Unit 2: Kinetic Model Evidence & Gas Laws

Understanding gas behavior through kinetic theory and mathematical relationships.

2.16 Kinetic Theory of Gases

The Kinetic Theory of Gases is a model that explains the behavior of gases based on the motion of their constituent particles. This theory assumes that gases consist of a large number of tiny particles in constant, random motion, and provides a molecular explanation for macroscopic gas properties like pressure, temperature, and volume.

Key Assumptions of the Kinetic Theory:
  • Gas particles are in constant, random, straight-line motion until they collide.
  • The volume of the gas particles themselves is negligible compared to the total volume of the gas.
  • There are no intermolecular forces between gas particles (ideal gas assumption).
  • Collisions between gas particles and with container walls are perfectly elastic (no energy is lost).
  • The average kinetic energy of gas particles is directly proportional to absolute temperature: $\overline{KE} = \frac{3}{2}kT$, where $k$ is Boltzmann's constant.
  • At any given temperature, different gases have the same average kinetic energy.
Explaining Gas Properties:
  • Pressure: Results from collisions of gas particles with container walls. More frequent or forceful collisions increase pressure.
  • Temperature: A measure of the average kinetic energy of gas particles. Higher temperature means faster-moving particles.
  • Volume: The space that gas particles occupy and move within. Gases expand to fill their containers completely.
Solved Examples:
  1. Explain why increasing temperature increases gas pressure in a fixed volume.
    Solution: According to kinetic theory, increasing temperature increases the average kinetic energy of gas particles. This means particles move faster and collide with container walls more frequently and with greater force, resulting in increased pressure.
  2. Why do gases expand when heated at constant pressure?
    Solution: When heated, gas particles gain kinetic energy and move faster. To maintain constant pressure (constant collision force per unit area), the volume must increase to reduce the frequency of collisions with the walls.
  3. Compare the average kinetic energies of $H_2$ and $O_2$ molecules at $25°C$.
    Solution: At the same temperature, all gases have the same average kinetic energy according to kinetic theory. Therefore, $H_2$ and $O_2$ molecules at $25°C$ have identical average kinetic energies, despite their different masses.
  4. Explain why gas pressure decreases when volume increases at constant temperature.
    Solution: When volume increases, gas particles have more space to move in. This reduces the frequency of collisions with container walls while maintaining the same collision force, resulting in decreased pressure.
  5. Why are collisions between gas particles considered elastic?
    Solution: Elastic collisions mean no kinetic energy is lost during particle collisions. This assumption explains why gases maintain their motion indefinitely and why temperature remains constant in a closed system without external energy input.

2.17 Brownian Motion & Diffusion

Brownian Motion is the random, zigzag movement of small particles suspended in a fluid (liquid or gas) as they are bombarded by the moving molecules of the fluid. This phenomenon, first observed by botanist Robert Brown in 1827, provides direct visual evidence for the kinetic theory of matter.

Characteristics of Brownian Motion:
  • Movement is random and unpredictable in direction.
  • Smaller particles exhibit more pronounced Brownian motion.
  • Higher temperatures increase the intensity of the motion.
  • Motion is continuous and never stops.
  • The motion is more noticeable in less viscous fluids.

Diffusion is the net movement of particles from a region of high concentration to a region of low concentration. This process occurs due to the random motion of particles and continues until equilibrium (uniform concentration) is reached. Diffusion provides evidence for particle motion and explains many everyday phenomena.

