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Review on the Properties of Gases

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Title: Review on the Properties of Gases


1
Review on the Properties of Gases
  • Kinetic Theory- particles are in constant motion
  • Increase in temperature increase movement of
    particles (increase kinetic energy)
  • Collisions cause pressure (collisions with walls
    of container and collisions with other molecules)

2
Review on Gases cont
  • Gases have no definite shape (they take the shape
    of their container)
  • Gas molecules are spread apart (not fixed in
    position)
  • Changing the kinetic energy of the particles will
    cause a significant change in volume (as
    discussed earlier)

3
Factors Affecting Gas Pressure
  • Number of Particles
  • What happens when you put air in your tire?
  • Pressure increases because you are increasing the
    number of gas particles which increases
    collisions.
  • If you double the number of gas particles then
    you will double the gas pressure
  • Volume of Gas/Container
  • What happens if you decrease the volume of a
    container of gas?increase the volume of the
    container?
  • Decreasing volume causes more collisions thus
    increasing pressureIncreasing volume decreases
    collision thus decreasing pressure.
  • Temperature
  • What happens if you increase the temperature of a
    gas?decrease it?
  • Increasing temperature will increase pressure due
    to the increase in kinetic energy. The particles
    are moving faster and hit the walls with more
    force and more frequently. Opposite is true for
    a decrease in temp

4
The Pressure-Volume Relationship Boyles Law
  • For a given mass of gas at a constant
    temperature, the volume of the gas varies
    inversely with pressure.
  • If you increase volume you decrease pressure.
  • If you decrease volume you increase pressure.

5
Boyles Law cont
  • The mathematical equation for Boyles Law is P1
    X V1 P2 X V2
  • P1 and V1 are the initial pressure and volume
    given
  • P2 and V2 are the final pressure and volume given
  • REMINDER!! Standard pressure is 1.0 atm or 760
    mmHg or 101 kPa or 76 cmHg

6
  • A high-altitude balloon contains 30.0 L of helium
    at standard temperature and pressure. What is
    the volume when the balloon rises to an altitude
    where the pressure is only 25.0 kPa? (assume
    temperature remains constant)
  • P1 101 kPa
  • V1 30.0 L
  • P2 25 kPa
  • V2 ?
  • 101 X 30 25 X V2 124 L
  • 25 25

7
The Temperature Volume Relationship Charles
Law
  • The volume of a fixed mass of gas is directly
    proportional to its Kelvin temperature if the
    pressure is kept constant.
  • As temperature increases volume increases
  • As temperature decreases volume decreases

8
Charles Law cont
  • The mathematical equation for Charles Law is
    V1/T1 V2/T2
  • Remember V1 and T1 are the initial and V2 and T2
    are the final
  • Temperature MUST be measured in Kelvin
  • To convert from C to K you add 273 to the Celsius
    temp
  • To convert from K to C you subtract 273 from
    Kelvin temp

9
The Temperature-Pressure Relationship
Gay-Lussacs Law
  • The pressure of a gas is directly proportional to
    the Kelvin temperature if the volume remains
    constant
  • The mathematical equation is represented by
    P1/T1 P2/T2

10
  • The gas left in a used aerosol can is at standard
    pressure and a temperature of 298 K. If this can
    is thrown into a fire, what is the pressure of
    the gas in kPa if the temperature of the aerosol
    can reaches 1201 K?
  • P1 101 kPa
  • T1 298 K
  • P2 ? kPa
  • T2 1201 K
  • 101 P2 101 X 1201 415 kPa
  • 298 1201 298

11
The Combined Gas Law
  • Boyles, Charles, and Gay-Lussacs Laws can be
    used together to form the Combined Gas Law
  • P1V1 P2V2
  • T1 T2

12
  • The Volume of a gas-filled balloon is 30.0 L at
    313 K and 153 kPa pressure. What volume will the
    balloon have at standard temperature and
    pressure?
  • V1 30L
  • T1 313K
  • P1 153 kPa
  • V2 ? L
  • T2 273 K (standard temperature)
  • P2 101 kPa
  • V2 30L X 153 kPa X 273K 39.5
  • 101.3 kPa X 313 K

13
Daltons Law of Partial Pressure
  • At constant volume and temperature, the total
    pressure exerted by a mixture of gases is equal
    to the sum of the partial pressures of the
    component gases.
  • This is represented by the equation
  • Ptotal P1 P2 P3

14
  • Air contains oxygen, nitrogen, carbon dioxide,
    and trace amounts of other gases. What is the
    partial pressure of oxygen if the total pressure
    is 101.3 kPa and the partial pressures of
    nitrogen, carbon dioxide, and other gases are
    79.10 kPa, 0.040 kPa, and 0.94 kPa respectively.
  • PN2 79.10 kPa
  • PCO2 0.040 kPa
  • Pothers 0.94 kPa
  • Ptotal 101.30 kPa
  • PO2 ? kPa
  • PO2 Ptotal (PN2 PCO2 Pothers)
  • 101.3 (79.1 .04 .94) 21.22 kPa

15
Ideal Gas Law
  • Based on Avogadros principl
  • -At equal pressures and temperature, equal
    volumes of gases contain the same number of
    molecules

16
  • One mole of any gas at STP will occupy the same
    volume as one mole of any other gas at STP
  • One mole of any gas at STP has a volume of 22.4
    liters
  • The molar volume of a gas at STP

17
Ideal Gas Equation
  • Uses the 3 variables of the combined gas law plus
    the molar volume

18
PV nRT
  • P pressure
  • V volume
  • n number of moles
  • R molar gas constant
  • T temperature

19
Molar Gas Constants
  • Derived from the molar volume of
  • a gas at STP
  • P standard pressure
  • 1 atm 760 mmHg 101.3 kPa
  • V 22.4 L
  • n one mole
  • T 273 K
  • R

20
Calculation of the Constant
  • When pressure is in atm
  • (1 atm)(22.4 L) (1 mole) R (273 K)
  • 22.4 atmL 273 mol K (R )
  • 0.0821 atm L/mol K R
  • When pressure is in kPa
  • (101.3 kPa)(22.4 L) (1 mole) R (273 K)
  • 2269 kPa L 273 mol K (R )
  • 8.31 kPa L/ mol K R

21
When pressure is in mmHG (Torr) (760
mmHg)(22.4 L) (1 mole) R (273 K) 17024
mmHg L 273 mol K (R ) 62.4 mmHg L/ mol K
R
22
What pressure is exerted by .306 moles of gas in
a 25.9 L container at 282 K ?-
P(25.9)(0.306mol)(0.0821atmL/mol K) (282 K)-
P (25.9L) 7.085 atm L- P 0.274 atm
23
What volume is occupied by 0.684
mol of a gas at 743 mmHg and 282 K? (743
mmHg)V 0.684 mol (62.4) (282K )
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