PhysicsISC Class 11 · Kinetic Theory of Gases

Gas Pressure and RMS Speed: What Temperature Means for Molecules

Updated 17 October 2026 · Fact-checked

In the kinetic theory the pressure of an ideal gas is p = ⅓ρc², where c is the rms speed, so c = √(3p/ρ) = √(3RT/M). The mean translational kinetic energy per molecule is (3/2)kT, which gives temperature its kinetic meaning.

Heating makes molecules faster, so c and mean kinetic energy (3/2)kT rise with T, and the harder wall hits give pressure p = ⅓ρc².

Illustration drawn for QuestionBuddy.

Understand Gas Pressure, RMS Speed and Temperature

Assumptions of the model. A gas is a large number of identical, point-like molecules in random motion; they obey Newton's laws, collide elastically with each other and the walls, exert no forces except during collisions, and occupy negligible volume compared with the container.

Pressure. Pressure comes from the change of momentum of molecules striking the walls. For N molecules of mass m in volume V: p = (1/3)(mN/V)c² = ⅓ρc², where c² is the mean of the squares of the speeds and c = √(c²) is the root mean square (rms) speed.

Meaning of temperature. Combining p = ⅓ρc² with pV = NkT gives ½mc² = (3/2)kT: the mean translational kinetic energy per molecule is proportional to the absolute temperature, whatever the gas. For a mole it is (3/2)RT. Heating raises the average kinetic energy; it does not depend on the pressure or the type of gas at a given T.

RMS speed. c = √(3kT/m) = √(3RT/M) = √(3p/ρ). At a given temperature, lighter molecules move faster: c ∝ 1/√M. At a given gas, c ∝ √T and, at a fixed temperature, it does not depend on pressure. The most probable and average speeds are slightly smaller than c, but their ratios are not needed here.

Counting molecules. From pV = NkT, N = pV/(kT). Useful constants (stated in problems): k = 1.38 × 10⁻²³ J K⁻¹, R = 8.31 J mol⁻¹ K⁻¹, N_A = 6.02 × 10²³ mol⁻¹. The total translational kinetic energy of a gas is (3/2)pV.

Gas Pressure, RMS Speed and Temperature: formulas and key facts

Pressure: $p = ⅓ ρ c^{2} ∣ p = ⅓ (N / V) m c^{2}$; c = rms speed.
Kinetic meaning of T: $½ m c^{2} = (3/2) k T$; Per molecule; per mole it is (3/2)RT.
RMS speed: $c = \sqrt (3 R T / M) ∣ c = \sqrt (3 p / ρ)$; M in kg mol⁻¹.
Equation of state: $p V = n R T = N k T$; N = number of molecules.

Gas Pressure, RMS Speed and Temperature in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Speed of oxygen molecules in a room

Suppose the air in a room is at 27 °C (300 K). Treat oxygen (M = 32 g mol⁻¹) as an ideal gas with R = 8.31 J mol⁻¹ K⁻¹.

Find the rms speed of an oxygen molecule.

Try it first, then show the solutionSolution

M = 0.032 kg mol⁻¹.
c = √(3RT/M) = √(3 × 8.31 × 300/0.032) = √(2.337 × 10⁵).
c ≈ 483 m s⁻¹.

Answer: About 483 m s⁻¹.

Memory hook: Convert molar mass to kg mol⁻¹ before using the formula.

Tricks to remember Gas Pressure, RMS Speed and Temperature

Short memory aids. Say each one aloud once, then use it on a question the same day.

Lighter, faster
At the same T, rms speed is inversely proportional to √M.
Use it when: Comparing hydrogen and oxygen.
T is average kinetic energy
Same T means same mean translational KE per molecule for all gases.
Use it when: Statements about two gases at equal temperature.

Gas Pressure, RMS Speed and Temperature: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Dependence of rms speed · 1 mark · MCQBoard level

The rms speed of the molecules of an ideal gas is directly proportional to:
(a) T
(b) √T
(c) 1/√T
(d) T²

Try it first, then show the solutionSolution

c = √(3kT/m), so c ∝ √T.