Factors Affecting Diffusion Rate:
  • Temperature: Higher temperature increases particle kinetic energy, leading to faster diffusion.
  • Particle Size/Mass: Smaller, lighter particles diffuse faster than larger, heavier ones.
  • Concentration Gradient: Greater concentration differences lead to faster diffusion rates.
  • Medium: Diffusion is fastest in gases, slower in liquids, and slowest in solids.
  • Distance: Diffusion rate decreases with increasing distance.
Solved Examples:
  1. Describe what you would observe when viewing pollen grains in water under a microscope.
    Solution: You would observe the pollen grains moving in random, jerky, zigzag patterns. This Brownian motion occurs because water molecules are constantly colliding with the pollen grains from all directions, causing the unpredictable movement that provides evidence for molecular motion.
  2. Explain why perfume scent spreads throughout a room.
    Solution: Perfume molecules undergo diffusion, moving from the high concentration area near the perfume bottle to areas of lower concentration throughout the room. The random motion of air molecules helps distribute the perfume molecules until they are evenly spread.
  3. Why does sugar dissolve faster in hot tea than in cold tea?
    Solution: In hot tea, both sugar and water molecules have higher kinetic energy, moving faster. This increased molecular motion accelerates the diffusion process of sugar molecules throughout the liquid, leading to faster dissolution.
  4. Compare the diffusion rates of hydrogen gas and carbon dioxide gas at the same temperature.
    Solution: Hydrogen gas ($H_2$, molar mass = 2 g/mol) will diffuse faster than carbon dioxide ($CO_2$, molar mass = 44 g/mol) because lighter molecules move faster at the same temperature. The diffusion rate is inversely related to molecular mass.
  5. Explain why Brownian motion is more pronounced at higher temperatures.
    Solution: At higher temperatures, fluid molecules have greater kinetic energy and move faster. This results in more frequent and forceful collisions with suspended particles, causing more vigorous and noticeable Brownian motion.

2.18 Graham's Law of Diffusion

Graham's Law of Diffusion, formulated by Scottish chemist Thomas Graham in 1846, describes the relationship between the rate of diffusion of gases and their molar masses. The law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Mathematical Expression:

For two gases A and B diffusing under the same conditions:

$\frac{\text{Rate of diffusion of A}}{\text{Rate of diffusion of B}} = \sqrt{\frac{M_B}{M_A}}$

Where $M_A$ and $M_B$ are the molar masses of gases A and B respectively.

Key Points:
  • Lighter gases diffuse faster than heavier gases.
  • The law applies to both diffusion and effusion (gas escaping through a small hole).
  • Temperature and pressure must be constant for accurate comparison.
  • The relationship is based on the kinetic theory assumption that average kinetic energy is the same for all gases at the same temperature.
Derivation from Kinetic Theory:

Since average kinetic energy is the same for all gases at the same temperature:

$$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2$$

Rearranging: $\frac{v_A}{v_B} = \sqrt{\frac{m_B}{m_A}} = \sqrt{\frac{M_B}{M_A}}$

Since diffusion rate is proportional to average speed, Graham's Law follows naturally.

Solved Examples:
  1. Calculate the relative rate of diffusion of hydrogen gas ($H_2$) compared to oxygen gas ($O_2$) at the same temperature.
    Solution: Using Graham's Law: $\frac{\text{Rate of }H_2}{\text{Rate of }O_2} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$
    Therefore, hydrogen diffuses 4 times faster than oxygen.
  2. If helium ($He$) effuses through a small hole at a rate of 12.0 mL/min, at what rate will carbon dioxide ($CO_2$) effuse under the same conditions?
    Solution: $\frac{\text{Rate of }He}{\text{Rate of }CO_2} = \sqrt{\frac{M_{CO_2}}{M_{He}}} = \sqrt{\frac{44}{4}} = \sqrt{11} = 3.32$
    Rate of $CO_2$ = $\frac{12.0}{3.32} = 3.61$ mL/min
  3. An unknown gas diffuses 0.68 times as fast as nitrogen gas ($N_2$). Calculate the molar mass of the unknown gas.
    Solution: Let the unknown gas be X. $\frac{\text{Rate of X}}{\text{Rate of }N_2} = 0.68 = \sqrt{\frac{M_{N_2}}{M_X}} = \sqrt{\frac{28}{M_X}}$
    Squaring both sides: $0.68^2 = \frac{28}{M_X}$
    $M_X = \frac{28}{0.68^2} = \frac{28}{0.462} = 60.6$ g/mol
  4. Explain why hydrogen gas escapes from balloons faster than helium gas.
    Solution: According to Graham's Law, the rate of effusion is inversely proportional to the square root of molar mass. Since hydrogen ($M = 2$ g/mol) has a lower molar mass than helium ($M = 4$ g/mol), it effuses $\sqrt{\frac{4}{2}} = \sqrt{2} = 1.41$ times faster, causing hydrogen balloons to deflate more quickly.
  5. Two gases are allowed to diffuse simultaneously. Gas A has a molar mass of 16 g/mol and gas B has a molar mass of 64 g/mol. If gas A travels 20 cm in a certain time, how far will gas B travel in the same time?
    Solution: $\frac{\text{Rate of A}}{\text{Rate of B}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2$
    Since gas A diffuses twice as fast, if A travels 20 cm, then B travels $\frac{20}{2} = 10$ cm in the same time.

