GAS LAWS CHEM 1 HONORS

Gas Laws Notes

GAS LAWS

KINETIC-MOLECULAR THEORY

• This theory describes the behavior of gases in terms of particles in motion.

• Objects in motion have kinetic energy.

• Gases consist of small particles that are separated from one another by empty space.

• Kinetic energy and temperature are related:
    - Definition: Temperature is a measure of the average kinetic energy of the particles.
    - Higher temperature means more kinetic energy, or faster moving molecules.

KINETIC-MOLECULAR THEORY (KMT) Simplified

• Gas particles are tiny with lots of space between themselves and their neighbors.

• They move in constant, random motion, based on the average temperature of the gas.

• When they eventually run into something along their path, they bounce off without losing energy.

COLLISIONS OF GAS PARTICLES

GASES

• Remember that the types of atoms present (composition) and their arrangement (structure/electron configuration) determine the chemical properties of matter.

• Unlike solids and liquids, gases have very similar physical properties despite having different compositions.
    - Example: Gold does not act like calcium, but oxygen gas acts like nitrogen gas.

• Mixtures: Impure substances consist of multiple “stuff” (chemical compounds) that can be physically separated but can retain all of the properties of their individual components.

BEHAVIOR OF GASES

• Low Density: The low density is from the empty space (volume) between the gas molecules.

• Compression & Expansion:
    - Gases can be compressed, causing the molecules to come closer to one another, similar to squeezing a pillow.
    - The gases will expand back to their original volume after compression.

• Diffusion & Effusion:
    - Diffusion: Random movement of molecules through a medium, occurs from an area of high concentration to low concentration.
    - Effusion: Process where gas escapes from an area of high pressure to low pressure.

DIFFUSION VS. EFFUSION

• Diffusion Example: A scent spreading throughout a room or people entering a theme park.

• Effusion Example: Air slowly leaking out of a tire or helium leaking out of a balloon.

GRAHAM’S LAW OF EFFUSION

• Definition: "At constant temperature and pressure, the rate of effusion is inversely proportional to the square root of their molar masses."

• Since this is the random escape of gas through a small opening, the size of the molecules will affect the rate of escape.
    - Smaller gases will effuse faster.

• Rate of Effusion of Gas A (velocity) / Rate of Effusion of Gas B (velocity) =
    $rac{ ext{Rate of Effusion Gas A}}{ ext{Rate of Effusion Gas B}} = rac{ ext{√Molar Mass Gas B}}{ ext{√Molar Mass Gas A}}$

PRACTICE PROBLEM

• Problem Statement: A certain gas takes three times as long (i.e., 3x slower) to effuse out as helium. Its molecular mass will be:
    - A = Helium; B = "a certain gas".
  
     $rac{v_A}{v_B} = rac{ ext{√Molar Mass Gas B}}{ ext{√Molar Mass Gas A}}$
     $3 = rac{ ext{√Molar Mass Gas B}}{ ext{√4}}$
     $ext{Molar Mass Gas B} = 36u$

PRACTICE PROBLEM

• Problem Statement: What is the time required for one-half of hydrogen to escape?
    - A = Hydrogen ($H_2$); B = Oxygen ($O_2$).
     
     $rac{ ext{Rate of Effusion Gas A}}{ ext{Rate of Effusion Gas B}} = rac{ ext{√Molar Mass Gas B}}{ ext{√Molar Mass Gas A}}$
     - If $x$ = 1, then:
    $ext{Rate of Effusion of Gas A} = 1$
    $ext{Rate of Effusion of Gas B} = x$
     
     $1 = rac{ ext{√32}}{ ext{√2}} <br>ightarrow x = rac{1}{4}$
     - Thus, if 1/2 of hydrogen escapes, then only 1/8 of the oxygen has also escaped.

PRACTICE PROBLEM

• Problem Statement: Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds, respectively. The molecular mass of A is 49u. What is the molecular mass of B?
     - A = A; B = B.
     - If A takes 2x longer than B, that means B is moving 2x faster.
     
