This MCQ module is based on: Kinetic Theory Pressure
TOPIC 20 OF 33
Kinetic Theory Pressure
🎓 Class 11
Physics
CBSE
Theory
Ch 12 – Kinetic Theory
⏱ ~14 min
🌐 Language: [gtranslate]
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Kinetic Theory Pressure
12.4 Kinetic Theory of an Ideal Gas — Postulates
Building on Bernoulli's idea (1738) and Maxwell-Boltzmann's machinery (1860–80), kinetic theory makes the following assumptions about an ideal gas:
Postulates of Kinetic Theory:
- A gas consists of a very large number of identical molecules in continuous random motion.
- The molecules occupy negligible volume compared to the volume of the container.
- There are no inter-molecular forces, except during the brief duration of a collision.
- All collisions (with each other and with the walls) are perfectly elastic.
- The duration of a collision is negligible compared to the time between collisions.
- Between collisions, molecules travel in straight lines at constant velocity (Newton's first law).
From these innocent-looking statements we can derive the macroscopic gas laws and the ideal gas equation — turning empirical regularities into predictions of mechanics.
12.5 Pressure of an Ideal Gas — Derivation
Consider a cubical box of side L (volume V = L³) containing N identical molecules, each of mass m. We compute the force per unit area on one wall (say, the wall at x = L, perpendicular to the x-axis).
Step 1: Momentum transfer in one collision
Take one molecule with velocity components (v_x, v_y, v_z). When it strikes the right wall, only its x-velocity reverses (elastic collision with rigid wall):
Δp_x = (−m v_x) − (m v_x) = −2 m v_x (change in molecule's x-momentum)
By Newton's third law, the wall absorbs +2 m v_x of x-momentum per collision.
Step 2: Frequency of hits
Between two successive hits on the right wall, the molecule must travel to the opposite wall and back — a round-trip distance 2L. The time between hits is:
Δt = 2L / v_x
The number of hits per second on the right wall by this molecule = v_x / (2L).
Step 3: Force on the wall
Force on the wall by this molecule = (momentum transferred per hit) × (hits per second):
f = (2 m v_x) × (v_x / 2L) = m v_x² / L
Summing over all N molecules and dividing by wall area A = L²:
P = F / A = (m / L³) Σ v_xᵢ² = (m N / V) ⟨v_x²⟩
Step 4: Use isotropy of motion
By symmetry, the molecular motion is equally likely in x, y, z directions:
⟨v_x²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⟨v²⟩ / 3
Substituting and using number density n = N/V:
Key result:
\[\boxed{\,P = \dfrac{1}{3}\, n\, m\, \langle v^2\rangle\, = \dfrac{1}{3}\,\rho\, \langle v^2\rangle\,}\]
where ρ = nm is the mass density of the gas. Pressure equals one-third of mass density times mean-square molecular speed.
This is one of the most beautiful results of physics: a macroscopic quantity (pressure measured in pascals) emerges from the microscopic motion of ~10²³ particles per cubic metre, each obeying Newton's laws. The averaging happens automatically because of the huge N.
12.5.1 An Equivalent Form
Multiplying numerator and denominator by ½:
P V = (2/3) × N × (½ m ⟨v²⟩) = (2/3) × (Total translational KE)
So pressure × volume = ⅔ of the total translational kinetic energy of the gas. We will use this in Part 3 to identify temperature.
| Macroscopic quantity | Microscopic origin |
|---|---|
| Pressure P | Average rate of momentum transfer to walls |
| Volume V | Region accessible to molecular motion |
| Total energy | Sum of translational (+ rotational/vibrational) KE |
| Density ρ | n × m (number density × mass per molecule) |
Worked Examples
Example 12.4: Pressure from molecular speeds
One mole of helium gas (m = 6.64 × 10⁻²⁷ kg per atom) is held in a 22.4-L box. The mean-square speed is ⟨v²⟩ = 1.95 × 10⁶ m²/s². Find the pressure.
n = N_A / V = 6.022 × 10²³ / (22.4 × 10⁻³) = 2.69 × 10²⁵ m⁻³.
P = (1/3) n m ⟨v²⟩ = (1/3)(2.69 × 10²⁵)(6.64 × 10⁻²⁷)(1.95 × 10⁶)
= (1/3)(0.1786)(1.95 × 10⁶) = 1.16 × 10⁵ Pa ≈ 1 atm. ✓
P = (1/3) n m ⟨v²⟩ = (1/3)(2.69 × 10²⁵)(6.64 × 10⁻²⁷)(1.95 × 10⁶)
= (1/3)(0.1786)(1.95 × 10⁶) = 1.16 × 10⁵ Pa ≈ 1 atm. ✓
Example 12.5: Mean-square speed from pressure
Air at 1.013 × 10⁵ Pa has density 1.29 kg/m³. Find the mean-square speed of an air molecule.
P = (1/3) ρ ⟨v²⟩ ⇒ ⟨v²⟩ = 3P/ρ = 3 × 1.013 × 10⁵ / 1.29 = 2.36 × 10⁵ m²/s².
√⟨v²⟩ ≈ 485 m/s — about 1.4 × the speed of sound, as we expect.
√⟨v²⟩ ≈ 485 m/s — about 1.4 × the speed of sound, as we expect.
Example 12.6: Hits per second on a wall
Consider a 1 m² wall of a 1 m³ box containing 2.69 × 10²⁵ N₂ molecules at 300 K. The mean ⟨v_x⟩ ≈ 290 m/s. Estimate hits/s on this wall.
