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The Gas Laws. Chemistry Dr. May. Gaseous Matter. Indefinite volume and no fixed shape Particles move independently of each other Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids. Avogadro’s Number.

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The gas laws

The Gas Laws

Chemistry

Dr. May


Gaseous matter
Gaseous Matter

  • Indefinite volume and no fixed shape

  • Particles move independently of each other

  • Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids


Avogadro s number
Avogadro’s Number

  • One mole of a gas contains Avogadro’s number of molecules

  • Avogadro’s number is

6.02 x 1023 or

602,000,000,000,000,000,000,000


Diatomic gas elements

Gas

Hydrogen (H2)

Nitrogen (N2)

Oxygen (O2)

Fluorine (F2)

Chlorine (Cl2)

Molar Mass

2 grams/mole

28 grams/mole

32 grams/mole

38 grams/mole

70 grams/mole

Diatomic Gas Elements


Inert gas elements

Gas

Helium

Neon

Argon

Krypton

Xenon

Radon

Molar Mass

4 grams/mole

20 grams/mole

40 grams/mole

84 grams/mole

131 grams/mole

222 grams/mole

Inert Gas Elements


Other important gases

Gas

Carbon Dioxide

Carbon Monoxide

Sulfur Dioxide

Methane

Ethane

Freon 14

Formula Molar Mass

CO2 44 g/mole

CO 28 g/mole

SO2 64 g/mole

CH4 16 g/mole

CH3CH3 30 g/mole

CF4 88 g/mole

Other Important Gases


One mole of oxygen gas o 2
One Mole of Oxygen Gas (O2)

  • Has a mass of 32 grams

  • Occupies 22.4 liters at STP

    • 273 Kelvins (0oC)

    • One atmosphere (101.32 kPa)(760 mm)

  • Contains 6.02 x 1023 molecules

    (Avogadro’s Number)


  • Mole of carbon dioxide co 2
    Mole of Carbon Dioxide (CO2)

    • Has a mass of 44 grams

    • Occupies 22.4 liters at STP

    • Contains 6.02 x 1023 molecules


    One mole of nitrogen gas n 2
    One Mole of Nitrogen Gas (N2)

    • Has a mass of 28 grams

    • Occupies 22.4 liters at STP

    • Contains 6.02 x 1023 molecules


    Mole of hydrogen gas h 2

    Mass

    Volume at STP

    Molecules

    2.0 grams

    22.4 liters

    6.02 x 1023

    Mole of Hydrogen Gas (H2)


    Standard conditions stp

    Molar Volume

    Standard

    Temperature

    Standard Pressure

    22.4 liters/mole

    0 oC

    273 Kelvins

    1 atmosphere

    101.32 kilopascals

    760 mm Hg

    Standard Conditions (STP)


    Gas law unit conversions

    liters  milliliters

    milliliters  liters

    o C  Kelvins

    Kelvins  o C

    mm  atm

    atm  mm

    atm  kPa

    kPa  atm

    Multiply by 1000

    Divide by 1000

    Add 273

    Subtract 273

    Divide by 760

    Multiply by 760

    Multiply by 101.32

    Divide by 101.32

    Gas Law Unit Conversions


    Charles law
    Charles’ Law

    • At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins

      V1 = V2

      T1 T2

      As the temperature goes up, the volume goes up


    Boyle s law
    Boyle’s Law

    • At constant temperature, the volume of a gas is inversely proportional to the pressure.

      P1V1 = P2V2

      As the pressure goes up, the volume goes down


    Combined gas law
    Combined Gas Law

    P1V1 = P2V2

    T1 T2

    • Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hg

    • Standard Temperature (T) is 273 K

    • Volume (V) is in liters, ml or cm3


    Charles law problem
    Charles’ Law Problem

    A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?

    1. Convert oC to Kelvins

    25 + 273 = 298 K

    38 + 273 = 311 K

    2. Insert into formula


    Charles law solution
    Charles’ Law Solution

    V1 = V2

    T1 T2

    V1 = 2 liters V2 = Unknown

    T1 = 298 K T2 = 311 K

    2 = V2

    298 311


    Charles law solution1
    Charles’ Law Solution

    2 = V2

    298 311

    298 V2 = (2) 311

    V2 = 622

    298

    V2 = 2.09 liters


    Charles law problem answer
    Charles’ Law Problem Answer

    A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?

    V2 = 2.09 liters


    Boyle s law problem
    Boyle’s Law Problem

    A balloon has a volume of 2.0 liters at

    743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?

    1. Convert pressure to the same units

    743  760 = .98 atm

    2. Insert into formula


    Boyle s law solution
    Boyle’s Law Solution

    P1V1 = P2V2

    P1 = 0.98 atm P2 = 2.5 atm

    V1 = 2.0 liters V2 = unknown

    0.98 (2.0) = 2.5 V2


    Boyle s law solution1
    Boyle’s Law Solution

    P1V1 = P2V2

    0.98 (2.0) = 2.5 V2

    V2= 0.98 (2.0)

    2.5

    V2 = 0.78 liters


    Boyle s law problem answer
    Boyle’s Law Problem Answer

    A balloon has a volume of 2.0 liters at

    743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?

    V2 = 0.78 liters


    Combined gas law problem
    Combined Gas Law Problem

    A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?

