Derive ideal gas equation PV=nRT from Boyle's and Charles' laws

Question

Starting from Boyle’s Law and Charles’ Law, derive the ideal gas equation $PV = nRT$ . Identify the universal gas constant $R$ and give its SI value.

Solution — Step by Step

Boyle’s Law (1662): At constant temperature (isothermal process) and fixed amount of gas, the pressure and volume of an ideal gas are inversely proportional.

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

PV = \text{constant} = k_1

Charles’ Law (1787): At constant pressure (isobaric process) and fixed amount of gas, the volume is directly proportional to the absolute temperature (in Kelvin).

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

\frac{V}{T} = \text{constant} = k_2

From Boyle’s Law: $V \propto \frac{1}{P}$ (at constant T)

From Charles’ Law: $V \propto T$ (at constant P)

Combining both (at fixed amount of gas):

V \propto \frac{T}{P}

V = k \cdot \frac{T}{P}

where $k$ is a proportionality constant that depends only on the amount of gas.

Avogadro’s Law: at constant $T$ and $P$ , equal volumes of gases contain equal numbers of moles. So volume is also proportional to number of moles $n$ :

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

Combining all three:

V \propto \frac{nT}{P}

V = R \cdot \frac{nT}{P}

where $R$ is the universal gas constant (same for all ideal gases).

Multiplying both sides by $P$ :

PV = nRT

This is the ideal gas equation (also called the ideal gas law or perfect gas equation).

Value of R: At STP (0°C = 273 K, 1 atm = 101.325 kPa), 1 mole of ideal gas occupies 22.4 L:

R = \frac{PV}{nT} = \frac{1 \text{ atm} \times 22.4 \text{ L}}{1 \text{ mol} \times 273 \text{ K}} = 0.0821\text{ L·atm·mol}^{-1}\text{K}^{-1}

In SI units: $R = 8.314\text{ J·mol}^{-1}\text{K}^{-1}$

Why This Works

The ideal gas equation is a beautiful synthesis of three empirical observations (Boyle, Charles, Avogadro) into one equation. Each law captures one independent variable’s effect on volume, and combining them gives the complete relationship.

Physically, the ideal gas model assumes: gas molecules have negligible size (point particles), and there are no intermolecular forces. Under these assumptions, the only energy is kinetic energy of molecules, which is directly proportional to temperature.

Real gases deviate from ideal behavior at high pressures (molecules are close together → intermolecular forces matter) and low temperatures (molecules slow down → intermolecular attractions become significant). The van der Waals equation corrects for these deviations: $(P + a/V^2)(V - b) = nRT$ .

Alternative Method

The ideal gas law can also be derived from the kinetic theory of gases. The kinetic theory derives $PV = \frac{1}{3}Nmv^2_{rms}$ from molecular motion principles. Combined with the thermodynamic result that the average kinetic energy of a molecule is $\frac{3}{2}k_BT$ , this gives $PV = Nk_BT = nRT$ (where $k_B$ is Boltzmann’s constant and $R = N_A k_B$ ).

This derivation is more fundamental — it derives the gas law from first principles rather than combining empirical observations.

For JEE and CBSE Class 11 chemistry, know both the derivation approach (combining Boyle + Charles + Avogadro) and the value of $R$ in different units. In problems: use $R = 0.082\text{ L·atm·mol}^{-1}\text{K}^{-1}$ when pressure is in atm and volume in litres. Use $R = 8.314\text{ J·mol}^{-1}\text{K}^{-1}$ when dealing with energy, work, or SI units. Temperature must always be in Kelvin (add 273.15 to Celsius). This conversion is one of the most common sources of error.

Common Mistake

The most common error in ideal gas calculations is using temperature in Celsius instead of Kelvin. $PV = nRT$ requires $T$ in absolute temperature (Kelvin). $0°C = 273\text{ K}$ , $25°C = 298\text{ K}$ , $100°C = 373\text{ K}$ . Using $T = 25$ instead of $T = 298$ gives a result that’s about 12× too large. Always write “T = 25°C = 298 K” explicitly before substituting. Another error: using $n$ = mass in grams instead of moles. Always compute $n = \text{mass}/\text{molar mass}$ first.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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