     $rac{ ext{Rate of Effusion of Gas A}}{ ext{Rate of Effusion of Gas B}} = rac{ ext{√Molar Mass Gas B}}{ ext{√Molar Mass Gas A}}$
     $1/2 = rac{ ext{√x}}{ ext{√49}} <br>ightarrow x = 12.25$

THE NATURE OF GASES

• All assumptions of the kinetic molecular theory components are based on four factors:
    - Number of Gas Particles: Quantity or the number of moles.
    - Pressure.
    - Temperature.
    - Volume of Gas Sample.

• All four factors work together to determine the behavior of the gas and when one changes, the others are also affected.

GAS PRESSURE

• Definition: Gas pressure is generated when gas molecules hit an object or surface.

• Equation for Pressure:
    $ext{Pressure} = rac{ ext{Force}}{ ext{Area}}$

• Earth's atmosphere is composed of gas molecules moving randomly and being pulled downward due to gravity, creating atmospheric pressure (air pressure) through collisions.

• Observation: Higher altitudes have less air pressure than sea-level due to lower density.

UNITS OF PRESSURE

• Measuring Instruments:
    - Barometers: Measure atmospheric pressure.
    - Manometer: Measures gas pressure in a closed container.

• SI Unit: Pascal (Pa) and often expressed in kilopascals (kPa).
    - Conversion Example: 1 kPa = 1,000 Pa

• Other units include:
    - Pounds per square inch (psi)
    - Millimeters of mercury (mmHg)
    - Torr (Torr)
    - Atmospheres (atm)

• At 0°C and sea level, the following equivalences hold:
    - $1.0 ext{ atm} = 760 ext{ mmHg}$
    - $760 ext{ Torr} = 101.3 ext{ kPa} = 101,300 ext{ Pa}$

PRACTICE PROBLEM

• Problem Statement: Find the force of air pressure acting on a baseball field tarp with an area of 100 ft × 100 ft at sea level.
     - Standard pressure: 1 atm, 101.3 kPa, 760 mm Hg, 14.7 lb/in².
     - Area conversion:
       $A = [100 ext{ ft} imes (1.0 ext{ m}/3.28 ext{ ft})]^2 = 929.5 ext{ m}^2$
     - Use Equation:
     $P = rac{F}{A} <br>ightarrow F = PA$
     $F = 101,300 ext{ Pa} imes 929.5 ext{ m}^2 = 94,158,982.7 ext{ N} <br>ightarrow 9.4 imes 10^7 ext{ N} ext{ or } ext{ ~10,000 tons}$

DALTON’S LAW OF PARTIAL PRESSURES

• Definition: When a mixture of gas is formed, each gas exerts pressure independently of the others, while the total pressure is the summation of each individual pressure:
    $P_ ext{total} = P_{ ext{gas 1}} + P_{ ext{gas 2}} + P_{ ext{gas 3}} + ext{ … }$

PRACTICE PROBLEM

• Problem Statement: A mixture of oxygen gas, carbon dioxide, and nitrogen gas has a total pressure of 0.97 atm. What is the partial pressure of oxygen gas, given that the partial pressure of carbon dioxide is 0.70 atm and nitrogen gas is 0.12 atm?
    - Setup:
    $P_ ext{total} = P_1 + P_2 + P_3$
    $0.97 ext{ atm} = P_{ ext{O}<em>2} + 0.70 ext{ atm} + 0.12 ext{ atm}$   $P</em>{ ext{O}_2} = 0.15 ext{ atm}$

BOYLE’S LAW

• Historian: Robert Boyle (1627-1691) studied the relationship between pressure and volume of a gas.

• Law Statement:
    - Doubling the pressure of a fixed amount of gas decreases its volume by one-half if kept at constant temperature.
    - Reducing the pressure by half results in a doubling of the volume.

• Conclusion: Pressure and volume are inversely proportional.