    1. Convert 25 oC to Kelvins 25 + 273 = 298 K

    2. Standard pressure is 101.32 kPa

    3. Standard temperature is 273 K

    4. Insert into formula


    Combined gas law solution
    Combined Gas Law Solution

    P1V1 = P2V2

    T1 T2

    P1 = 98 kPa P2 = 101.32 kPa

    V1 = 2.0 liters V2 = unknown

    T1 = 298 K T2 = 273 K


    Combined gas law solution1
    Combined Gas Law Solution

    P1V1 = P2V2

    T1 T2

    98 (2.0) = 101.32 V2

    298 273

    (298) (101.32) V2 = (273) (98) (2.0)


    Combined gas law solution2
    Combined Gas Law Solution

    P1V1 = P2V2

    T1 T2

    (298) (101.32) V2 = (273) (98) (2.0)

    V2= 273 (98) (2.0)

    (298) (101.32)


    Combined gas law solution3
    Combined Gas Law Solution

    P1V1 = P2V2

    T1 T2

    V2= 273 (98) (2.0)

    (298) (101.32)

    V2=53508=1.77 liters

    30193


    Combined gas law problem answer
    Combined Gas Law Problem Answer

    A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?

    V2 = 1.77 liters


    Combined gas law v 2
    Combined Gas Law – V2

    P1V1 = P2V2

    T1 T2

    P1V1T2 = P2V2T1

    P1V1T2= V2

    P2T1


    The end
    The End

    • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology

    • Please send suggestions and comments to rmay@nccvt.k12.de.us


    The ideal gas law

    The Ideal Gas Law

    Chemistry

    Dr. May


    Kinetic molecular theory
    Kinetic Molecular Theory

    Molecules of an ideal gas

    • Are dimensionless points

    • Are in constant, straight-line motion

    • Have kinetic energy proportional to their absolute temperature

    • Have elastic collisions

    • Exert no attractive or repulsive forces on each other


    Ideal gas law
    Ideal Gas Law

    PV = nRT

    P = pressure in kilopascals (kPa) or atmospheres (atm)

    V = volume in liters

    n = moles

    T = temperature in Kelvins

    R = universal gas constant


    Ideal gas law pv nrt

    Pressure (P)

    Volume (V)

    Moles (n)

    Temperature (T)

    The universal gas constant (R)

    Atm or kPa

    Always liters

    Moles

    Kelvins

    0.0821 ( P in atm) or

    8.3 (P in kPa)

    Ideal Gas Law: PV = nRT


    Universal gas constant
    Universal Gas Constant

    R = 0.0821 if P = atmospheres

    R = 8.3 if P = kilopascals

    R = PV

    nT


    Deriving r for p in atmospheres
    Deriving R for P in Atmospheres

    R = PV

    nT

    Assume n = 1 mole of gas

    Standard P = 1atmosphere

    Standard V = molar volume = 22.4 liters

    Standard T = 273 Kelvins


    R value when p is in atmospheres
    R Value When P Is In Atmospheres

    R = PV

    nT

    R = (1) (22.4)

    (1) 273

    R = 0.0821atm Liters

    mole Kelvins


    Deriving r for p in kilopascals
    Deriving R For P In Kilopascals

    R = PV

    nT

    Assume n = 1 mole of gas

    Standard P = 101.32 kilopascals

    Standard V = molar volume = 22.4 liters

    Standard T = 273 Kelvins


    R value when p is in kilopascals
    R Value When P Is In Kilopascals

    R = PV

    nT

    R = (101.32) (22.4)

    (1) 273

    R = 8.3kPa Liters

    mole Kelvins


    Ideal gas law pressure
    Ideal Gas Law - Pressure

    PV = nRT

    P = nRT

    V

    Solves for pressure when moles, temperature, and volume are known


    Ideal gas law volume
    Ideal Gas Law - Volume

    PV = nRT

    V = nRT

    P

    Solves for volume when moles, temperature, and pressure are known


    Ideal gas law temperature
    Ideal Gas Law - Temperature

    PV = nRT

    T = PV

    nR

    Solves for temperature when moles, pressure, and volume are known


    Ideal gas law moles
    Ideal Gas Law - Moles

    PV = nRT

    n = PV

    RT

    Solves for moles when pressure, temperature, and volume are known


    Ideal gas law problem
    Ideal Gas Law Problem

    What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?

    • V = 2.3 liters

    • P = 1.2 atmospheres

    • T = 25 oC = 298 Kelvins

    • R = 0.0821 since P is in atms.

    • Find moles (n), then grams


    Ideal gas law solution moles
    Ideal Gas Law Solution (moles)

    PV = nRT

    1.2 (2.3) = n (0.0821) (298)

    n = 1.2 ( 2.3) = 0.11 moles

    (0.0821) (298)


    Ideal gas law solution grams
    Ideal Gas Law Solution (Grams)

    Grams = moles x molecular weight (MW)

    • Moles = 0.11

    • Molecular Weight of N2 = 28 g/mole

    • Grams = 0.11 x 28 = 3.1 grams


    Ideal gas law answer
    Ideal Gas Law Answer

    What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?

    The answer is 0.11 moles and

    3.1 grams


    The end1
    The End

    • This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology

    • Please send suggestions and comments to rmay@nccvt.k12.de.us