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CBSE Class 11 Physics · 10 questions · 24 marks
Kinetic theory explains the pressure, temperature and specific heats of a gas by treating it as a swarm of tiny particles in ceaseless random motion. Once you accept a handful of simple assumptions, a few lines of mechanics deliver the gas laws that experimenters had found by trial. This chapter also tells you what temperature really measures at the microscopic level, and why a diatomic gas needs more heat than a monatomic one for the same rise in temperature.
The average translational kinetic energy of a molecule of an ideal gas depends on:
Answer
The absolute temperature only — kinetic theory gives the average translational kinetic energy per molecule as (3/2)kT, an expression containing no reference whatever to the mass of the molecule or the nature of the gas. At the same temperature a hydrogen molecule and an oxygen molecule carry equal average translational energy; the light hydrogen molecule simply achieves it by moving four times faster, since (1/2)mv² is fixed while m differs.
The value of γ = C_p/C_v for a rigid diatomic gas at ordinary temperatures is:
Answer
1.40 — a rigid diatomic molecule has f = 5 degrees of freedom, three translational and two rotational, since rotation about the bond axis carries negligible moment of inertia. Then γ = 1 + 2/f = 1 + 2/5 = 1.4. The value 1.67 belongs to a monatomic gas with f = 3, and 1.33 to a non-linear polyatomic gas with f = 6.
If the absolute temperature of a gas is increased four times, its root mean square molecular speed becomes:
Answer
Twice the original value — since v_rms = √(3RT/M), the speed varies as the square root of the absolute temperature. Multiplying T by 4 multiplies v_rms by √4 = 2. A common error is to assume the speed scales directly with T; it is the kinetic energy, not the speed, that is proportional to T.
According to kinetic theory, the pressure exerted by an ideal gas of density ρ is:
Answer
(1/3)ρv_rms² — the factor of one-third arises because molecular motion is random and shared equally among the three perpendicular directions, so only one-third of the mean square speed contributes to the momentum transferred to any one wall. The expression (2/3)ρv_rms² would be twice the kinetic energy per unit volume divided by... in fact P = (2/3) × (kinetic energy per unit volume), which is the same statement written differently, but as a formula in ρ and v_rms the correct coefficient is 1/3.
Assertion (A): At the same temperature, hydrogen molecules move faster on average than oxygen molecules. Reason (R): The root mean square speed of a gas molecule is inversely proportional to the square root of its molar mass.
Answer
Both A and R are true and R is the correct explanation of A — from v_rms = √(3RT/M), at a fixed temperature v_rms ∝ 1/√M. Hydrogen has M = 2 g/mol and oxygen M = 32 g/mol, a ratio of 16, so hydrogen molecules move √16 = 4 times faster on average. This is also the reason light gases such as hydrogen and helium leak out of the Earth's upper atmosphere over geological time while heavier gases are retained.
Calculate the root mean square speed of nitrogen molecules at 300 K. Take the molar mass of nitrogen as 28×10⁻³ kg/mol and R = 8.3 J/mol·K.
Answer
v_rms = √(3RT/M) = √((3 × 8.3 × 300)/(28×10⁻³)) Numerator: 3 × 8.3 × 300 = 7470. Dividing: 7470/(28×10⁻³) = 7470/0.028 = 2.667×10⁵. v_rms = √(2.667×10⁵) ≈ 516 m/s. So nitrogen molecules at room temperature move at roughly 5.2×10² m/s, comfortably faster than the speed of sound in air, which is what you would expect since sound is carried by these very molecules.
A sealed rigid cylinder holds an ideal gas at 27 °C and a pressure of 1.5×10⁵ Pa. It is heated until the pressure reaches 2.5×10⁵ Pa. (a) Find the final temperature in °C. (b) By what factor does the average kinetic energy of a molecule change?
Answer
(a) The cylinder is rigid, so the volume and the number of molecules are fixed. From PV = NkT with V and N constant, P ∝ T: P₁/T₁ = P₂/T₂. T₁ = 27 + 273 = 300 K. T₂ = T₁ × (P₂/P₁) = 300 × (2.5×10⁵/1.5×10⁵) = 300 × (5/3) = 500 K. In Celsius: 500 − 273 = 227 °C. (b) The average kinetic energy per molecule is (3/2)kT, which is directly proportional to the absolute temperature. Ratio = T₂/T₁ = 500/300 = 5/3 ≈ 1.67. So the average molecular kinetic energy increases by a factor of about 1.67, the same factor as the pressure — which makes sense, since at fixed volume the pressure is just a measure of the kinetic energy density.
