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#Boyle's law

13 public questions tagged with this topic.

A gas at 2 atm and 300 K occupies 20 litres. If the pressure is increased to 4 atm at constant temperature, what is the

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2 atm, V₁ = 20 litres, P₂ = 4 atm.V₂ = (P₁ V₁)/(P₂) = (2 × 20)/(4) = 10 litres. Substituting values gives 10 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

A gas at 1.5 atm and 300 K has a volume of 18 litres. If the pressure decreases to 0.75 atm at constant temperature, wha

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1.5 atm, V₁ = 18 litres, P₂ = 0.75 atm.V₂ = (P₁ V₁)/(P₂) = (1.5 × 18)/(0.75) = 36 litres. Substituting values gives 36 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

A gas at 3 atm and 600 K has a volume of 15 litres. If the pressure decreases to 1.5 atm at constant temperature, what i

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 3 atm, V₁ = 15 litres, P₂ = 1.5 atm.V₂ = (P₁ V₁)/(P₂) = (3 × 15)/(1.5) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

A gas at 4 atm and 500 K has a volume of 20 litres. If the pressure increases to 8 atm at constant temperature, what is

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 4 atm, V₁ = 20 litres, P₂ = 8 atm.V₂ = (P₁ V₁)/(P₂) = (4 × 20)/(8) = 10 litres. Substituting values gives 10 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

A gas at 2 atm and 400 K has a volume of 6 litres. If the pressure decreases to 1 atm at constant temperature, what is t

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2 atm, V₁ = 6 litres, P₂ = 1 atm.V₂ = (P₁ V₁)/(P₂) = (2 × 6)/(1) = 12 litres. Substituting values gives 12 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

A gas at 1 atm and 300 K has a volume of 10 litres. If the volume is reduced to 5 litres at constant temperature, what i

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1 atm, V₁ = 10 litres, V₂ = 5 litres.P₂ = (P₁ V₁)/(V₂) = (1 × 10)/(5) = 2 atm . Substituting values gives 2.0 atm, which matches expected kinetic theory result, confirming mean free path λ

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

A gas at 1 atm and 273 K has a volume of 8 litres. If the pressure increases to 4 atm at constant temperature, what is t

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1 atm, V₁ = 8 litres, P₂ = 4 atm.V₂ = (P₁ V₁)/(P₂) = (1 × 8)/(4) = 2 litres. Substituting values gives 2 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

A gas at 3 atm and 400 K has a volume of 15 litres. If the pressure drops to 1.5 atm at constant temperature, what is th

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 3 atm, V₁ = 15 litres, P₂ = 1.5 atm.V₂ = (P₁ V₁)/(P₂) = (3 × 15)/(1.5) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

What happens to the volume of an ideal gas if its pressure is doubled while the temperature remains constant?

According to Boyle’s Law (Section 10.4), for an ideal gas at constant temperature, P V = constant . If pressure doubles ( P_2 = 2P_1 ), then V_2 = V_1 / 2, so the volume halves.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Kinetic Theory (Latest NCERT 2026-27), Topic: RMS speed v_rms ∝ √T, temperature dependence, ratio v₂/v₁ = √(T₂/T₁) and calculation. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page.

Which gas law relates pressure and volume when temperature is held constant?

Boyle’s Law describes the inverse relationship between pressure and volume of an ideal gas at constant temperature (PV = constant). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Boyle’s Law. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

What happens to the pressure of an ideal gas if its volume is halved and temperature is kept constant?

Boyle’s Law (Section 10.4) states PV = constant at constant temperature. If V2 = V1/2, then P2 = 2P1, so the pressure doubles. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It doubles. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.