Chapter 13: Kinetic Theory

Introduction

Overview

• Gases consist of rapidly moving atoms or molecules.
• Kinetic theory explains gas behavior by neglecting inter-atomic forces.
• Developed in 19ᵗʰ century by Maxwell, Boltzmann, and others.
• Provides molecular interpretation of pressure and temperature.
• Consistent with gas laws and Avogadro’s hypothesis.
• Explains specific heat capacities of gases.

Solved Examples

Ex 1

Explain why kinetic theory neglects inter-atomic forces in gases.

Step 1: Recognize gas molecules are far apart.

Step 2: Inter-atomic forces are negligible due to large distances.

Step 3: Hence, gas behavior can be modeled by free moving particles.

Practice Set

• Q1: What does kinetic theory assume about gas molecules?
  • A. Stationary particles
  • B. Rapidly moving particles
  • C. Strongly interacting particles
  • D. Fixed positions
• Q2: Who contributed to the development of kinetic theory?
  • A. Dalton and Avogadro
  • B. Maxwell and Boltzmann
  • C. Newton and Einstein
  • D. Curie and Feynman
• Q3: Kinetic theory explains which of the following?
  • A. Molecular weight of gases
  • B. Molecular interpretation of pressure and temperature
  • C. Chemical reactions in gases
  • D. Atomic structure

Answer Key

• Q1: B - Gas molecules are assumed to be rapidly moving particles.
• Q2: B - Maxwell and Boltzmann developed kinetic theory.
• Q3: B - Kinetic theory explains pressure and temperature molecularly.

Memory Aids

• Kinetic theory = gas molecules in rapid motion + negligible forces.
• Maxwell & Boltzmann = kinetic theory pioneers.
• Pressure & temperature explained molecularly.

Molecular Nature of Matter

Key Concepts

• Feynman’s atomic hypothesis: Atoms are tiny particles in constant motion, attracting at a distance and repelling when very close.
• Dalton’s Atomic Theory:
  • First law: Fixed mass proportion in compounds.
  • Second law: Mass ratios of elements in multiple compounds are small integers.
• Gay Lussac’s Law: Volumes of gases combining chemically are in small integer ratios.
• Avogadro’s Hypothesis: Equal volumes of gases at same temperature and pressure contain equal molecules.
• Mean Free Path: Average distance a molecule travels without collision.

Solved Examples

Ex 1

Calculate mean free path if molecule diameter d = 3 × 10⁻¹⁰ m and number density n = 2.5 × 10²⁵ m⁻³.

Step 1: Use formula \( \lambda = \frac{1}{\sqrt{2} \pi d^{2} n} \)

Step 2: Substitute values \( \lambda = \frac{1}{\sqrt{2} \pi (3 \times 10^{-10})^{2} \times 2.5 \times 10^{25}} \)

Step 3: Calculate \( \lambda \approx 1.06 \times 10^{-7} \) m

Practice Set

• Q1: According to Avogadro’s hypothesis, equal volumes of gases at same T and P have?
  • A. Equal mass
  • B. Equal number of molecules
  • C. Equal density
  • D. Equal pressure
• Q2: Gay Lussac’s law relates to?
  • A. Mass ratios in compounds
  • B. Volume ratios of combining gases
  • C. Molecular speeds
  • D. Atomic masses
• Q3: Mean free path depends on?
  • A. Molecular diameter and number density
  • B. Temperature only
  • C. Pressure only
  • D. Volume only

Answer Key

• Q1: B - Equal volumes contain equal molecules at same T and P.
• Q2: B - Gay Lussac’s law states volume ratios of gases combine in small integers.
• Q3: A - Mean free path depends on molecular size and density.

Memory Aids

• Avogadro = equal volume = equal molecules.
• Gay Lussac = volume ratios small integers.
• Mean free path \( \lambda = \frac{1}{\sqrt{2} \pi d^{2} n} \).

