Chapter 12: Kinetic Theory Case Study Questions | physics Class 11 CBSE

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

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A research team is studying the behaviour of an ideal gas inside a sealed metallic container of fixed volume. They heat the gas from 27°C to 127°C while continuously monitoring the pressure using a sensor attached to the wall. The container is rigid and no gas escapes during the experiment. The team knows from kinetic theory that gas molecules are in incessant random motion and that pressure arises from the momentum transferred by molecules to the walls. They record that the pressure increases proportionally with the rise in absolute temperature. They also note that the average kinetic energy of each molecule changes during the process. The experiment is designed to verify the kinetic interpretation of temperature and the gas laws derived from kinetic theory.

Question 1: State the gas law that governs the behaviour of the gas in this experiment. Write the mathematical relation between pressure and temperature for this process.

Since the container is sealed and rigid, volume and number of moles are constant; Gay-Lussac's law applies: P/T = constant.
From the ideal gas equation PV = µRT with fixed V and µ, pressure P is directly proportional to absolute temperature T: P₁/T₁ = P₂/T₂.
This is a direct consequence of kinetic theory: P = (1/3)nmv̄² and ½mv̄² = (3/2)kBT, so P ∝ T at constant n.

Question 2: Calculate the ratio of final pressure to initial pressure when the gas is heated from 27°C to 127°C at constant volume.

Initial absolute temperature T₁ = 27 + 273 = 300 K; final temperature T₂ = 127 + 273 = 400 K.
At constant volume, P₁/T₁ = P₂/T₂, so P₂/P₁ = T₂/T₁ = 400/300 = 4/3.
The final pressure is 4/3 times the initial pressure, meaning the pressure increases by approximately 33.3%.

Question 3: Using the kinetic interpretation of temperature, calculate the percentage increase in the rms speed of the gas molecules when the gas is heated from 27°C to 127°C.

The rms speed vrms = √(3kBT/m), so vrms ∝ √T.
vrms₂/vrms₁ = √(T₂/T₁) = √(400/300) = √(4/3) = 2/√3 ≈ 1.1547.
Percentage increase = (vrms₂ − vrms₁)/vrms₁ × 100 = (1.1547 − 1) × 100 ≈ 15.47%, so the rms speed increases by approximately 15.5% even though the absolute temperature increased by 33.3%, illustrating the square-root dependence of speed on temperature.

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Case Set 2

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A chemistry student is performing an experiment on gaseous diffusion. Two gases, hydrogen (H₂, molar mass 2 u) and oxygen (O₂, molar mass 32 u), are released simultaneously from opposite ends of a long horizontal tube at the same temperature of 27°C and 1 atm pressure. The student expects the two gases to meet somewhere along the tube after diffusing toward each other. According to kinetic theory, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (Graham's law of diffusion). The student wishes to locate where the two gases will first meet, and also to compare their rms speeds and average kinetic energies at the same temperature.

Question 1: At the same temperature, compare the average kinetic energy per molecule of H₂ and O₂. Justify your answer using kinetic theory.

The average kinetic energy per molecule of any ideal gas equals (3/2)kBT and depends only on absolute temperature, not on the nature or mass of the gas.
Since both gases are at the same temperature T = 300 K, the average kinetic energy per molecule is identical for H₂ and O₂: KE = (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
This is the kinetic interpretation of temperature: temperature is a measure of average translational kinetic energy, making it independent of molecular identity.

Question 2: Calculate the ratio of rms speeds of H₂ to O₂ at 27°C and explain which gas diffuses faster.

vrms = √(3RT/M), so vrms(H₂)/vrms(O₂) = √(M_O₂/M_H₂) = √(32/2) = √16 = 4.
H₂ molecules have rms speed 4 times greater than O₂ molecules at the same temperature.
Since diffusion rate ∝ vrms ∝ 1/√M (Graham's law), H₂ diffuses 4 times faster than O₂, and the two gases will meet much closer to the oxygen end of the tube.