• Mathematical Formulation:
    - $P_i V_i = P_f V_f$

PRACTICE PROBLEM

• Problem Statement: A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0 L volume is 210 kPa, what will be the pressure at 2.5 L?
     - Application of Boyle's Law:
    $P_i V_i = P_f V_f <br>ightarrow 210 ext{ kPa} imes 4.0 ext{ L} = P_f imes 2.5 ext{ L}$
    $P_f = 335 ext{ kPa} ext{ or } 340 ext{ kPa}$

PRACTICE PROBLEM

• Problem Statement: If an ideal gas ends with 1.9 L at 3.7 atm, then what was its original volume if the starting pressure was 5.6 atm?
     - Application of Boyle's Law:
    $P_i V_i = P_f V_f <br>ightarrow 5.6 ext{ atm} imes V_i = 3.7 ext{ atm} imes 1.9 ext{ L}$
    $V_i = 1.255 ext{ L} ext{ or } 1.3 ext{ L}$

PRESSURE-VOLUME RELATIONSHIP

• Table showing relationship of pressure and volume:
      - P1V1 = P2V2 = P3V3 = 100 L kPa
        - (P1, V1): P1 = 100 kPa, V1 = 1.0 L
        - (P2, V2): P2 = 50 kPa, V2 = 2.0 L
        - (P3, V3): P3 = 200 kPa, V3 = 0.5 L

TEMPERATURE SCALES

• Developed temperature scales allow for the comparison of measurements.
    - Centigrade (Celsius): A scale based on properties of water, devised in 1741 by Swedish physicist Anders Celsius.
    - Fahrenheit: Named after German physicist Gabriel D. Fahrenheit.
    - Kelvin Scale: Absolute temperature scale, named after Lord Kelvin, associated with thermal dynamics.

TEMPERATURE

• Kelvin and Centigrade are metric, Kelvin is the standard metric to avoid negative temperatures which imply negative kinetic energy.

• Equations:
     $T_K = T_{℃} + 273$
     $T_{℉} = rac{9}{5} T_{℃} + 32.0$

• When the Kelvin temperature doubles, the average kinetic energy also doubles.

CHARLES’S LAW

• Historian: French physicist Jacques Charles (1746-1823) studied the relationship between volume and temperature.

• Law Statement: Temperature and Volume are directly proportional when pressure is held constant.

• Equations:
     $rac{V_i}{T_i} = rac{V_f}{T_f} ext{ (Temperatures must be converted to Kelvin!)}$

CHARLES’S LAW - SIMPLIFIED

• Explained using kinetic-molecular theory: As temperature increases, gas molecules move faster, striking the container walls with more force (and more often), which increases volume, pushing the walls back.

PRACTICE PROBLEM

• Problem Statement: A gas sample at 40.0°C occupies a volume of 2.32 L. If the temperature is raised to 75.0°C, what will the volume be, assuming constant pressure?
    - Convert Temperatures:
     - $T1 = 40.0°C + 273 = 313 K$
     - $T2 = 75.0°C + 273 = 348 K$
    - Solution:
     $rac{2.32 L}{313 K} = rac{V2}{348 K} <br>ightarrow V2 = 2.58 L$

PRACTICE PROBLEM

• Problem Statement: If a gas is held at constant pressure, what is the initial temperature at 1.63 L when the gas ends up at a state of 0.89 L and 335 K?
     - Application of Charles's Law:
    $rac{1.63 L}{T_i} = rac{0.89 L}{335 K} <br>ightarrow T_i = 613.5 K ext{ or } 340°C$

GAY-LUSSAC’S LAW

• Definition: If Boyle’s Law deals with pressure and volume while Charles’s deals with temperature and volume, Gay-Lussac’s Law states that raising the temperature of a gas will increase its pressure.

• Kinetic-Molecular Theory Explanation: Higher temperatures equate to more kinetic energy, causing faster-moving particles and resulting in more frequent/harder collisions, thus creating higher pressure.

GAY-LUSSAC’S LAW

• Pressure and Temperature are directly proportional if volume is held constant.

