1. “KMT and the Behavior of Gases” adapted from Stephen L. Cotton

2. The Nature of Gases Kinetic refers to motion Kinetic refers to motion kinetic energy – the energy an object has because of its motion kinetic energy – the energy an object has because of its motion The kinetic theory states that the tiny particles in all forms of matter are in constant motion! The kinetic theory states that the tiny particles in all forms of matter are in constant motion!

3. KMT – Kinetic Molecular Theory 1. All matter is composed of very tiny particles 1. All matter is composed of very tiny particles 2. Particles of matter are continually moving 2. Particles of matter are continually moving 3. The collisions of these particles are “elastic” (no loss of energy) 3. The collisions of these particles are “elastic” (no loss of energy)

4. The Nature of Gases Three basic assumptions of the kinetic theory as it applies to gases: Three basic assumptions of the kinetic theory as it applies to gases: #1. The particles in a gas are considered to be small, hard spheres with insignificant volume #1. The particles in a gas are considered to be small, hard spheres with insignificant volume #2. The motion of the particles in a gas are rapid, constant and random #3. All collisions between particles in a gas are perfectly elastic.

5. - Page 385 Top

6. The Nature of Gases (no volume or shape) Gas Pressure –the force exerted by a gas per unit surface area of an object Gas Pressure –the force exerted by a gas per unit surface area of an object The result of simultaneous collisions of billions of rapidly moving particles. The result of simultaneous collisions of billions of rapidly moving particles. No particles present? Then there cannot be any collisions, and thus no pressure – called a vacuum No particles present? Then there cannot be any collisions, and thus no pressure – called a vacuum

7. The Nature of Gases Atmospheric pressure results from the collisions of air molecules with objects Atmospheric pressure results from the collisions of air molecules with objects Air exerts pressure on earth because gravity holds the particles in the air from Earth’s atmosphere. Air exerts pressure on earth because gravity holds the particles in the air from Earth’s atmosphere. Barometer is the measuring device for atmospheric pressure, which is dependent upon weather & altitude Barometer is the measuring device for atmospheric pressure, which is dependent upon weather & altitude

8. Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure Torricelli

9. An Early Barometer 760 mm Hg = 101.3 kPa = 1 atmosphere

10. The Nature of Gases For gases, it is important to relate measured values to standards For gases, it is important to relate measured values to standards Standard values are defined as a temperature of 0 o C and a pressure of 101.3 kPa, or 1 atm Standard values are defined as a temperature of 0 o C and a pressure of 101.3 kPa, or 1 atm This is called Standard Temperature and Pressure, or STP This is called Standard Temperature and Pressure, or STP

11. The Nature of Gases Absolute zero (0 K, or –273 o C) is the temperature at which the motion of particles theoretically ceases Absolute zero (0 K, or –273 o C) is the temperature at which the motion of particles theoretically ceases Kelvin = °C + 273 Kelvin = °C + 273 °C = Kelvin – 273 °C = Kelvin – 273 °C = (°F – 32) x.555 °C = (°F – 32) x.555

12. The Nature of Gases The Kelvin temperature scale reflects a direct relationship between temperature and average kinetic energy The Kelvin temperature scale reflects a direct relationship between temperature and average kinetic energy Particles of He gas at 200 K have twice the average kinetic energy as particles of He gas at 100 K Particles of He gas at 200 K have twice the average kinetic energy as particles of He gas at 100 K

13. The Nature of Liquids Liquid particles are also in motion. Liquid particles are also in motion. a phase of a substance that has a definite volume but no definite shape a phase of a substance that has a definite volume but no definite shape

14. The Nature of Liquids vaporization vaporization – conversion of a liquid to a gas by adding heat evaporation evaporation – conversion of a liquid to a gas at room temperature

15. The Nature of Liquids Evaporation of a liquid in a closed container is somewhat different Evaporation of a liquid in a closed container is somewhat different vapor pressurea measure of the force exerted by a gas above a liquid vapor pressure – a measure of the force exerted by a gas above a liquid An increase in the temperature of a liquid increases the vapor pressure. A decrease in the temperature decreases the vapor pressure.

16. The Nature of Liquids The boiling point (bp) the temperature at which the vapor pressure of a liquid is just equal to the external pressure. The boiling point (bp) the temperature at which the vapor pressure of a liquid is just equal to the external pressure.

17. Section 13.2 The Nature of Liquids Normal bp of water = 100 o C Normal bp of water = 100 o C However, in Denver = 95 o C, since Denver is 1600 m above sea level and average atmospheric pressure is about 85.3 kPa (Recipe adjustments?) However, in Denver = 95 o C, since Denver is 1600 m above sea level and average atmospheric pressure is about 85.3 kPa (Recipe adjustments?) In pressure cookers, which reduce cooking time, water boils above 100 o C due to the increased pressure In pressure cookers, which reduce cooking time, water boils above 100 o C due to the increased pressure

18. - Page 394 Not Boiling Normal Boiling Point @ 101.3 kPa = 100 o C Boiling, but @ 34 kPa = 70 o C

19. - Page 394 Questions: a. 60 o C b. about 20 kPac. about 30 kPa

20. The Behavior of Gases

21. Properties of Gases Compressibility: a measure of how much the volume of matter decreases under pressure. Compressibility: a measure of how much the volume of matter decreases under pressure. Gases are easily compressed because of the space between the particles. (remember KMT) Gases are easily compressed because of the space between the particles. (remember KMT)

22. 3 Factors Affecting Gas Pressure 1. Amount of Gas: by adding gas you increase the particles and number of collisions so the pressure increases, and vise versa. 1. Amount of Gas: by adding gas you increase the particles and number of collisions so the pressure increases, and vise versa.

