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Kinetic Theory of Gases: JEE Main Physics formulas and notes

Kinetic theory is a short, formula-driven chapter: JEE Mains usually asks 1 to 2 questions, nearly always from the speeds (rms, mean, most probable), degrees of freedom with C(p), C(v) and γ, mean free path, and gas mixtures. Most are solved in under a minute from the formulas on the Formulas tab. About 10 hours is enough.

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Formulas with solved examples

Gas laws and basics

Ideal gas equation

How to apply. Convert to SI units and kelvin first. Use PV = nRT when moles are given and P = nkT when molecules are counted. Two states of the same amount of gas: (P₁V₁)/(T₁) = (P₂V₂)/(T₂).

Example. 2 mol of gas at 400 K in 10 L (0.01 m³): P = 2 × 8.31 × 400 ÷ 0.01 = 6.65 × 10⁵ Pa.

Boyle’s, Charles’s and Gay-Lussac’s laws

How to apply. Each law is the ideal gas equation with two quantities held fixed. Name the quantity that is constant, and write only the ratio that applies.

Example. 4 L at 2 atm compressed isothermally to 1 L: P = 8 atm. 6 L at 300 K heated at constant P to 450 K: V = 9 L. A rigid vessel at 2 atm and 300 K heated to 400 K: P = 2.67 atm.

Avogadro number, Boltzmann constant and molar volume

How to apply. Use k for one molecule and R for one mole. The mass of a molecule is the molar mass divided by Avogadro’s number.

Example. Molecules per cm³ at 273 K and 1 atm: n = P ÷ kT = 1.013 × 10⁵ ÷ (1.38 × 10⁻²³ × 273) = 2.69 × 10²⁵ per m³ = 2.69 × 10¹⁹ per cm³.

Density of a gas

How to apply. Put P, M (in kg/mol) and T into ρ = PM/RT. Compare two gases or states by taking the ratio.

Example. Nitrogen (M = 0.028 kg/mol) at 300 K and 1.013 × 10⁵ Pa: ρ = 0.028 × 1.013 × 10⁵ ÷ (8.31 × 300) = 1.14 kg/m³.

Dalton’s law of partial pressures and mixtures

How to apply. Treat each gas as if it were alone in the vessel, then add the pressures. The ratio of partial pressures is the ratio of moles (same V and T).

Example. 1 mol of H₂ and 1 mol of He in 10 L at 300 K: each gas gives 2.49 × 10⁵ Pa, so P(total) = 4.99 × 10⁵ Pa.

Connected vessels and leaking gas

How to apply. Write n = PV/RT for each part, equate the total number of moles before and after, and keep P equal in the connected parts.

Example. Vessel A (2 L, 3 atm) joined to vessel B (3 L, 1 atm) at the same T: P = (6 + 3) ÷ 5 = 1.8 atm. A rigid vessel heated from 300 K to 400 K with the valve open keeps (300)/(400) = 75% of its gas.

Kinetic theory and pressure

Assumptions of the kinetic theory

How to apply. Use these when a question asks what is true for an ideal gas or what causes pressure. Pressure comes from momentum given to the walls by collisions; it is the same in every direction in a vessel.

Example. Because there are no forces between the molecules, an ideal gas has no intermolecular potential energy: its internal energy depends only on the kinetic energy of its molecules (translation and, for diatomic gases, rotation).

Pressure of a gas

How to apply. The factor (1)/(3) comes from the three directions. Use ρ = (mass of gas) ÷ V. At constant volume and mass, P ∝ v(rms)², so doubling the speeds multiplies P by 4.

Example. 10²³ molecules of mass 5 × 10⁻²⁶ kg, v(rms) = 500 m/s, in 1 L: P = (1)/(3) × 10²³ × 5 × 10⁻²⁶ × 500² ÷ 10⁻³ = 4.17 × 10⁵ Pa.

Pressure and kinetic energy per unit volume

How to apply. Use this when the energy is given instead of the speed. The total translational KE of N molecules is E = N × (½mv(rms)²).

Example. A gas has 300 J of translational KE in 2 L: P = (2)/(3) × 300 ÷ 0.002 = 1.0 × 10⁵ Pa.

Speed of sound and v(rms)

How to apply. The speed of sound in a gas is a fixed fraction of the rms speed of its molecules. It is the same relation for any ideal gas, with its own γ.

Example. Air: v(rms) = √(3 × 10⁵ ÷ 1.2) = 500 m/s, γ = 1.4, so v(sound) = 500 × √(1.4 ÷ 3) = 342 m/s.

Temperature and molecular speeds

Kinetic meaning of temperature

How to apply. The mean KE depends only on T, not on the type of gas or on the pressure. So at the same T a heavy molecule moves slower, not with a smaller energy.