Behaviour of Gases

Gas Laws and Constants

• At low temperature and high pressure, PV = K T, where K = N k_B.
• k_B (Boltzmann constant) = 1.38 × 10⁻²³ J K⁻¹, same for all gases.
• N = number of molecules; if P, V, T are same, N is same for all gases.
• Avogadro number N_A = 6.02 × 10²³ molecules in 22.4 L of any gas.
• Ideal gas equation: PV = BCRT, where BC = number of moles, R = N_A k_B = 8.314 J mol⁻¹ K⁻¹.
• BC = M / M_0 = N / N_A, where M = mass, M_0 = molar mass.
• Alternate form: PV = k_B N T or P = k_B n T, where n = number density.
• Real gases behave ideally at low pressure and high temperature.
• Pressure formula: P = (M / M_0)(RT / V) = C1 RT / M_0, where C1 = mass density.
• Ideal gas is a theoretical model; no real gas is perfectly ideal.

Solved Examples

• Ex 1: Calculate pressure of 2 moles of ideal gas at 300 K occupying 10 L.
  • Use PV = BCRT
  • P = BCRT / V = (2)(8.314)(300) / 0.01 = 498840 Pa

Practice Set

• Q1: Avogadro number is
  • A) 6.02 × 10²⁰
  • B) 6.02 × 10²³
  • C) 1.38 × 10⁻²³
  • D) 8.314
• Q2: For an ideal gas, PV / BC T is
  • A) Variable with pressure
  • B) Constant and equals R
  • C) Zero
  • D) Depends on volume
• Q3: Pressure of 1 mole of gas at 273 K and volume 22.4 L is
  • A) 1 atm
  • B) 2 atm
  • C) 0.5 atm
  • D) 22.4 atm

Answer Key

• Q1: B; Avogadro number is 6.02 × 10²³ molecules/mol.
• Q2: B; For ideal gas, PV / BC T = R constant.
• Q3: A; Using PV = BCRT, P = (1)(0.0821)(273)/22.4 ≈ 1 atm.

Memory Aids

• Ideal gas constant R = 8.314 J/mol·K or 0.0821 L·atm/mol·K.
• Avogadro number = 6.02 × 10²³ molecules/mol.
• Use PV = BCRT for ideal gases.

Boyle’s Law

Key Points

• At constant temperature and moles, PV = constant.
• Pressure varies inversely with volume: P B1 1/V.
• Valid for ideal gases at low pressure and high temperature.
• Experimental P–V curves for steam agree well with Boyle’s law at high T and low P.

Solved Examples

• Ex 1: If volume of gas is halved at constant T, pressure
  • Initial: P_1 V_1 = constant
  • Final: P_2 V_2 = P_1 V_1
  • Since V_2 = V_1/2, P_2 = 2 P_1

Practice Set

• Q1: If volume doubles at constant T, pressure
  • A) Doubles
  • B) Halves
  • C) Remains same
  • D) Quadruples
• Q2: For a gas at constant T, if pressure is 3 atm at 2 L, volume at 6 atm is
  • A) 1 L
  • B) 0.67 L
  • C) 3 L
  • D) 4 L
• Q3: A gas at 1 atm and 4 L is compressed to 2 L at constant T. New pressure is
  • A) 0.5 atm
  • B) 1 atm
  • C) 2 atm
  • D) 4 atm

Answer Key

• Q1: B; Pressure inversely proportional to volume.
• Q2: B; P_1 V_1 = P_2 V_2 1 D3 D3 0.67 L.
• Q3: C; P_2 = P_1 V_1 / V_2 = 1 D7 4 / 2 = 2 atm.

Memory Aids

• Boyle’s law: P V = constant at constant T.
• Pressure and volume are inversely related.
• Use P_1 V_1 = P_2 V_2 for calculations.

Charles’ Law

Key Points

• At constant pressure, volume is proportional to absolute temperature: V B1 T.
• Expressed as V / T = constant at constant P.
• Valid for ideal gases at low pressure and high temperature.
• Experimental T–V curves for CO₂ agree well with Charles’ law.

Solved Examples

• Ex 1: Volume of gas at 300 K is 2 L. Find volume at 600 K at constant pressure.
  • V_1 / T_1 = V_2 / T_2
  • V_2 = V_1 T_2 / T_1 = 2 D7 600 / 300 = 4 L

Practice Set

• Q1: If temperature doubles at constant pressure, volume
  • A) Halves
  • B) Doubles
  • C) Remains same
  • D) Quadruples
• Q2: Volume of gas is 3 L at 300 K. Volume at 450 K is
  • A) 2 L
  • B) 3 L
  • C) 4.5 L
  • D) 6 L
• Q3: A gas at 400 K has volume 5 L. Find volume at 200 K at constant pressure.
  • A) 2.5 L
  • B) 10 L
  • C) 5 L
  • D) 1 L

Answer Key

• Q1: B; Volume proportional to temperature.
• Q2: C; V_2 = 3 D7 450 / 300 = 4.5 L.
• Q3: A; V_2 = 5 D7 200 / 400 = 2.5 L.