2.19 Boyle's, Charles' & Gay-Lussac's Laws

The gas laws describe the mathematical relationships between pressure, volume, temperature, and amount of gas. These empirical laws, discovered through experimentation, can be explained using kinetic theory and form the foundation for understanding gas behavior.

1. Boyle's Law (Pressure-Volume Relationship)

At constant temperature and amount of gas, the pressure of a gas is inversely proportional to its volume.

$P \propto \frac{1}{V} \quad \text{or} \quad PV = \text{constant}$
$P_1V_1 = P_2V_2$

  • Discovered by Robert Boyle in 1662.
  • As volume increases, pressure decreases proportionally.
  • Graph of P vs V is a hyperbola; P vs 1/V is a straight line.
2. Charles' Law (Volume-Temperature Relationship)

At constant pressure and amount of gas, the volume of a gas is directly proportional to its absolute temperature.

$$V \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant}$$
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

  • Discovered by Jacques Charles in 1787.
  • Temperature must be in Kelvin (K).
  • Graph of V vs T is a straight line that extrapolates to absolute zero.
  • At $0 K$ (absolute zero), the volume theoretically becomes zero.
3. Gay-Lussac's Law (Pressure-Temperature Relationship)

At constant volume and amount of gas, the pressure of a gas is directly proportional to its absolute temperature.

$$P \propto T \quad \text{or} \quad \frac{P}{T} = \text{constant}$$
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

  • Discovered by Joseph Gay-Lussac in 1808.
  • Also known as Amonton's Law (Guillaume Amontons, 1702).
  • Temperature must be in Kelvin (K).
  • Graph of P vs T is a straight line through the origin.
Solved Examples:
  1. A gas occupies 2.5 L at 1.2 atm pressure. What volume will it occupy at 2.4 atm if temperature remains constant?
    Solution: Using Boyle's Law: $P_1V_1 = P_2V_2$
    $(1.2 \text{ atm})(2.5 \text{ L}) = (2.4 \text{ atm})(V_2)$
    $V_2 = \frac{(1.2)(2.5)}{2.4} = 1.25 \text{ L}$
  2. A balloon contains 3.0 L of gas at $27°C$. What will be its volume at $127°C$ if pressure remains constant?
    Solution: Using Charles' Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
    First convert to Kelvin: $T_1 = 27 + 273 = 300 \text{ K}$, $T_2 = 127 + 273 = 400 \text{ K}$
    $\frac{3.0}{300} = \frac{V_2}{400}$
    $V_2 = \frac{(3.0)(400)}{300} = 4.0 \text{ L}$
  3. A gas has a pressure of 2.5 atm at $25°C$. What will be the pressure at $125°C$ if volume remains constant?
    Solution: Using Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
    Convert to Kelvin: $T_1 = 25 + 273 = 298 \text{ K}$, $T_2 = 125 + 273 = 398 \text{ K}$
    $\frac{2.5}{298} = \frac{P_2}{398}$
    $P_2 = \frac{(2.5)(398)}{298} = 3.34 \text{ atm}$
  4. Explain Boyle's Law using kinetic theory.
    Solution: When volume decreases at constant temperature, gas particles have less space to move. This increases the frequency of collisions with container walls while maintaining the same collision force (constant temperature), resulting in increased pressure. The inverse relationship occurs because $P \times V =$ constant collision frequency $\times$ constant force.
  5. A gas sample at STP (Standard Temperature and Pressure: 0°C, 1 atm) occupies 22.4 L. What volume will it occupy at 2 atm and 273°C?
    Solution: This requires both Boyle's and Charles' Laws. We can solve step by step:
    First, pressure change (Boyle's): $V_{intermediate} = \frac{P_1V_1}{P_2} = \frac{(1)(22.4)}{2} = 11.2 \text{ L}$
    Then, temperature change (Charles'): $T_1 = 273 \text{ K}$, $T_2 = 273 + 273 = 546 \text{ K}$
    $V_2 = \frac{V_{intermediate} \times T_2}{T_1} = \frac{11.2 \times 546}{273} = 22.4 \text{ L}$