Hits/s ≈ (1/4) × n × ⟨v⟩ × A — the famous flux formula. Approximately:
≈ (1/4)(2.69 × 10²⁵)(470 m/s)(1 m²) ≈ 3.16 × 10²⁷ hits/second.
A truly enormous number — yet they average to a steady macroscopic pressure.
≈ (1/4)(2.69 × 10²⁵)(470 m/s)(1 m²) ≈ 3.16 × 10²⁷ hits/second.
A truly enormous number — yet they average to a steady macroscopic pressure.
Interactive: Pressure from Molecular Speed L3 Apply
Adjust gas density ρ and the rms speed v_rms to see the pressure from P = (1/3) ρ v_rms².
Activity 12.2 — Pressure-by-Drumming Demo
L4 Analyse
Predict: If a hundred small marbles are dropped on a flat metal sheet rapidly, will the sheet feel a steady force, or jolts? How can a smooth gas pressure arise from discrete collisions?
- Take a metal tray (acts as the "wall") and a fistful of dried peas or small ball-bearings.
- From a height of ~30 cm, sprinkle the peas continuously onto the tray for 5 seconds. Listen and feel.
- Now reduce the rate of sprinkling but use heavier marbles. Compare.
Observation: A continuous fast stream feels like a smooth push — the discrete impacts blur into a steady force. Heavier projectiles deliver more momentum per hit. This is exactly how 10²⁵ molecular hits per m² produce steady gas pressure: P = (1/3) ρ⟨v²⟩.
Competency-Based Questions
A cubical box of side 0.10 m contains 10²³ N₂ molecules. The molar mass of N₂ is 28 g/mol, so each molecule has mass m = 4.65 × 10⁻²⁶ kg. The mean-square speed is ⟨v²⟩ = 2.5 × 10⁵ m²/s².
Q1. L1 Remember List any three postulates of kinetic theory.
(i) Gas consists of large number of identical molecules in random motion. (ii) Molecular volume is negligible compared to container volume. (iii) Collisions are perfectly elastic and instantaneous.
Q2. L2 Understand Why does only the x-component of velocity matter in computing pressure on a wall perpendicular to x-axis?
Only v_x reverses on collision with that wall (elastic). The y and z components are unchanged because the wall is smooth and perpendicular to x — they exert no force along x. Hence only v_x contributes to momentum transfer onto the wall.
Q3. L3 Apply Compute the pressure inside the box.
n = N/V = 10²³ / (10⁻³) = 10²⁶ m⁻³. P = (1/3) n m ⟨v²⟩ = (1/3)(10²⁶)(4.65 × 10⁻²⁶)(2.5 × 10⁵) = 3.88 × 10⁵ Pa ≈ 3.8 atm.
Q4. L4 Analyse If we doubled the box volume but kept N and ⟨v²⟩ unchanged, what happens to P? Justify from the formula.
P = (1/3)(N/V)m⟨v²⟩ — doubling V halves n, so P halves. This is Boyle's law, derived microscopically.
Q5. L5 Evaluate A student says "the formula P = (1/3) ρ ⟨v²⟩ shows that heavier gases at the same speed exert more pressure." Comment.
Correct: at the same number density n and same ⟨v²⟩, larger m means larger ρ = nm and larger P. But at the same temperature, heavier molecules move slower (⟨v²⟩ ∝ T/m), so the two effects cancel and PV = NkT is independent of m. The student's statement is true only if speed (not temperature) is held fixed — a subtle but important distinction.
Assertion-Reason Questions
Assertion (A): The pressure of a gas in a container is uniform throughout.
Reason (R): Molecular motion is isotropic in equilibrium.
Answer: A. Isotropy means ⟨v_x²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩, so all walls receive the same momentum flux, and pressure is uniform.
Assertion (A): If gas molecules collided inelastically with the walls, gas pressure would steadily decrease.
Reason (R): Energy lost to walls would lower the kinetic energy of the gas over time.
Answer: A. Inelastic wall collisions would drain KE from the gas, decreasing ⟨v²⟩ and hence P. Elasticity is required for steady-state pressure.
Assertion (A): The factor 1/3 in P = (1/3) n m ⟨v²⟩ comes from averaging over three spatial directions.
Reason (R): Pressure is a vector quantity.
Answer: C. A is true (the 1/3 = ⟨v_x²⟩/⟨v²⟩ from isotropy). R is false — pressure is a scalar (force per unit area, regardless of direction in equilibrium).
Frequently Asked Questions - Kinetic Theory Pressure
What is the main concept covered in Kinetic Theory Pressure?
In NCERT Class 11 Physics Chapter 12 (Kinetic Theory), "Kinetic Theory Pressure" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Kinetic Theory Pressure useful in real-life applications?
Real-life applications of Kinetic Theory Pressure from NCERT Class 11 Physics Chapter 12 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Kinetic Theory Pressure?
Key formulas in Kinetic Theory Pressure (NCERT Class 11 Physics Chapter 12 Kinetic Theory) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 12?
NCERT Class 11 Physics Chapter 12 (Kinetic Theory) is structured so each part builds on the previous one. Kinetic Theory Pressure connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Kinetic Theory Pressure?
CBSE board questions from Kinetic Theory Pressure typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Kinetic Theory Pressure lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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