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I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids I will compare types of intermolecular forces.

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Presentation on theme: "I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids I will compare types of intermolecular forces."— Presentation transcript:

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2 I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids I will compare types of intermolecular forces I will explain how kinetic energy and intermolecular forces combine to determine the state of a substance I will describe the role of energy in phase changes I will use the kinetic-molecular theory to explain the physical properties of gases, liquids and solids I will compare types of intermolecular forces I will explain how kinetic energy and intermolecular forces combine to determine the state of a substance I will describe the role of energy in phase changes States of Matter

3 13.1 Gases I will use the Kinetic-molecular theory to explain the behavior of gases I will describe how mass affects the rates of diffusion and effusion I will explain how gas pressure is measured and calculate the partial pressure of a gas Vocabularykinetic-molecular theoryelastic collision temperatureGraham’s Law of effusiondiffusion Pressurebarometerpascalatmosphere Dalton’s law of partial pressures

4 Gases Substances that are gases at room temperature usually display similar physical preperties despite their different compositions. Why is there so little variation in behavior among gases?

5 Kinetic- Molecular Theory

6 Describes the behavior of gases in terms of particles in motion Makes several assumptions about gas particles: Size Motion Energy

7 Gas Assumptions Particle Size Separated by empty space Volume of particle = small Volume of empty space = large NO significant attractive or repulsive forces Particle Motion Constant random motion Move in a STRAIGHT line Collide with walls or other particles Collisions are elastic (NO kinetic energy is lost, just transferred)

8 Gas Assumptions Cont... Particle Energy Affected by mass and velocity KE = ½ mv 2 Single gas Particles have same mass Particles have different velocity Particles have different KE Temperature = measure of the average KE of the particles in a sample of matter At a given temperature, ALL gases have the SAME average KE

9 Gas Behavior

10 Low Density Compressible Expandable Diffuse Effuse

11 Gas Behavior Low Density Density = mass/volume Large space between gas particles Fewer gas molecules than solid or liquid molecules in the same volume

12 Gas Behavior Compression & Expansion Large amount of empty space between gas particles Allows particles to be squished into a smaller volume Stop squishing; random motion of particles fills the available space, expands to original volume

13 Gas Behavior Diffusion The movement of one material through another material Gas particles have no significant forces of attraction Particles can slide past each other Mix until evenly distributed Particles diffuse from areas of high concentration to areas of low concentration Rate of diffusion Depends on mass of particles Lighter = faster Heavier = slower

14 Gas Behavior

15 Related to diffusion When a gas escapes through a tiny opening Ex. Puncture a balloon or tire Inverse relationship between effusion rates and molar mass Effusion

16 Gas Behavior Graham’s Law of Effusion The rate of effusion for a gas is inversely proportional to the square root of its molar mass Also applies to rate of diffusion Set up a proportion to compare the diffusion rates of 2 gases

17 Finding a Ratio of Diffusion Rates Let’s Try!

18 Practice Problems 1.RN 2 /RNe = 0.849 2.RCO/RCO 2 = 1.25 3.2.5mol/min 1.RN 2 /RNe = 0.849 2.RCO/RCO 2 = 1.25 3.2.5mol/min 1.Calculate the ratio of effusion rates for nitrogen (N 2 ) and Neon (Ne). 1.Calculate the ratio of diffusion rates for carbon monoxide (CO) and carbon dioxide (CO 2 ) 1.What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.6 mol/min? Use to find ratios

19 Gas Pressure

20 Force per unit area Gas particles exert pressure when they collide with: The walls of their container Each other Exert pressure in ALL directions (because particles move in all directions) Individual particles can only exert little pressure Many particles colliding can exert substantial pressure pressure of gases is what keeps our tires inflated, makes our basketballs bounce, makes hairspray come out of the can, etc. Pressure

21 Gas Pressure Pressure increases when temperature increases because the molecules are moving with greater speed and colliding against the sides of their containers more often. Therefore, the pressure inside that container is greater, because there are more collisions.

22 Gas Pressure Atmospheric Pressure Air pressure at Earth’s surface Equal to: Pressure exerted by 1 kg mass on a square centimeter Varies by elevation Mountains = less Sea Level = more

23 Gas Pressure Measuring Air Pressure Barometer Measures atmospheric pressure Mercury in it is always about 760mm Exact amount determined by 2 forces Gravity- downward force Air pressure- upward force (air presses on surface of Hg) Air pressure varies because of: Changes in air temperature Changes in humidity Increase in air pressure = Hg rises Decrease in air pressure = Hg falls

24 Gas Pressure Measuring Enclosed Gas Pressure Manometer Measures pressure of enclosed gas The difference in height of the mercury in the 2 arms is used to calculate the pressure of the gas in the flask