2.20 Combined Gas Law

The Combined Gas Law brings together Boyle's Law, Charles' Law, and Gay-Lussac's Law into a single equation. It describes the relationship between pressure, volume, and temperature for a fixed amount of gas when more than one variable changes simultaneously.

Mathematical Expression:

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
Where:

  • $P_1, V_1, T_1$ = initial pressure, volume, and temperature
  • $P_2, V_2, T_2$ = final pressure, volume, and temperature
  • Temperature must always be in Kelvin (K)
  • Pressure and volume can be in any units as long as they're consistent

Special Cases:
  • Constant Temperature ($T_1 = T_2$): Reduces to Boyle's Law: $P_1V_1 = P_2V_2$
  • Constant Pressure ($P_1 = P_2$): Reduces to Charles' Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
  • Constant Volume ($V_1 = V_2$): Reduces to Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Applications:
  • Calculating gas behavior in weather systems
  • Design of pressure vessels and containers
  • Understanding respiratory physiology
  • Scuba diving calculations for gas mixtures
  • Industrial gas processing and storage
Solved Examples:
  1. A gas sample occupies 5.0 L at 2.0 atm and $27°C$. What volume will it occupy at 1.5 atm and $127°C$?
    Solution: Using the Combined Gas Law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
    Given: $P_1 = 2.0$ atm, $V_1 = 5.0$ L, $T_1 = 27 + 273 = 300$ K
    $P_2 = 1.5$ atm, $T_2 = 127 + 273 = 400$ K
    $V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{(2.0)(5.0)(400)}{(1.5)(300)} = \frac{4000}{450} = 8.89$ L
  2. A weather balloon contains 10.0 L of helium at sea level (1.0 atm, $15°C$). At an altitude where pressure is 0.5 atm and temperature is $-40°C$, what will be the balloon's volume?
    Solution: Given: $P_1 = 1.0$ atm, $V_1 = 10.0$ L, $T_1 = 15 + 273 = 288$ K
    $P_2 = 0.5$ atm, $T_2 = -40 + 273 = 233$ K
    $V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{(1.0)(10.0)(233)}{(0.5)(288)} = \frac{2330}{144} = 16.2$ L
  3. A gas cylinder contains gas at 25.0 atm and $20°C$. If the cylinder is heated to $100°C$, what will be the new pressure? (Volume remains constant)
    Solution: Since volume is constant, we use Gay-Lussac's Law (special case of Combined Gas Law):
    $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
    $T_1 = 20 + 273 = 293$ K, $T_2 = 100 + 273 = 373$ K
    $P_2 = \frac{P_1T_2}{T_1} = \frac{(25.0)(373)}{293} = 31.8$ atm
  4. A scuba tank contains air at 200 atm and $25°C$. If the tank cools to $5°C$ underwater, what will be the new pressure?
    Solution: Using Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
    $T_1 = 25 + 273 = 298$ K, $T_2 = 5 + 273 = 278$ K
    $P_2 = \frac{P_1T_2}{T_1} = \frac{(200)(278)}{298} = 186.6$ atm
  5. A gas occupies 2.0 L at STP (0°C, 1 atm). What pressure is needed to compress it to 0.5 L at $50°C$?
    Solution: Using Combined Gas Law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
    $P_1 = 1$ atm, $V_1 = 2.0$ L, $T_1 = 273$ K
    $V_2 = 0.5$ L, $T_2 = 50 + 273 = 323$ K
    $P_2 = \frac{P_1V_1T_2}{V_2T_1} = \frac{(1)(2.0)(323)}{(0.5)(273)} = \frac{646}{136.5} = 4.73$ atm

2.21 Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures, formulated by John Dalton in 1801, states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. Each gas in the mixture behaves independently and exerts pressure as if it alone occupied the entire volume.

