Gas laws — Boyle, Charles, Gay-Lussac, Avogadro, ideal gas equation connection

Question

How are the individual gas laws (Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s) connected? Show how combining them gives the ideal gas equation $PV = nRT$ .

Gas Law Connection Map

flowchart TD
    A["Boyle's Law: PV = constant (at constant T, n)"] --> E["Combine all four"]
    B["Charles's Law: V/T = constant (at constant P, n)"] --> E
    C["Gay-Lussac's Law: P/T = constant (at constant V, n)"] --> E
    D["Avogadro's Law: V/n = constant (at constant T, P)"] --> E
    E --> F["PV = nRT — The Ideal Gas Equation"]
    F --> G["R = 8.314 J/mol/K = 0.0821 L atm/mol/K"]

Solution — Step by Step

At constant temperature and amount of gas:

P \propto \frac{1}{V} \quad \Rightarrow \quad PV = k_1

If you compress a gas (decrease V), its pressure increases. The P-V graph is a rectangular hyperbola (isotherm).

Practical application: When you push a syringe plunger, the air inside compresses and pressure increases — that is Boyle’s law.

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

At constant pressure and amount of gas:

V \propto T \quad \Rightarrow \quad \frac{V}{T} = k_2

Here $T$ must be in Kelvin (absolute temperature). As temperature increases, gas expands.

\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(at constant P and n)}

This law predicts that at $0 \text{ K}$ ( $-273.15°\text{C}$ ), the volume would become zero — which is the theoretical basis for absolute zero.

At constant volume and amount of gas:

P \propto T \quad \Rightarrow \quad \frac{P}{T} = k_3

As you heat a gas in a rigid container, pressure increases. This is why pressure cookers have safety valves — heating the sealed container increases pressure.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \text{(at constant V and n)}

At constant temperature and pressure:

V \propto n \quad \Rightarrow \quad \frac{V}{n} = k_4

Equal volumes of all gases at the same T and P contain equal number of molecules. At STP (273.15 K, 1 atm), one mole of any ideal gas occupies 22.4 L.

From the four laws:

V \propto \frac{1}{P} \times T \times n

V = \frac{nRT}{P}

\boxed{PV = nRT}

where $R$ is the universal gas constant:

$R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$ (SI units)
$R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$ (CGS)
$R = 2 \text{ cal mol}^{-1}\text{K}^{-1}$ (approximate)

Why This Works

The ideal gas equation is not a separate law — it is what you get when you combine the four individual gas laws into one. Each law holds one or two variables constant and examines the relationship between the remaining variables. The ideal gas equation captures all four relationships simultaneously.

When solving gas law problems, always check which variables are constant and which are changing. If T and n are constant, use Boyle’s law. If P and n are constant, use Charles’s law. If V and n are constant, use Gay-Lussac’s. For complex problems where multiple variables change, go directly to $PV = nRT$ or the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ .

Quick Reference Table

Law	Relationship	Constant	Graph Shape
Boyle’s	$PV = k$	T, n	Hyperbola (P vs V)
Charles’s	$V/T = k$	P, n	Straight line through origin (V vs T in K)
Gay-Lussac’s	$P/T = k$	V, n	Straight line through origin (P vs T in K)
Avogadro’s	$V/n = k$	T, P	Straight line through origin (V vs n)

Common Mistake

The most common error: using Celsius instead of Kelvin in gas law calculations. Charles’s law and Gay-Lussac’s law require absolute temperature (Kelvin). Using Celsius gives absurd results — for instance, at $0°\text{C}$ , $V/T$ would appear to be infinite. Always convert: $T(\text{K}) = T(°\text{C}) + 273.15$ . JEE and NEET examiners deliberately give temperatures in Celsius to catch this mistake.