• Equations:
     $rac{P_i}{T_i} = rac{P_f}{T_f} ext{(Temperatures must be in Kelvin!)}$

PRACTICE PROBLEM

• Problem Statement: The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank?
     - Convert Temperatures:
     - $T1 = 22.0°C + 273 = 295 K$
     - $T2 = 60.0°C + 273 = 333 K$
    - Application of Gay-Lussac's Law:
     $rac{3.20 atm}{295 K} = rac{P2}{333 K} <br>ightarrow P2 = 3.61 atm$

PRACTICE PROBLEM

• Problem Statement: Find the final temperature of a gas that was at 77.0°C and 105 kPa, if the pressure changes to 89.0 kPa.
     - Application of Gay-Lussac's Law:
     $rac{105 kPa}{350 K} = rac{89.0 kPa}{Tf}$
     - Solution: $Tf = 296.6 K ext{ or } 23.7°C$

COMBINED GAS LAW

• Definition: A combination of all three previous gas laws.

• Mathematical Expression:
     $rac{P_1 V_1}{T_1} = rac{P_2 V_2}{T_2}$

• Note: Temperature must be in Kelvin; pressure and volume can be any unit as long as they’re consistent on both sides.

PRACTICE PROBLEM

• Problem Statement: A closed system has 1.5 L of gas at 1.1 atm and a temperature of 26°C. What is the final temperature when the pressure increases to 1.8 atm and the volume finishes at 1.9 L?
     - Convert temperature:
     $T = 26°C + 273 = 299 K$
     
     $rac{1.1 atm imes 1.5 L}{299 K} = rac{1.8 atm imes 1.9 L}{Tf} <br>ightarrow Tf = 619.7 K ext{ or } 350°C$

STP (Standard Temperature and Pressure)

• Definition: Pressures and temperatures vary due to differences in altitudes, climates, etc. To standardize laboratory experiments, scientists have agreed upon standard conditions for gas experiments.
    - Standard Temperature = 0°C or 273 K.
    - Standard Pressure = 1.0 atm or 101.325 kPa (760 mmHg).

AVOGADRO’S PRINCIPLE

• Definition: Particles making up a gas mixture will vary in mass and, therefore, size. Amadeo Avogadro (1776-1856) studied the influence of the quantity of molecules/moles on pressure, volume, and temperature.

• Principle Statement: When any gas or gas mixture has equal volumes, temperatures, and pressures, the quantity of gas is always the same.
    - $1.0 mole = 6.02 imes 10^{23} molecules/atoms$
    - At STP, 1.0 mole of gas or gas mixture always occupies 22.4 L.

PRACTICE PROBLEM

• Problem Statement: A gas has a volume of 3.72 L at STP. Find the number of particles.
     - Calculation:
     $3.72 L imes rac{1 ext{ mol}}{22.4 L} imes rac{6.02 imes 10^{23} ext{ molecules}}{1 ext{ mol}} = 9.99 imes 10^{22} ext{ or } 1.00 imes 10^{23} ext{ molecules}$

AVOGADRO’S LAW

• Definition: The quantity of gas present has a direct relationship with the volume it occupies (directly proportional).
    - Mathematical Expression:
     $rac{V_1}{n_1} = rac{V_2}{n_2}$
     - Where n is the number of moles of gas present in (mol).

PRACTICE PROBLEM

• Problem Statement: A balloon is filled with 0.0920 mol of helium gas at 1.90 L. If another 0.0210 mol is added, while temperature and pressure are held constant, what will the final volume be?
     - Calculation:
     $rac{1.90 ext{ L}}{0.0920 ext{ mol}} = rac{V_2}{0.0920 + 0.0210} <br>ightarrow V_2 = 2.33 ext{ L}$

IDEAL GAS LAW

• Definition: The laws of Avogadro, Boyle, Charles, and Gay-Lussac can be combined into a single mathematical statement.

