How do billions of randomly moving molecules create the smooth pressure we measure? Kinetic theory builds a bridge between the chaotic microscopic world and the ordered macroscopic gas laws — one of physics' most elegant derivations.
💥Derive the pressure of an ideal gas: pV = ⅓Nm<c²>
📐Define and calculate rms speed c_rms = √<c²>
🌡️Show that mean KE = ½m<c²> = 3kT/2, linking temperature to molecular motion
📊Describe and interpret the Maxwell-Boltzmann speed distribution
🔺Explain how the distribution shifts with temperature
🔗Link kinetic theory to the ideal gas equation pV = NkT
Deriving Gas Pressure from Molecular Motion
Consider N identical molecules, each of mass m, in a cubic box of side length L (volume V = L³). We derive the pressure on one wall.
Consider one molecule with velocity component u in the x-direction:
It hits the right wall with momentum mu, bounces elastically, so the change in momentum = 2mu (toward wall is positive).
Time between successive hits on the same wall = 2L/u (travels to far wall and back).
Force on wall from this molecule: F = Δp/Δt = 2mu/(2L/u) = mu²/L
Summing over all N molecules: F_total = (m/L)Σuᵢ² = (m/L)N⟨u²⟩
Pressure on the wall: p = F/A = (m/L)N⟨u²⟩/L² = Nm⟨u²⟩/L³ = Nm⟨u²⟩/V
By symmetry, ⟨u²⟩ = ⟨v²⟩ = ⟨w²⟩ = ⟨c²⟩/3 (where ⟨c²⟩ is mean square speed in 3D).
pV = ⅓Nm⟨c²⟩
Mean square speed ⟨c²⟩: The average of the squares of all molecular speeds. Note: ⟨c²⟩ ≠ ⟨c⟩² — the mean square speed is NOT the square of the mean speed.
RMS Speed and Mean Kinetic Energy
The root-mean-square (rms) speed c_rms is defined as:
c_rms = √⟨c²⟩
This is the square root of the mean square speed. It is the most useful measure of molecular speed because it appears directly in kinetic energy and pressure equations.
Comparing pV = ⅓Nm⟨c²⟩ with pV = NkT (ideal gas equation):
⅓Nm⟨c²⟩ = NkT → ½m⟨c²⟩ = 3kT/2
Mean KE per molecule = ½m⟨c²⟩ = 3kT/2
This is a profound result: temperature is a direct measure of the mean kinetic energy of molecules. The mean KE depends only on T, not on the type of gas. At the same temperature, molecules of different masses have the same mean KE but different rms speeds.
From ½m⟨c²⟩ = 3kT/2: solving for rms speed:
c_rms = √(3kT/m) = √(3RT/M)
where M is the molar mass (kg mol⁻¹). Lighter molecules have higher rms speeds at the same temperature.
Gas
Molar mass M (g mol⁻¹)
c_rms at 300 K (m s⁻¹)
Hydrogen H₂
2.0
1934
Helium He
4.0
1368
Nitrogen N₂
28
517
Oxygen O₂
32
484
Carbon dioxide CO₂
44
412
Maxwell-Boltzmann Speed Distribution
Individual molecules have a wide range of speeds — they continuously exchange energy through collisions. The Maxwell-Boltzmann distribution f(c) describes the fraction of molecules with speeds between c and c + dc.
Key features of the distribution curve (f(c) vs c):
Most probable speed c_p: at the peak of the distribution. This is the speed of the largest fraction of molecules. c_p = √(2kT/m) = √(2RT/M)
Mean speed ⟨c⟩: slightly higher than c_p. ⟨c⟩ = √(8kT/πm)
RMS speed c_rms: highest of the three. c_rms = √(3kT/m)
Ordering: c_p < ⟨c⟩ < c_rms always.
Effect of increasing temperature:
The peak shifts to the right (higher most probable speed)
The peak height decreases (the distribution broadens)
The area under the curve remains constant (total number of molecules fixed)
More molecules have high speeds (right-hand tail extends further)
The high-speed tail of the Maxwell-Boltzmann distribution is crucial in chemistry: it represents the molecules with enough energy to overcome activation energy barriers. This is why reaction rates increase dramatically with small temperature rises — the area of the high-energy tail grows disproportionately.