(a) List the main assumptions of the kinetic theory of gases. (b) Derive the expression P = (1/3)(mN/V)v̄² for the pressure exerted by an ideal gas on the walls of its container. (c) Use the result to deduce that the average translational kinetic energy per molecule is (3/2)kT, and hence show that Avogadro's law follows.
Answer
(a) Assumptions of kinetic theory: 1. A gas consists of a very large number of identical molecules in constant random motion. 2. The size of a molecule is negligible compared with the average distance between molecules, so the molecules themselves occupy no appreciable volume. 3. Molecules exert no force on one another except during collisions; there is no potential energy of interaction, so all the internal energy is kinetic. 4. Collisions between molecules, and between a molecule and the wall, are perfectly elastic, so kinetic energy and momentum are conserved. 5. The duration of a collision is negligible compared with the time between collisions. 6. Between collisions the molecules travel in straight lines obeying Newton's laws. (b) Derivation of the pressure relation. Consider N molecules, each of mass m, in a cubical box of side L, so the volume is V = L³. Take one molecule with velocity components v_x, v_y and v_z. Momentum change at one wall: the molecule strikes the wall perpendicular to the x-axis and rebounds elastically, so its x-velocity reverses from +v_x to −v_x. The change in momentum of the molecule is Δp = (−mv_x) − (mv_x) = −2mv_x, so the momentum imparted to the wall is 2mv_x. Time between successive impacts on the same wall: the molecule must travel to the opposite wall and back, a distance 2L, at speed v_x, so Δt = 2L/v_x. Force exerted on the wall by this one molecule: F = momentum transferred per unit time = 2mv_x/(2L/v_x) = mv_x²/L. Total force from all N molecules: F_total = (m/L)(v_x₁² + v_x₂² + … + v_xN²) = (mN/L) × (mean of v_x²). Pressure on this wall of area L²: P = F_total/L² = (mN/L³) × (mean of v_x²) = (mN/V) × (mean of v_x²). Now use the randomness of the motion. For each molecule v² = v_x² + v_y² + v_z², and since no direction is preferred, the averages of v_x², v_y² and v_z² are all equal. Hence mean of v_x² = (1/3) v̄², where v̄² is the mean square speed. Substituting: P = (1/3)(mN/V)v̄². Since mN/V is just the density ρ, this is also P = (1/3)ρv̄², and since mN is the total mass, PV = (1/3)mN v̄². (c) Average kinetic energy and Avogadro's law. Rewrite the result as PV = (2/3) × N × (1/2)m v̄² = (2/3) N E, where E = (1/2)m v̄² is the average translational kinetic energy of one molecule. Compare with the experimental ideal gas equation PV = NkT: (2/3) N E = N k T E = (3/2) kT. So the average translational kinetic energy of a molecule depends on the absolute temperature alone. This gives temperature a clear microscopic meaning: it is a direct measure of the mean translational kinetic energy of the molecules, and at T = 0 K that motion would cease. Avogadro's law: consider two gases at the same pressure P, the same volume V and the same temperature T. From PV = (2/3)N₁E₁ and PV = (2/3)N₂E₂, we get N₁E₁ = N₂E₂. But equal temperature means E₁ = E₂ = (3/2)kT. Therefore N₁ = N₂. That is, equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules, which is exactly Avogadro's law, here deduced rather than assumed.
(a) State the law of equipartition of energy. (b) Use it to obtain C_v, C_p and γ for a monatomic gas, a rigid diatomic gas and a non-linear polyatomic gas. (c) A mixture contains 3 moles of helium (monatomic) and 2 moles of oxygen (rigid diatomic). Find the molar specific heat at constant volume of the mixture and its effective γ. Take R = 8.3 J/mol·K.