Mathematical Expression:

$P_{total} = P_1 + P_2 + P_3 + ... + P_n$
Where $P_1, P_2, P_3, ..., P_n$ are the partial pressures of individual gases.

Partial Pressure Calculation:

The partial pressure of a gas in a mixture can be calculated using:
$P_i = X_i \times P_{total}$
Where $X_i$ is the mole fraction of gas i:
$X_i = \frac{n_i}{n_{total}} = \frac{\text{moles of gas i}}{\text{total moles of all gases}}$

Key Concepts:
  • Partial Pressure: The pressure that a gas would exert if it alone occupied the entire volume at the same temperature.
  • Mole Fraction: The ratio of moles of one component to the total moles in the mixture.
  • Independent Behavior: Each gas molecule behaves as if other gas molecules don't exist.
  • The sum of all mole fractions in a mixture equals 1.0.
Applications:
  • Calculating gas compositions in atmospheric studies
  • Medical applications (oxygen therapy, anesthesia)
  • Scuba diving gas mixtures (nitrox, trimix)
  • Industrial gas separation processes
  • Collection of gases over water (water vapor pressure correction)
Gas Collection Over Water:

When gases are collected over water, the total pressure includes water vapor pressure:

$P_{total} = P_{gas} + P_{H_2O(vapor)}$
Therefore: $P_{gas} = P_{total} - P_{H_2O(vapor)}$

Solved Examples:
  1. A mixture contains 2.0 mol of nitrogen, 1.0 mol of oxygen, and 0.5 mol of argon at a total pressure of 1.5 atm. Calculate the partial pressure of each gas.
    Solution: Total moles = 2.0 + 1.0 + 0.5 = 3.5 mol
    Mole fractions: $X_{N_2} = \frac{2.0}{3.5} = 0.571$, $X_{O_2} = \frac{1.0}{3.5} = 0.286$, $X_{Ar} = \frac{0.5}{3.5} = 0.143$
    Partial pressures: $P_{N_2} = 0.571 \times 1.5 = 0.857$ atm
    $P_{O_2} = 0.286 \times 1.5 = 0.429$ atm, $P_{Ar} = 0.143 \times 1.5 = 0.214$ atm
    Check: 0.857 + 0.429 + 0.214 = 1.500 atm ✓
  2. Air is approximately 78% nitrogen and 21% oxygen by volume. If atmospheric pressure is 760 mmHg, what are the partial pressures of nitrogen and oxygen?
    Solution: For gases, volume percentage equals mole percentage.
    $P_{N_2} = 0.78 \times 760 = 593$ mmHg
    $P_{O_2} = 0.21 \times 760 = 160$ mmHg
    (The remaining 1% is mainly argon and other gases: 7 mmHg)
  3. Hydrogen gas is collected over water at $25°C$ and 750 mmHg total pressure. If the vapor pressure of water at $25°C$ is 23.8 mmHg, what is the pressure of the dry hydrogen gas?
    Solution: Using Dalton's Law for gas collection over water:
    $P_{H_2} = P_{total} - P_{H_2O(vapor)} = 750 - 23.8 = 726.2$ mmHg
  4. A gas mixture has partial pressures of 0.30 atm COâ‚‚, 0.15 atm Oâ‚‚, and 0.25 atm Nâ‚‚. What is the mole fraction of each gas?
    Solution: First find total pressure: $P_{total} = 0.30 + 0.15 + 0.25 = 0.70$ atm
    Mole fractions: $X_{CO_2} = \frac{0.30}{0.70} = 0.429$
    $X_{O_2} = \frac{0.15}{0.70} = 0.214$, $X_{N_2} = \frac{0.25}{0.70} = 0.357$
    Check: 0.429 + 0.214 + 0.357 = 1.000 ✓
  5. A scuba tank contains a nitrox mixture (nitrogen and oxygen) at 200 atm total pressure. If the oxygen makes up 32% by volume, what are the partial pressures of nitrogen and oxygen?
    Solution: $P_{O_2} = 0.32 \times 200 = 64$ atm
    $P_{N_2} = 0.68 \times 200 = 136$ atm
    Check: 64 + 136 = 200 atm ✓
    This mixture is safer for deep diving as it reduces nitrogen narcosis risk.

