Concept of Pressure Kt

8 MCQs3 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 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 13, page 5) 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.

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 pressure exerted by an ideal gas in a container is due to:

MCQ 2Easy RecallPractice

For an ideal gas, the average translational kinetic energy per molecule depends on:

MCQ 3Easy RecallPractice

The SI unit of the Boltzmann constant k is:

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)

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:

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)

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:

MCQ 8CalculationPractice

An ideal gas at 400 K has RMS speed v. The temperature at which the RMS speed becomes v/2 is:

Quick recall before you leave

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. ---

Before solving, remember these

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, p. 5

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.5

Test yourself on this topic with real past-paper questions:

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Avogadro Number Degrees of Freedom

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Concept of Pressure Kt

Lesson

Practice MCQs

Quick recall before you leave

PV = nRT = NkT

v_rms = √(3RT/M)

⟨KE⟩ = (3/2)kT

Worked Example

Before solving, remember these

Pressure from kinetic theory

Formulas

Average translational KE per molecule

Valid when

Cv from degrees of freedom

Valid when

Ideal gas equation

Valid when

Mean free path of gas molecule

Valid when

RMS speed of gas molecules

Valid when

Exam Traps & Common Mistakes

v_rms vs T: linear vs sqrt scaling

When it triggers

How to avoid

Scales v_rms linearly with T instead of √T.

Correction

Past Year Questions

How NEET usually asks this

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

Common distractors

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

Common distractors

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

Common distractors

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

Common distractors

Sources

Free NEET study resources