Example. At 300 K: (3)/(2)kT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J = 0.039 eV; per mole (3)/(2)RT = 3740 J.

Root-mean-square speed

How to apply. Use 3RT/M with M in kg per mole (0.032 for O₂, 0.002 for H₂). Equal T gives equal KE, so the ratio of speeds is the inverse ratio of √M.

Example. O₂ at 300 K: v(rms) = √(3 × 8.31 × 300 ÷ 0.032) = 483 m/s. H₂ at 300 K: √(3 × 8.31 × 300 ÷ 0.002) = 1934 m/s.

Mean speed and most probable speed

How to apply. The three speeds are in a fixed ratio for a given gas and temperature, so find one and the others follow. The order v(mp) < v(avg) < v(rms) always holds.

Example. N₂ at 300 K: v(rms) = 517 m/s, v(avg) = 517 × 0.921 = 476 m/s, v(mp) = 517 × 0.816 = 422 m/s.

Dependence of speeds on T and M

How to apply. Write the ratio with the square root and cancel what does not change. Use kelvin: a change from 27 °C to 54 °C is 300 K to 327 K, not a doubling.

Example. The speeds double when T goes from 300 K to 1200 K (927 °C). H₂ and O₂ at the same T: v(H₂) ÷ v(O₂) = √(32 ÷ 2) = 4. O₂ has the speed of H₂ at 300 K when T = 300 × 16 = 4800 K.

Speed of a mixture of gases

How to apply. Do not average the speeds; average the squares of the speeds, weighted by the number of molecules (this is not the same as averaging the energies: at one T every gas has the same mean KE, but different v²). The heavier gas is slower at the same T.

Example. 1 mol of H₂ and 1 mol of He at 300 K: v² (H₂) = 3.74 × 10⁶, v² (He) = 1.87 × 10⁶, so the number-averaged v(rms) = √(2.80 × 10⁶) = 1675 m/s. (Using √(3P/ρ) = √(3RT/M(mix)) with M(mix) = 3 g/mol instead gives 1579 m/s; use that if the question gives the pressure and the density of the mixture.)

Escape of gases from a planet

How to apply. Compare v(rms) of the gas with the escape speed. This is why the Earth has little hydrogen and the Moon has no atmosphere. Higher T and smaller M give a greater chance of escape.

Example. The temperature at which H₂ would have v(rms) = 11.2 km/s: T = 0.002 × (11200)² ÷ (3 × 8.31) = 1.0 × 10⁴ K. For O₂: 1.6 × 10⁵ K.

Equipartition and heat capacities

Degrees of freedom and the law of equipartition

How to apply. Count f from the shape of the molecule: a single atom has no rotation about its own axis that counts, a line has two rotations, a non-linear molecule has three. Use f = 5 for a diatomic gas unless vibration is stated.

Example. Helium: f = 3, nitrogen: f = 5, ammonia and water vapour: f = 6, carbon dioxide (linear): f = 5 (without vibration).

Internal energy of an ideal gas

How to apply. Use ΔU = nC(v)ΔT even if the volume or pressure changes. Because U = (f)/(2)PV, the change in internal energy can also be read from the change in PV.

Example. 2 mol of N₂ at 300 K: U = 2.5 × 2 × 8.31 × 300 = 1.25 × 10⁴ J.

C(v), C(p) and γ for an ideal gas

How to apply. Work out f first, then every other quantity follows. Choose C(p) when the pressure is constant and C(v) when the volume is constant.

Example. Heating 1 mol of O₂ by 10 K at constant pressure: Q = 3.5 × 8.31 × 10 = 291 J (at constant volume it takes 208 J).

Vibration and the temperature dependence of C(v)

How to apply. Use f = 5 for a diatomic gas, unless the problem mentions a vibrating molecule (then f = 7) or very low temperature. Always read the data given in the question.

Example. A diatomic gas with vibration: C(v) = 3.5R = 29.1 J/mol K and γ = 9 ÷ 7 = 1.29; without vibration γ = 7 ÷ 5 = 1.4.

Internal energy and heat capacity of a mixture

How to apply. Add the energies of the separate gases (they are at the same T). For γ add C(p) and C(v) separately and divide; do not average the two values of γ.

Example. 1 mol of He and 2 mol of O₂ at 300 K: U = (3 + 2 × 5) ÷ 2 × 8.31 × 300 = 1.62 × 10⁴ J. 2 mol of monatomic with 1 mol of diatomic: C(v) = 1.83R, C(p) = 2.83R, γ = 17 ÷ 11 = 1.55.

