PhysicsCBSE Class 11 · Kinetic Theory

Pressure of an Ideal Gas, Temperature and RMS Speed Class 11

Updated 17 October 2026 · Fact-checked

The kinetic theory gives the pressure of an ideal gas as P = ⅓nmv²rms = ⅓ρv²rms and shows that the mean translational kinetic energy per molecule is ½mv²rms = (3/2)k_BT, so v_rms = √(3k_BT/m) = √(3RT/M). The pressure of a mixture is the sum of the partial pressures (Dalton), and PV = nRT = Nk_BT.

Understand Pressure of an Ideal Gas, Temperature and RMS Speed

Assumptions. A gas consists of a very large number of identical molecules in random motion, the molecules are point-like (their size is negligible compared with their separation), they exert no forces on each other except during elastic collisions, and collisions with the walls are elastic. The time of a collision is negligible compared with the time between collisions.

Pressure. By computing the momentum delivered to a wall by molecules, $P = \frac{1}{3} nm \overline{v^{2}} = \frac{1}{3} ρ v_{r m s}^{2}$ where $n$ is the number density. Equivalently, $P V = \frac{2}{3} N (\frac{1}{2} m \overline{v^{2}})$ : pressure is two-thirds of the translational kinetic energy per unit volume.

Temperature. Comparing with $P V = N k_{B} T$ :
$ $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T, v_{r m s} = \frac{3 k _{B} T}{m} = \frac{3 R T}{M} = \frac{3 P}{ρ} .$ $
Temperature is proportional to the mean translational kinetic energy of a molecule, and is the same for all gases in thermal equilibrium; at a given $T$ , lighter molecules move faster ( $v_{r m s} \propto 1/ M$ ). The total translational kinetic energy of $n$ moles is $\frac{3}{2} n R T$ . The mean kinetic energy does not depend on the pressure, volume or mass of the gas. Doubling $T$ multiplies $v_{r m s}$ by $2$ .

Ideal gas laws. $P V = n R T = N k_{B} T$ with $R = 8.314 J mol^{- 1} K^{- 1}$ , $k_{B} = R / N_{A} = 1.38 \times 1 0^{- 23}$ J/K, $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ . From it follow Boyle's law, Charles' law and Gay-Lussac's law. Dalton's law: $P = P_{1} + P_{2} + \dots$ with $P_{i} = n_{i} R T / V$ for a non-reacting mixture. Real gases approach ideal behaviour at low pressure and high temperature.

Number density. $n = \frac{P}{k _{B} T}$ , and the typical molecular spacing is about $n^{- 1/3}$ .

Pressure of an Ideal Gas, Temperature and RMS Speed: formulas and key facts

Pressure: $P = \frac{1}{3} nm \overline{v^{2}} = \frac{1}{3} ρ v_{r m s}^{2}$; n is the number per unit volume.
Mean kinetic energy: $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$; Per molecule.
RMS speed: $v_{r m s} = \frac{3 R T}{M} = \frac{3 P}{ρ}$; M in kg/mol.
Ideal gas law: $P V = n R T = N k_{B} T$; R = 8.314 J mol⁻¹ K⁻¹.
Dalton's law: $P = P_{1} + P_{2} + \dots, P_{i} = \frac{n _{i} R T}{V}$; Non-reacting mixture.
Number density: $n = \frac{P}{k _{B} T}$; Molecules per m³.

How to solve Pressure of an Ideal Gas, Temperature and RMS Speed questions

Work in SI units, with the molar mass in kg/mol and T in kelvin; find n from PV = nRT before anything else.

1Convert T to kelvin, V to m³, P to Pa and M to kg/mol.
2Use PV = nRT to get n (or N = nN_A).
3For speeds, use v_rms = √(3RT/M) or √(3P/ρ).
4For energies, use (3/2)k_BT per molecule or (3/2)RT per mole.
5For mixtures, compute each partial pressure from its own n or use the mole fraction.
6Check ratios: v_rms ∝ √(T/M).

Pressure of an Ideal Gas, Temperature and RMS Speed 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.

A helium balloon

Suppose a balloon holds helium (M = 4.0 × 10⁻³ kg/mol) at 300 K.

Find the rms speed of its atoms.

Try it first, then show the solutionSolution

v_rms = √(3RT/M) = √(3 × 8.314 × 300/4.0 × 10⁻³).
= √(1.871 × 10⁶) ≈ 1368 m/s.

Answer: About 1.37 km/s.

Memory hook: Light atoms move fast.

Air in a room

Suppose air in a room is at 27 °C.

