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Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
Chapter 10.3 : The Gas Laws
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Chapter 10.3 : The Gas Laws

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  • 1. The Gas lawsChapter 10.3
    Objectives:
    Use the kinetic-molecular theory to explain the relationships between gas volume, temperature, and pressure.
    Use Boyle’s law to calculate volume-pressure changes at constant temperature.
    Use Charles’s law to calculate volume-temperature changes at constant pressure.
    Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.
    Use the combined gas law to calculate volume-temperature-pressure changes.
    Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.
  • 2. The Gas Laws
    Simple mathematical relationships between
    Volume
    Temperature
    Pressure
    Amount of gas
    Gas Law Program:
    Shows the relationship of all four of the above on gases
    http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
    Constant :
    Volume and amount of gas
    Shows:
    Change in pressure and temperature
  • 3. Boyle’s Law
    States – volume of a fixed mass of gas varies inversely with the pressure at constant temperature.
    Constant temperature and amount of gas
    If you double volume, pressure is cut in half
    If you cut volume in half, pressure doubles
  • 4. Boyle’s Law
    Pressure caused by
    Moving molecules hitting container walls
    Speed of particles (force) and number of collisions
    Both increase pressure
    Mathematically:
    Volume-Pressure Data for Gas Sample
    1
    Volume Pressure P x V
    (mL) (atm)
    V
    k
    =
    PV
    or
    =
    k
    P
    k is constant
    1200 0.5 600
    P is pressure
    600 1.0 600
    V is volume
    300 2.0 600
    200 3.0 600
    150 4.0 600
    120 5.0 600
    100 6.0 600
    Interactive graph
  • 5. Boyle’s Law
    Boyle’s Law Equation:
    P1V1
    =
    k
    P2V2
    =
    k
    So:
    P1V1
    =
    P2V2
    Sample Problem 1
    A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant.
    P1
    =
    0.947 atm
    P1V1
    =
    P2V2
    V1
    =
    150. mL
    (0.947 atm)
    (150. mL)
    P2
    P2
    V2
    =
    P2
    =
    0.987 atm
    0.987 atm
    P1V1
    V2
    =
    ?
    =
    V2
    =
    144 mL
    P2
  • 6. Charles’s Law
    Volume-temperature Relationship
    When using temperature, Must use absolute zero, Kelvin Temperature scale
    Temperature -273.15oC is absolute zero
    All molecular movement would stop.
    K = oC + 273
    212
    100
    373
    So : How many Kelvin is 10 oC?
    K = oC + 273
    = 10oC + 273
    32
    0
    273
    = 283 K
    Fahrenheit
    Celsius
    Kelvin
    Temperature Scales
  • 7. Charles’s Law
    Volume-Temperature Data for Gas Sample
    Temperature Kelvin Volume V/T
    (oC) (K) (mL) (mL/K)
    273 546 1092 2
    100 373 746 2
    10 283 566 2
    1 274 548 2
    0 273 546 2
    -1 272 544 2
    -73 200 400 2
    -173 100 200 2
    -223 50 100 2
  • 8. Charles’s Law
    States that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.
    Constant Pressure and amount of gas
    If you double temperature, the volume will double
    If you cut the temperature in half, the volume will be half as much.
    V
    V
    k
    =
    or
    kT
    =
    T
    k is constant
    T is temperature
    V is volume
    Charles’s Law Equation:
    V1
    V2
    =
    k
    =
    k
    T1
    T2
    So:
    V2
    V1
    =
    T2
    T1
  • 9. Charles’s Law
    Sample Problem 2
    A sample of neon gas occupies a volume of 752 mL at 25oC. What volume will the gas occupy at 50oC if the pressure remains constant?
    V2
    V1
    V1
    =
    752 mL
    T2
    =
    T2
    x
    x
    T2
    T1
    T1
    =
    25oC
    + 273 =
    298 K
    V2
    =
    ?
    V1
    T2
    (752 mL)
    (323 K)
    T2
    =
    50oC
    + 273 =
    323 K
    V2
    =
    =
    T1
    298 K
    **Always convert to Kelvin!!
    =
    815 mL
  • 10. Gay-Lussac’s Law
    Pressure-temperature Relationship
    Increasing temperature, increases the speed of the gas particles
    Thus, more collisions with the container walls
    Causing an increase in pressure
  • 11. Gay-Lussac’s Law
    States that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.
    Constant volume and amount of gas
    If you double temperature, pressure doubles
    If you cut temperature in half, pressure is also cut in half
    P
    P
    k
    =
    or
    kT
    =
    T
    k is constant
    T is temperature
    P is pressure
    P1
    P2
    =
    k
    =
    k
    T1
    T2
    So:
    P2
    P1
    =
    T1
    T2
  • 12. Gay-Lussac’s Law
    Sample Problem 3
    The gas in an aerosol can is at a pressure of 3.00 atm at 25oC. Directions on the can warn the user not to keep the can in a place where the temperature exceed 52oC. What would the gas pressure in the can be at 52oC?
    P1
    P2
    P1
    =
    3.00 atm
    T2
    =
    T2
    x
    x
    T2
    T1
    T1
    =
    25oC
    + 273 =
    298 K
    P2
    =
    ?
    P1
    T2
    (3.00 atm)
    (325 K)
    T2
    =
    52oC
    + 273 =
    325 K
    P2
    =
    =
    T1
    298 K
    **Always convert to Kelvin!!
    =
    3.27 atm
  • 13. The Combined Gas Law
    Sample Problem 4
    A helium-filled balloon has a volume of 50.0 L at 25oC and 1.08 atm. What volume will it have at 0.855 atm and 10.oC?
    P1V1
    P2V2
    P1
    1.08 atm
    =
    T2
    =
    x
    T2
    x
    T2
    T1
    T1
    =
    25oC
    + 273 =
    298 K
    V1
    =
    50.0 L
    P1V1
    T2
    P2V2
    =
    P2
    =
    0.855 atm
    T1
    P2
    P2
    T2
    =
    10.oC
    + 273 =
    283 K
    P1V1
    T2
    V2
    =
    ?
    V2
    =
    T1
    P2
    (1.08 atm)
    (283 K)
    (50.0 L)
    =
    (298 K)
    (0.855 atm)
    60.0 L
    =
  • 14. Dalton’s Law of Partial Pressures
    States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
    Partial pressure: pressure of each gas in a mixture
    PT = p1 + p2 + p3 + ……
    PT= Total Pressure
    p1 + p2 + p3 = partial pressures
  • 15. Dalton’s Law of Partial Pressures
    Gas collected by water displacement.
    Must include the pressure exerted by water vapor
    PT = p1 + p2 + p3 + ……
    So: Patm = pgas + pH2O
  • 16. Dalton’s Law of Partial Pressures
    Sample Problem 5
    Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0oC, respectively. What was the partial pressure of the oxygen collected?
    Patm = pO2 + pH2O
    Patm = 731.0 torr
    PO2 = ?
    PH2O = 17.5 torr (from appendix in table A-8, pg. 899)
    pO2 = Patm - pH2O
    pO2 = 731.0 torr – 17.5 torr
    = 713.5 torr

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