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 Class 11 Physics, Ch. 12, p. 256RMS Speed Gas
8 MCQs4 revision cards9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 25 Sep 2026
RMS Speed Gas, explained for NEET
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 12, page 250), 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.
Can you answer these RMS Speed Gas 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:
Show answer and why every option is right or wrong
Answer: B. From v_rms = √(3RT/M), the RMS speed is proportional to the square root of absolute temperature (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because v_rms ∝ √T, not T². There is no squared-temperature dependence in the kinetic theory expression for RMS speed.
Why C is wrong: C is wrong because v_rms ∝ √T, not T. This is the core trap — treating the dependence as linear instead of square-root (trap: v_rms vs T linear vs sqrt scaling).
Why D is wrong: D is wrong because v_rms increases with temperature, not decreases. An inverse relationship would imply molecules slow down when heated.
MCQ 2Easy RecallPractice
The SI unit of molar mass M used in the formula v_rms = √(3RT/M) is:
Show answer and why every option is right or wrong
Answer: D. The formula v_rms = √(3RT/M) requires M in kg/mol for dimensional consistency, since R is in J mol⁻¹ K⁻¹ and 1 J = 1 kg m² s⁻² (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because g/mol is the commonly tabulated unit but must be converted to kg/mol before substitution. Using g/mol without conversion makes M numerically 1000 times too large, so v_rms comes out smaller by a factor of √1000.
Why B is wrong: B is wrong because kg is a unit of mass, not molar mass. Molar mass has dimensions of mass per amount of substance.
Why C is wrong: C is wrong because amu (atomic mass unit) is used for individual atomic/molecular masses, not for the molar-mass variable in the gas-constant form of v_rms.
MCQ 3Easy RecallPractice
To double the RMS speed of gas molecules, the absolute temperature must be increased by a factor of:
Show answer and why every option is right or wrong
Answer: B. Since v_rms ∝ √T, doubling v_rms requires T to increase by a factor of 2² = 4 (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because this assumes v_rms ∝ T (linear scaling). The actual dependence is v_rms ∝ √T, so doubling T only increases v_rms by √2, not 2 (trap: v_rms vs T linear vs sqrt scaling).
Why C is wrong: C is wrong because increasing T by √2 would increase v_rms by (√2)^(1/2) = 2^(1/4) ≈ 1.19, not 2.
Why D is wrong: D is wrong because a factor of 8 in T would give v_rms increasing by √8 = 2√2 ≈ 2.83, overshooting the target of 2.
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:
Show answer and why every option is right or wrong
Answer: D. At the same T, v_rms ∝ 1/√M. The ratio is √(M_O₂/M_H₂) = √(32/2) = √16 = 4, so v_rms(H₂) = 4v₀ (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because different gases at the same temperature have different RMS speeds unless they have the same molar mass. Hydrogen is 16 times lighter than oxygen.
Why B is wrong: B is wrong because this inverts the relationship — lighter molecules move faster, not slower. v_rms(H₂) > v_rms(O₂).
Why C is wrong: C is wrong because this uses the molar mass ratio directly (32/2 = 16) instead of its square root. v_rms ∝ 1/√M, not 1/M.
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:
Show answer and why every option is right or wrong
Answer: C. Convert to kelvin: T₁ = 300 K, T₂ = 600 K. The ratio v₂/v₁ = √(T₂/T₁) = √(600/300) = √2 (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because this uses °C values directly without converting to kelvin, and also treats the dependence as linear. Both errors compound.
Why B is wrong: B is wrong because this uses °C values (327/27) instead of kelvin values (600/300). Temperature ratios in kinetic theory must always use absolute temperature (trap: forgetting °C to K conversion).
Why D is wrong: D is wrong because this assumes v_rms ∝ T (linear scaling). Since T doubled (300 K → 600 K), linear scaling would give 2, but the correct √T dependence gives √2 (trap: v_rms vs T linear vs sqrt scaling).
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?
Show answer and why every option is right or wrong
Answer: A. v_rms doubles (400/200 = 2). Since v_rms ∝ √T, the new temperature T₂ = T × (v₂/v₁)² = T × 4 = 4T (NCERT Class 11 Physics, Chapter 12, page 250).
Why B is wrong: B is wrong because doubling T only increases v_rms by √2 ≈ 1.41, not by 2. This is the linear-scaling trap (trap: v_rms vs T linear vs sqrt scaling).
Why C is wrong: C is wrong because reducing temperature would decrease v_rms, not increase it. This answer goes in the wrong direction entirely.
Why D is wrong: D is wrong because T√2 gives v_rms increasing by (√2)^(1/2) = 2^(1/4) ≈ 1.19, far short of the required factor of 2.
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:
Show answer and why every option is right or wrong
Answer: C. v_rms triples, so v₂/v₁ = 3. Since v_rms ∝ √T, T₂ = T₁ × (v₂/v₁)² = 300 × 9 = 2700 K (NCERT Class 11 Physics, Chapter 12, page 250).
Why A is wrong: A is wrong because 900 K = 300 × 3, which uses linear scaling (v_rms ∝ T). The correct square-root dependence requires squaring the speed ratio: 3² = 9, giving 300 × 9 = 2700 K (trap: v_rms vs T linear vs sqrt scaling).
Why B is wrong: B is wrong because 100 K = 300/3, which inverts the relationship. Increasing speed requires increasing temperature, not decreasing it.
Why D is wrong: D is wrong because 300√3 ≈ 520 K would give v_rms increasing by √(√3) = 3^(1/4) ≈ 1.32, not 3. This confuses the square root with the fourth root.
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:
Show answer and why every option is right or wrong
Answer: A. Setting v_rms equal: √(3R × 200/M) = √(3R × T_Y/4M). Squaring: 200/M = T_Y/4M, so T_Y = 800 K. The heavier gas requires 4 times the temperature to match the lighter gas's RMS speed (NCERT Class 11 Physics, Chapter 12, page 250).
Why B is wrong: B is wrong because equal temperatures give equal v_rms only for gases of equal molar mass. Since gas Y is 4 times heavier, it moves slower at the same temperature.
Why C is wrong: C is wrong because 50 K = 200/4, which inverts the mass relationship. A heavier gas needs a higher temperature (not lower) to achieve the same v_rms as a lighter gas.
Why D is wrong: D is wrong because 400 K = 200 × √4 takes the square root of the mass ratio. Since v_rms ∝ √(T/M), equal speeds need T ∝ M itself, giving T_Y = 200 × 4 = 800 K.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
RMS Speed Gas: quick recall before you leave
How do you solve a RMS Speed Gas question? A 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.
---
What to remember before solving RMS Speed Gas questions
Which RMS Speed Gas formulas do you need for NEET?
1 formula — click to collapse
RMS speed of gas molecules
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
Where do students lose marks on RMS Speed Gas?
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
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.
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.
More in Kinetic Theory: 4 formulas · 3 question patterns from its other lessons.
RMS Speed Gas questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
13295°C
23097 K
3223 K
4669°C
NTA Answer: Option 1(final)
How does NEET ask about RMS Speed Gas?
1 recurring pattern from past papers — click to collapse
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 12, p.250
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →