This MCQ module is based on: Molecular Behaviour Gas Laws
TOPIC 19 OF 33
Molecular Behaviour Gas Laws
🎓 Class 11
Physics
CBSE
Theory
Ch 12 – Kinetic Theory
⏱ ~14 min
🌐 Language: [gtranslate]
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Molecular Behaviour Gas Laws
12.1 Introduction — From Macroscopic to Microscopic
In Chapter 11 we treated thermodynamics in purely macroscopic terms — pressure, volume, temperature, internal energy. Kinetic theory takes a complementary view: it explains all those macroscopic observations starting from the assumption that a gas is a swarm of tiny molecules in incessant random motion. Robert Boyle (1661), and later Maxwell and Boltzmann in the 19th century, built this picture into one of the most successful theories of physics.
Kinetic theory gives a molecular interpretation of pressure, temperature, and the gas laws. It also predicts specific heats, viscosity, diffusion and conduction — all in terms of just a few molecular parameters.
12.2 Molecular Nature of Matter
Richard Feynman remarked that the discovery "matter is made of atoms" is among the most significant in the history of science. The modern atomic theory is credited to John Dalton (~1808), who used it to explain the laws of definite and multiple proportions. Avogadro then sharpened the picture with his hypothesis: equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
A typical molecule has size ~ 10⁻¹⁰ m (an Angstrom). At standard temperature and pressure (STP), 1 mole of any gas (6.022 × 10²³ molecules) occupies about 22.4 litres. The mean separation between molecules in a gas at STP is roughly 10 times the molecular diameter — gases are mostly empty space.
Inter-molecular forces are short-range and important in solids and liquids; in gases the molecules are so far apart that we can treat them as non-interacting except during instantaneous collisions. This single simplification is what makes gas behaviour so much easier to describe than that of solids and liquids.
12.3 Behaviour of Gases — The Gas Laws
Long before kinetic theory, careful experiments produced three empirical gas laws. They are what kinetic theory must reproduce.
12.3.1 Boyle's Law (T constant)
At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.
\[PV = \text{constant} \quad \text{(at constant }T, n)\]
12.3.2 Charles' Law (P constant)
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
\[\dfrac{V}{T} = \text{constant} \quad \text{(at constant }P, n)\]
12.3.3 Avogadro's Hypothesis
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Combining these three relations gives the ideal gas equation:
PV = nRT (n = number of moles, R = 8.314 J·mol⁻¹·K⁻¹)
Equivalently in terms of total molecules N and the Boltzmann constant k_B = R/N_A = 1.38 × 10⁻²³ J/K:
PV = N k_B T
12.3.4 Dalton's Law of Partial Pressures
For a mixture of non-reacting ideal gases, each gas acts independently. The total pressure equals the sum of the partial pressures each gas would exert if it occupied the volume alone:
P_total = P₁ + P₂ + P₃ + … (Dalton's law)
| Constant | Symbol | Value (SI) |
|---|---|---|
| Universal gas constant | R | 8.314 J·mol⁻¹·K⁻¹ |
| Boltzmann constant | k_B | 1.381 × 10⁻²³ J·K⁻¹ |
| Avogadro number | N_A | 6.022 × 10²³ mol⁻¹ |
| Molar volume at STP (273.15 K, 1 atm) | V_m | 22.4 × 10⁻³ m³ |
| Atmospheric pressure | P_atm | 1.013 × 10⁵ Pa |
Worked Examples
Example 12.1: Number of molecules at STP
How many molecules of an ideal gas are present in 1 cm³ at STP (T = 273 K, P = 1.013 × 10⁵ Pa)?
N = PV/(k_B T) = (1.013 × 10⁵ × 1 × 10⁻⁶)/(1.381 × 10⁻²³ × 273)
= 0.1013 / (3.77 × 10⁻²¹) ≈ 2.69 × 10¹⁹ molecules.
This is the Loschmidt number n₀ — the molecular density of an ideal gas at STP.
= 0.1013 / (3.77 × 10⁻²¹) ≈ 2.69 × 10¹⁹ molecules.
This is the Loschmidt number n₀ — the molecular density of an ideal gas at STP.
Example 12.2: Boyle's law compression
A bubble of air rises from the bottom of a lake (10.3 m deep, T constant). At the bottom its volume is 1.0 cm³. Atmospheric pressure = 1.013 × 10⁵ Pa, ρ_water = 10³ kg/m³, g = 9.8 m/s². Find its volume when it reaches the surface.
P_bottom = P_atm + ρgh = 1.013 × 10⁵ + 10³ × 9.8 × 10.3 ≈ 2.02 × 10⁵ Pa.
P_top = P_atm = 1.013 × 10⁵ Pa.
By Boyle's law (constant T): P₁V₁ = P₂V₂. So V₂ = (P₁/P₂)V₁ = (2.02/1.013) × 1.0 ≈ 2.0 cm³ — the bubble doubles in size.
P_top = P_atm = 1.013 × 10⁵ Pa.