Memory Aids

• Charles’ law: V / T = constant at constant pressure.
• Volume changes directly with absolute temperature.
• Use V_1 / T_1 = V_2 / T_2 for calculations.

Dalton’s Law of Partial Pressures

Key Points

• Total pressure of gas mixture = sum of partial pressures: P_total = P_1 + P_2 + P_3 + 26.
• At constant T and V, P_i = BC_i R T / V.
• Total pressure: P_total = (R T / V)(BC_1 + BC_2 + 26) = (R T / V) BC_{total}.
• Partial pressure of i-th gas: P_i = P_{total} (BC_i / BC_{total}), where mole fraction = BC_i / BC_{total}.

Solved Examples

• Ex 1: In a mixture of 2 moles of gas A and 3 moles of gas B at 300 K and volume 10 L, find partial pressure of gas A.
  • Total moles = 5
  • Total pressure P = BC R T / V = 5 D7 8.314 D7 300 / 0.01 = 1.245 D7 10^6 Pa
  • Partial pressure of A: P_A = P BC_A / BC_{total} = 1.245 D7 10^6 D7 2/5 = 4.98 D7 10^5 Pa

Practice Set

• Q1: Total pressure of mixture of 1 mole gas A and 2 moles gas B is 3 atm. Partial pressure of gas A is
  • A) 1 atm
  • B) 2 atm
  • C) 3 atm
  • D) 0.5 atm
• Q2: Mole fraction of gas B in mixture of 3 moles A and 2 moles B is
  • A) 0.4
  • B) 0.6
  • C) 0.3
  • D) 0.5
• Q3: In a gas mixture, partial pressure of gas C is 2 atm and mole fraction is 0.25. Total pressure is
  • A) 0.5 atm
  • B) 8 atm
  • C) 2 atm
  • D) 4 atm

Answer Key

• Q1: A; Partial pressure = total pressure × mole fraction = 3 × 1/3 = 1 atm.
• Q2: A; Mole fraction = 2 / (3+2) = 0.4.
• Q3: B; Total pressure = partial pressure / mole fraction = 2 / 0.25 = 8 atm.

Memory Aids

• Dalton’s law: P_total = 26 P_i.
• Partial pressure proportional to mole fraction.
• Use P_i = P_{total} BC_i / BC_{total}.

Kinetic Theory of an Ideal Gas

Basic Concepts

• Gas molecules move randomly in all directions inside a container.
• Collisions with container walls exert force, creating pressure.
• Intermolecular distances are much larger than molecule size.
• Collisions are perfectly elastic; no energy loss.
• Molecules possess only kinetic energy.
• Gas density is uniform throughout the container.

Solved Examples

• Ex 1: Calculate pressure exerted by N molecules each of mass m moving with velocity component \(v_x\) in a cube of side L and face area A.
  Solution:
  Change in momentum \(\Delta p_x = 2mv_x\)
  Time between collisions \(\Delta t = \frac{2L}{v_x}\)
  Force by one molecule \(F = \frac{\Delta p_x}{\Delta t} = \frac{mv_x^2}{L}\)
  Total force \(F_{avg} = \frac{N m v_x^2}{L}\)
  Pressure \(P = \frac{F_{avg}}{A} = \frac{N m v_x^2}{A L}\)

Practice Set

• Q1: Which assumption is NOT part of kinetic theory?
  A) Molecules are in constant random motion
  B) Collisions are perfectly elastic
  C) Molecules attract each other continuously
  D) Energy is purely kinetic
  
• Q2: Pressure exerted by gas is proportional to:
  A) Average kinetic energy of molecules
  B) Volume of container
  C) Number of container walls
  D) Molecular size
  
• Q3: For gas molecules with average speed squared \(\overline{v^2}\), mass m, number N, and volume V, pressure P is:
  A) \(P = \frac{N m \overline{v^2}}{3V}\)
  B) \(P = \frac{3 N m \overline{v^2}}{V}\)
  C) \(P = \frac{N m \overline{v^2}}{V}\)
  D) \(P = \frac{N m \overline{v^2}}{6V}\)

Answer Key

• Q1: C - Molecules do not exert continuous attraction; forces act only during collisions.
• Q2: A - Pressure depends on average kinetic energy (motion) of molecules.
• Q3: A - Pressure formula from kinetic theory is \(P = \frac{N m \overline{v^2}}{3V}\).