Knowledge Check (25 Questions)

Answer: 1) Gas particles are in constant random motion, 2) Particle volume is negligible compared to container volume, 3) No intermolecular forces exist, 4) Collisions are perfectly elastic, 5) Average kinetic energy is proportional to absolute temperature.

Answer: Brownian motion is the random, zigzag movement of small particles suspended in a fluid due to bombardment by fluid molecules. It provides direct visual evidence for the kinetic theory of matter.

Answer: $CH_4$ will diffuse faster because it has a lower molar mass. The rate ratio is $\sqrt{\frac{44}{16}} = 1.66$, so methane diffuses 1.66 times faster than carbon dioxide.

Answer: At constant temperature, the pressure of a gas is inversely proportional to its volume. Mathematical expression: $P_1V_1 = P_2V_2$.

Answer: Using Boyle's Law: $V_2 = \frac{P_1V_1}{P_2} = \frac{(1.5)(4.0)}{3.0} = 2.0$ L.

Answer: At constant pressure, volume is directly proportional to absolute temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. Temperature must be in Kelvin because the relationship is proportional - at 0 K, volume theoretically becomes zero.

Answer: Convert to Kelvin: $T_1 = 300$ K, $T_2 = 400$ K. Using Charles' Law: $V_2 = \frac{V_1T_2}{T_1} = \frac{(2.5)(400)}{300} = 3.33$ L.

Answer: At constant volume, pressure is directly proportional to absolute temperature: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$. We use it when analyzing pressure changes in rigid containers due to temperature changes.

Answer: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Answer: Using Combined Gas Law: $V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{(2.0)(3.0)(348)}{(1.5)(298)} = 4.67$ L.

Answer: The total pressure of a gas mixture equals the sum of the partial pressures of all component gases: $P_{total} = P_1 + P_2 + P_3 + ...$

Answer: Mole fraction is the ratio of moles of one component to total moles: $X_i = \frac{n_i}{n_{total}}$. It equals the ratio of partial pressure to total pressure.

Answer: Total moles = 2.0 mol. Mole fraction of $N_2 = \frac{0.5}{2.0} = 0.25$. Partial pressure = $0.25 \times 2.0 = 0.5$ atm.

Answer: Temperature, particle mass/size, concentration gradient, medium (gas/liquid/solid), and distance.

Answer: Rate ratio = $\frac{8.0}{2.0} = 4 = \sqrt{\frac{M_B}{M_A}}$. Therefore, $\frac{M_B}{M_A} = 16$. Gas B is 16 times heavier than gas A.

Answer: Higher temperature increases average kinetic energy, making particles move faster. This increases both collision frequency and force with container walls, resulting in higher pressure.

Answer: Hydrogen has very low molar mass (2 g/mol), making it highly buoyant. According to Graham's Law, it also diffuses quickly, providing good lift efficiency despite safety concerns.

Answer: The collected gas is saturated with water vapor. By Dalton's Law, total pressure = gas pressure + water vapor pressure. We subtract to find the pressure of just the collected gas.

Answer: Partial pressure of $O_2 = 0.21 \times 1$ atm $= 0.21$ atm or 160 mmHg.

Answer: Average kinetic energy is directly proportional to absolute temperature, so it also doubles when temperature (in Kelvin) doubles.

Answer: Total pressure = 0.4 + 0.6 = 1.0 atm. Mole fraction of $CO_2 = \frac{0.4}{1.0} = 0.4$.

Answer: Rate ratio = $\sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$. Helium diffuses 3.16 times faster than argon.

Answer: Perfume molecules undergo diffusion, moving from high concentration (near the source) to low concentration areas due to random molecular motion, eventually reaching equilibrium.

Answer: Using Combined Gas Law: $P_2 = \frac{P_1V_1T_2}{V_2T_1} = \frac{(1)(V_1)(546)}{(0.5V_1)(273)} = 4.0$ atm.

Answer: Gas particles are far apart with large empty spaces between them. When compressed, these spaces decrease. Liquids and solids have particles already close together with minimal empty space.
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