Formula
Pressure from kinetic theory
P = (1/3) n m v_rms² = (1/3) ρ v_rms², where v_rms is root-mean-square molecular speed. Pressure is proportional to mean molecular kinetic energy.
-- NCERT Class 11 Physics, Ch. 12, p. 250Concept of Pressure Kinetic Theory
8 MCQs3 revision cards9-step worked example
Source: NCERT Kinetic TheoryOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Concept of Pressure Kinetic Theory, explained for NEET
The trap first: when temperature doubles, students reflexively double v_rms. That costs marks. The RMS speed scales as √T, not T — so doubling temperature multiplies v_rms by √2 ≈ 1.414, not 2. This single confusion is the highest-frequency distractor in NEET kinetic-theory pressure questions.
What pressure actually is, microscopically. Gas molecules slam into container walls. Each collision transfers momentum. Pressure is the net momentum transfer per unit area per unit time across all molecules. The kinetic theory derivation (NCERT Class 11 Physics, Chapter 12, page 248) starts from a single molecule bouncing inside a cube and sums over N molecules to give:
P = (1/3)(N/V)m·v²_rms
Rewriting with the ideal gas law PV = NkT, we get:
(1/2)m·v²_rms = (3/2)kT
This is the microscopic meaning of temperature: T is a direct measure of average translational kinetic energy per molecule. It is independent of the gas species — all ideal gases at the same T have the same average translational KE.
Bridge to NEET. Questions on this topic test two things: (1) can you connect PV = nRT to the molecular picture, and (2) do you handle the √T dependence of v_rms correctly? The ideal gas equation appears in PYQ 2024 and 2025; the v_rms scaling appeared in PYQ 2023.
Watch out: v_rms = √(3RT/M). To double v_rms, you need T to quadruple (factor of 4), not double. Write the ratio v₂/v₁ = √(T₂/T₁) every time — it prevents the linear-scaling slip.
Can you answer these Concept of Pressure Kinetic Theory 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 pressure exerted by an ideal gas in a container is due to:
Show answer and why every option is right or wrong
Answer: C. Kinetic theory derives pressure from the rate of momentum transfer when molecules collide elastically with the container walls (NCERT Class 11 Physics, Chapter 12, page 248). No intermolecular forces are considered in the ideal gas model.
Why A is wrong: A is wrong because the ideal gas model neglects gravitational attraction between molecules; pressure arises purely from wall collisions, not intermolecular gravity.
Why B is wrong: B is wrong because ideal gas assumptions treat molecules as non-interacting point particles; there are no repulsive forces in the model.
Why D is wrong: D is wrong because the kinetic theory model does not invoke electrostatic interactions; molecules are treated as neutral hard spheres.
MCQ 2Easy RecallPractice
For an ideal gas, the average translational kinetic energy per molecule depends on:
Show answer and why every option is right or wrong
Answer: D. ⟨KE⟩ = (3/2)kT — the average translational KE per molecule depends only on the absolute temperature T, not on the molecular species or mass (NCERT Class 11 Physics, Chapter 12, page 249).
Why A is wrong: A is wrong because the formula ⟨KE⟩ = (3/2)kT contains no mass term; molar mass affects individual molecular speed but not the average translational KE at a given T.
Why B is wrong: B is wrong because although PV = NkT links pressure and volume to temperature, the average KE per molecule is determined solely by T, not independently by P and V.
Why C is wrong: C is wrong because molar mass does not appear in ⟨KE⟩ = (3/2)kT; heavier molecules move slower but carry the same average translational KE at the same temperature.
MCQ 3Easy RecallPractice
The SI unit of the Boltzmann constant k is:
Show answer and why every option is right or wrong
Answer: B. k = R/Nₐ. Since R has units J·mol⁻¹·K⁻¹ and Nₐ has units mol⁻¹, k has units J·K⁻¹ (NCERT Class 11 Physics, Chapter 12, page 249).
Why A is wrong: A is wrong because J·mol⁻¹·K⁻¹ is the unit of the universal gas constant R, not the Boltzmann constant k (which is R divided by Avogadro's number, removing the mol⁻¹).
Why C is wrong: C is wrong because J·kg⁻¹·K⁻¹ is the unit of specific heat capacity, not Boltzmann's constant.
Why D is wrong: D is wrong because Pa·m³·mol⁻¹ is dimensionally equivalent to J·mol⁻¹, which does not match k = R/Nₐ = J·K⁻¹.
MCQ 4Direct ApplicationPractice
The RMS speed of oxygen molecules at temperature T is v. The RMS speed of hydrogen molecules at the same temperature T is: (Molar mass: O₂ = 32 g/mol, H₂ = 2 g/mol)
Show answer and why every option is right or wrong
Answer: A. v_rms = √(3RT/M). At the same T, v_rms ∝ 1/√M. Ratio = √(M_O₂/M_H₂) = √(32/2) = √16 = 4. So v_rms(H₂) = 4v (NCERT Class 11 Physics, Chapter 12, page 249).
Why B is wrong: B is wrong because it assumes RMS speed is independent of molar mass; v_rms ∝ 1/√M, so lighter molecules move faster at the same temperature.
Why C is wrong: C is wrong because it inverts the ratio — this would mean heavier H₂ moves slower, but H₂ is lighter than O₂; the correct scaling gives H₂ a higher speed by factor √(32/2) = 4.
Why D is wrong: D is wrong because it uses a linear mass ratio (32/2 = 16) instead of the square-root ratio; v_rms ∝ 1/√M, not 1/M.
MCQ 5Direct ApplicationPractice
An ideal gas is at temperature 300 K. Its RMS speed is v₀. To increase the RMS speed to 2v₀, the gas must be heated to:
Show answer and why every option is right or wrong
Answer: A. v_rms ∝ √T. To double v_rms: (2v₀/v₀)² = T₂/T₁ → T₂ = 4 × 300 = 1200 K. The common trap (trap: vrms t linear vs sqrt) is picking 600 K by assuming linear scaling.