Answer
(a) Law of equipartition of energy: in thermal equilibrium at temperature T, the total energy of a system is shared equally among all its active degrees of freedom, each degree of freedom carrying an average energy of (1/2)kT per molecule, or equivalently (1/2)RT per mole. A degree of freedom is any independent coordinate needed to specify the energy of a molecule. Each translational and each rotational mode contributes one such term of the form (1/2)mv² or (1/2)Iω², so each carries (1/2)kT. A vibrational mode is different: it has both a kinetic and a potential term, so it carries kT in all, that is 2 × (1/2)kT. (b) Specific heats from degrees of freedom. General method: internal energy per mole U = (f/2)RT, where f is the number of degrees of freedom. Then C_v = dU/dT = (f/2)R, C_p = C_v + R = (f/2 + 1)R by Mayer's relation, and γ = C_p/C_v = 1 + 2/f. Monatomic gas (helium, argon): only 3 translational degrees of freedom, f = 3. C_v = (3/2)R = 12.45 J/mol·K, C_p = (5/2)R = 20.75 J/mol·K, γ = 5/3 ≈ 1.67. Rigid diatomic gas (oxygen, nitrogen at ordinary temperatures): 3 translational + 2 rotational, f = 5. Rotation about the line joining the two atoms is ignored because the moment of inertia about that axis is negligible. C_v = (5/2)R = 20.75 J/mol·K, C_p = (7/2)R = 29.05 J/mol·K, γ = 7/5 = 1.40. Non-linear polyatomic gas (water vapour, methane): 3 translational + 3 rotational, f = 6. C_v = 3R = 24.9 J/mol·K, C_p = 4R = 33.2 J/mol·K, γ = 4/3 ≈ 1.33. (c) The mixture. For a mixture, the internal energy is simply the sum of the internal energies of the components, so the molar specific heat is the mole-weighted average: C_v(mix) = (n₁C_v₁ + n₂C_v₂)/(n₁ + n₂). Helium: n₁ = 3, C_v₁ = (3/2)R = 1.5 × 8.3 = 12.45 J/mol·K. Oxygen: n₂ = 2, C_v₂ = (5/2)R = 2.5 × 8.3 = 20.75 J/mol·K. C_v(mix) = (3 × 12.45 + 2 × 20.75)/(3 + 2) = (37.35 + 41.50)/5 = 78.85/5 = 15.77 J/mol·K. Mayer's relation holds for the mixture too, since both components are ideal: C_p(mix) = C_v(mix) + R = 15.77 + 8.3 = 24.07 J/mol·K. γ(mix) = C_p/C_v = 24.07/15.77 ≈ 1.53. This lies between 1.67 and 1.40 as expected, and closer to the helium value because helium is the more abundant component.
Read the following and answer the questions that follow: A laboratory flask of volume 5×10⁻³ m³ contains an ideal monatomic gas at a pressure of 2.0×10⁵ Pa and a temperature of 300 K. Take R = 8.3 J/mol·K, k = 1.38×10⁻²³ J/K and N_A = 6.02×10²³ /mol. (a) Find the number of moles and the number of molecules of gas in the flask. (b) Find the total translational kinetic energy of all the molecules. (c) The gas is now heated at constant volume to 600 K. State what happens to the pressure, to v_rms and to the mean free path, giving a reason in each case. (d) A student claims that at 600 K every molecule in the flask is moving at exactly v_rms. Explain why this is wrong.
Answer
(a) Number of moles from PV = nRT: n = PV/(RT) = (2.0×10⁵ × 5×10⁻³)/(8.3 × 300) = 1000/2490 ≈ 0.402 mol. Number of molecules: N = n N_A = 0.402 × 6.02×10²³ ≈ 2.42×10²³ molecules. (b) Total translational kinetic energy. Each molecule has an average of (3/2)kT, so E_total = N × (3/2)kT = (3/2) × 2.42×10²³ × 1.38×10⁻²³ × 300. First: 2.42×10²³ × 1.38×10⁻²³ ≈ 3.34. Then: (3/2) × 3.34 × 300 = 1.5 × 1002 = 1503 J. So the molecules together carry about 1.5×10³ J. (The same figure follows more quickly from E = (3/2)nRT = 1.5 × 0.402 × 8.3 × 300 ≈ 1501 J, the small difference being rounding.) (c) Heating from 300 K to 600 K at constant volume: Pressure: at fixed V and N, P ∝ T, so the pressure doubles to 4.0×10⁵ Pa. The molecules strike the walls both harder and more often. v_rms: since v_rms ∝ √T, doubling the temperature multiplies v_rms by √2 ≈ 1.41, an increase of about 41%, not 100%. Mean free path: λ = 1/(√2 πd²n) depends only on the number density n and the molecular diameter d. The flask is sealed and rigid, so neither N nor V has changed and n is unaltered. The mean free path is therefore unchanged. The molecules cover the same average distance between collisions, but they cover it faster, so the collision frequency rises by the factor √2. (d) The claim is wrong because v_rms is a statistical average, not a value shared by every molecule. At any instant the molecules have a wide spread of speeds described by the Maxwell distribution: a few are nearly stationary, a few move far faster than v_rms, and the most probable speed v_mp is actually smaller than v_rms, with v_mp < v_avg < v_rms. Constant collisions keep reshuffling the energy among molecules, so an individual molecule's speed changes from moment to moment. What stays fixed at a given temperature is the shape of the distribution, and v_rms is simply the square root of the mean of the squared speeds — a convenient single number that summarises it, since it is the speed that reproduces the correct total kinetic energy.
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