MHT-CET Physics · Kinetic Theory of Gases

Pressure and R.M.S. Speed From Molecular Motion

Molecules striking the walls give a pressure P = ⅓ρ⟨v²⟩, which is two thirds of the kinetic energy per unit volume; since that energy is fixed by temperature, the r.m.s. speed is √(3RT/M), rising as √T and falling as 1/√M.

All Kinetic Theory of Gases notes MHT-CET Physics strategy

Why this matters

36 PYQs, 8 of them HARD. Thirteen are about pressure itself — the assumptions of kinetic theory, pressure as two thirds of the kinetic energy density, the number of molecules from their energy, the mean free path. Twenty scale the r.m.s. speed with temperature or molar mass, and include the speed of sound. Three chain it with an adiabatic expansion. Three cards.

Concept 1 of 3: Pressure as Two Thirds of the Kinetic Energy Density

Kinetic theory assumes identical point molecules that exert no force except in perfectly elastic collisions, so momentum and kinetic energy are both conserved. Their collisions with the WALLS — not with each other — make the pressure P = ⅓ρ⟨v²⟩ = ⅓(Nm/V)⟨v²⟩. Since the kinetic energy per unit volume is ½ρ⟨v²⟩, P = (2/3)(E/V). So pressure is proportional to the mean of the square of the speed. Halving each molecule's mass and doubling its speed doubles mv², and with it the pressure. From PV = NkT and E = (3/2)kT per molecule, the number of molecules is N = 3PV/(2E). The mean free path, the average distance between collisions, depends on how many molecules share each unit volume: at constant volume it does not change with temperature.

Definition

Assumptions: identical molecules, elastic collisions, no force except in collision; pressure from wall collisions.
$P = \tfrac{1}{3}\rho\langle v^2\rangle = \tfrac{2}{3}\dfrac{E}{V}$ (7500 J in 10 L ⇒ $5\times10^5$ Pa).
$P \propto m\langle v^2 \rangle$ : m halved, v doubled ⇒ 2P.
Molecules from energy: $N = \dfrac{3PV}{2E}$ .
Mean free path $\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$ : unchanged when heated at constant volume.
$T \propto$ mean square velocity.

Kinetic pressure

P = \frac{1}{3}\rho\langle v^2\rangle = \frac{2}{3}\frac{E}{V}

Worked example

A 1 L vessel holds gas at 1.5 × 10⁵ Pa. Its total translational kinetic energy?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q41Hard

Example 1 · Kinetic Theory of Gases · Kinetic Theory — Pressure, RMS Speed, and Temperature

An ideal gas in a container of volume 500 cc is at a pressure of

2 \times 10^5

N/m

^2

. The average kinetic energy of each molecule is

6 \times 10^{-21}

J. The number of gas molecules in the container is

Writing P = ⅓ of the energy density

P = ⅓ρv², but the kinetic energy density is ½ρv². So P is TWO thirds of the energy per unit volume.

Heating a gas at constant volume changes its mean free path

The mean free path depends on molecules per unit volume. In a rigid vessel that number does not change, so neither does the mean free path.

Drill 12 more on pressure as two thirds of the kinetic energy density

Concept 2 of 3: R.M.S. Speed, Temperature and Molar Mass

Equal average energy for all molecules at one temperature means lighter molecules move faster: v_rms = √(3RT/M) = √(3kT/m). Quadruple the kelvin temperature and the speed doubles; an isothermal compression leaves it alone. Two gases have the same r.m.s. speed when T/M is the same. Sound travels at √(γRT/M), a similar form with γ in place of 3, so its speed ratio between two gases at the same temperature is √(γ₁M₂/γ₂M₁).

Definition

$v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$ ; $v^2/T$ constant; isothermal ⇒ unchanged.
Doubling v needs 4T: −68 °C ⇒ 547 °C; 27 °C ⇒ 927 °C.
Same speed: $\dfrac{T_1}{M_1} = \dfrac{T_2}{M_2}$ (He at 57 °C ↔ O₂ at 2640 K).
At constant P, doubling v means 4T and so 4V.
Sound: $v = \sqrt{\dfrac{\gamma RT}{M}}$ (H₂ against He ⇒ $\dfrac{\sqrt{42}}{5}$ ).

R.M.S. speed

v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}

Worked example

Hydrogen molecules (M = 2 g/mol) at 300 K: r.m.s. speed?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q18Moderate

Example 2 · Kinetic Theory of Gases · Kinetic Theory — Pressure, RMS Speed, and Temperature

The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at

57^{\circ}C

is (molecular masses of oxygen and helium are 32 and 4 respectively.)

Scaling speed with temperature in °C

−68 °C is 205 K; doubling the speed needs 820 K = 547 °C. Multiplying −68 by 4 gives nonsense.

Scaling speed with T instead of √T

v_rms ∝ √T. Four times the temperature doubles the speed; the options offer 4x as well.

Drill 19 more on r.m.s. speed, temperature and molar mass

Concept 3 of 3: R.M.S. Speed Through an Adiabatic Expansion

To cut the r.m.s. speed k times, the temperature must fall k² times. In an adiabatic expansion TV^(γ−1) is constant, so the volume must grow by (k²)^(1/(γ−1)). With γ = 1.5 that exponent is 2, and the volume grows by k⁴: halving the speed needs 16 times the volume, cutting it three times needs 81.

Definition

Speed ÷ k ⇒ T ÷ k² ⇒ $V \times k^{2/(\gamma - 1)}$ .
γ = 1.5: speed ÷ 2 ⇒ V × 16; ÷ 3 ⇒ V × 81; ÷ 4 ⇒ V × 256.

Adiabatic cooling

\frac{V_2}{V_1} = \left(\frac{T_1}{T_2}\right)^{1/(\gamma - 1)}, \qquad \frac{T_1}{T_2} = \left(\frac{v_1}{v_2}\right)^2

Worked example

A gas with γ = 1.5 expands adiabatically until its r.m.s. speed halves. By what factor has its volume grown?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q8Hard

Example 3 · Kinetic Theory of Gases · Kinetic Theory — Pressure, RMS Speed, and Temperature

An ideal gas expands adiabatically

(\gamma = 1.5)

. To reduce the r.m.s velocity of the molecules 3 times, the gas has to be expanded

Using the speed ratio as the temperature ratio

Speed goes as √T, so halving the speed quarters the temperature. Feeding 2 instead of 4 into the adiabatic relation gives 4 instead of 16.

Drill 2 more on r.m.s. speed through an adiabatic expansion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Pressure as Two Thirds of the Kinetic Energy Density
Kinetic pressure
$P = \frac{1}{3}\rho\langle v^2\rangle = \frac{2}{3}\frac{E}{V}$
R.M.S. Speed, Temperature and Molar Mass
R.M.S. speed
$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}$
R.M.S. Speed Through an Adiabatic Expansion
Adiabatic cooling
$\frac{V_2}{V_1} = \left(\frac{T_1}{T_2}\right)^{1/(\gamma - 1)}, \qquad \frac{T_1}{T_2} = \left(\frac{v_1}{v_2}\right)^2$

Watch out for (5)

Writing P = ⅓ of the energy density
→ Pressure as Two Thirds of the Kinetic Energy Density
Heating a gas at constant volume changes its mean free path
→ Pressure as Two Thirds of the Kinetic Energy Density
Scaling speed with temperature in °C
→ R.M.S. Speed, Temperature and Molar Mass
Scaling speed with T instead of √T
→ R.M.S. Speed, Temperature and Molar Mass
Using the speed ratio as the temperature ratio
→ R.M.S. Speed Through an Adiabatic Expansion

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