• Formula:
     $PV = nRT$
    - Where:
      - P = pressure in atm
      - V = volume in L
      - n = moles of gas (mol)
      - R = constant, 0.08206 L∙atm/mol∙K
      - T = temperature in K

PRACTICE PROBLEM

• Problem Statement: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon. Then, find the mass of argon in the sample.
     - Use Ideal Gas Law:
    $1.0 atm imes 56.2 L = n imes 0.08206 L∙atm/mol∙K imes 273 K$
     - Solving gives $n = 2.51 ext{ mol}$
     - Mass Calculation:
     $2.51 ext{ mol} imes 40 g/ ext{mol} = 100.4 g ext{ or } 1.00 × 10^{2} g ext{ Ar}$

PRACTICE PROBLEM

• Problem Statement: Determine the volume occupied by 2.34 grams of carbon dioxide gas at STP.
     - Molar Mass Calculation:
     - $2.34 g ext{ CO}_2 imes rac{1 ext{ mol CO}_2}{44 g CO_2} = 0.0532 ext{ mol CO}_2$
     - Use Ideal Gas Law:
     $PV = nRT <br>ightarrow 1.0 atm imes V = 0.0532 imes 0.08206 imes 273 K$
     - Thus, $V = 1.19 L$

IDEAL VS. REAL GASES

• Ideal Gases:
    - Follow the Kinetic Molecular Theory (KMT)
    - Take up no space and follow all gas laws.

• Real Gases:
    - Encounter real-life conditions
    - Take up space/volume (albeit tiny) and interact with one another through electron cloud reactivity (bonding).

• Most gases behave as ideal gases under standard conditions but deviate under very cold temperatures and extremely high pressures, causing particle interaction.

HISTORY OF SCIENCE: GAS LAWS

• Timeline of significant contributions to gas law development:
    - 1650-1750: Boyle's and Charles's laws.
    - 1800: Gay-Lussac’s law.
    - 1811: Avogadro’s Particle Number Theory.
    - Various historical milestones, including the signing of the U.S. Constitution and Latin American independence movements during the 18th and 19th centuries.

GAS LAW CALCULATIONS

• Important important formulas related to gas laws:
    - Ideal Gas Law: $PV = nRT$
    - Dalton’s Law of Partial Pressures: $P_T = P_A + P_B$
    - Charles’s Law: $rac{T_1}{T_2} = rac{V_1}{V_2}$
    - Boyle’s Law: $P_1V_1 = P_2V_2$
    - Gay-Lussac’s Law: $rac{P_1}{P_2} = rac{T_1}{T_2}$
    - Combined Gas Law: $rac{P_1V_1}{T_1} = rac{P_2V_2}{T_2}$
    - Avogadro’s Law: Add or remove gases accordingly.
    - R = 0.0821 L atm / mol K
    - Pressure Conversions: 1 atm = 760 mm Hg = 101.3 kPa

• Bernoulli’s Principle: Fast moving fluids create low pressure.

• Density Equations: $rac{T_1}{D_1} = rac{T_2}{D_2}$

• Graham’s Law of Diffusion vs. Effusion.

DENSITY OF GASES

• Density Definition: The formula for density of any substance:
    - $ext{Density} = rac{ ext{mass}}{ ext{volume}}$

• For gas samples, mass remains constant, but changing pressure/temperature can result in a change in volume.

• As mass is constant, it can be set to 1, leading to relationships:
    - $D1 imes V1 = D2 imes V2$

• Volume is also linked to temperature and pressure variability.

• Final Equation:
    - $P_1 T_1 D_1 = P_2 T_2 D_2$

DENSITY AND THE IDEAL GAS LAW

• Formula Combining Density with Ideal Gas Law:
    - $P = rac{DRT}{W}$
    - Where W stands for molar mass.

GAS STOICHIOMETRY

• Reactions’ coefficients reflect molar amounts of reactants/products and also indicate volume ratios, as per Avogadro's principle.

• For reactions, knowing a balanced chemical equation and at least one gas volume is crucial to determine other volumes.

CALCULATIONS INVOLVING ONLY VOLUME

• Example Reaction:
    - $CH_4 (g) + 2 O_2 (g) → CO_2 (g) + 2 H_2O (g)$
    - Volume ratios in reaction: 2.0 L of oxygen required to react with 1.0 L of methane; producing 1.0 L of carbon dioxide and 2.0 L of water.