By Boyle's law (constant T): P₁V₁ = P₂V₂. So V₂ = (P₁/P₂)V₁ = (2.02/1.013) × 1.0 ≈ 2.0 cm³ — the bubble doubles in size.
Example 12.3: Heating a sealed tyre
A tyre is inflated to 2.5 × 10⁵ Pa at 27 °C. After driving on a hot road its temperature rises to 57 °C. Assuming volume is unchanged, find the new pressure.
At constant V (Gay-Lussac form): P/T = const ⇒ P₂ = P₁ × T₂/T₁ = 2.5 × 10⁵ × (330/300) = 2.75 × 10⁵ Pa. A 30 K rise produces ~10% pressure increase — relevant to tyre safety.
Interactive: Ideal Gas Law Calculator L3 Apply
Set the moles n, temperature T and volume V; the applet computes the pressure P from PV = nRT.
Activity 12.1 — Squeezing a Sealed Syringe
L3 Apply
Predict: If you seal a syringe with your finger and push the plunger down to half-volume, what happens to the pressure inside?
- Take a 10-mL plastic syringe; pull plunger to 10 mL with the tip open. Now seal the tip with your finger.
- Push the plunger from 10 mL to 5 mL (volume halves) at room temperature.
- Note the resistance you feel — that resistance is the increased pressure pushing back on the plunger.
- Release; observe the plunger spring back to nearly its original position.
Observation: Halving V at constant T doubles P (Boyle's law). When you release, P drops back and V returns. This is a direct demonstration of PV = const at constant T. The thumb-tip force on a 1-cm² plunger at 2 atm is about 10 N — easily felt.
Competency-Based Questions
A flexible balloon containing 2.0 L of helium at 27 °C and 1.0 atm is taken from a warm classroom into a cold storage room at −13 °C. The balloon adjusts so that the internal pressure remains 1.0 atm.
Q1. L1 Remember State Avogadro's hypothesis.
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
Q2. L2 Understand Why is the ideal gas law more accurate at low pressures and high temperatures?
At low P the molecules are far apart, so the volume occupied by molecules themselves is negligible compared to V. At high T the kinetic energy dominates the inter-molecular potential energy. Both conditions push the gas towards the "no interactions, no size" idealisation.
Q3. L3 Apply Find the new volume of the balloon in the storage room.
Charles' law (P const): V₂ = V₁ × T₂/T₁ = 2.0 × (260/300) = 1.73 L.
Q4. L4 Analyse A mixture of 0.10 mol N₂ and 0.20 mol O₂ is placed in a 5.0 L vessel at 300 K. Find the partial pressure of O₂ and the total pressure. (R = 8.314 J/mol·K)
P_O₂ = n_O₂ RT/V = (0.20)(8.314)(300)/(5.0×10⁻³) = 9.98 × 10⁴ Pa ≈ 1.0 atm. Similarly P_N₂ ≈ 0.5 atm. Total P = 1.5 atm.
Q5. L5 Evaluate A student claims a real gas at very high pressure can have PV/(nT) less than R. Is this consistent with experiments? Justify.
Yes — at intermediate pressures, attractive inter-molecular forces (the "a/V²" term in van der Waals' equation) lower the effective pressure, so PV/(nT) drops below R before rising again at very high P (where the finite size correction "−b" dominates). Fig 12.2 shows this dip in CO₂.
Assertion-Reason Questions
Assertion (A): The ideal gas law is universal — at given P, T, V, all gases contain the same number of moles.
Reason (R): The constant R has the same numerical value for every ideal gas.
Answer: A. The universality of R (= 8.314 J/mol·K) is exactly what makes the ideal gas law work for any gas, embodying Avogadro's hypothesis.
Assertion (A): Boyle's law fails for a real gas near its liquefaction temperature.
Reason (R): Inter-molecular forces become non-negligible near liquefaction.
Answer: A. Both true; the inter-molecular forces are the direct cause of deviation from PV = nRT.
Assertion (A): At STP, one mole of any ideal gas occupies 22.4 L.
Reason (R): The Boltzmann constant equals R/N_A.
Answer: B. Both A and R are true, but R simply re-defines k_B; the 22.4 L value comes from V = nRT/P at T=273 K, P=1.013×10⁵ Pa, not directly from R/N_A.
Frequently Asked Questions - Molecular Behaviour Gas Laws
What is the main concept covered in Molecular Behaviour Gas Laws?
In NCERT Class 11 Physics Chapter 12 (Kinetic Theory), "Molecular Behaviour Gas Laws" 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 Molecular Behaviour Gas Laws useful in real-life applications?
Real-life applications of Molecular Behaviour Gas Laws 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 Molecular Behaviour Gas Laws?
Key formulas in Molecular Behaviour Gas Laws (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. Molecular Behaviour Gas Laws 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 Molecular Behaviour Gas Laws?
CBSE board questions from Molecular Behaviour Gas Laws 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 Molecular Behaviour Gas Laws 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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