Question 3: If the tube is 1 m long and both gases start diffusing simultaneously from opposite ends, estimate the point along the tube (measured from the H₂ end) where they first meet. Justify using Graham's law.

By Graham's law, the ratio of distances covered in the same time is equal to the ratio of diffusion rates, which equals the ratio of rms speeds: d_H₂/d_O₂ = vrms(H₂)/vrms(O₂) = 4.
Since d_H₂ + d_O₂ = 1 m and d_H₂ = 4 × d_O₂: 4d_O₂ + d_O₂ = 1 m, so d_O₂ = 0.2 m and d_H₂ = 0.8 m.
The gases first meet at 0.8 m from the H₂ end (or 0.2 m from the O₂ end), reflecting the fact that H₂ travels 4 times farther than O₂ in the same elapsed time.

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Case Set 3

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A physics teacher demonstrates the law of equipartition of energy using three sealed identical flasks at the same temperature and pressure: flask A contains helium (monatomic), flask B contains nitrogen (diatomic, treated as rigid rotator), and flask C contains a polyatomic gas with 2 vibrational modes. The teacher explains that each quadratic energy term in the molecular energy expression contributes ½kBT at temperature T. She asks students to calculate the total internal energy per mole of each gas and the molar specific heat at constant volume, then to predict which gas would show the greatest temperature rise if the same amount of heat is supplied at constant volume to one mole of each gas.

Question 1: State the law of equipartition of energy and calculate the average energy per molecule at temperature T for the gas in flask A (helium).

The law of equipartition of energy states that in thermal equilibrium at temperature T, the total energy is equally distributed among all energy modes, with each quadratic term contributing ½kBT.
Helium is monatomic with 3 translational degrees of freedom. Average energy per molecule = 3 × ½kBT = (3/2)kBT.
At T = 300 K, this equals (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J per helium atom.

Question 2: Calculate the molar specific heat at constant volume (Cv) for nitrogen in flask B and the polyatomic gas in flask C. Compare the two values.

For nitrogen (rigid diatomic, f = 5 degrees): U = (5/2)RT per mole, so Cv(N₂) = (5/2)R = 2.5 × 8.314 = 20.79 J mol⁻¹ K⁻¹.
For the polyatomic gas with f = 3 translational + 3 rotational + 2 vibrational modes: each vibrational mode contributes kBT (two quadratic terms), so total energy = (3/2 + 3/2 + 2)RT = 5RT per mole and Cv = 5R = 41.57 J mol⁻¹ K⁻¹.
The polyatomic gas has Cv = 5R, more than double that of nitrogen (5/2)R, because it stores energy across more degrees of freedom, requiring more heat to raise its temperature by 1 K.

Question 3: If 500 J of heat is supplied at constant volume to one mole of each gas, calculate the temperature rise in each flask and identify which gas shows the greatest rise. (R = 8.31 J mol⁻¹ K⁻¹)

ΔT = Q/(µCv). For helium: Cv = (3/2)R = 12.47 J mol⁻¹ K⁻¹; ΔT = 500/12.47 ≈ 40.1 K.
For nitrogen: Cv = (5/2)R = 20.78 J mol⁻¹ K⁻¹; ΔT = 500/20.78 ≈ 24.1 K. For polyatomic: Cv = 5R = 41.55 J mol⁻¹ K⁻¹; ΔT = 500/41.55 ≈ 12.0 K.
Helium shows the greatest temperature rise (≈40.1 K) because it has the fewest degrees of freedom; all the heat goes into only 3 translational modes, giving the largest ΔT per joule of heat supplied.