PRACTICE PROBLEM

• Problem Statement: What volume of oxygen gas is required for the complete combustion of 4.00 L of propane gas (C3H8)?
     - Reaction: $C_3H_8 (g) + 5 O_2 (g) → 3 CO_2 (g) + 4 H_2O (g)$
     - Calculation:
     $4.00 L ext{ C}_3H_8 imes rac{5.0 L ext{ O}_2}{1.0 L ext{ C}_3H_8} = 20.0 L ext{ O}_2$

PRACTICE PROBLEM - GAS STOICHIOMETRY (VOLUME-MASS)

• Example of Synthesis Reaction:

• $N_2 (g) + 3 H_2 (g) → 2 NH_3 (g)$
     - If 5.00 L of nitrogen reacts completely, how many grams of ammonia are produced?
     - Calculation:
   $5.00 L ext{ N}_2 imes rac{2.0 L ext{ NH}_3}{1.0 L ext{ N}_2} = 10.0 L ext{ NH}_3$
     - Use Ideal Gas Law:
     $PV = nRT <br>ightarrow 3.00 atm imes 10.0 L = n imes 0.08206 L∙atm/mol∙K imes 298 K$
     - Resulting in $n = 1.2267 mol ext{ NH}_3$
     - Conversion to mass:
     $n imes rac{17 g}{1 mol} = 20.9 g ext{ NH}_3$

PRACTICE PROBLEMS

• Problem: Determine the volume of hydrogen gas needed to react completely with 6.00 L of oxygen gas to form water.
    - Reaction: $2 H_2 (g) + O_2 (g) → 2 H_2O (g)$
    - Hence:
     $6.00 L ext{ O}_2 imes rac{2.0 L ext{ H}_2}{1.0 L ext{ O}_2} = 12.0 L ext{ H}_2$

• Problem: Ammonium nitrate ($NH_4NO_3$) decomposes into dinitrogen oxide and water. Given that 0.100 L of $N_2O$ is acquired at STP, how much ammonium nitrate was used?
     - $NH_4NO_3 (g) <br>ightarrow N_2O (g) + 2 H_2O (g)$
     - $0.100 L ext{ N}_2O imes rac{1.0 L ext{ NH}_4NO_3}{1.0 L ext{ N}_2O} = 0.100 L ext{ NH}_4NO_3$
     - Using Ideal Gas Law again, $PV = nRT$
     - $1.0 atm imes 0.100 L = n imes 0.08206 L∙atm/mol∙K imes 273 K$
     - Resulting in $n = 0.00446 mol ext{ NH}_4NO_3$
     - Lastly, conversion gives:
     $0.00446 mol imes rac{80 g}{1 mol} = 0.357 g ext{ NH}_4NO_3$

• Problem: Determine the amount of water vapor produced at 1.00 atm and 200°C through complete combustion of 10.5 L of methane gas (CH4).
     - $10.5 L ext{ CH}_4 imes rac{2 L ext{ H}_2O}{1 L ext{ CH}_4} = 21.0 L ext{ H}_2O$
     - Using Ideal Gas Law to find moles:
     $PV=nRT <br>ightarrow 1.0 atm imes 21.0 L = n imes 0.08206 L∙atm/mol∙K imes 473 K$
     - Resultant: $n = 0.541 mol ext{ H}_2O$

NEED TO KNOW

• Kinetic Molecular Theory

• Properties of Gases (state of matter)

• Units:
    - Pressure (atm, kPa, Torr, mmHg)
    - Volume (L, mL)
    - Temperature (K)
    - Standard Temperature & Pressure (1.0 atm & 273 K or 0.0°C)

• Concepts to Differentiate:
    - Ideal Gases vs. Real Gases
    - Gas Laws:
      - Dalton’s Law of Partial Pressures
      - Graham’s Law of Effusion
      - Boyle’s Law – Constant Temperature
      - Charles’s Law – Constant Pressure
      - Gay-Lussac’s Law – Constant Volume
      - Combined Gas Law
      - Ideal Gas Law (including molar quantities)
      - Avogadro’s Principle

Study Guide

KINETIC-MOLECULAR THEORY

• Describes the behavior of gases in terms of particles in motion.

• Objects in motion have kinetic energy.

• Gases consist of small particles that are separated by empty space.

• Kinetic energy and temperature are related:
    - Temperature: Measure of average kinetic energy of particles. Higher temperature means more kinetic energy (faster molecules).

