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Chapter 11 Gas Laws

Chapter 11 Gas Laws. Elements that exist as gases at 25 0 C and 1 atmosphere. Physical Characteristics of Gases. Gases assume the volume and shape of their containers. Gases are the most compressible state of matter.

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Chapter 11 Gas Laws

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  1. Chapter 11 Gas Laws

  2. Elements that exist as gases at 250C and 1 atmosphere

  3. Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids.

  4. Force Area Barometer What is Pressure? Pressure = Pressure is Measured in Units of: Pascals (Pa), millimeters of mercury, (mmHg), torr (torr), atmosphere (atm), and pounds of square inch (psi)

  5. STP To compare volumes of gases, one must know temperature and pressure at which the volumes are measured. Scientists have agreed upon standards conditions of exactly 1 atm pressure and 0 degrees celsius (273K).

  6. Under STP: 1 atmosphere of pressure = 760 mmHg 1atm = 760 torr 1 atm = 101.325kPa 1atm = 14.7psi

  7. The average atmospheric pressure at sea level is 0°C. Basically the weight of air at sea level is equal to 1 atm. What happens to atm pressure as you move further away from sea level? 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

  8. A Barometer is an instrument used to measure atmospheric pressure

  9. Convert Units of Pressure: Example #1: Express 0.830 atm in units of mmHg and kPa 0.830 atm x 760 mmHg = 631 mmHg 1 atm 0.830 atm x 101.325 kPa = 84.1 kPa 1 atm

  10. Dalton’s Law of Partial Pressures In a mixture of gases, the total pressure is calculated by adding the pressure of each individual gas in the mixture. V and T are constant P1 P2 Ptotal= P1 + P2

  11. Pg. 367 #1-4 • 1. Force / Area • 2. mmHg, torr, kPa, psi, atm • 3. STP = 273K and 1 atm • 4. a) 151.98 kPa x 1 atm= 1.5003 atm 101.3kPa b) 456 torr x 1 atm = 6.00 x 10-1 atm 760 torr

  12. Boyle’s Law

  13. What is Boyle’s Law? • Boyle’s Law states that when a gas is under pressure it takes up less space: • Boyles Law tells us about the relationship between the volume of a gas and its pressure at a constant temperature. The higher the pressure, the smaller the volume.

  14. VIDEO

  15. Volume vs. PressureInverse Relationship

  16. Sample Problem: • A deep sea diver is working at a depth where the pressure is 3.0 atmospheres. He is breathing out air bubbles. The volume of each air bubble is 2 cm3. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

  17. Figure out which variables the problems deals with (Volume and Pressure only) Use the Combined Gas Law Formula (Refer to reference sheet) Ignore the Temperature in the Formula P1V1= P2V2 T1 T2 BOYLE’S LAW

  18. How we work this out: • This is the formula we are going to use: • Formula first: P1 x V1 = P2 x V2 (Refer to reference Page) • Given:= P1 = 3.0 atm • V1 = 2 cm3 • P2 = 1.0 atm • Unknown: V2

  19. Here’s what you should have calculated P1 x V1 = P2 x V2 3 atm x 2 cm3 = 1 atm x V2 6 = 1 (V2) V2= 6 cm3

  20. Charles’ Law The Temperature-Volume Relationship

  21. Charles’ Law • French chemist Jacques Charles discovered that the volume of a gas at constant pressure changes with temperature. • Temperature and volume are directly related……As the temperature of the gas increases, so does its volume, and vie versa

  22. VIDEO

  23. V T Volume vs. TemperatureDirect Relationship

  24. Example • If the temperature of a given amount of gas is doubled, for example, its volume will also double (as long as pressure remains unchanged).

  25. Charles’ Law Example Problem: A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Given: V1 = 752 mL T1 = 25°C + 273 = 298K T2 = 50°C + 273 = 323K

  26. Charles’ Law • Determine which variables are in the problem. • Delete the variable that is not in the problem. (Pressure) That means that pressure remains constant. • Use the Combined Gas Law to extrapolate the right equation to solve this problem (reference sheet) P1V1= P2V2 T1 T2

  27. Equation Used: V1 = V2 T1 T2 752 mL = V2 298 K 323K 298K = 242896mL V2 = 815 mL

  28. Gay-Lussac’s Law

  29. What is Gay-Lussac’s Law? • Gay-Lussac’s Law states that when a gas is under constant volume, there is a direct relationship between temperature and pressure. • The higher the temperature, the higher the pressure.

  30. Gay-Lussac’s Law: P and T Gay-Lussac’s law states that • The pressure exerted by a gas is directly related to the Kelvin temperature. • V is constant. • An increase in temperature increases the pressure of a gas. 30

  31. PRESSURE VS. TEMP

  32. Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18 ˚C. What is the new pressure when the temperature is increased to 62 ˚C? Which variable is constant? 32

  33. Used the Combined Gas Law When you have problems that give you volume and pressure. Delete Volume in the Equation P1 V1= P2 V2 T1 T2 Gay-Lussac’s Law

  34. This formula is used when the volume is constant. P1 = P2 T1 T2 Gay-Lussac’s Law

  35. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK

  36. Ideal Gas Law • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

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