Work and heat in processes of an ideal gas (brief)

How to apply. Quote the first law, then choose the process. In an adiabatic compression the temperature rises; in an isothermal change all the heat goes to work.

Example. 1 mol of diatomic gas heated at constant pressure by 50 K: W = 8.31 × 50 = 416 J and Q = 3.5 × 8.31 × 50 = 1454 J. Compressed adiabatically to V ÷ 32: T becomes 32^0.4 = 4 times and P becomes 32^1.4 = 128 times.

Specific heat per unit mass

How to apply. JEE gives the specific heat per kg or asks for it. Divide the molar value by the molar mass in kg/mol. Q = m c ΔT.

Example. Helium (M = 0.004 kg/mol): c(v) = 1.5 × 8.31 ÷ 0.004 = 3116 J/kg K, c(p) = 2.5 × 8.31 ÷ 0.004 = 5194 J/kg K.

Solids: Dulong and Petit

How to apply. Use this when a solid is treated by the law of equipartition: it gives the molar specific heat of a monatomic solid.

Example. C = 3 × 8.31 = 24.9 J/mol K; for one mole U = 3RT = 7480 J at 300 K.

Mean free path

Mean free path

How to apply. Use P = nkT to move between n, P and T. When the temperature is raised at constant pressure, n falls and λ increases. If only the container volume is fixed, λ stays the same.

Example. N₂, d = 3 × 10⁻¹⁰ m, 300 K and 10⁵ Pa: λ = 1.38 × 10⁻²³ × 300 ÷ (1.414 × π × 9 × 10⁻²⁰ × 10⁵) = 1.0 × 10⁻⁷ m. At constant P a rise from 300 K to 600 K doubles λ.

Collision frequency and time between collisions

How to apply. Collisions per second is the speed divided by the distance travelled between collisions. Use the average speed (not the rms) for z.

Example. For the N₂ case above: v(avg) = 476 m/s, so z = 476 ÷ 1.04 × 10⁻⁷ = 4.6 × 10⁹ per second.

Distribution, diffusion and real gases

Maxwell distribution of speeds (qualitative)

How to apply. Use the picture: increasing T shifts the peak to a higher speed and flattens it, but the area stays the same. Do not use the formula for a calculation; use the ratios of speeds.

Example. Raising T from 300 K to 1200 K moves v(mp) to twice its value, and the peak of the curve falls to half its height (the area is fixed).

Diffusion and effusion (Graham’s law)

How to apply. Write the ratio of the rates as the square root of the inverse ratio of molar masses. For the same volume to escape the time ratio is the reverse.

Example. Rate of H₂ to rate of O₂ = √(32 ÷ 2) = 4. Helium escapes √(16 ÷ 4) = 2 times faster than methane (M = 16).

Brownian motion

How to apply. The motion needs no outside cause: it comes from the random collisions of molecules. The particle is far heavier than a molecule, so it is much slower.

Example. A dust particle of mass 10⁻¹⁵ kg at 300 K: v(rms) = √(3 × 1.38 × 10⁻²³ × 300 ÷ 10⁻¹⁵) = 3.5 × 10⁻³ m/s.

Real gases and the van der Waals equation

How to apply. Mention this when a question asks when the ideal gas law fails: the molecules have a size and they attract. Both corrections become small when the gas is dilute and hot.

Example. CO₂ (a = 0.364 Pa m⁶/mol², b = 4.27 × 10⁻⁵ m³/mol), 1 mol in 1 L at 300 K: P = 8.31 × 300 ÷ (10⁻³ − 4.27 × 10⁻⁵) − 0.364 ÷ 10⁻⁶ = 2.24 × 10⁶ Pa, against 2.49 × 10⁶ Pa for an ideal gas.

Most asked question types

How do I solve “Ideal gas equation and gas laws” questions in JEE Main?

A gas changes P, V or T, or is heated or compressed with one quantity fixed. Find the final value of one of them. 1. Name what is constant (T, P or V). 2. Write (P₁V₁)/(T₁) = (P₂V₂)/(T₂). 3. Use kelvin. Example: 6 L at 300 K heated at constant pressure to 450 K gives 9 L.

How do I solve “Mass, moles and number of molecules” questions in JEE Main?

A mass of gas, a volume at STP or a number of molecules is given. Find n, the molecular mass or the number of molecules. 1. n = m/M or V/22.4 L at STP. 2. N = n N(A). 3. Mass of one molecule = M/N(A). Example: 1 cm³ of gas at 273 K and 1 atm has 2.69 × 10¹⁹ molecules.

How do I solve “Density of a gas and mixtures of gases” questions in JEE Main?

