Rms Speed Gas

8 MCQs2 revision cards9-step worked example

Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap first: when temperature doubles, students reflexively double v_rms. That loses you 4 marks before you blink. The RMS speed does not scale linearly with temperature — it scales with the square root.

The formula. From kinetic theory (NCERT Class 11 Physics, Chapter 13, page 6), the root-mean-square speed of gas molecules is:

v_rms = √(3RT/M) = √(3kT/m)

where R is the gas constant (8.314 J mol⁻¹ K⁻¹), T is absolute temperature in kelvin, M is molar mass in kg/mol, k is Boltzmann's constant (1.38 × 10⁻²³ J/K), and m is the mass of a single molecule in kg.

What the formula actually says. v_rms ∝ √T and v_rms ∝ 1/√M. Two consequences that NEET tests directly:

Temperature scaling: to double v_rms, you must quadruple T (since √4 = 2). Tripling v_rms requires T to increase by a factor of 9.
Mass dependence: lighter molecules move faster at the same temperature. Hydrogen (M = 2 × 10⁻³ kg/mol) has a higher v_rms than oxygen (M = 32 × 10⁻³ kg/mol) at identical T.

Molar mass unit trap. The formula requires M in kg/mol, not g/mol. Forgetting to convert (e.g., using 32 instead of 0.032 for O₂) inflates v_rms by a factor of √1000 ≈ 31.6 — an obviously wrong answer, but under exam pressure, students pick the distractor that matches their miscalculation.

NEET bridge. Questions typically give an initial temperature and ask for the new temperature (or new v_rms) after a stated change. The pattern: set up the ratio v₂/v₁ = √(T₂/T₁), square both sides, solve. No calculator needed — the arithmetic is designed to come out clean.

Watch-out: always convert °C to K before substituting. Using 27 instead of 300 K is a common route to the wrong option.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The RMS speed of molecules of an ideal gas is proportional to:

MCQ 2Easy RecallPractice

The SI unit of molar mass M used in the formula v_rms = √(3RT/M) is:

MCQ 3Easy RecallPractice

To double the RMS speed of gas molecules, the absolute temperature must be increased by a factor of:

MCQ 4Direct ApplicationPractice

The RMS speed of oxygen molecules (M = 32 × 10⁻³ kg/mol) at 300 K is v₀. The RMS speed of hydrogen molecules (M = 2 × 10⁻³ kg/mol) at the same temperature is:

MCQ 5Direct ApplicationPractice

The temperature of an ideal gas is increased from 27°C to 327°C. The ratio of the new RMS speed to the original RMS speed is:

MCQ 6Direct ApplicationPractice

The RMS speed of gas molecules at temperature T is 200 m/s. At what temperature will the RMS speed become 400 m/s?

MCQ 7CalculationPractice

An ideal gas is at temperature 300 K. If the RMS speed of its molecules must increase to 3 times the original value, the final temperature is:

MCQ 8CalculationPractice

Two gases X and Y have molar masses M and 4M respectively. If gas X is at 200 K and gas Y is at T_Y, and both have the same RMS speed, then T_Y is:

Quick recall before you leave

Worked Example

Pattern: NEET pattern: rms speed temp scaling (PYQ 2023, medium difficulty, multi-step)

1
Given
The RMS speed of molecules of a gas at temperature T₁ = 127°C is v₁. We need the temperature at which the RMS speed becomes 2v₁.
2
Required
Find T₂ such that v_rms = 2v₁.
3
Concept
v_rms ∝ √T for a given gas (fixed M). The ratio of RMS speeds at two temperatures gives v₂/v₁ = √(T₂/T₁), where T must be in kelvin.
4
Formula
v₂/v₁ = √(T₂/T₁) Squaring: T₂ = T₁ × (v₂/v₁)²
5
Substitution
First, convert temperature: T₁ = 127 + 273 = 400 K. The speed ratio: v₂/v₁ = 2. T₂ = 400 × (2)² = 400 × 4
6
Calculation
T₂ = 1600 K Converting back (if required): T₂ = 1600 − 273 = 1327°C. Note: the factor 2 (speed ratio) and the squaring exponent 2 are exact integers. The addition of 273 for °C-to-K conversion is an exact defined offset. These do not limit significant figures.
7
Final answer
T₂ = 1600 K (or 1327°C).
8
Common trap
The linear-scaling error: a student who assumes v_rms ∝ T (instead of √T) would compute T₂ = 400 × 2 = 800 K. This is exactly half the correct answer and is a high-frequency distractor in NEET papers on this pattern.
9
Similar NEET-style question
The RMS speed of nitrogen molecules at 27°C is v. At what temperature will the RMS speed become v√3? *Setup:* T₁ = 300 K, speed ratio = √3. T₂ = 300 × (√3)² = 300 × 3 = 900 K = 627°C. ---

Before solving, remember these

Formula

Kinetic interpretation of temperature

Average translational KE per molecule: <½ m v²> = (3/2) k T. So v_rms = √(3kT/m) = √(3RT/M). Temperature is a measure of microscopic kinetic energy.