KINETIC-MOLECULAR THEORY (KMT) Simplified

• Gas particles are tiny and have a lot of space between them.

• Move in constant, random motion based on average temperature.

• When they collide, they bounce off without losing energy.

COLLISIONS OF GAS PARTICLES

GASES

• The chemical properties of matter depend on the atoms present and their arrangement.

• Gases have similar physical properties despite different compositions (e.g., oxygen gas vs. nitrogen gas).

• Mixtures consist of multiple substances that can be separated but retain individual properties.

BEHAVIOR OF GASES

• Low Density: Due to empty space between gas molecules.

• Compression & Expansion: Gases can be compressed and will return to original volume when allowed to expand.

• Diffusion & Effusion:
    - Diffusion: Movement of molecules from high to low concentration.
    - Effusion: Gas escapes from high pressure to low pressure.

DIFFUSION VS. EFFUSION

• Diffusion Example: Scent spreading in a room.

• Effusion Example: Air leaking from a tire.

GRAHAM’S LAW OF EFFUSION

• Definition: Rate of effusion is inversely proportional to the square root of their molar masses. Smaller gases effuse faster.

• Formula:
    $\frac{\text{Rate of Effusion Gas A}}{\text{Rate of Effusion Gas B}} = \frac{\sqrt{Molar Mass Gas B}}{\sqrt{Molar Mass Gas A}}$

PRACTICE PROBLEMS

Problem 1: Molar Mass Calculation

• A certain gas takes three times as long to effuse as helium.

• Calculate molecular mass:
    $\frac{v_A}{v_B} = \frac{\sqrt{Molar Mass Gas B}}{\sqrt{4}} \rightarrow Molar Mass Gas B = 36u$

Problem 2: Time for Gas Escape

• Find time for half of hydrogen to escape:
    $\frac{Rate_{A}}{Rate_{B}} = \frac{\sqrt{32}}{\sqrt{2}} \rightarrow x = \frac{1}{4}$

Problem 3: Mole Calculations

• Two gases A (49u) and B have diffused in 20 and 10 seconds. Find molecular mass of B.

• Using the rate relationship, calculate molecular mass of B.

THE NATURE OF GASES

• Influenced by factors: Number of particles, pressure, temperature, volume. All interrelated.

GAS PRESSURE

• Definition: Pressure generated when gas molecules hit an object.

• Equation:
    $\text{Pressure} = \frac{\text{Force}}{\text{Area}}$

• Higher altitudes reduce air pressure.

UNITS OF PRESSURE

• Instruments: Barometers measure atmospheric pressure; manometers measure gas pressure in closed containers.

• Common Units: Pascal (Pa), kilopascal (kPa), mmHg, Torr, atm.
    - $1.0 atm = 760 mmHg = 101.3 kPa$

PRACTICE PROBLEMS

Problem 4: Force of Air Pressure

• Find force acting on a tarp at 100 ft x 100 ft.
    $F = PA = 101300 imes 929.5 = 94,158,982.7 N$

DALTON’S LAW OF PARTIAL PRESSURES

• Definition: Total pressure equals the sum of individual pressures of gas ingredients.
    $P_{total} = P_{gas 1} + P_{gas 2} + P_{gas 3} + …$

PRACTICE PROBLEM

Problem 5: Partial Pressure Calculation

• Calculate partial pressure of oxygen:
    $P_{total} = 0.97 atm = P_{O_2} + 0.70 atm + 0.12 atm <br>ightarrow P_{O_2} = 0.15 atm$

BOYLE’S LAW

• Law Statement: Pressure inversely related to volume at constant temperature.
    $P_i V_i = P_f V_f$

PRACTICE PROBLEMS

Problem 6: Helium Gas Pressure

• Helium compressed from 4.0 L to 2.5 L at 210 kPa initial pressure:
    $P_f = 335 kPa$

Problem 7: Original Volume Calculation

• Gas with 1.9 L at 3.7 atm; find original volume at 5.6 atm:
    $V_i = 1.255 L$

PRESSURE-VOLUME RELATIONSHIP

• Examples demonstrating P-V relationships within various conditions like constant kPa and L.

