n): Number of moles. 1 mole = $6.022 \times 10^{23}$ molecules (Avogadro’s number).

Standard conditions: STP (Standard Temperature and Pressure) = 0°C and 1 atm. At STP, 1 mole of ideal gas occupies 22.4 L.

Boyle’s Law — Pressure and Volume

Statement: At constant temperature, the pressure of a given mass of gas is inversely proportional to its volume.

P \propto \frac{1}{V} \quad (\text{at constant } T, n)

PV = \text{constant}

P_1V_1 = P_2V_2

Why this makes physical sense: When you compress a gas (decrease V), molecules are packed into a smaller space. They collide with the walls more frequently → pressure increases. At constant temperature, the average kinetic energy (and hence speed) of molecules is unchanged.

Graph: P vs V is a hyperbola. P vs 1/V is a straight line through the origin.

Example: A diver’s lungs. Deep underwater, pressure is high → volume of air in lungs is smaller. This is why scuba divers must breathe from pressurised tanks.

P_1V_1 = P_2V_2 \quad (\text{constant } T, n)

Valid for ideal gases. Real gases deviate at high pressures.

Charles’s Law — Temperature and Volume

Statement: At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.

V \propto T \quad (\text{at constant } P, n)

\frac{V}{T} = \text{constant}

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

Why this makes physical sense: Higher temperature → molecules move faster → they hit the walls harder and more frequently → if pressure is to remain constant, the container must expand to allow the same number of hits per unit area → volume increases.

Absolute zero: Charles’s law predicts that at T = 0 K (−273°C), volume would be zero. This defines absolute zero — the temperature at which molecular motion theoretically stops. In reality, gases liquefy/solidify before reaching 0 K.

Graph: V vs T (Kelvin) is a straight line through the origin. V vs T (Celsius) is a straight line intersecting the T-axis at −273°C.

The most common error in Charles’s law problems: using Celsius temperature instead of Kelvin. Always convert first! $T_K = T_C + 273$ . This is the single most tested source of errors in gas law numerical problems.

Gay-Lussac’s Law — Pressure and Temperature

Statement: At constant volume, pressure is directly proportional to absolute temperature.

P \propto T \quad (\text{at constant } V, n)

\frac{P}{T} = \text{constant}

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Real-life application: Pressure cookers. At constant volume (sealed pot), heating increases temperature → pressure increases → water boils above 100°C → food cooks faster.

Car tyre pressure is higher after a long drive — temperature increases at constant volume (rigid tyre), so pressure rises.

Avogadro’s Law — Amount and Volume

Statement: At constant temperature and pressure, equal volumes of all gases contain equal numbers of molecules.

V \propto n \quad (\text{at constant } T, P)

\frac{V_1}{n_1} = \frac{V_2}{n_2}

This is a remarkable result — it doesn’t matter whether the gas is H₂ (lightest) or UF₆ (very heavy). At the same T and P, equal volumes contain equal molecules. It means molecular size is negligible compared to intermolecular distances in a gas.

Avogadro’s number: $N_A = 6.022 \times 10^{23}$ molecules/mol.

Molar volume at STP: 22.4 L/mol (IUPAC: 22.7 L/mol at 0°C and 1 bar).

The Ideal Gas Law — Combining All Three

Combining Boyle’s, Charles’s, and Avogadro’s laws:

PV = nRT

Where:

P = pressure (Pa, atm, or bar — be consistent)
V = volume (m³ or L)
n = moles of gas
R = universal gas constant
T = temperature (KELVIN ONLY)

Value of R:

$R = 8.314$ J mol⁻¹ K⁻¹ (when P in Pa, V in m³)
$R = 0.0821$ L atm mol⁻¹ K⁻¹ (when P in atm, V in L)
$R = 8.314$ Pa m³ mol⁻¹ K⁻¹ = 8.314 J mol⁻¹ K⁻¹

Use R = 0.0821 L·atm/(mol·K) for most chemistry problems with pressures in atm.

Combined Gas Law (for fixed amount of gas)

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

This is used when P, V, and T all change simultaneously for the same amount of gas.

Partial Pressures — Dalton’s Law

When a mixture of gases that don’t react with each other is in a container:

Dalton’s Law of Partial Pressures: Total pressure = sum of partial pressures of each component.

P_{total} = P_1 + P_2 + P_3 + \ldots

The partial pressure of each gas is the pressure it would exert if it alone occupied the entire container.

P_i = x_i \times P_{total}

where $x_i = n_i / n_{total}$ is the mole fraction of component $i$ .

Application: Scuba diving gas mixtures, blood gas pressures, industrial gas processing.

CBSE Class 11 and JEE frequently test Dalton’s law with “wet gas” problems: when gas is collected over water, the gas is saturated with water vapour. Total pressure = partial pressure of collected gas + vapour pressure of water. Subtract the vapour pressure of water to get the dry gas pressure. This is a standard 3-mark numerical question.

Deviations from Ideal Behaviour

Real gases deviate from the ideal gas law at:

High pressures (molecules are close together; intermolecular forces matter)
Low temperatures (molecules are slow; intermolecular attractions cause condensation)
High molecular masses (larger molecules have stronger London dispersion forces)

The van der Waals equation corrects for these deviations:

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

where $a$ = correction for intermolecular attractions, $b$ = correction for actual volume of molecules.

A gas behaves most ideally at high temperature and low pressure (molecules far apart, kinetic energy dominates over attractions).

