The gas laws form the quantitative backbone of States of Matter and are the most directly tested topic in NEET numerical problems.

Boyle's Law

At constant temperature and moles: $P \propto 1/V$, so $PV = k$ (a constant). Equivalently, $P_1V_1 = P_2V_2$. The PV vs. P plot is a horizontal line (PV is constant); the P vs. V plot is a rectangular hyperbola. The key NEET trap: Boyle's law applies only at constant temperature. A compression that raises temperature violates the law's conditions.

Charles's Law

At constant pressure: $V \propto T$ (in Kelvin). $V_1/T_1 = V_2/T_2$. The V vs. T(K) plot is a straight line through the origin. Extrapolating to $V = 0$ gives $T = 0$ K (absolute zero = −273.15°C), which is the thermodynamic basis for the Kelvin scale. The trap: using Celsius instead of Kelvin — a classic calculation error.

Gay-Lussac's Law

At constant volume: $P \propto T$ (in Kelvin). $P_1/T_1 = P_2/T_2$. This law explains why sealed gas containers (like aerosol cans) are dangerous when heated — pressure rises proportionally with absolute temperature.

Avogadro's Law

At constant T and P: $V \propto n$. At STP (0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L (molar volume). This is the basis for molar mass determination from gas density: $M = dRT/P$.

Ideal Gas Equation and Numerical Strategy

$PV = nRT$. NEET numerical protocol: (1) Convert temperature to Kelvin ($T_K = T_C + 273$). (2) Choose the correct value of R to match the pressure unit: $R = 0.0821\ \text{L·atm/(mol·K)}$ for atm pressure; $R = 8.314\ \text{J/(mol·K)}$ for Pa or SI units. (3) Convert mass to moles: $n = w/M$. (4) Solve for the unknown.

Worked example: Find the volume of 4 g of $CH_{4}$ at 300 K and 1 atm. $n = 4/16 = 0.25\ \text{mol}$; $V = nRT/P = (0.25 \times 0.0821 \times 300)/1 = 6.16\ \text{L}$

Dalton's Law — Application to Collected Gases

When a gas is collected over water, the total pressure includes the vapor pressure of water: $P_\text{dry gas} = P_\text{total} - P_\text{water vapor}$. Example: gas collected over water at 25°C (where P_\text{H_2O} = 23.8\ \text{mmHg}) at $P_\text{total} = 760\ \text{mmHg}$ → $P_\text{dry gas} = 736.2\ \text{mmHg}$.

Graham's Law — Calculation Strategy

$r_1/r_2 = \sqrt{M_2/M_1}$. Equivalently, $r \propto 1/\sqrt{M}$. The most common NEET form: given two gases with known $M_1$, $M_2$, find the ratio of rates; or given one rate ratio, find unknown $M$. Critical step: the heavier gas always has the slower rate. Never invert the mass inside the square root — $r_1/r_2 = \sqrt{M_2/M_1}$ means if gas 1 has smaller $M$, it diffuses faster (larger $r_1$).

Molar Mass from Gas Density

$M = \frac{dRT}{P}$

At STP: $M = 22.4 \times d$ (where $d$ is in g/L). This is a one-step calculation frequently tested in NEET.

Relative Density and Vapour Density

Vapour density (VD) = $M/2$ for a diatomic or polyatomic gas relative to $H_{2}$. NEET often states VD and asks for molar mass: $M = 2 \times VD$.