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## GASES

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**The Kinetic-Molecular Theory for Ideal Gases**The kinetic-molecular theory describes the behavior of IDEAL gases in terms of particles in motion. Real gases do not obey the kinetic molecular theory.**The Kinetic-Molecular Theory for Ideal Gases**1. Gas particles are much smaller than the spaces between them. The gas particles themselves have virtually no (negligible) volume.**The Kinetic-Molecular Theory for Ideal Gases**2. Gas particles are in constant, random motion. Particles move in a straight line until they collide with other particles or with the walls of their container.**The Kinetic-Molecular Theory for Ideal Gases**3. Gas particles do not attract or repel each other. Therefore ideal gases would never condense to form liquids.**The Kinetic-Molecular Theory for Ideal Gases**4. No kinetic energy is lost when gas particles collide with each other or with the walls of their container. Collisions are perfectly elastic.**The Kinetic-Molecular Theory for Ideal Gases**5. All gases have the same average kinetic energy at a given temperature.**Gas Solubility**Gases are less soluble in warm liquids than in cooler liquids.**Gas Solubility**Gases are more soluble when under pressure.**Vapor Pressure**The vapor pressure of water increases as the temperature increases.**STP**• STP stands for standard temperature and pressure.**Standard Pressure**• Standard pressure is 1atm (atmosphere) which is equal to 760 mm Hg, 760 torr, or 101.3 kPa (kilopascals).**Standard Temperature**• Standard temperature is 273 Kelvin.**Standard Atmospheric Pressure**1) Perform the following pressure conversions. a) 144 kPa = _____ atm (1.42) b) 795 mm Hg = _____ atm (1.05)**Standard Atmospheric Pressure**Perform the following pressure conversions. c) 669 torr = ______ kPa (89.2) d) 1.05 atm = ______ mm Hg (798)**Standard Atmospheric Pressure**• Air pressure at higher altitudes, such as on a mountaintop, is slightly lower than air pressure at sea level.**Standard Atmospheric Pressure**• Air pressure is measured using abarometer.**Boyle’s Law**• Boyle’s lawstates that the pressure and volume of a gas at constant temperature are inversely proportional. • Inversely proportional means as one goes up the other goes down.**1 atm**4 Liters**As the pressure on a gas increases, the volume decreases.**2 atm 2 Liters**Boyle’s Law**• The P-V graph for Boyle’s law results in a hyperbola because pressure and volume are inversely proportional.**Boyle’s Law**• P x V = K (K is some constant) P1 V1 = P2 V2**Example**• A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume?**Example**• First, make sure the pressure or volume units in the question match. A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? THEY DO!**Example**P1 • A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? V1 = P2 V2 V2 1.0 atm (25 L) 1.5 atm V2 = 17 L**Problem 2**• A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to 43 L? P2 = 2.2 atm**Problem 3**• A gas is collected in a 242 cm3 container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at standard pressure? V2 = 209 cm3**Problem 4**• A gas is collected in a 24.2 L container. The pressure of the gas in the container is determined to be 756 mm Hg. What is the pressure of this gas if the volume increases to 30.0 L? P2 = 610. mm Hg**Charles’ Law**• Charles’ Law states that the volume of a gas is directly proportional to the Kelvin temperature if the pressure is held constant. • Directly proportional means that as one goes up, the other goes up as well.**Charles’ Law**• The V-T graph for Charles’ law results in a straight line because volume and temperature are directly proportional.**Charles’ Law**• V / T = K (K is some constant) V1 V2 = T1 T2**Charles’ Law**• In any gas law problem involving temperature, temperature must be in Kelvin. K = °C + 273**Example**• What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure?**Example**• First, make sure the volume units in the question match. • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? THEY DO!**Example**• Second, make sure to convert degrees Celsius to Kelvin. • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? °C 25 K = + 273 K = 298 K**Example**V1 • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? V2 2.5 L 4.1 L = T1 T2 298 K T2 = 489 K**Problem 5**• What is the final volume of a gas that starts at 8.3 L and 17 ºC and is heated to 96 ºC? V2 = 11 L**Problem 6**• A 225 cm3 volume of gas is collected at 57 ºC. What volume would this sample of gas occupy at standard temperature? V2 = 186 cm3**Problem 7**• A 225 cm3 volume of gas is collected at 42 ºC. If the volume is decreased to 115 cm3, what is the new temperature? T2 = 161 K**Gay-Lusaac’s Law**• Gay-Lusaac’ Law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume is held constant.**Gay-Lusaac’s Law**• At higher temperatures, the particles in a gas have greater kinetic energy. • They move faster and collide with the walls of the container more often and with greater force, so the pressure rises.**Gay-Lusaac’s Law**• The P-T graph for Gay-Lusaac’s law results in a straight line because pressure and temperature are directly proportional.**Gay-Lussac’s Law**• P / T = K (K is some constant) P1 P2 = T1 T2**Example**• What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? Volume remains constant.**Example**• First, make sure the pressure units in the question match. • What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? There is only one pressure unit!**Example**• Second, make sure to convert degrees Celsius to Kelvin. • What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? °C 25 K = + 273 K = 298 K