Memory Aids

• Pressure \(P = \frac{1}{3} \rho \overline{v^2}\) relates pressure to density and average speed squared.
• Collisions are elastic → no energy loss.
• Gas molecules move randomly → kinetic energy only.

Kinetic Interpretation of Temperature

Key Points

• Internal energy \(E = N \times \frac{1}{2} m \overline{v^2}\) is purely kinetic for ideal gases.
• From ideal gas law and kinetic theory: \(PV = \frac{2}{3} E\).
• Average kinetic energy per molecule: \(\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T\).
• Temperature \(T\) is proportional to average kinetic energy.
• Independent of pressure, volume, or gas type.
• Dalton's law: total pressure of gas mixture \(P = (n_1 + n_2 + ...) k_B T\).
• Root mean square speed: \(v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M_0}}\).

Solved Examples

• Ex 1: Find average kinetic energy of a molecule at temperature T.
  Solution:
  \(KE_{avg} = \frac{3}{2} k_B T\)
• Ex 2: Calculate \(v_{rms}\) for oxygen (Molar mass \(M_0 = 32 \times 10^{-3} kg/mol\)) at 300 K.
  Solution:
  \(v_{rms} = \sqrt{\frac{3 R T}{M_0}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} \approx 480 m/s\)

Practice Set

• Q1: Average kinetic energy of a molecule is proportional to:
  A) Pressure
  B) Volume
  C) Absolute temperature
  D) Molecular mass
  
• Q2: Internal energy of an ideal gas depends on:
  A) Pressure and volume
  B) Temperature only
  C) Number of molecules and temperature
  D) Volume only
  
• Q3: Root mean square speed \(v_{rms}\) increases with:
  A) Increasing molar mass
  B) Decreasing temperature
  C) Increasing temperature
  D) Increasing pressure
  

Answer Key

• Q1: C - Average kinetic energy \(\propto T\).
• Q2: C - Internal energy depends on number of molecules and temperature.
• Q3: C - \(v_{rms} \propto \sqrt{T}\), increases with temperature.

Memory Aids

• \(E = \frac{3}{2} N k_B T\) links energy and temperature.
• \(v_{rms} = \sqrt{\frac{3RT}{M_0}}\) for quick speed calculation.
• Temperature measures average kinetic energy.

Maxwellian Distribution of Molecular Speeds

Essentials

• Molecular speeds vary due to collisions but distribution remains steady.
• Maxwell's distribution function:
  \(dN = 4 \pi N \left(\frac{M}{2 \pi R T}\right)^{3/2} v^2 e^{-\frac{M v^2}{2 R T}} dv\)
• Fraction of molecules with speed between \(v\) and \(v + dv\) is constant over time.
• Graph plots fraction \(\frac{1}{N} \frac{dN}{dv}\) vs speed \(v\).
• Key speeds:
  - Most probable speed \(v_{mp} = \sqrt{\frac{2 k_B T}{m}} = \sqrt{\frac{2 R T}{M_0}}\)
  - Average speed \(v_{av} = \sqrt{\frac{8 k_B T}{\pi m}} = \sqrt{\frac{8 R T}{\pi M_0}}\)
  - Root mean square speed \(v_{rms} > v_{av} > v_{mp}\)
• Area under curve = 1 (total fraction).
• Curve is asymmetric; average speed differs from most probable speed.