Linking Kinetic Theory to Gas Properties
From pV = ⅓Nm⟨c²⟩ and ½m⟨c²⟩ = 3kT/2, the total internal energy U of an ideal monatomic gas is:
U = N × (3kT/2) = (3/2)NkT = (3/2)nRT
This means for an ideal gas, U depends only on temperature. This is consistent with the first law: in an isothermal process (ΔT = 0), ΔU = 0 and Q = −W.
This shows pressure increases with both number density (more molecules) and temperature (faster molecules → harder/more frequent impacts).
⚠️ The kinetic theory derivation assumes: (1) molecules move randomly in all directions; (2) elastic collisions; (3) no intermolecular forces between collisions; (4) molecular volume is negligible. These are identical to the ideal gas assumptions.
Why ⟨u²⟩ = ⟨c²⟩/3: By the isotropy of molecular motion (no preferred direction), the mean square velocity components in x, y and z are equal: ⟨u²⟩ = ⟨v²⟩ = ⟨w²⟩. Since ⟨c²⟩ = ⟨u²⟩ + ⟨v²⟩ + ⟨w²⟩ = 3⟨u²⟩, we get ⟨u²⟩ = ⟨c²⟩/3.
Calculate the rms speed of nitrogen molecules (M = 28 × 10⁻³ kg mol⁻¹) at 300 K and at 1200 K.
4Note: T × 4 → c_rms × 2. This confirms c_rms ∝ √T.
At 300 K: c_rms = 517 m s⁻¹; at 1200 K: c_rms = 1034 m s⁻¹. Doubling rms speed requires quadrupling temperature.
A container holds 3.0 × 10²³ nitrogen molecules (m = 4.65 × 10⁻²⁶ kg each) with mean square speed 5.0 × 10⁵ m² s⁻². The container volume is 2.0 × 10⁻³ m³. Calculate (a) the pressure, (b) the temperature.
4p = 6975 / (6.0×10⁻³) = 6975/0.006 = 1.163 × 10⁶ Pa ≈ 1.16 × 10⁶ Pa
5Temperature: ½m⟨c²⟩ = 3kT/2 → T = m⟨c²⟩/(3k) = (4.65×10⁻²⁶ × 5.0×10⁵)/(3 × 1.38×10⁻²³) = 2.325×10⁻²⁰/(4.14×10⁻²³) = 561.6 K ≈ 562 K
p = 1.16 × 10⁶ Pa; T ≈ 562 K (289°C)
Compare the rms speeds of hydrogen (H₂, M = 2.0 × 10⁻³ kg mol⁻¹) and oxygen (O₂, M = 32 × 10⁻³ kg mol⁻¹) at the same temperature. What ratio of temperatures would be needed for oxygen molecules to have the same rms speed as hydrogen at 300 K?
1Since c_rms ∝ √(1/M) at constant T: ratio = c_H₂/c_O₂ = √(M_O₂/M_H₂) = √(32/2) = √16 = 4
2Hydrogen molecules are 4 times faster than oxygen at the same T.
3For O₂ at T₂ to equal H₂ at 300 K: √(3RT₂/M_O₂) = √(3R×300/M_H₂)
H₂ is 4 × faster than O₂ at same T. O₂ needs T = 4800 K to match H₂ at 300 K.
Estimate the mean kinetic energy of an air molecule at room temperature (20°C). If air has mean molar mass ~29 g mol⁻¹, find the rms speed of air molecules.
Mean KE ≈ 6.1 × 10⁻²¹ J; c_rms ≈ 502 m s⁻¹ (~1800 km h⁻¹ — much faster than sound!)
Q1. In the derivation of gas pressure, why is the change in momentum of a molecule bouncing off a wall equal to 2mu (not mu)?
Q2. What does the expression ½m⟨c²⟩ = 3kT/2 tell us about temperature?
Q3. A gas temperature is doubled (in Kelvin). By what factor does the rms speed change?
Q4. On a Maxwell-Boltzmann distribution, which speed is highest?
Q5. The mean KE of a gas molecule at 300 K is approximately:
Challenge 1. Using pV = ⅓Nm⟨c²⟩ and pV = NkT, derive the expression for mean kinetic energy per molecule. Then show that the total internal energy of n moles of a monatomic ideal gas is U = (3/2)nRT.