Find the average translational kinetic energy per molecule (k_B = 1.38 × 10⁻²³ J/K).

Try it first, then show the solutionSolution

T = 300 K; K = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 300.
= 6.2 × 10⁻²¹ J.

Answer: 6.2 × 10⁻²¹ J.

Memory hook: The same for every gas at that temperature.

Tricks to remember Pressure of an Ideal Gas, Temperature and RMS Speed

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

T/M under the root
v_rms ∝ √(T/M): same ratio, same speed.
Use it when: Comparing gases.
Same T, same KE
All gases at the same T have the same mean translational KE per molecule.
Use it when: Hydrogen vs oxygen.
Two thirds
PV = (2/3) × total translational KE.
Use it when: Pressure from energy.
Mole fractions
Partial pressure = mole fraction × total pressure.
Use it when: Dalton's law.

Pressure of an Ideal Gas, Temperature and RMS Speed: solved examples

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

Example 1: Number of molecules · 2 marks · Very short answerBoard level

Find the number of moles and the number of molecules in 10 L of an ideal gas at 1.0 × 10⁵ Pa and 27 °C (R = 8.314 J/mol K, N_A = 6.022 × 10²³ mol⁻¹).

Try it first, then show the solutionSolution

n = PV/RT = 1.0 × 10⁵ × 10⁻²/(8.314 × 300) = 0.40 mol. (1)
N = nN_A = 0.401 × 6.022 × 10²³ ≈ 2.4 × 10²³. (1)

Answer: 0.40 mol; 2.4 × 10²³ molecules.

Example 2: RMS speed and kinetic energy · 3 marks · Short answerBoard level

Find the rms speed of helium atoms (M = 4.0 × 10⁻³ kg/mol) at 400 K and the mean translational kinetic energy per atom (k_B = 1.38 × 10⁻²³ J/K).

Try it first, then show the solutionSolution

v_rms = √(3RT/M) = √(3 × 8.314 × 400/4.0 × 10⁻³). (1)
= √(2.494 × 10⁶) ≈ 1.58 × 10³ m/s. (1)
K = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 400 = 8.3 × 10⁻²¹ J. (1)

Answer: About 1.58 km/s; 8.3 × 10⁻²¹ J.

Example 3: Why light gases escape · 3 marks · Short answerHarder board level

At what temperature would the rms speed of helium atoms equal the Earth's escape speed of 11.2 km/s (M = 4.0 × 10⁻³ kg/mol)? Comment on its relevance to the Earth's atmosphere.

Try it first, then show the solutionSolution

v² = 3RT/M gives T = Mv²/3R = 4.0 × 10⁻³ × (1.12 × 10⁴)²/(3 × 8.314). (1)
T = 5.02 × 10⁵/24.94 ≈ 2.0 × 10⁴ K. (1)
At ordinary temperatures the rms speed is far below this, but the fast tail of the distribution plus the low mass lets light gases like helium and hydrogen escape over long times, unlike heavier molecules. (1)

Answer: About 2.0 × 10⁴ K.

Example 4: Mixture of gases · 4 marks · Long answerHarder board level

A 10 L vessel at 300 K holds 2.0 g of hydrogen (M = 2 g/mol) and 8.0 g of helium (M = 4 g/mol). Find the total pressure, the partial pressure of each gas and the rms speed of each.

Try it first, then show the solutionSolution

n(H₂) = 1.0 mol, n(He) = 2.0 mol; total 3.0 mol. (1)
P = nRT/V = 3 × 8.314 × 300/10⁻² = 7.48 × 10⁵ Pa. (1)
Partial pressures: H₂ = 2.49 × 10⁵ Pa, He = 4.99 × 10⁵ Pa (mole fractions 1/3 and 2/3). (1)
v_rms(H₂) = √(3RT/0.002) = 1934 m/s; v_rms(He) = √(3RT/0.004) = 1368 m/s. (1)

Answer: 7.48 × 10⁵ Pa; 2.49 × 10⁵ and 4.99 × 10⁵ Pa; 1934 m/s and 1368 m/s.

Example 5: Molecular picture of argon · 4 marks · Long answerHOTS / competency

Argon (M = 0.040 kg/mol) is at 1.0 × 10⁵ Pa and 300 K (k_B = 1.38 × 10⁻²³ J/K). (a) Find its density. (b) Find v_rms in two ways and show that they agree. (c) Find the number density and the average spacing of atoms, n^(−1/3).

