Questions on the kinetic theory model and its derivation

Question 1

State two assumptions of the kinetic theory model of an ideal gas.

Accepted Answer

Any two of the following, one mark each (2): molecules are in constant random motion; collisions with the walls and each other are perfectly elastic; the volume of the molecules is negligible compared with the container; the time of collisions is negligible; and there are no forces between molecules except during collisions.

Question 2

Calculate the average kinetic energy of a gas molecule at 300 K, using mean KE = (3/2)kT (k = 1.38 × 10⁻²³ J K⁻¹).

Accepted Answer

Mean KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 (1) Mean KE = 6.21 × 10⁻²¹ J (1)

Question 3

State the kinetic theory equation relating pressure, volume and molecular speed.

Accepted Answer

pV = ⅓Nm(c_rms)² (1), where N is the number of molecules, m the mass of one molecule and c_rms the root-mean-square speed (1).

Question 4

Smoke particles suspended in air are viewed through a microscope. Describe what is observed, and state what this observation is evidence for.

Accepted Answer

The smoke particles move continually in random, jerky paths (Brownian motion) (1). This is evidence that air consists of molecules in rapid random motion, bombarding the smoke particles unevenly (1).

Question 5

State what is meant by the root-mean-square speed of the molecules of a gas.

Accepted Answer

The square root of the mean of the squares of the molecular speeds (1).

Question 6

At one instant, four gas molecules have speeds of 300 m s⁻¹, 400 m s⁻¹, 500 m s⁻¹ and 600 m s⁻¹. Calculate their root-mean-square speed.

Accepted Answer

Mean square speed = (300² + 400² + 500² + 600²)/4 = 2.15 × 10⁵ m² s⁻² (1) c_rms = √(2.15 × 10⁵) = 464 m s⁻¹ (1)

Question 7

Calculate the total kinetic energy of one mole of an ideal gas at 400 K, using R = 8.31 J mol⁻¹ K⁻¹.

Accepted Answer

Each molecule carries (3/2)kT on average, so a mole carries (3/2)RT (1) = 1.5 × 8.31 × 400 (1) = 4986 J ≈ 5.0 × 10³ J (1)

Question 8

A nitrogen molecule has a mass of 4.7 × 10⁻²⁶ kg. Calculate its root-mean-square speed at 300 K, using ½m(c_rms)² = (3/2)kT.

Accepted Answer

c_rms = √(3kT/m) (1) = √((3 × 1.38 × 10⁻²³ × 300)/(4.7 × 10⁻²⁶)) (1) c_rms = 514 m s⁻¹ (1)

Question 9

A gas of density 1.2 kg m⁻³ has molecules with an rms speed of 500 m s⁻¹. Calculate its pressure, using p = ⅓ρ(c_rms)².

Accepted Answer

p = ⅓ρ(c_rms)² (1) = (1/3) × 1.2 × 500² (1) p = 1.00 × 10⁵ Pa (1)

Question 10

At the same temperature, explain which has the greater root-mean-square speed: a light molecule or a heavy one.

Accepted Answer

The light molecule (1). At a given temperature all molecules have the same average kinetic energy (3/2)kT, so ½m(c_rms)² is the same (1); a smaller mass m therefore requires a larger c_rms, since c_rms ∝ 1/√m (1).

Question 11

A nitrogen molecule of mass 4.65 × 10⁻²⁶ kg strikes a container wall head-on at 480 m s⁻¹ and rebounds along the same line at the same speed. Calculate the change in momentum of the molecule. The molecule returns and strikes the same wall once every 1.2 ms. Go on to calculate the average force it exerts on the wall.

Accepted Answer

Δp = 2mv = 2 × 4.65 × 10⁻²⁶ × 480 (1) Δp = 4.46 × 10⁻²³ kg m s⁻¹ (1) F = Δp/t = 4.46 × 10⁻²³/(1.2 × 10⁻³) = 3.7 × 10⁻²⁰ N (1)

Question 12

A container holds a mixture of helium (molecular mass 6.6 × 10⁻²⁷ kg) and argon (molecular mass 6.6 × 10⁻²⁶ kg) at the same temperature. State the ratio of the mean kinetic energy of a helium molecule to that of an argon molecule, and calculate the ratio c_rms(helium)/c_rms(argon).

Accepted Answer

The mean kinetic energies are equal (ratio 1), because mean KE = (3/2)kT depends only on temperature (1)
Equal ½m(c_rms)² gives c_rms ∝ 1/√m, so the ratio = √(m_Ar/m_He) = √10 (1)
Ratio = 3.2 (1)

Question 13

A rigid container of gas is sealed and kept at constant temperature. Explain, with reference to one assumption of the kinetic theory model, why the pressure does not decrease over time.