Answer: (b) √T

Example 2: RMS speed of a gas · 3 marks · Short answerBoard level

Find the rms speed of oxygen molecules (M = 32 g mol⁻¹) at 27 °C. (R = 8.31 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

T = 300 K, M = 0.032 kg mol⁻¹. (0.5 mark)
c = √(3RT/M) = √(3 × 8.31 × 300/0.032). (1 mark)
3 × 8.31 × 300 = 7479; 7479/0.032 = 2.337 × 10⁵. (1 mark)
c = 483 m s⁻¹. (0.5 mark)

Answer: 483 m s⁻¹

Example 3: Matching speeds · 3 marks · Short answerHarder board level

At what temperature will the rms speed of hydrogen (M = 2 g mol⁻¹) equal that of oxygen (M = 32 g mol⁻¹) at 300 K? Also compare their rms speeds at the same temperature.

Try it first, then show the solutionSolution

c ∝ √(T/M); equal speeds need T/M equal. (1 mark)
T_H = 300 × 2/32 = 18.75 K. (1 mark)
At the same T: c_H/c_O = √(32/2) = 4. (1 mark)

Answer: 18.75 K; hydrogen molecules move 4 times faster than oxygen at the same temperature.

Example 4: Kinetic energy of molecules · 4 marks · Short answerHarder board level

Find the average translational kinetic energy of a molecule at 27 °C in joule and electron-volt (1 eV = 1.6 × 10⁻¹⁹ J), and the total translational kinetic energy of one mole (k = 1.38 × 10⁻²³ J K⁻¹, R = 8.31 J mol⁻¹ K⁻¹).

Try it first, then show the solutionSolution

Per molecule: (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300. (1 mark)
= 6.21 × 10⁻²¹ J. (1 mark)
In eV: 6.21 × 10⁻²¹/1.6 × 10⁻¹⁹ = 0.039 eV. (1 mark)
Per mole: (3/2)RT = 1.5 × 8.31 × 300 = 3740 J. (1 mark)

Answer: 6.2 × 10⁻²¹ J (0.039 eV); 3740 J mol⁻¹.

Example 5: Molecules in a vessel · 5 marks · Long answerHOTS / competency

A 2.0 L vessel of an ideal gas is at a pressure of 1.01 × 10⁵ Pa and a temperature of 300 K. The density of the gas is 1.2 kg m⁻³. Find (i) the rms speed, (ii) the rms speed at 600 K, (iii) the number of molecules, and (iv) the total translational kinetic energy.

Try it first, then show the solutionSolution

(i) c = √(3p/ρ) = √(3 × 1.01 × 10⁵/1.2) = √(2.525 × 10⁵) = 502 m s⁻¹. (1 mark)
(ii) c ∝ √T, so c = 502 × √2 = 710 m s⁻¹. (1 mark)
(iii) N = pV/kT = 1.01 × 10⁵ × 2 × 10⁻³/(1.38 × 10⁻²³ × 300) = 202/4.14 × 10⁻²¹. (1 mark)
N = 4.9 × 10²² molecules. (1 mark)
(iv) K = (3/2)pV = 1.5 × 202 = 303 J. (1 mark)

Answer: 502 m s⁻¹; 710 m s⁻¹; 4.9 × 10²² molecules; 303 J.

Where students lose marks in Gas Pressure, RMS Speed and Temperature

The slips examiners see most often, and what to write instead.

Mistake: Molar mass in grams
Using M = 32 instead of 0.032 kg mol⁻¹.
Do this: Convert to kg mol⁻¹ before using c = √(3RT/M).
Usually costs: 2 marks.
Mistake: Celsius temperature
Substituting 27 instead of 300.
Do this: Use kelvin in every gas formula.
Usually costs: 2 marks.
Mistake: Saying rms speed depends on pressure
Claiming c rises with p at constant T.
Do this: At fixed T, c depends only on T and M.
Usually costs: 1 mark.