Try it first, then show the solutionSolution

(a) ρ = PM/RT = 10⁵ × 0.040/(8.314 × 300) = 1.60 kg/m³. (1)
(b) √(3P/ρ) = √(3 × 10⁵/1.60) = 433 m/s; √(3RT/M) = √(3 × 8.314 × 300/0.040) = 433 m/s. (1)
(c) n = P/k_BT = 10⁵/(1.38 × 10⁻²³ × 300) = 2.4 × 10²⁵ m⁻³. (1)
Spacing = n^(−1/3) = 3.5 × 10⁻⁹ m, many times larger than an atom, as the model assumes. (1)

Answer: (a) 1.60 kg/m³; (b) 433 m/s both ways; (c) 2.4 × 10²⁵ m⁻³ and 3.5 nm.

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

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

Mistake: Using the molar mass in g/mol in v_rms.
Unit mismatch.
Do this: M must be in kg/mol.
Usually costs: The answer mark.
Mistake: Using °C in the gas law.
Not converting to kelvin.
Do this: T(K) = t(°C) + 273.
Usually costs: Whole numerical.
Mistake: Saying hotter gases of the same kind have the same speed.
Missing the T-dependence.
Do this: v_rms ∝ √T.
Usually costs: Concept mark.
Mistake: Believing heavier molecules have more kinetic energy at the same T.
Confusing speed and kinetic energy.
Do this: Mean KE depends only on T.
Usually costs: Reasoning mark.

Exam tips

State "k_B = R/N_A" and the value you use.
Convert all quantities to SI before substituting.
Show v_rms in two forms when asked to verify.
For mixtures, state Dalton's law first.
Keep three significant figures.

Practice questions on Pressure of an Ideal Gas, Temperature and RMS Speed

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 · MCQHarder board level
Gas X (molar mass 4 g/mol) at 200 K and gas Y (16 g/mol) at 800 K are compared. The ratio v_rms(X) : v_rms(Y) is
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4
Hint
Compare T/M.
Answer
(a) 1 : 1, since T/M = 50 for both.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Gas molecules do not lose kinetic energy in collisions with the walls of the container.
Reason (R): The collisions of the molecules with the walls are perfectly elastic.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Elastic collisions conserve KE.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the pressure of 2 mol of an ideal gas at 300 K in a vessel of volume 5 L (R = 8.314 J/mol K).
Hint
P = nRT/V.
Answer
P = 2 × 8.314 × 300/5 × 10⁻³ = 9.98 × 10⁵ Pa ≈ 1.0 × 10⁶ Pa.
Q4 · 3 marks · Short answerHarder board level
A gas at a pressure of 2.0 × 10⁵ Pa has a density of 2.4 kg/m³. Find the rms speed of its molecules and, if its molar mass is 0.032 kg/mol, its temperature.
Hint
v² = 3P/ρ; T = Mv²/3R.
Answer
v = √(3 × 2 × 10⁵/2.4) = 500 m/s; T = 0.032 × 2.5 × 10⁵/(3 × 8.314) ≈ 321 K.
Q5 · 4 marks · Long answerHOTS / competency
Vessel A holds 0.5 mol of hydrogen at 300 K and vessel B (same volume) holds 0.5 mol of oxygen (M = 32 g/mol) at 600 K. Compare (a) their pressures, (b) the mean kinetic energy per molecule, (c) the total translational kinetic energy and (d) the rms speeds of the molecules (M of hydrogen = 2 g/mol).
Hint
Same n and V.
Answer
(a) P ∝ T: P_B/P_A = 2. (b) K ∝ T: 2. (c) Same n, so total K ratio is 2. (d) v ∝ √(T/M): v_B/v_A = √((600/32)/(300/2)) = √0.125 = 0.354.

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 formula?Show answer
⅓ρv²rms.
2.Mean KE per molecule?Show answer
(3/2)k_BT.
3.v_rms?Show answer
√(3RT/M).
4.Dalton's law?Show answer
Total pressure is the sum of partial pressures.
5.k_B?Show answer
1.38 × 10⁻²³ J/K.
6.Number density?Show answer
P/k_BT.

Pressure of an Ideal Gas, Temperature and RMS Speed: frequently asked questions

What does temperature measure in kinetic theory?

The mean translational kinetic energy of a molecule: K = (3/2)k_BT.

Why do lighter molecules move faster at the same temperature?

Because ½mv² is the same, a smaller m means a larger v.

Is v_rms the same as the mean speed?

No; the rms speed is slightly higher than the mean speed.

When does a real gas behave ideally?

At low pressure and high temperature, when molecular size and forces are negligible.

What happens to pressure if the rms speed doubles at constant volume?

It becomes four times, since P ∝ v².