-- NCERT, p. 6

Formulas

5 formulas — click to collapse

Average translational KE per molecule

⟨ K E ⟩ = \frac{3}{2} k T

Microscopic interpretation of temperature: T is direct measure of average translational kinetic energy.

Symbol	Quantity	SI Unit
k	Boltzmann constant	J/K
T	absolute temperature	K

Valid when

Translational degrees of freedom only
Ideal gas

Cv from degrees of freedom

C_{v} = \frac{f}{2} R

Each quadratic DoF contributes (1/2)R to molar Cv. Mono: f=3, Cv=3R/2; di-rigid: f=5, Cv=5R/2; poly-rigid: f=6, Cv=3R.

Symbol	Quantity	SI Unit
Cv	molar specific heat	J/mol/K
f	degrees of freedom	-
R	gas constant	J/mol/K

Valid when

Equipartition holds (temperature high enough)
Quadratic energy modes

Ideal gas equation

P V = n R T = N k T

Fundamental equation of state of ideal gas relating pressure, volume, temperature.

Symbol	Quantity	SI Unit
P	pressure	Pa
V	volume	m^3
n	moles	mol
R	8.314	J/mol/K
N	molecule count	-
k	Boltzmann 1.38e-23	J/K
T	temp	K

Valid when

Gas obeys ideal gas approximation (low pressure, high temperature relative to phase transitions)

Mean free path of gas molecule

λ = \frac{1}{2 π n d ^{2}}

Average distance between successive molecular collisions.

Symbol	Quantity	SI Unit
lambda	mean free path	m
n	number density	1/m^3
d	molecular diameter	m

Valid when

Hard-sphere model
Equilibrium gas

RMS speed of gas molecules

v_{rms} = \frac{3 R T}{M} = \frac{3 k T}{m}

Root-mean-square molecular speed; depends on T and molar mass M (or molecular mass m).

Symbol	Quantity	SI Unit
R	gas constant	J/mol/K
T	temp	K
M	molar mass	kg/mol
k	Boltzmann	J/K
m	molecular mass	kg

Valid when

Ideal gas
Maxwell-Boltzmann distribution

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

2 items — click to collapse

Trap

v_rms vs T: linear vs sqrt scaling

Category: Similar Terms

Student treats v_rms ∝ T instead of √T. Doubling T does NOT double v_rms; it multiplies by √2.

When it triggers

Question asks for new v_rms after T change.

How to avoid

v_rms = √(3RT/M). v_rms ∝ √T. To double v_rms, T must quadruple.

Mistake

Scales v_rms linearly with T instead of √T.

Root cause: formula misuse

Correction

v_rms = √(3RT/M), so v_rms ∝ √T. To double v_rms, T must quadruple (factor of 4). Common error: assume doubling T doubles v_rms.

Past Year Questions

6 questions from NEET 2020, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: 100 [Given, R= Jmol–1K–1, and molecular mass of O2 = 32, 1 atm pressure = 1.01 × 105 N/m] 12

10.156 kg

20.125 kg

30.144 kg

40.116 kg

NTA Answer: Option 4(final)

NEET 2024Revised key

The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P1, P2 and P3 compared with those of Charles’s law represented as dotted lines. Then the correct relation is:

1P3 > P2 > P1

2P1 > P3 > P2

3P2 > P1 > P3

4P1 > P2 > P3

NTA Answer: Option 4(revised_final)

NEET 2023

The temperature of a gas is –50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

13295°C

23097 K

3223 K

4669°C

NTA Answer: Option 1(final)

NEET 2022

The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

15.6 m3

25.6 × 106 m3

35.6 × 103 m3

45.6 × 10–3 m3

NTA Answer: Option 1(final)

NEET 2020

The average thermal energy for a mono-atomic gas 16. U — is : (k is Boltzmann constant and T, absolute B ( U =k B U =T) temperature) 7 7

12 k B T 1 1

22 k B T 3 3

32 k B T 5 5

42 k B T

NTA Answer: Option 3(final)

NEET 2020

, fl d 23. The mean free path for a gas, with molecular fl n , U diameter d and number density n can be expressed — as : 1 1

12 n 2 2 d 2 1 1

22 n d 1 1

32 n d2 1 1

42 n 2 d 2

NTA Answer: Option 3(final)

How NEET usually asks this

4 recurring patterns from past papers — click to collapse

Average translational KE per molecule = (3/2)kT (monoatomic). Total via DoF.

Direct ApplicationEasy

Common distractors

uses 3 2 RT not 3 2 kT

Confuses per-mole RT with per-molecule kT

PV = nRT. Given two of (P, V, T, n) and changes, find missing.

Multi StepEasy

Common distractors

forgets temperature conversion

Mixes °C with K

Mean free path scaling with n, d. lambda = 1/(sqrt(2)pin*d^2).

InterpretationEasy

Common distractors

ignores d squared

Treats d linearly

Find new T given v_rms scales by factor k. v_rms ∝ sqrt(T), so T_new = T*k^2.

Multi StepMedium

Common distractors

uses linear scaling

Treats v_rms ∝ T not sqrt(T)

Sources

NCERT refs: Class 11 Physics Chapter 13, p.6

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Mean Free Path

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