Solved Examples

• Ex 1: Calculate most probable speed for nitrogen (Molar mass \(M_0 = 28 \times 10^{-3} kg/mol\)) at 300 K.
  Solution:
  \(v_{mp} = \sqrt{\frac{2 R T}{M_0}} = \sqrt{\frac{2 \times 8.314 \times 300}{0.028}} \approx 422 m/s\)
• Ex 2: Order the speeds: \(v_{mp}, v_{av}, v_{rms}\).
  Solution:
  \(v_{rms} > v_{av} > v_{mp}\)

Practice Set

• Q1: Which speed is the highest?
  A) Most probable speed
  B) Average speed
  C) Root mean square speed
  D) All equal
  
• Q2: The Maxwell distribution curve area under between speeds \(v_1\) and \(v_2\) gives:
  A) Total number of molecules
  B) Fraction of molecules with speeds in \(v_1\) to \(v_2\)
  C) Average speed
  D) Pressure
  
• Q3: The speed at which maximum molecules move is:
  A) Average speed
  B) Root mean square speed
  C) Most probable speed
  D) None
  

Answer Key

• Q1: C - Root mean square speed is highest.
• Q2: B - Area under curve between speeds gives fraction of molecules.
• Q3: C - Most probable speed corresponds to peak of distribution.

Memory Aids

• Order speeds: \(v_{rms} > v_{av} > v_{mp}\).
• Maxwell distribution shows speed variation in gases.
• Area under curve = 1 (total molecules fraction).

Law of Equipartition of Energy

Fundamental Principle

• In thermal equilibrium, total energy is equally divided among all energy modes.
• Each degree of freedom has average energy \( \frac{1}{2}kT \), where \( k \) is Boltzmann constant and \( T \) is temperature in Kelvin.
• Translational and rotational degrees of freedom contribute \( \frac{1}{2}kT \) each.
• Each vibrational mode contributes \( kT \) (kinetic + potential energy).

Energy of Monoatomic Molecules

• Only translational motion considered.
• Energy per molecule: \( E = \frac{1}{2}m\overline{v_x^2} + \frac{1}{2}m\overline{v_y^2} + \frac{1}{2}m\overline{v_z^2} = \frac{3}{2}kT \).

Energy of Diatomic Molecules (No Vibration)

• Translational + rotational energy considered.
• Energy per molecule: \( E = E_t + E_r = \frac{1}{2}m(v_x^2 + v_y^2 + v_z^2) + \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 = \frac{5}{2}kT \).

Energy Including Vibrational Mode (Diatomic)

• Vibrational energy: \( E_v = \frac{1}{2}m\left(\frac{dy}{dt}\right)^2 + \frac{1}{2}ky^2 \).
• Total energy: \( E = E_t + E_r + E_v \).

Solved Examples

• Ex 1: Calculate average kinetic energy of a monoatomic gas molecule at temperature \( T = 300\,K \).
  • Use \( E = \frac{3}{2}kT \).
  • Substitute \( k = 1.38 \times 10^{-23} J/K \), \( T = 300 K \).
  • \( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} J \).
• Ex 2: Find total energy per molecule of a diatomic gas (no vibration) at \( 400\,K \).
  • Use \( E = \frac{5}{2}kT \).
  • \( E = \frac{5}{2} \times 1.38 \times 10^{-23} \times 400 = 1.38 \times 10^{-20} J \).

Practice Set

• Q1: What is the average energy per degree of freedom at temperature \( T \)?
  • A) \( kT \)
  • B) \( \frac{1}{2}kT \)
  • C) \( 2kT \)
  • D) \( \frac{3}{2}kT \)
• Q2: The total energy of a monoatomic gas molecule at temperature \( 500\,K \) is?
  • A) \( \frac{3}{2}k \times 500 \)
  • B) \( \frac{5}{2}k \times 500 \)
  • C) \( k \times 500 \)
  • D) \( 2k \times 500 \)
• Q3: For a diatomic molecule including vibrational mode, total energy per molecule is?
  • A) \( \frac{5}{2}kT \)
  • B) \( \frac{7}{2}kT \)
  • C) \( \frac{3}{2}kT \)
  • D) \( 3kT \)

Answer Key

• Q1: B) \( \frac{1}{2}kT \) — Each degree of freedom contributes half \( kT \) energy.
• Q2: A) \( \frac{3}{2}k \times 500 \) — Monoatomic gas energy is \( \frac{3}{2}kT \).
• Q3: D) \( 3kT \) — Vibrational mode adds \( kT \) to \( \frac{5}{2}kT \), total \( 3kT \).

Memory Aids

• Each degree of freedom = \( \frac{1}{2}kT \).
• Monoatomic gas energy = \( \frac{3}{2}kT \) (3 translational).
• Diatomic gas (no vibration) = \( \frac{5}{2}kT \) (3 translational + 2 rotational).
• Vibrational mode adds \( kT \) (kinetic + potential).