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Case Set 4

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An engineer is designing a gas-filled thermal insulator using the concept of mean free path. She knows that in highly evacuated chambers, heat transfer by convection and conduction through the gas is minimised because the mean free path of gas molecules becomes comparable to or larger than the dimensions of the chamber. She uses the formula l = 1/(√2 π n d²) for the mean free path, where n is the number density and d is the molecular diameter. She works with air molecules (d ≈ 2 × 10⁻¹⁰ m) and varies the pressure at a fixed temperature of 300 K to study how the mean free path changes. At standard pressure (n = 2.7 × 10²⁵ m⁻³), the mean free path is about 2.9 × 10⁻⁷ m.

Question 1: State the formula for mean free path and explain the physical significance of each symbol in the context of kinetic theory.

Mean free path l = 1/(√2 π n d²), where n is the number of molecules per unit volume (number density) and d is the diameter of the gas molecule.
The √2 factor arises from accounting for the relative motion of all molecules (average relative velocity = √2 × average speed), making the expression more accurate than the simpler estimate l = 1/(π n d²).
Physically, l is the average distance a molecule travels in a straight line before colliding with another molecule; a larger n or larger d means more frequent collisions and a shorter mean free path.

Question 2: If the pressure in the insulating chamber is reduced to 1/100 of atmospheric pressure at the same temperature, calculate the new mean free path and compare it with the chamber dimension of 1 cm.

At constant temperature, from P = nkBT, reducing pressure by factor 100 reduces n by factor 100: n_new = 2.7 × 10²⁵/100 = 2.7 × 10²³ m⁻³.
New mean free path: l_new = l_old × 100 = 2.9 × 10⁻⁷ × 100 = 2.9 × 10⁻⁵ m = 29 µm.
The mean free path (29 µm) is still much smaller than the chamber dimension (1 cm = 10⁴ µm); for effective thermal insulation, the pressure must be reduced further until l exceeds the chamber dimensions.

Question 3: At what pressure (in Pa) must the air in the chamber be reduced so that the mean free path equals the chamber dimension of 1 cm? (kB = 1.38 × 10⁻²³ J K⁻¹, T = 300 K, d = 2 × 10⁻¹⁰ m)

Set l = 0.01 m: 0.01 = 1/(√2 π n d²), so n = 1/(√2 π × 0.01 × 4 × 10⁻²⁰) = 1/(1.776 × 10⁻²¹) = 5.63 × 10²⁰ m⁻³.
Pressure P = nkBT = 5.63 × 10²⁰ × 1.38 × 10⁻²³ × 300 = 5.63 × 10²⁰ × 4.14 × 10⁻²¹ ≈ 2.33 Pa.
This is approximately 2.33 Pa, about 1/43,000 of atmospheric pressure, showing that very high vacuum is needed for the mean free path to reach centimetre scale—consistent with the operating conditions of Thermos flasks and vacuum-insulated panels.

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Case Set 5

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A student is studying the molecular nature of water by comparing its properties in liquid and vapour states. She is given that the density of liquid water is 1000 kg m⁻³ and the density of water vapour at 100°C and 1 atm is 0.6 kg m⁻³. She assumes the density of a single water molecule equals the bulk density of liquid water (since molecules are closely packed in the liquid). Using Avogadro's number NA = 6 × 10²³ mol⁻¹ and the molar mass of water M = 18 g mol⁻¹, she attempts to estimate the radius of a water molecule and the average separation between molecules in the vapour phase.

Question 1: Estimate the mass and volume of a single water molecule using the given data. Hence find the radius of a water molecule.

Mass of one molecule = molar mass/NA = 0.018/(6 × 10²³) = 3 × 10⁻²⁶ kg.
Volume of one molecule = mass/density = (3 × 10⁻²⁶)/(1000) = 3 × 10⁻²⁹ m³.
Setting volume = (4/3)πr³: r³ = (3 × 3 × 10⁻²⁹)/(4π) ≈ 7.16 × 10⁻³⁰ m³, so r ≈ 1.93 × 10⁻¹⁰ m ≈ 2 Å, confirming the atomic scale of water molecules.

