Chapter 12: Kinetic Theory Application Questions | physics Class 11 CBSE

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Chapter 12: Kinetic Theory Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A mountaineer at high altitude notices that it takes longer to boil water than at sea level. Using kinetic theory, explain why the boiling point of water decreases at high altitude.

At high altitude, atmospheric pressure is lower because there are fewer air molecules per unit volume (lower number density n), directly reducing the air pressure from P = nkBT.
Boiling occurs when the vapour pressure of water equals the surrounding atmospheric pressure. At lower pressure, this equilibrium is achieved at a lower temperature.
From kinetic theory, vrms = √(3kBT/m); at lower pressure (and thus lower boiling temperature T), water molecules have lower average kinetic energy when they escape from the liquid surface.
This means food cooked in boiling water at altitude receives less thermal energy per unit time, so it takes longer to cook even though water is visibly boiling—a practical consequence of the pressure-temperature relationship of gases.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

A scuba diver breathes compressed air from a tank. Explain, using Dalton's law of partial pressures, why breathing high-pressure air at great depths can cause nitrogen narcosis.

In a scuba tank, air is a mixture of nitrogen and oxygen. By Dalton's law, the total pressure is the sum of partial pressures: P_total = P_N₂ + P_O₂, each gas acting independently.
At depth, the ambient water pressure is much higher than at the surface, so the diver must breathe air at this elevated pressure to prevent lung collapse.
The partial pressure of nitrogen in the compressed air increases proportionally with depth; at great depths, the elevated partial pressure of nitrogen causes it to dissolve in greater quantities in the blood and tissues.
Excess dissolved nitrogen affects nerve function, producing nitrogen narcosis (an intoxication-like state), illustrating how Dalton's law has critical physiological implications—each gas in a mixture behaves as if it alone occupies the volume at its partial pressure.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

With reference to the kinetic theory of gases, explain why a tyre inflated in the morning at 27°C has higher pressure in the afternoon when the temperature rises to 47°C. Calculate the new pressure if initial pressure is 2 × 10⁵ Pa.

When temperature rises, the average kinetic energy per molecule (½ mv̄² = (3/2)kBT) increases, so molecules strike the tyre walls with greater speed and momentum, increasing the pressure.
For a sealed tyre at constant volume, the amount of gas (µ) is fixed, so from PV = µRT: P/T = constant, giving P₁/T₁ = P₂/T₂.
With T₁ = 300 K and T₂ = 320 K: P₂ = P₁ × (T₂/T₁) = 2 × 10⁵ × (320/300) = 2 × 10⁵ × 1.067 ≈ 2.13 × 10⁵ Pa.
This increase in pressure is a direct consequence of the kinetic interpretation of temperature: higher T means higher vrms and thus greater momentum transferred per wall collision per unit time, which is the microscopic basis of the macroscopic pressure rise.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at 12°C to the surface at 35°C. Find the volume of the bubble at the surface. (P₀ = 1.013 × 10⁵ Pa, ρ_water = 1000 kg m⁻³, g = 10 m/s²)

Pressure at 40 m depth: P₁ = P₀ + ρgh = 1.013 × 10⁵ + 1000 × 10 × 40 = 5.013 × 10⁵ Pa. Pressure at surface: P₂ = 1.013 × 10⁵ Pa.
Temperatures: T₁ = 285 K, T₂ = 308 K. Number of moles in the bubble is constant (no gas escapes), so apply the combined gas law: P₁V₁/T₁ = P₂V₂/T₂.
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 1.0 × (5.013/1.013) × (308/285) = 1.0 × 4.95 × 1.081 ≈ 5.35 cm³.
The bubble grows more than fivefold because both the large pressure drop (dominant effect) and the temperature rise contribute to the expansion—a practical illustration of the ideal gas equation combining Boyle's and Charles' laws simultaneously.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A perfume bottle is opened in one corner of a room. Although perfume vapour molecules travel at hundreds of metres per second, it takes minutes for the scent to reach the other corner. Explain this observation using the concept of mean free path.