Exam tips

Convert °C to K and g mol⁻¹ to kg mol⁻¹ first.
State the assumptions of the kinetic theory when asked for the derivation.
Use ratio methods (c ∝ √(T/M)) to avoid arithmetic.
Give the constants k, R, N_A that you use.

Practice questions on Gas Pressure, RMS Speed and Temperature

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At a given temperature, which gas molecules have the highest average translational kinetic energy?
(a) hydrogen
(b) oxygen
(c) nitrogen
(d) all the same
Hint
KE = (3/2)kT for any gas.
Answer
(d) all the same
Q2 · 2 marks · Short answerBoard level
A gas of density 0.09 kg m⁻³ is at a pressure of 97 200 Pa. Find the rms speed of its molecules.
Hint
c = √(3p/ρ).
Answer
c = √(3 × 97 200/0.09) = √(3.24 × 10⁶) = 1800 m s⁻¹.
Q3 · 3 marks · Short answerHarder board level
By what factor must the kelvin temperature of a gas be raised to double its rms speed? If the gas is at 27 °C, what is the new temperature in °C?
Hint
c ∝ √T.
Answer
T must be multiplied by 4. 300 K → 1200 K = 927 °C.
Q4 · 3 marks · Short answerHarder board level
How many molecules are there in 1 cm³ of an ideal gas at 1.01 × 10⁵ Pa and 273 K? (k = 1.38 × 10⁻²³ J K⁻¹)
Hint
N = pV/kT.
Answer
N = 1.01 × 10⁵ × 10⁻⁶/(1.38 × 10⁻²³ × 273) = 0.101/3.77 × 10⁻²¹ = 2.7 × 10¹⁹.
Q5 · 3 marks · Short answerHarder board level
Find the ratio of the rms speeds of hydrogen (M = 2) and helium (M = 4) at the same temperature, and of the same gas at 100 K and 400 K.
Hint
c ∝ √(T/M).
Answer
c_H/c_He = √(4/2) = 1.41. For the same gas, c₄₀₀/c₁₀₀ = √4 = 2.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a sealed cylinder contains helium (M = 4 g mol⁻¹) at 300 K (R = 8.31 J mol⁻¹ K⁻¹).
(i) Find the rms speed of its atoms. (1 mark)
(ii) Find the mean translational kinetic energy per atom (k = 1.38 × 10⁻²³ J K⁻¹). (1 mark)
(iii) Find the temperature at which the rms speed would equal 11.2 km s⁻¹. (1 mark)
(iv) Explain why very light gases tend to leak away from a planet's atmosphere more easily than heavy ones. (1 mark)
Hint
c = √(3RT/M); T = Mc²/3R.
Answer
(i) c = √(3 × 8.31 × 300/0.004) = 1367 m s⁻¹. (ii) 6.2 × 10⁻²¹ J. (iii) T = 0.004 × (11 200)²/(3 × 8.31) = 2.0 × 10⁴ K. (iv) At the same temperature, lighter molecules move faster, so a larger share of them reach speeds near the escape velocity.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Pressure of a gas in terms of density and rms speed?Show answer
p = ⅓ρc².
2.Mean KE per molecule?Show answer
(3/2)kT.
3.RMS speed in terms of T and M?Show answer
√(3RT/M).
4.Does c depend on the pressure at constant temperature?Show answer
No.

Gas Pressure, RMS Speed and Temperature: frequently asked questions

Why do all molecules not have the same speed?

Collisions constantly exchange energy, so the molecules have a spread of speeds; the rms speed is a particular average of the distribution.

Does the average velocity of the molecules equal the rms speed?

No. The average velocity is zero in a gas at rest, because molecules move equally in all directions. The rms speed is not zero.

Is temperature defined for a single molecule?

No. Temperature is a property of a large collection; for a single molecule only its kinetic energy has meaning.