Mean Free Path

Concept and Formula

• Mean free path (\(\lambda\)) is the average distance a gas molecule travels between collisions.
• In time \(t\), a molecule moves distance \(vt\) and collides with molecules in a cylindrical volume \(V = \pi d^2 vt\), where \(d\) is molecular diameter.
• Number density \(n = \frac{N}{V}\), where \(N\) is number of molecules in volume \(V\).
• Number of collisions in volume \(V\) is \(N = n \pi d^2 vt\).
• Mean free path formula ignoring molecular motion: \[ \lambda = \frac{vt}{N} = \frac{1}{n \pi d^2} \]
• Exact formula considering all molecular motion: \[ \lambda = \frac{1}{\sqrt{2} n \pi d^2} \]

Solved Examples

• Ex 1: Calculate \(\lambda\) if \(n = 2.5 \times 10^{25} \text{ m}^{-3}\) and \(d = 3 \times 10^{-10} \text{ m}\).
  • Use \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\)
  • \(\lambda = \frac{1}{\sqrt{2} \times 2.5 \times 10^{25} \times \pi \times (3 \times 10^{-10})^2} \approx 1.06 \times 10^{-7} \text{ m}\)

Practice Set

• Q1: Mean free path depends inversely on
  • A) Number density \(n\)
  • B) Molecular diameter \(d\)
  • C) Temperature
  • D) Pressure
• Q2: If molecular diameter doubles, mean free path becomes
  • A) Half
  • B) One-fourth
  • C) Double
  • D) Unchanged
• Q3: For gas with \(n = 3 \times 10^{25} \text{ m}^{-3}\), \(d = 2 \times 10^{-10} \text{ m}\), find \(\lambda\).
  • A) \(1.88 \times 10^{-7} \text{ m}\)
  • B) \(2.5 \times 10^{-7} \text{ m}\)
  • C) \(3.5 \times 10^{-7} \text{ m}\)
  • D) \(4.0 \times 10^{-7} \text{ m}\)

Answer Key

• Q1: A — Mean free path \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\) inversely depends on number density \(n\).
• Q2: B — \(\lambda \propto \frac{1}{d^2}\), doubling \(d\) reduces \(\lambda\) by factor 4.
• Q3: A — Using formula, \(\lambda = \frac{1}{\sqrt{2} \times 3 \times 10^{25} \times \pi \times (2 \times 10^{-10})^2} \approx 1.88 \times 10^{-7} \text{ m}\).

Memory Aids

• Mean free path \(\lambda\) inversely proportional to \(n\) and \(d^2\).
• Exact formula includes \(\sqrt{2}\) in denominator.
• Think "collision cylinder" volume \(\pi d^2 vt\) for collisions.

Avogadro’s Number

Definition and Value

• Avogadro’s number is the number of units (atoms, molecules, ions, electrons) in one mole of a substance.
• Value: \(6.022140857 \times 10^{23}\) units/mole.
• Used to relate microscopic particles to macroscopic amounts.

Solved Examples

• Ex 1: Number of molecules in 2 moles of oxygen gas.
  • Number = \(2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}\) molecules.

Practice Set

• Q1: Avogadro’s number represents
  • A) Number of atoms in 1 gram
  • B) Number of units in 1 mole
  • C) Number of molecules in 1 liter
  • D) Number of ions in 1 mole
• Q2: Number of molecules in 0.5 mole of water is
  • A) \(3.01 \times 10^{23}\)
  • B) \(6.02 \times 10^{23}\)
  • C) \(1.20 \times 10^{24}\)
  • D) \(9.03 \times 10^{23}\)
• Q3: Number of atoms in 3 moles of helium gas is
  • A) \(1.81 \times 10^{24}\)
  • B) \(6.02 \times 10^{23}\)
  • C) \(3.00 \times 10^{23}\)
  • D) \(9.03 \times 10^{23}\)

Answer Key

• Q1: B — Avogadro’s number is units per mole.
• Q2: A — \(0.5 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}\) molecules.
• Q3: A — \(3 \times 6.022 \times 10^{23} = 1.81 \times 10^{24}\) atoms.

Memory Aids

• Avogadro’s number = \(6.022 \times 10^{23}\) units/mole.
• Relates microscopic particles to macroscopic moles.
• Use to convert moles ↔ number of particles.