Perfume vapour molecules travel at high speeds (~300–500 m/s) comparable to the speed of sound, but the mean free path in air at STP is only about 2.9 × 10⁻⁷ m.
Each molecule undergoes roughly 10⁹ collisions per second, and each collision deflects it in a random direction. The molecule follows a highly tortuous random-walk path rather than moving in a straight line toward the other end of the room.
The net displacement in a random walk grows only as the square root of the number of steps (steps being the mean free path), so diffusion is far slower than ballistic motion—it takes many minutes to cross a room by diffusion.
This is why the mean free path is crucial: it determines the scale of each random step. A smaller mean free path (denser gas, larger molecules) would result in even slower diffusion despite high individual molecular speeds.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to the kinetic theory, explain why a gas at high temperature and low pressure behaves more like an ideal gas than at low temperature and high pressure.

Ideal gas behaviour requires that intermolecular forces be negligible. At low pressure, molecules are widely spaced (large interatomic distances), so intermolecular attractions and repulsions are very weak and can be ignored.
At high temperature, molecules have large kinetic energies (½ mv̄² = (3/2)kBT >> potential energy of interaction), so even when they come close, the interaction does not significantly alter their trajectories.
At high pressure, molecules are compressed into a small volume; they spend more time near each other, and intermolecular forces become significant, causing deviations from PV = µRT (as seen in the P-V graph for steam in the chapter).
At low temperature, molecular kinetic energies are comparable to intermolecular potential energies; attractions dominate, clustering occurs, and the gas may even liquefy—far from ideal behaviour.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm at 27°C. After some oxygen is withdrawn, the gauge pressure drops to 11 atm and temperature drops to 17°C. Estimate the mass of oxygen withdrawn. (R = 8.31 J mol⁻¹ K⁻¹, M_O₂ = 32 u)

Initial state: absolute pressure P₁ = 16 atm = 16 × 1.013 × 10⁵ Pa, T₁ = 300 K, V = 30 × 10⁻³ m³. Moles initially: µ₁ = P₁V/(RT₁) = (16 × 1.013 × 10⁵ × 30 × 10⁻³)/(8.31 × 300) ≈ 19.48 mol.
Final state: P₂ = 12 atm = 12 × 1.013 × 10⁵ Pa, T₂ = 290 K. Moles remaining: µ₂ = P₂V/(RT₂) = (12 × 1.013 × 10⁵ × 30 × 10⁻³)/(8.31 × 290) ≈ 15.09 mol.
Moles of oxygen withdrawn: Δµ = µ₁ − µ₂ ≈ 19.48 − 15.09 = 4.39 mol.
Mass withdrawn = Δµ × M = 4.39 × 32 ≈ 141 g. This calculation relies on the ideal gas equation applied at each state to find the molar content, with gauge pressure converted to absolute by adding atmospheric pressure (1 atm).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

Estimate the total number of air molecules in a room of volume 25 m³ at 27°C and 1 atm pressure. (kB = 1.38 × 10⁻²³ J K⁻¹, P = 1.013 × 10⁵ Pa)

From the ideal gas equation in the form PV = NkBT, solve for N: N = PV/(kBT).
Substituting: N = (1.013 × 10⁵ × 25) / (1.38 × 10⁻²³ × 300) = (2.5325 × 10⁶) / (4.14 × 10⁻²¹).
N ≈ 6.12 × 10²⁶ molecules. This is approximately 10³ times Avogadro's number, confirming the enormous number of molecules in even a modest volume of gas at ordinary conditions.
This large number is the statistical basis for the reliability of macroscopic laws: pressure, temperature, and other bulk properties are averages over ~10²⁶ molecules, making fluctuations from the mean utterly negligible in practice.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic) with partial pressure ratio P_Ne : P_O₂ = 3:2. Estimate the ratio of (i) number of molecules and (ii) mass densities. (M_Ne = 20.2 u, M_O₂ = 32 u)

