Topic 3.5: Kinetic Molecular Theory Worksheet

A1. (a) The particles’ own volume is negligible beside the container’s volume; the particles move constantly, randomly, in straight lines between collisions; the particles exert no attractions or repulsions on each other except during collisions; all collisions are perfectly elastic, so total kinetic energy is conserved; the average kinetic energy of the particles is directly proportional to the kelvin temperature. (b) The constant random motion postulate: moving particles collide with the walls, each collision transfers momentum, and the accumulated impacts are the pressure. (c) The no intermolecular forces postulate: condensation needs attractions to pull particles into a liquid, and the ideal model sets those attractions at exactly zero. (d) The other particle must leave more slowly, carrying less kinetic energy, so that the pair’s total is unchanged; that is the perfectly elastic collision postulate. One particle can gain only what the other loses.

A2. (a) It depends on the absolute (kelvin) temperature alone. It does not depend on the identity of the gas, its molar mass, its pressure, or how much gas is present. (b) They are equal, per mole and per particle: at the same temperature every ideal gas has the same average kinetic energy, and 400 K fixes that value for helium and xenon alike, even though a xenon atom weighs about 33 times a helium atom. (c) The average per particle is identical, since both flasks sit at 300 K. The totals differ: the 3 mol flask holds three times as many particles each carrying the same average, so its total kinetic energy is three times the 1 mol flask’s. (d) Celsius zero is not a true zero of kinetic energy, so Celsius ratios are meaningless for energy: 27 °C to 327 °C is a ratio of about 12 in Celsius but only 600 K to 300 K, a factor of 2, on the scale the proportionality actually uses.

A3. (a) The number of particles (the fraction of particles) travelling at each speed. (b) The total number of particles in the sample. (c) The most probable speed, the speed at which the largest number of particles travel. (d) On heating, the peak moves to the right, to higher speeds, and the curve becomes lower and broader. What does not change is the area under the curve: the sample is sealed, so the particle count, and therefore the area, is the same.

A4. (a) Effusion is the escape of gas particles through a tiny hole into a vacuum, one particle at a time through an opening too small for bulk flow. Diffusion is the gradual spreading and mixing of one gas through another (or through space) driven by random motion, with no hole involved. (b) Graham’s law. The deciding property is molar mass: helium atoms (4.003 g/mol) are far lighter than nitrogen molecules (28.02 g/mol), so at the same temperature they move faster and slip through the balloon’s microscopic pores at √(28.02 ÷ 4.003) ≈ 2.65 times the rate. (c) The molecules are continually colliding with air molecules, and each collision knocks them onto a new straight run in a new direction. The path is a zigzag of very short runs, so the net displacement grows far more slowly than the instantaneous speed suggests. (d) Yes. Diffusion is driven by the same molecular speeds that set effusion rates, so the lighter gas diffuses faster by the same √(M₂/M₁) comparison; only the setting differs, mixing through a gas rather than escaping through a hole.

A5. (a) In the classical model it is the temperature at which average translational kinetic energy reaches zero and particle motion reaches its minimum: the floor of the temperature scale, the zero from which the kelvin scale is built, rather than an arbitrary reference point like the freezing of water. (b) 27 °C = 300.15 K, about 300 K; 327 °C = 600.15 K, about 600 K. (c) Average kinetic energy is proportional to the kelvin temperature, so it changes by 600.15 ÷ 300.15 ≈ 2.00: it doubles. (d) The student ratioed Celsius readings, whose zero is not the zero of kinetic energy; only kelvin ratios carry the proportionality.

B1. (a) (3/2) × 8.314 × 300 = 3,741 J per mole, about 3.74 kJ/mol. (b) 3,741 ÷ (6.022 × 10²³) ≈ 6.21 × 10⁻²¹ J per particle. (c) (3/2) × 8.314 × 600 = 7,483 J per mole, about 7.48 kJ/mol. That is exactly twice part (a), as it must be: the temperature doubled and average kinetic energy is directly proportional to it.

B2. (a) uᵣₘₛ = √(3 × 8.314 × 300 ÷ 0.03200) = √233,831 ≈ 484 m/s. (b) Speed scales with the square root of temperature: √(150 ÷ 300) = √0.5 ≈ 0.707, so uᵣₘₛ = 484 × 0.707 ≈ 342 m/s (direct substitution gives the same 342 m/s). (c) The average kinetic energy per mole halves, from 3.74 kJ to 1.87 kJ, because it is directly proportional to T. There is no contradiction: energy follows T itself, speed follows √T, and √(1/2) ≈ 0.707 is exactly the factor that halves ½mv².

B3. (a) rate and speed ratios run as √(M₂/M₁): √(44.01 ÷ 2.016) = √21.83 ≈ 4.67. Hydrogen’s root-mean-square speed is about 4.67 times carbon dioxide’s. (b) uᵣₘₛ(CO₂) = 1,920 ÷ 4.67 ≈ 411 m/s. (c) At one temperature the average ½mv² is the same for both gases, so v² must stand in the inverse mass ratio, and v therefore stands in the inverse square root of the mass ratio. The mass ratio is about 22, but speeds answer to √22 ≈ 4.7, not to 22 itself.