Solved Examples

Easy — CBSE Level

Q: A gas occupies 500 mL at 27°C. What volume will it occupy at 127°C, pressure being constant?

Solution: Using Charles’s Law: $V_1/T_1 = V_2/T_2$ .

$T_1 = 27 + 273 = 300$ K; $T_2 = 127 + 273 = 400$ K.

$V_2 = V_1 \times T_2/T_1 = 500 \times 400/300 = 666.7$ mL.

Medium — JEE Main Level

Q: At 27°C, a gas in a container has pressure 1.2 atm. If the container is heated to 127°C at constant volume, what is the new pressure?

Solution: Gay-Lussac’s Law: $P_1/T_1 = P_2/T_2$ .

$T_1 = 300$ K, $T_2 = 400$ K.

$P_2 = P_1 \times T_2/T_1 = 1.2 \times 400/300 = 1.6$ atm.

Hard — JEE Main Level

Q: A mixture contains 8 g of O₂ and 14 g of N₂ at total pressure 0.9 atm. Find the partial pressure of O₂.

Solution:

Moles of O₂ = 8/32 = 0.25 mol. Moles of N₂ = 14/28 = 0.5 mol. Total moles = 0.75 mol.

Mole fraction of O₂ = 0.25/0.75 = 1/3.

Partial pressure of O₂ = (1/3) × 0.9 = 0.3 atm.

Exam-Specific Tips

CBSE Class 11: Boyle’s law, Charles’s law, Gay-Lussac’s law, Avogadro’s law, and combined gas law are individually tested. Know which variable is held constant for each law. Numerical problems with unit conversions are common.

JEE Main: Ideal gas law (PV = nRT), Dalton’s partial pressures, mole fractions, and van der Waals equation corrections. Combined gas law problems where you must identify which variables change.

NEET: Conceptual questions about which law applies (constant T? constant P? constant V?) and straightforward numerical problems at the CBSE level.

Common Mistakes to Avoid

Mistake 1: Using Celsius temperature in gas law equations. Always convert to Kelvin: $T_K = T_C + 273$ . Using $T_C = 27°C$ instead of $T_K = 300 K$ gives completely wrong answers.

Mistake 2: Confusing Boyle’s law with Charles’s law. Boyle’s: $P \times V = const$ (inverse relationship, constant T). Charles’s: $V/T = const$ (direct relationship, constant P). In Boyle’s: if P doubles, V halves. In Charles’s: if T doubles, V doubles.

Mistake 3: Using R = 8.314 when pressure is in atm and volume in litres. Use R = 0.0821 L·atm/(mol·K) when working in atm and litres. Using the wrong R value gives answers off by a factor of about 100.

Mistake 4: In Dalton’s law problems with gas collected over water: forgetting to subtract water vapour pressure. The gas collected is WET — it’s a mixture of the desired gas and water vapour. $P_{gas} = P_{total} - P_{water}$ .

Mistake 5: Saying that equal volumes of gases at the same conditions have the same mass. Equal volumes (at same T, P) have the same NUMBER of molecules (Avogadro’s law), but not the same mass — mass depends on molar mass.

Practice Questions

1. At STP, what is the volume of 3 moles of an ideal gas?

At STP (0°C = 273K, 1 atm), 1 mole of ideal gas = 22.4 L. So 3 moles = 3 × 22.4 = 67.2 L. Alternatively using PV = nRT: V = nRT/P = 3 × 0.0821 × 273 / 1 = 67.2 L.

2. What is the molar mass of a gas if 5 g of it occupies 2.8 L at STP?

At STP, 22.4 L contains 1 mole. So 2.8 L contains 2.8/22.4 = 0.125 mol. Molar mass = mass/moles = 5/0.125 = 40 g/mol. (This is argon, Ar!)

3. Why does a balloon deflate when taken from a warm room to outdoors on a cold day?

Charles’s law: at constant pressure, volume is proportional to absolute temperature. When temperature decreases outdoors, the kinetic energy of gas molecules decreases → molecules hit the balloon walls less frequently/hard → the balloon contracts to maintain equilibrium pressure with the surroundings. Volume decreases with temperature.

4. Calculate the pressure of 0.5 mol of gas at 27°C in a 10 L container.

PV = nRT. P = nRT/V = (0.5 × 0.0821 × 300) / 10 = 12.315 / 10 = 1.23 atm.

FAQs

Q: Why is absolute zero (0 K) theoretically the coldest possible temperature? At absolute zero, all molecular motion would cease — molecules have zero kinetic energy. Removing kinetic energy beyond zero is impossible. In practice, we can get very close to 0 K (experiments have reached 10⁻¹⁰ K) but never reach it exactly (Third Law of Thermodynamics).

Q: What is meant by saying a gas is “ideal”? An ideal gas has two assumptions: (1) gas molecules have negligible volume compared to the container; (2) there are no intermolecular attractions or repulsions. Real gases approximate ideal behaviour at low pressures (molecules far apart) and high temperatures (kinetic energy dominates over attractions). He and H₂ are closest to ideal among real gases.

Q: What is the compressibility factor Z? $Z = PV/(nRT)$ for a real gas. For an ideal gas, Z = 1 always. For real gases: Z > 1 means the gas is harder to compress than ideal (repulsions dominate at very high pressures). Z < 1 means the gas is easier to compress than ideal (attractions dominate at moderate pressures).