Question 2: Calculate the ratio of the volume occupied by water vapour to that of liquid water for the same mass. What does this tell you about the fraction of molecular volume to total volume in the vapour?

For the same mass, density_liquid/density_vapour = 1000/0.6 ≈ 1667, so the vapour occupies about 1667 times more volume than the same mass of liquid.
In liquid water, molecules are closely packed, so the fraction of molecular volume to total volume ≈ 1 (nearly all volume is occupied by molecules).
In vapour, this fraction is reduced by the same factor: molecular volume/total volume ≈ 1/1667 ≈ 6 × 10⁻⁴, showing that in the vapour state, molecules occupy only 0.06% of the total volume—mostly empty space.

Question 3: Estimate the average distance between water molecules in the vapour phase and compare it with the molecular diameter. What does this comparison tell you about the validity of the ideal gas approximation for water vapour?

The volume increases by a factor of ~1667, so the average distance between molecules increases by 1667^(1/3) ≈ 11.8. Average separation ≈ 11.8 × 2r ≈ 11.8 × 4 Å ≈ 47 Å ≈ 40 Å (as given in the chapter's Example 12.3).
The molecular diameter is 2r ≈ 4 Å, so the average intermolecular separation in vapour (≈40 Å) is about 10 times the molecular diameter.
Since molecules are so far apart relative to their own size, intermolecular forces are negligible, and water vapour at 100°C and 1 atm can be treated as approximately ideal—consistent with the chapter's discussion that real gases approximate ideal behaviour at high temperature and low pressure.

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Chapter 12: Kinetic Theory | Case Study Questions

Passage

Question 1: State the gas law that governs the behaviour of the gas in this experiment. Write the mathematical relation between pressure and temperature for this process.

Question 2: Calculate the ratio of final pressure to initial pressure when the gas is heated from 27°C to 127°C at constant volume.

Question 3: Using the kinetic interpretation of temperature, calculate the percentage increase in the rms speed of the gas molecules when the gas is heated from 27°C to 127°C.

Passage

Question 1: At the same temperature, compare the average kinetic energy per molecule of H₂ and O₂. Justify your answer using kinetic theory.

Question 2: Calculate the ratio of rms speeds of H₂ to O₂ at 27°C and explain which gas diffuses faster.

Question 3: If the tube is 1 m long and both gases start diffusing simultaneously from opposite ends, estimate the point along the tube (measured from the H₂ end) where they first meet. Justify using Graham's law.

Passage

Question 1: State the law of equipartition of energy and calculate the average energy per molecule at temperature T for the gas in flask A (helium).

Question 2: Calculate the molar specific heat at constant volume (Cv) for nitrogen in flask B and the polyatomic gas in flask C. Compare the two values.

Question 3: If 500 J of heat is supplied at constant volume to one mole of each gas, calculate the temperature rise in each flask and identify which gas shows the greatest rise. (R = 8.31 J mol⁻¹ K⁻¹)

Passage

Question 1: State the formula for mean free path and explain the physical significance of each symbol in the context of kinetic theory.

Question 2: If the pressure in the insulating chamber is reduced to 1/100 of atmospheric pressure at the same temperature, calculate the new mean free path and compare it with the chamber dimension of 1 cm.

Question 3: At what pressure (in Pa) must the air in the chamber be reduced so that the mean free path equals the chamber dimension of 1 cm? (kB = 1.38 × 10⁻²³ J K⁻¹, T = 300 K, d = 2 × 10⁻¹⁰ m)

Passage

Question 1: Estimate the mass and volume of a single water molecule using the given data. Hence find the radius of a water molecule.

Question 2: Calculate the ratio of the volume occupied by water vapour to that of liquid water for the same mass. What does this tell you about the fraction of molecular volume to total volume in the vapour?

Question 3: Estimate the average distance between water molecules in the vapour phase and compare it with the molecular diameter. What does this comparison tell you about the validity of the ideal gas approximation for water vapour?

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