B4. (a) rate(He) ÷ rate(N₂) = √(28.02 ÷ 4.003) = √7.00 ≈ 2.65. Helium effuses about 2.65 times as fast. (b) For a fixed amount, time is inversely proportional to rate, so time(N₂) = 30 × 2.65 ≈ 79 s. (c) Rate measures amount per unit time, so a gas that effuses at 1/2.65 of the rate needs 2.65 times as long to pass the same amount: the time ratio is the rate ratio inverted, √(M₁/M₂) instead of √(M₂/M₁).

B5. (a) The unknown passes the same amount in twice the time, so its rate is half of methane’s: rate(unknown) ÷ rate(CH₄) = 30 ÷ 60 = 0.500. (b) Graham’s law: 0.500 = √(16.04 ÷ M), so 0.250 = 16.04 ÷ M and M = 16.04 ÷ 0.250 = 64.16, about 64.2 g/mol. Equivalently, halving the rate means quadrupling the mass: 16.04 × 4 = 64.16 g/mol. (c) SO₂ at 64.07 g/mol matches the calculated 64.2 g/mol within rounding; CO₂ (44.01) and Cl₂ (70.90) are both well away from it.

B6. (a) The 1,200 K curve’s peak sits further right: its particles hold four times the average kinetic energy and twice the root-mean-square speed. (b) The areas are equal. The sample is sealed, the particle count is unchanged, and area under the curve is that count. (c) The 1,200 K peak is lower. The same area spread over a wider range of speeds cannot stand as tall; the 300 K curve, compressed into a narrower band, peaks higher. (d) Average kinetic energy scales with T: 1,200 ÷ 300 = 4, so it grows by a factor of 4. Root-mean-square speed scales with √T: √4 = 2, so it doubles.

C1. (a) rate(NH₃) ÷ rate(HCl) = √(36.46 ÷ 17.03) = √2.141 ≈ 1.46. Ammonia diffuses about 1.46 times as fast as hydrogen chloride. (b) In any given time the gases travel distances in the rate ratio, 1.46 to 1, sharing the 1.00 m between them. HCl’s share is 1.00 × (1 ÷ 2.46) ≈ 0.41 m, so the ring forms about 0.41 m from the HCl end (about 0.59 m from the NH₃ end). (c) The midpoint would require equal distances in equal times, which means equal rates. Hydrogen chloride molecules are heavier (36.46 against 17.03 g/mol), move more slowly at the same temperature, and so cover less of the tube before the two fronts meet; the meeting point is dragged towards the HCl end. (d) Both gases’ speeds rise by the same √T factor, so both rates increase in the same proportion, the ring forms sooner, but the ratio of rates is unchanged and the ring forms in exactly the same position, about 0.41 m from the HCl end.

C2. (a) Average kinetic energy per mole = (3/2)RT: in X, (3/2) × 8.314 × 300 = 3,741 J, about 3.74 kJ/mol; in Y, (3/2) × 8.314 × 600 = 7,483 J, about 7.48 kJ/mol. The argon holds twice the average kinetic energy per mole, because its temperature is twice as high, and the gas’s identity plays no part. (b) uᵣₘₛ(He at 300 K) = √(3 × 8.314 × 300 ÷ 0.004003) = √1,869,248 ≈ 1,367 m/s. uᵣₘₛ(Ar at 600 K) = √(3 × 8.314 × 600 ÷ 0.03995) = √374,598 ≈ 612 m/s. The helium particles move faster on average, by a factor of about 2.2. (c) The containers have the same volume and the same number of moles, so by PV = nRT the pressure follows the temperature: Y’s pressure is twice X’s. In particle terms, argon’s collisions strike the walls with twice the average energy, and that outweighs helium’s more frequent, weaker impacts. (d) Average kinetic energy is ½mv², a product of mass and speed squared. An argon atom carries about ten times the mass of a helium atom, so it can hold twice the kinetic energy while moving at well under half the speed: the doubled temperature raises the energy each particle carries, but the heavy particle banks that energy in its mass rather than in speed.

C3. (a) Rearranged, M = 3RT ÷ uᵣₘₛ² = (3 × 8.314 × 300) ÷ 300² = 7,483 ÷ 90,000 = 0.0831 kg/mol, which is 83.1 g/mol. (b) Kr at 83.80 g/mol is the close match; Xe (131.29) and Rn (222) are far too heavy to give so high a speed at 300 K. (c) Speed scales with √T: √(1,200 ÷ 300) = 2, so uᵣₘₛ = 300 × 2 = 600 m/s. (d) (3/2) × 8.314 × 300 = 3,741 J, about 3.74 kJ per mole. Average kinetic energy is set by the kelvin temperature alone: at 300 K every ideal gas, whatever its particles weigh, has the same average per mole. The three candidates differ in speed, not in average energy.