TEMPERATURE SCALES

• Scales: Celsius, Fahrenheit, Kelvin (absolute).

• Use conversions for calculations:
    $T_K = T_{℃} + 273$

CHARLES’S LAW

• Direct Proportionality: Temperature and volume at constant pressure.
    $\frac{V_i}{T_i} = \frac{V_f}{T_f}$

PRACTICE PROBLEM

Problem 8: Temperature and Volume Relationship

• Calculate volume at raised temperatures for a gas sample.

GAY-LUSSAC’S LAW

• Relationship of temperature and pressure held constant:
    $\frac{P_i}{T_i} = \frac{P_f}{T_f}$

COMBINED GAS LAW

• Combines all gas laws:
    $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

STP (Standard Temperature and Pressure)

• Standard conditions: STP = 0°C or 273 K, 1.0 atm.

AVOGADRO’S PRINCIPLE AND LAW

• Direct relationship between gas quantity and volume at equal conditions.
    - $1 mole = 6.02 \times 10^{23} molecules$ at STP = 22.4L.

IDEAL GAS LAW

• Definition: Combines various gas laws into one equation:
    $PV = nRT$
    - Where P, V, n, R, T represent pressure, volume, moles, gas constant, and temperature respectively.

IDEAL VS. REAL GASES

• Gases behavior as ideal under specific conditions and deviate under extremes.

Practice Test

Definitions

1. Kinetic Molecular Theory: Describes the behavior of gases in terms of particles in motion; explains how temperature is related to kinetic energy.

2. Diffusion: The movement of molecules from an area of high concentration to an area of low concentration.

3. Effusion: The process by which gas escapes through a tiny opening into a vacuum or lower-pressure area.

4. Graham's Law of Effusion: States that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

5. Dalton’s Law of Partial Pressures: The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture.

6. Boyle’s Law: States that pressure and volume are inversely proportional for a given amount of gas at constant temperature: $P_i V_i = P_f V_f$.

7. Charles’s Law: Volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure: $\frac{V_i}{T_i} = \frac{V_f}{T_f}$.

8. Gay-Lussac’s Law: Pressure of a given mass of gas varies directly with the absolute temperature when the volume remains constant: $\frac{P_i}{T_i} = \frac{P_f}{T_f}$.

9. Ideal Gas Law: Combines Boyle’s, Charles’s, and Avogadro’s laws: $PV = nRT$.

10. Avogadro’s Principle: Equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

Practice Problems

1. Molar Mass Calculation: A certain gas takes three times as long to effuse as helium, which has a molar mass of 4 u. Determine the molar mass of the gas using Graham's Law.
      - Solution: $\frac{v_A}{v_B} = \frac{\sqrt{Molar\ Mass\ Gas\ B}}{\sqrt{4}}$. Find the molecular mass.

2. Partial Pressure Calculation: A sample of a gas mixture contains 0.70 atm of carbon dioxide and 0.12 atm of nitrogen. If the total pressure is 0.97 atm, what is the partial pressure of oxygen?
      - Solution: Use Dalton’s Law: $P_{total} = P_{O_2} + 0.70 atm + 0.12 atm$.

3. Volume Calculation: A gas occupies a volume of 2.50 L at 1.0 atm. If the pressure is increased to 2.0 atm, what will the volume be at constant temperature?
      - Solution: Use Boyle’s Law: $P_i V_i = P_f V_f$ to find the new volume.

4. Temperature Change Problem: A gas is held in a container at 1.5 atm and a temperature of 300 K. If the temperature is increased to 600 K, what will be the new pressure, assuming volume is constant?
      - Solution: Use Gay-Lussac's Law: $\frac{P_i}{T_i} = \frac{P_f}{T_f}$.

5. Efficiency of Effusion: If gas A (molar mass 16 g/mol) effuses twice as fast as gas B, calculate the molar mass of gas B using Graham's Law.
      - Solution: $\frac{v_A}{v_B} = \frac{\sqrt{Molar\ Mass\ B}}{\sqrt{16}}$.