✓ From pV = ⅓Nm⟨c²⟩ and pV = NkT: equating: NkT = ⅓Nm⟨c²⟩. Dividing both sides by N: kT = ⅓m⟨c²⟩. Multiplying both sides by 3/2: (3/2)kT = ½m⟨c²⟩. Therefore mean KE per molecule = ½m⟨c²⟩ = (3/2)kT. QED. Total internal energy: for N molecules, U = N × ½m⟨c²⟩ = N × (3/2)kT = (3/2)NkT. Since N = nN_A and k = R/N_A: U = (3/2)(nN_A)(R/N_A)T = (3/2)nRT. QED. Note: the factor of 3 comes from 3 degrees of translational freedom for a monatomic molecule. Diatomic molecules also have rotational degrees of freedom, giving U = (5/2)nRT at room temperature.
Challenge 2. The escape velocity from Earth is 11.2 km s⁻¹. Calculate the temperature at which the rms speed of hydrogen (H₂, M = 2.0 × 10⁻³ kg mol⁻¹) would equal the escape velocity. Does this explain why Earth has very little hydrogen in its atmosphere?
✓ c_rms = v_escape: √(3RT/M) = v_escape → T = Mv²/(3R) = (2.0×10⁻³ × (11200)²)/(3 × 8.314) = (2.0×10⁻³ × 1.2544×10⁸)/24.942 = (2.5088×10⁵)/24.942 = 10 058 K ≈ 10 000 K. At room temperature (~300 K), c_rms(H₂) ≈ 1934 m s⁻¹ — well below escape velocity. However, the Maxwell-Boltzmann distribution has a high-speed tail. Even at 300 K, a small fraction of H₂ molecules have speeds exceeding 11.2 km s⁻¹. These molecules escape to space over geological timescales (this is called Jeans escape). Over billions of years, Earth has lost most of its primordial hydrogen this way. Jupiter and Saturn (much more massive, higher escape velocity ~60 km s⁻¹) have retained their hydrogen. This also explains why helium (light, fast) is rare in Earth's atmosphere despite being produced by radioactive decay.
Challenge 3. Show that the pressure of an ideal gas can be written as p = (1/3)ρ⟨c²⟩ where ρ is the gas density. Then use this to estimate the rms speed of air molecules (density 1.29 kg m⁻³ at STP, pressure 1.013 × 10⁵ Pa) without using molecular masses.
✓ From pV = ⅓Nm⟨c²⟩: p = (Nm/V)(⟨c²⟩/3) = ρ⟨c²⟩/3. (Since Nm = total mass and V = volume, Nm/V = ρ.) QED. Rearranging for c_rms: c_rms = √(3p/ρ) = √(3 × 1.013×10⁵ / 1.29) = √(3.039×10⁵ / 1.29) = √(2.356×10⁵) = 485 m s⁻¹. Comparison with our earlier result (502 m s⁻¹ for M = 29 g mol⁻¹): the small difference arises from using STP density (0°C) vs 20°C. This method is experimentally accessible — you only need to measure pressure and density, without knowing individual molecular masses.
Challenge 4 (Synoptic — Chemistry link). The activation energy for a certain reaction is E_a = 8.0 × 10⁻²⁰ J. At 300 K, what fraction of molecules have kinetic energy exceeding E_a? (Use Boltzmann factor: fraction ∝ e^(−E_a/kT).) How many times more molecules exceed E_a at 310 K? Explain the significance for reaction rate.
✓ At 300 K: E_a/kT = 8.0×10⁻²⁰ / (1.38×10⁻²³ × 300) = 8.0×10⁻²⁰ / 4.14×10⁻²¹ = 19.32. Fraction ∝ e^(−19.32) = 4.04×10⁻⁹. At 310 K: E_a/kT = 8.0×10⁻²⁰ / (1.38×10⁻²³ × 310) = 8.0×10⁻²⁰ / 4.278×10⁻²¹ = 18.69. Fraction ∝ e^(−18.69) = 7.64×10⁻⁹. Ratio: 7.64/4.04 = 1.89 — nearly double! A 10 K rise doubles the fraction of molecules with E > E_a, roughly doubling the reaction rate. This is the basis of the Arrhenius equation and explains the rule of thumb that reaction rates double for every 10°C rise. Physically: even though the mean KE only increases by ~3% (3k×10/2 per molecule), the exponential tail of the Maxwell-Boltzmann distribution means the high-energy fraction is extremely sensitive to temperature.