Accepted Answer

Collisions of molecules with the walls (and with each other) are perfectly elastic, so no kinetic energy is lost in collisions (1). The molecules therefore keep the same mean square speed, the rate of change of momentum at the walls is unchanged, and the pressure stays constant (1).

Question 14

For an ideal gas, the mean translational kinetic energy of one molecule is plotted against thermodynamic temperature. The graph is a straight line through the origin. State the quantity the gradient of the line represents and calculate its value, then explain what it shows that the same line is obtained for every ideal gas (k = 1.38 × 10⁻²³ J K⁻¹).

Accepted Answer

Mean kinetic energy = (3/2)kT, so the gradient is 3k/2 (1)
Gradient = 1.5 × 1.38 × 10⁻²³ = 2.07 × 10⁻²³ J K⁻¹ (1)
One line for every gas shows that the mean translational kinetic energy per molecule is set by the absolute temperature alone, and does not depend on which gas it is or on the mass of its molecules (1)

Question 15

An oxygen molecule in air has a mass of 5.3 × 10⁻²⁶ kg. Calculate (a) its mean kinetic energy at 290 K and (b) its root-mean-square speed.

Accepted Answer

(a) Mean KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 290 (1)
Mean KE = 6.0 × 10⁻²¹ J (1)
(b) c_rms = √(3kT/m) = √((3 × 1.38 × 10⁻²³ × 290)/(5.3 × 10⁻²⁶)) (1)
c_rms = 476 m s⁻¹ (1)

Question 16

A gas is at 300 K. Calculate the temperature to which it must be raised to double the root-mean-square speed of its molecules.

Accepted Answer

c_rms ∝ √T (1) Doubling c_rms therefore needs four times the absolute temperature (1) T = 4 × 300 = 1200 K (1)

Question 17

Explain how the kinetic theory model accounts for the pressure a gas exerts on its container.

Accepted Answer

Molecules move randomly and repeatedly collide with the walls (1). Each collision changes a molecule's momentum, so by Newton's laws the wall exerts a force on it and it exerts an equal force back (1). The many collisions per second over the wall area produce a steady average pressure (1).

Question 18

A cubical box of side L contains N identical molecules, each of mass m, in random motion. Starting from the change in momentum when one molecule collides elastically with a wall, derive the equation pV = ⅓Nm(c_rms)².

Accepted Answer

One molecule moving at velocity component v_x towards a wall rebounds elastically, a change in momentum of 2mv_x (1). It returns after travelling 2L, so it strikes the wall every 2L/v_x seconds, giving an average force 2mv_x/(2L/v_x) = mv_x²/L (1). Summing over all N molecules and replacing v_x² by its mean gives a total force Nm(v_x²)_mean/L (1). The motion is random, so the mean square speed is shared equally among three directions: (v_x²)_mean = ⅓(c_rms)² (1). Pressure = force/L² and V = L³, so pV = ⅓Nm(c_rms)² (1).

Question 19

The escape speed from the Earth is 11 km s⁻¹. Calculate the temperature at which hydrogen molecules (mass 3.3 × 10⁻²⁷ kg) would have a root-mean-square speed equal to the escape speed (k = 1.38 × 10⁻²³ J K⁻¹). The temperature of the upper atmosphere is only a few hundred kelvin, yet it contains almost no hydrogen. Suggest why.

Accepted Answer

From ½m(c_rms)² = (3/2)kT: T = m(c_rms)²/(3k) (1)
T = 3.3 × 10⁻²⁷ × (1.1 × 10⁴)²/(3 × 1.38 × 10⁻²³) = 9.6 × 10³ K (1)
The molecular speeds are spread about c_rms, so at a few hundred kelvin a small fraction of hydrogen molecules still move faster than the escape speed (1)
These fast molecules escape, and over a long time the hydrogen is lost (1)

Question 20

A cylinder holds 0.020 m³ of helium, an ideal gas, at a pressure of 2.5 × 10⁵ Pa. Explain why the internal energy of the helium is entirely kinetic, and calculate this internal energy.

Accepted Answer

An ideal gas has no intermolecular forces (except in collisions), so there is no potential-energy contribution to the internal energy (1)
Total kinetic energy = N × (3/2)kT = (3/2)pV, using pV = NkT (1)
U = 1.5 × 2.5 × 10⁵ × 0.020 = 7500 J (1)