25 Gas Pressure Units of Pressure Comparison of Pressure Units UnitCompared with 1 atm Compared with 1 KPa Kilopascal (kPa) 1atm = 101.3 kPa mm Hg1 atm = 760 mm Hg1 kPa = 7.501 mm Hg torr1 atm = 760 torr1 kPa = 7.501 torr psi1 atm = 14.7 psi1 kPa = 0.145 psi atm1 kPa = 0.009 869 atm SI Unit = Pascal (Pa) Atmosphere = atm (used to report air pressure)

26 Gas Laws

27 Dalton’s Law of Partial Pressures Gay-Lussac’s Law Boyle’s Law Charles’s Law Ideal Gas Law Combined Gas Law

28 Dalton’s Law of Partial Pressures P total = P1 + P2 + P3 +…Pn The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. Partial pressure = the portion of the total pressure contributed by a single gas Depends on # of moles of gas Size of container Temperature of the mixture Does NOT depend on identity of gas (at a given temperature and pressure, the partial pressure of 1 mole of ANY gas is the same)

29 Dalton’s Law of Partial Pressures Practice Problem A mixture of oxygen (O 2 ), carbon dioxide (CO 2 ), and nitrogen (N 2 ) has a total pressure of 0.97 atm. What is the partial pressure of O 2, if the partial pressure of CO 2 is 0.70 atm and the partial pressure of N 2 is 0.12 atm? PO 2 = 0.97 atm – 0.70 atm -0.12 atm PO 2 = 0.15 atm

30 Boyle’s Law P 1 V 1 = P 2 V 2 The volume of a given amount of gas held at a constant temperature varies inversely with pressure As pressure increases, volume decreases and vice versa

31 Boyle’s Law Practice Problem 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 the pressure be at 2.5 L? 210 kPa (4.0 L) = P 2 (2.5 L) P 2 = 340 kPa

32 Charles’s Law V 1 V 2 T 1 = T 2 The volume of a given mass of a gas is directly proportional to its Kelvin temperature at constant pressure. When Kelvin temperature increases, volume increases and vice versa T k = 273 + T c

33 Charles’s Law Practice Problem A gas sample at 40 degrees C occupies a volume of 2.32 L. If the temperature is raised to 75.0 degrees C, what will the volume be, assuming the pressure remains constant? 40.0 degrees C + 273 = 313K 75 degrees C + 273 = 348K 2.32 L = V2 313 K348 K V 2 = 2.58 L

34 Gay-Lussac’s Law P 1 P 2 T 1 = T 2 The pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant. As Kelvin temperature increases, pressure increases and vice versa

35 Gay-Lussac’s Law Practice Problem The pressure of a gas in a tank is 3.20 atm at 22.0 degrees C. If the temperature rises to 60.0 degrees C, what will be the gas pressure in the tank? 22.0 degrees C + 273 = 295 K 60.0 degrees C + 273 = 333 K 3.20 atm = P2 295 K 333 K P 2 = 3.61 atm

36 The Combined Gas Law P1V1 = P2V2 T1 T2 The relationship among pressure, volume, and temperature of a fixed amount of gas. Pressure increases, volume decreases Pressure increases, temperature increases Volume increases, temperature increases

37 The Combined Gas Law Practice Problem A gas at 110 kPa and 30.0 degrees C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 degrees C and the pressure increased to 440 kPa, what is the new volume? 30.0 degrees C + 273 = 303 K 80.0 degrees C + 273 = 353 K 110(2.00) = 440(V2) 303 353 V 2 = 0.58 L

38 Ideal Gas One whose particles Take up NO space Have NO intermolecular attractive forces Follows ideal gas laws under ALL conditions of temperature and pressure

39 Real Gas NO gas in the real world is truly ideal particles have some volume some attractive forces Real gases deviate most from ideal behavior at low temperatures and high pressures Examples: Liquid nitrogen is used to store biological tissues at low temps Increased pressure allows a larger mass of propane to fit into a smaller volume for easier transport

40 Ideal Gas Law PV = nRT Describes the physical behavior of an ideal gas in terms of the pressure, volumes, temperature, and number of moles of gas present. R = ideal gas constant Numerical Values of the Gas Constant, R Units of RNumerical R Value Units of PUnits of VUnits of TUnits of n L atm/ mol K0.0821atmLKmol L kPa/ mol K8.314kPaLKmol L mm Hg / mol K 62.4mm HgLKmol

41 Ideal Gas Law Calculate the number of moles of gas contained in a 3.0 L vessel at 3.00 x 10 2 K with a pressure of 1.50 atm. V = 3.0 LT = 3.00 x 10 2 K P = 1.50 atmR = 0.0821 L atm/ mol K n = ? 1.50 (3.0) = n (0.0821)(3.00 x 10 2 K) n = 0.18 mol

42 Density of a Gas D = MP RT Practice Problem What is the density of a gas at STP that has a molar mass of 44.0 g/mol? STP = 273 K, 1 atm D = 44.0(1) 0.0821(273) D = 1.96 g/L


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