Why B is wrong: B is wrong because 600 K assumes v_rms ∝ T (linear scaling). Since v_rms ∝ √T, doubling v_rms requires quadrupling T, not doubling it (trap: v_rms vs T linear vs sqrt scaling).
Why C is wrong: C is wrong because 900 K = 3 × 300, which would give v_rms = √3 · v₀ ≈ 1.73v₀, not 2v₀.
Why D is wrong: D is wrong because halving T to 150 K would reduce v_rms by factor 1/√2, not increase it.
MCQ 6Direct ApplicationPractice
Two moles of an ideal gas occupy a volume of 0.050 m³ at a pressure of 1.0 × 10⁵ Pa. The temperature of the gas is approximately: (R = 8.314 J·mol⁻¹·K⁻¹)
Show answer and why every option is right or wrong
Answer: B. PV = nRT → T = PV/(nR) = (1.0 × 10⁵ × 0.050)/(2 × 8.314) = 5000/16.628 ≈ 301 K. Straightforward ideal gas law application (NCERT Class 11 Physics, Chapter 12, page 245).
Why A is wrong: A is wrong because 150 K results from using n = 4 (doubling the moles) or an arithmetic slip; the correct denominator is 2 × 8.314.
Why C is wrong: C is wrong because 602 K results from using n = 1 instead of n = 2 in PV = nRT; the question specifies two moles.
Why D is wrong: D is wrong because 1204 K results from using n = 0.5 or misplacing a factor of 4; careful substitution into T = PV/(nR) gives ~301 K.
MCQ 7CalculationPractice
The RMS speed of gas molecules at 27°C is 500 m/s. If the temperature is raised to 327°C, the new RMS speed is:
Show answer and why every option is right or wrong
Answer: C. Convert to kelvin: T₁ = 300 K, T₂ = 600 K. v_rms ∝ √T → v₂ = v₁ × √(T₂/T₁) = 500 × √(600/300) = 500 × √2 ≈ 707 m/s. The multi-step: temperature conversion first, then ratio application.
Why A is wrong: A is wrong because 1000 m/s assumes v_rms ∝ T (linear scaling). Since v_rms ∝ √T, doubling the absolute temperature multiplies v_rms by √2, not 2 (trap: v_rms vs T linear vs sqrt scaling).
Why B is wrong: B is wrong because 250√2 comes from halving v₁ instead of multiplying by √2; the ratio √(T₂/T₁) is a multiplicative factor applied to the original speed.
Why D is wrong: D is wrong because 500 m/s implies no change in speed, which would require the same temperature; T has doubled from 300 K to 600 K.
MCQ 8CalculationPractice
An ideal gas at 400 K has RMS speed v. The temperature at which the RMS speed becomes v/2 is:
Show answer and why every option is right or wrong
Answer: D. v_rms ∝ √T. (v/2)/v = √(T₂/400) → 1/4 = T₂/400 → T₂ = 100 K. Two steps: set up the ratio, then square both sides.
Why A is wrong: A is wrong because 200 K assumes v_rms ∝ T; halving v_rms requires T to become one-quarter (not one-half), since v_rms ∝ √T (trap: v_rms vs T linear vs sqrt scaling).
Why B is wrong: B is wrong because T₂ = 50 K would give v_rms = v × √(50/400) = v × √(1/8) = v/(2√2), not v/2.
Why C is wrong: C is wrong because 800 K would increase v_rms by factor √(800/400) = √2, not decrease it to v/2.
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Concept of Pressure Kinetic Theory: quick recall before you leave
How do you solve a Concept of Pressure Kinetic Theory question? A worked example
Pattern: NEET pattern: rms speed temp scaling (PYQ 2023, medium difficulty, negative-marking risk medium)
- 1
Given
The RMS speed of nitrogen molecules (N₂, M = 28 g/mol = 0.028 kg/mol) at temperature T₁ = 300 K is v₁. The temperature is raised until the RMS speed triples (v₂ = 3v₁).
- 2
Required
Find T₂, the new temperature.
- 3
Concept
RMS speed scales as the square root of absolute temperature: v_rms ∝ √T (at constant molar mass). This means the ratio of speeds equals the square root of the ratio of temperatures.
- 4
Formula
v_rms = √(3RT/M)
Therefore: v₂/v₁ = √(T₂/T₁) - 5
Substitution
3v₁/v₁ = √(T₂/300)
3 = √(T₂/300) - 6
Calculation
Squaring both sides:
9 = T₂/300
T₂ = 9 × 300 = 2700 K
Note on exact constants: the factor 3 (the speed multiplier) and 300 K (given temperature) are problem-defined exact values. They do not limit significant figures in the answer. - 7
Final answer
T₂ = 2700 K
To triple the RMS speed, the absolute temperature must increase by a factor of 9 (= 3²), not 3. - 8
Common trap
The linear-scaling trap (trap: vrms t linear vs sqrt): a student who assumes v_rms ∝ T would answer T₂ = 3 × 300 = 900 K. This is the most common distractor for this pattern. Always write the ratio and square it.
- 9
Similar NEET-style question
"The RMS speed of helium atoms at 200 K is u. At what temperature will the RMS speed become u√3?" → Set up u√3/u = √(T₂/200), square: 3 = T₂/200, T₂ = 600 K.
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What to remember before solving Concept of Pressure Kinetic Theory questions
More in Kinetic Theory: 2 exam traps and mistakes · 5 formulas · 4 question patterns from its other lessons.
Concept of Pressure Kinetic Theory questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 12, p.248
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.
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