Gases are mixed in a vessel, or the density of a gas at a given P and T is asked. 1. ρ = PM/RT. 2. Dalton: add the partial pressures. 3. Use the mean molar mass for a mixture. Example: 1 mol of H₂ and 1 mol of He in 10 L at 300 K: P = 4.99 × 10⁵ Pa.

How do I solve “Pressure from molecular speeds” questions in JEE Main?

The speeds, the mass or the density of the gas is given. Find the pressure or the rms speed. 1. P = ⅓ρv(rms)² or P = ⅓(mN/V)v(rms)². 2. Find v(rms) = √(3P/ρ). 3. At fixed V and N: P ∝ v(rms)². Example: P = 10⁵ Pa and ρ = 1.2 kg/m³ give v(rms) = 500 m/s.

How do I solve “Pressure from kinetic energy” questions in JEE Main?

The total or the mean KE of molecules is given along with the volume. Find the pressure, or the reverse. 1. P = ⅔E/V. 2. E = N(3/2)kT = (3/2)nRT. 3. Compare two cases by ratio. Example: 300 J of translational KE in 2 L gives 10⁵ Pa.

How do I solve “Mean KE of a molecule and temperature” questions in JEE Main?

The temperature of a gas or the energy of a molecule is given. Find the KE, in joule or eV, or the temperature at which a given KE is reached. 1. KE = (3/2)kT per molecule. 2. Use kelvin. 3. 1 eV = 1.6 × 10⁻¹⁹ J. Example: At 300 K, KE = 6.2 × 10⁻²¹ J = 0.039 eV.

How do I solve “rms, average and most probable speeds” questions in JEE Main?

A gas at a given temperature is described. Find v(rms), v(avg) or v(mp), or use their ratio. 1. v(rms) = √(3RT/M) with M in kg/mol. 2. Ratio √3 : √(8/π) : √2. 3. Check v(mp) < v(avg) < v(rms). Example: N₂ at 300 K: 517, 476 and 422 m/s.

How do I solve “Change of speed with temperature or molar mass” questions in JEE Main?

T or the gas is changed, and a speed is compared. Find the new speed or the temperature at which two speeds are equal. 1. v ∝ √(T/M). 2. Write the ratio of the two cases. 3. Use kelvin. Example: O₂ has the speed of H₂ at 300 K when T = 4800 K.

How do I solve “Degrees of freedom and internal energy” questions in JEE Main?

The type of molecule is given, or a mixture. Find U, the number of degrees of freedom or the energy per molecule. 1. Count f: 3, 5 or 6. 2. U = (f/2)nRT. 3. Add for a mixture. Example: 2 mol of N₂ at 300 K: U = 1.25 × 10⁴ J.

How do I solve “Specific heats and γ” questions in JEE Main?

The type of gas or a mixture is given. Find C(v), C(p), γ or the heat for a rise of temperature. 1. C(v) = fR/2, C(p) = C(v) + R. 2. γ = 1 + 2/f. 3. For a mixture add n C(v) and n C(p). Example: 2 mol of monatomic gas and 1 mol of diatomic gas: γ = 17/11 = 1.55.

How do I solve “Heat, work and ΔU at constant P, V or in an adiabatic change” questions in JEE Main?

A gas is heated at constant P, constant V, or compressed without heat. Find Q, W, ΔU or the final T. 1. Isobaric W = nRΔT, Q = nC(p)ΔT. 2. Isochoric W = 0. 3. Adiabatic: TV^γ − 1 = constant. Example: 1 mol of diatomic gas, 50 K rise at constant P: W = 416 J, Q = 1454 J.

How do I solve “Mean free path and collision frequency” questions in JEE Main?

The size of a molecule, T, P or n is given or changed. Find λ, the collision frequency, or the change in λ. 1. λ = kT/(√2 π d² P). 2. λ ∝ T/P, and constant at fixed V. 3. z = v(avg)/λ. Example: N₂ with d = 3 × 10⁻¹⁰ m at 300 K and 10⁵ Pa: λ = 1.0 × 10⁻⁷ m.

How do I solve “Maxwell speed distribution (qualitative)” questions in JEE Main?

A graph of the number of molecules against speed is shown for different T or gases. Identify the curves or compare the speeds. 1. Peak at v(mp). 2. Higher T: peak lower and to the right. 3. Area under the curve is the same. Example: T from 300 K to 1200 K doubles v(mp).

How do I solve “Graham’s law and escape of gases” questions in JEE Main?

Two gases diffuse or effuse through a hole, or the escape of a gas from a planet is considered. 1. Rate ∝ 1/√M. 2. Time for the same volume ∝ √M. 3. Compare v(rms) with the escape speed. Example: H₂ diffuses 4 times as fast as O₂.

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Simulations for Kinetic Theory of Gases

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