Both gases obey PV = µRT. At the same V and T: P_Ne/P_O₂ = µ_Ne/µ_O₂ = 3/2. Since µ = N/NA, we get N_Ne/N_O₂ = 3/2.
The ratio of number of molecules of neon to oxygen is 3:2—directly equal to the partial pressure ratio, because each mole contains the same number of molecules (Avogadro's number).
For mass densities: ρ = (µM)/V, so ρ_Ne/ρ_O₂ = (µ_Ne × M_Ne)/(µ_O₂ × M_O₂) = (3/2) × (20.2/32.0) = (3 × 20.2)/(2 × 32.0) = 60.6/64.0 ≈ 0.947.
So neon has a slightly lower mass density than oxygen in this mixture, even though neon has a higher mole fraction, because its molecular mass (20.2 u) is significantly less than O₂ (32 u).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

Estimate the average thermal energy of a helium atom at (i) room temperature (27°C), (ii) the surface of the Sun (6000 K), and (iii) the core of a star (10⁷ K). (kB = 1.38 × 10⁻²³ J K⁻¹)

Average thermal energy per molecule of a monatomic ideal gas = (3/2)kBT. At T = 300 K (room temperature): E = (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J ≈ 0.039 eV.
At T = 6000 K (surface of Sun): E = (3/2) × 1.38 × 10⁻²³ × 6000 = 1.24 × 10⁻¹⁹ J ≈ 0.78 eV—about 20 times the room temperature value.
At T = 10⁷ K (stellar core): E = (3/2) × 1.38 × 10⁻²³ × 10⁷ = 2.07 × 10⁻¹⁶ J ≈ 1290 eV ≈ 1.29 keV.
At stellar core temperatures, thermal energies are in the keV range—sufficient to overcome the Coulomb barrier between nuclei and initiate nuclear fusion reactions. This shows how the kinetic interpretation of temperature connects directly to astrophysics and nuclear physics.

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Chapter 12: Kinetic Theory | Application Questions

A mountaineer at high altitude notices that it takes longer to boil water than at sea level. Using kinetic theory, explain why the boiling point of water decreases at high altitude.

A scuba diver breathes compressed air from a tank. Explain, using Dalton's law of partial pressures, why breathing high-pressure air at great depths can cause nitrogen narcosis.

With reference to the kinetic theory of gases, explain why a tyre inflated in the morning at 27°C has higher pressure in the afternoon when the temperature rises to 47°C. Calculate the new pressure if initial pressure is 2 × 10⁵ Pa.

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at 12°C to the surface at 35°C. Find the volume of the bubble at the surface. (P₀ = 1.013 × 10⁵ Pa, ρ_water = 1000 kg m⁻³, g = 10 m/s²)

A perfume bottle is opened in one corner of a room. Although perfume vapour molecules travel at hundreds of metres per second, it takes minutes for the scent to reach the other corner. Explain this observation using the concept of mean free path.

With reference to the kinetic theory, explain why a gas at high temperature and low pressure behaves more like an ideal gas than at low temperature and high pressure.

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm at 27°C. After some oxygen is withdrawn, the gauge pressure drops to 11 atm and temperature drops to 17°C. Estimate the mass of oxygen withdrawn. (R = 8.31 J mol⁻¹ K⁻¹, M_O₂ = 32 u)

Estimate the total number of air molecules in a room of volume 25 m³ at 27°C and 1 atm pressure. (kB = 1.38 × 10⁻²³ J K⁻¹, P = 1.013 × 10⁵ Pa)

A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic) with partial pressure ratio P_Ne : P_O₂ = 3:2. Estimate the ratio of (i) number of molecules and (ii) mass densities. (M_Ne = 20.2 u, M_O₂ = 32 u)

Estimate the average thermal energy of a helium atom at (i) room temperature (27°C), (ii) the surface of the Sun (6000 K), and (iii) the core of a star (10⁷ K). (kB = 1.38 × 10⁻²³ J K⁻¹)

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