23. 2. Volume of Gas: by increasing the volume you increase the space that the particles can move in. Thus the pressure decreases as the number of collisions decreases, and vice versa. 2. Volume of Gas: by increasing the volume you increase the space that the particles can move in. Thus the pressure decreases as the number of collisions decreases, and vice versa.

24. 3. Temperature: as the temperature increases the kinetic energy of the particles increases and they hit the walls of the container and each other with more energy, increasing the pressure, and vice versa. 3. Temperature: as the temperature increases the kinetic energy of the particles increases and they hit the walls of the container and each other with more energy, increasing the pressure, and vice versa. Warm temp.Hot temp.Cold temp.

25. The Gas Laws Boyle’s Law: If the temperature is constant, as the pressure of a gas increases, the volume decreases. Boyle’s Law: If the temperature is constant, as the pressure of a gas increases, the volume decreases. P 1 V 1 = P 2 V 2

27. Pressure and Volume ExperimentPressureVolume P x V (atm) (L) (atm x L) 18.0 2.0 16 24.04.0_____ 32.08.0_____ 41.016_____ Boyle's Law P x V = k (constant) when T remains constant T remains constant P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L P1V1 = P2V2 = k Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes. new vol. old vol. x Pfactor new P old P x Vfactor new vol. old vol. x Pfactor new P old P x Vfactor V2 = V1 x P1 P2 = P1 x V1 P2 V2 P2 V2

28. P and V Changes P1P1 P2P2 V1V1 V2V2

29. Boyle's Law The pressure of a gas is inversely related to the volume when T does not change The pressure of a gas is inversely related to the volume when T does not change Then the PV product remains constant Then the PV product remains constant P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L

30. PV Calculation Prepare a data table DATA TABLE Initial conditionsFinal conditions P 1 = 50 mm HgP 2 = 200 mm Hg V 1 = 1.6 LV 2 = ? ?

31. 1)If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? 1)If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be?

32. 2)I have added 15 L of air to a balloon at sea level (1.0 atm). If I take the balloon with me to Denver, where the air pressure is 0.85 atm, what will the new volume of the balloon be? 2)I have added 15 L of air to a balloon at sea level (1.0 atm). If I take the balloon with me to Denver, where the air pressure is 0.85 atm, what will the new volume of the balloon be?

33. 3)I’ve got a car with an internal volume of 12,000 L. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.0 atm to 1.4 atm? 3)I’ve got a car with an internal volume of 12,000 L. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.0 atm to 1.4 atm?

34. Charles Law: As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant. Charles Law: As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.

35. When you use temperature in ANY gas law you must change the temperature into Kelvin When you use temperature in ANY gas law you must change the temperature into Kelvin Kelvin = ºC + 273 Kelvin = ºC + 273

36. Find the volume of 250 mL of a gas at 25 °C if the temperature is dropped to 10 °C. Find the volume of 250 mL of a gas at 25 °C if the temperature is dropped to 10 °C.

37. Gay-Lussac’s Law: As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant. Gay-Lussac’s Law: As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.

38. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 10 3 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 10 3 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm.

39. The Combined Gas Law: all three gas laws combined. The Combined Gas Law: all three gas laws combined.

40. Convert 250 mL of a gas at 50 °C and 650 mm Hg to STP Convert 250 mL of a gas at 50 °C and 650 mm Hg to STP

41. A gas starts at 1 atm, 30 °C and 1 L. What is the final temperature if the final amounts are 5 atm and 0.5 L. A gas starts at 1 atm, 30 °C and 1 L. What is the final temperature if the final amounts are 5 atm and 0.5 L.

42. Ideal Gas Law: includes the number of particles (n= moles) and the ideal gas constant (related to pressure). Ideal Gas Law: includes the number of particles (n= moles) and the ideal gas constant (related to pressure).

43. 1 mole = 6.02 x 10 23 particles 1 mole = 6.02 x 10 23 particles R = (22.4 L x 1 atm/1 mol x 273 K) = 0.0821 L x atm/mol x K R = (22.4 L x 1 atm/1 mol x 273 K) = 0.0821 L x atm/mol x K

44. How many moles of a gas will you have if 1 liter of gas is collected 28.0 °C and 1.7 atmospheres pressure. How many moles of a gas will you have if 1 liter of gas is collected 28.0 °C and 1.7 atmospheres pressure.

45. R = (22.4 L x 760 mmHg/1 mol x 273 K) = 62.4 L x mmHg/mol x K R = (22.4 L x 760 mmHg/1 mol x 273 K) = 62.4 L x mmHg/mol x K R = (22.4 L x 1 atm/1 mol x 273 K) = 0.0821 L x atm/mol x K R = (22.4 L x 1 atm/1 mol x 273 K) = 0.0821 L x atm/mol x K R = (22.4 L x 101.3 kPa/1 mol x 273 K) = 8.31 L x kPa/mol x K R = (22.4 L x 101.3 kPa/1 mol x 273 K) = 8.31 L x kPa/mol x K