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The Gaseous State of Matter Chapter 12

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## The Gaseous State of Matter Chapter 12

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**The Gaseous State of MatterChapter 12**Hein and Arena Version 1.1**Chapter Outline**12.2 The KineticMolecularTheory 12.9Combined Gas Laws 12.10Dalton’s Law of PartialPressures 12.3 Measurement of Pressure ofGases 12.11 Avogadro’s Law 12.4 Dependence of Pressure on Number of Molecules and Temperature 12.12Mole-Mass-VolumeRelationships of Gases 12.13Density of Gases 12.5 Boyle’s Law 12.14 Ideal Gas Equation 12.6Charles’ Law 12.15 Gas Stoichiometry 12.7Gay Lussac’s Law 12.8Standard Temperature andPressure 12.16 Real Gases**The Kinetic-Molecular Theory**• KMT is based on the motions of gas particles. • A gas that behaves exactly as outlined by KMT is known as an ideal gas. • While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.**Principle Assumptions of the KMT**• Gases consist of tiny subatomic particles. • The distance between particles is large compared with the size of the particles themselves. • Gas particles have no attraction for one another.**Principle Assumptions of the KMT**• Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container. • No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic.**Principle Assumptions of the KMT**• The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.**mH**= 2 vH 4 2 2 = mO = 32 1 vO 2 2 Kinetic Energy • All gases have the same average kinetic energy at the same temperature. • As a result lighter molecules move faster than heavier molecules.**Stopcock closed No diffusion occurs**Stopcock open Diffusion occurs Diffusion The ability of two or more gases to mix spontaneously until they form a uniform mixture.**Effusion**A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.**Graham’s Law of Effusion**The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.**The pressure resulting from the collisions of gas molecules**with the walls of the balloon keeps the balloon inflated.**The pressure exerted by a gas depends on the**• Number of gas molecules present. • Temperature of the gas. • Volume in which the gas is confined.**A tube of mercury is inverted and placed in a dish of**mercury. Mercury Barometer The barometer is used to measure atmospheric pressure.**Average Composition of Dry Air**Gas Volume Percent Gas Volume Percent N2 78.08% He 0.0005% O2 20.95% CH4 0.0002% Ar 0.93% Kr 0.0001% CO2 0.033% Xe, H2, and N2O Trace Ne 0.0018%**Dependence of Pressure on Number of Molecules and**Temperature • Pressure is produced by gas molecules colliding with the walls of a container. • At a specific temperature and volume, the number of collisions depends on the number of gas molecules present. • For an ideal gas the number of collisions is directly proportional to the number of gas molecules present.**The pressure exerted by a gas is directly proportional to**the number of molecules present. V = 22.4 LT = OoC**Dependence of Pressure on Temperature**• The pressure of a gas in a fixed volume increases with increasing temperature. • When the pressure of a gas increases, its kinetic energy increases. • The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.**Increased pressure is due to more frequent and more**energetic collisions of the gas molecules with the walls of the container at the higher temperature. Lower T Lower P Higher T Higher P The pressure of a gas in a fixed volume increases with increasing temperature.**Boyle’s Law**At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).**Graph of pressure versus volume. This shows the inverse PV**relationship of an ideal gas.**An 8.00 L sample of N2 is at a pressure of 500 torr. What**must be the pressure to change the volume to 3.00 L? (T is constant). Method A. Conversion Factors Step 1. Determine whether volume is being increased or decreased. Initial volume = 8.00 L Final volume = 3.00 L volume decreases pressure increases**An 8.00 L sample of N2 is at a pressure of 500 torr. What**must be the pressure to change the volume to 3.00 L? (T is constant). Step 2. Multiply the original pressure by a ratio of volumes that will result in an increase in pressure. new pressure = original pressure x ratio of volumes**P1 = 500 torr**V1 = 8.00 L P2 = ? V2 = 3.00 L An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant). Method B. Algebraic Equation Step 1. Organize the given information:**An 8.00 L sample of N2 is at a pressure of 500 torr. What**must be the pressure to change the volume to 3.00 L? (T is constant). Step 2. Write and solve the equation for the unknown.**An 8.00 L sample of N2 is at a pressure of 500 torr. What**must be the pressure to change the volume to 3.00 L? (T is constant). Step 3. Put the given information into the equation and calculate.**Charles’ Law**• If a given volume of any gas at 0oC is cooled by 1oC the volume of the gas decreases by . • If a given volume of any gas at 0oC is cooled by 20oC the volume of the gas decreases by . Absolute Zero on the Kelvin Scale**If a given volume of any gas at 0oC is cooled by 273oC the**volume of the gas decreases by . Absolute Zero on the Kelvin Scale • -273oC (more precisely –273.15oC) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.**Charles’ Law**At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.**Effect of temperature on the volume of a gas. Pressure is**constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.**A 255 mL sample of nitrogen at 75oC is confined at a**pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC? Method A. Conversion Factors Step 1. Change oC to K: oC + 273 = K 75oC + 273 = 348 K 250oC + 273 = 523 K**A 255 mL sample of nitrogen at 75oC is confined at a**pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC? Step 2: Multiply the original volume by a ratio of Kelvin temperatures that will result in an increase in volume:**V1 = 255 mL**T1 = 75oC = 348 K V2 = ? T2 = 250oC = 523 K A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC? Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same):**A 255 mL sample of nitrogen at 75oC is confined at a**pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC? Step 2. Write and solve the equation for the unknown:**V1 = 255 mL**T1 = 75oC = 348 K V2 = ? T2 = 250oC = 523 K A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC? Step 3. Put the given information into the equation and calculate:**Gay-Lussac’s Law**The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.**At a temperature of 40oC an oxygen container is at a**pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? Method A. Conversion Factors Step 1. Change oC to K: oC + 273 = K 40oC + 273 = 313 K 100oC + 273 = 373 K Determine whether temperature is beingincreased or decreased. temperature increases pressure increases**At a temperature of 40oC an oxygen container is at a**pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:**At a temperature of 40oC an oxygen container is at a**pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? A temperature ratio greater than 1 will increase the pressure**P1 = 21.5 atm**T1 = 40oC = 313 K P2 = ? T2 = 100oC = 373 K At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same):**At a temperature of 40oC an oxygen container is at a**pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? Step 2. Write and solve the equation for the unknown:**P1 = 21.5 atm**T1 = 40oC = 313 K P2 = ? T2 = 100oC = 373 K At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen? Step 3. Put the given information into the equation and calculate:**STP**Standard Conditions Standard Temperature and Pressure Standard Temperature and Pressure Selected common reference points of temperature and pressure. 273.15 K or 0.00oC 1 atm or 760 torr or 760 mm Hg**Combined Gas Laws**• A combination of Boyle’s and Charles’ Law. • Used when pressure and temperature change at the same time. • Solve the equation for any one of the 6 variables