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Properties of Gases & The gas Laws
Chapter 14 Section 1 & 2 Properties of Gases & The gas Laws
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Measurable Properties of Gases
Pressure (P) Temperature (T) Volume (V) Moles (n) Units atm, Pa (SI unit), mmHg, torr K (SI unit), ºC L, mL, cm3, m3 moles, mol Instrument barometer, manometer thermometer graduated cylinder NA Meaning force exerted over area by colliding gas particles amount of average kinetic energy of particles space occupied by particles, not the volume of particles 1 mol = 6.02×1023 particles
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Properties of Gases Compressible Have mass
Gas particles always in motion Gas particles exert pressure when they run into a wall Take up any shape and size of container – diffuse (= to spread out) Are described with four variables: the amount of gas (n), volume (V), pressure (P), and temperature (T) ** variable = something you can change and represented by a letter
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Pressure vs. Volume (Boyle’s Law)
Describe the picture.
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Boyle’s Law Equations P1∙V1 = P2∙V2
“1” and “2” refer to two different sets of condition Match the unit for each variable The volume is inversely proportional to the pressure The volume increases (decreases) as the pressure decreases (increases) The temperature and amount of gas must remain unchanged for this law to work
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Bolye’s Law (Pressure-Volume Relationship)
hyperbola
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Example A given sample of gas occupies 523 mL at 760 torr. The pressure is increased to 1.97 atm, while the temperature remains the same. What is the new volume of the gas?
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Practices on Boyle’s Law
1) A flask containing 155 cm3 of hydrogen gas was collected under a pressure of 22.5 kPa. What pressure would have been required for the volume of the gas to have been 90.0 cm3?
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2) A sample of oxygen gas has a volume of 150
2) A sample of oxygen gas has a volume of 150. mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains the constant?
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3) A gas has a pressure of 1. 26 atm and occupies a volume of 7. 40 L
3) A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume 2.93 L, what will its pressure be, assuming constant temperature?
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Pressure-Volume Data (Boyle’s Law)
Pressure(kPa) Volume (L) PV(kPa×L) 150 0.334 50.1 200 0.250 50.0 250 0.200 300 0.167
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Charles’S Law (Temperature – Volume Relationship)
Volume (mL) Temperature (K) V/T (mL/K) 748 373 2.01 567 283 2.00 402 200 199 100 1.99
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Charles’s Law (Temperature-Volume Relationship)
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Charles’S Law 1) 2) The volume and temperature are directly proportional 3) The volume of gas increases (decreases) as the temperature increases (decreases) 4) The temperature MUST be in kelvin (K = C + 273) 5) Must keep the pressure and the amount of gas unchanged for this law to work
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Graph of Charles’s Law (Temperature-Volume Relationship)
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Example A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath. What, at −78.5°C, is its volume, assuming the pressure remains constant?
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Practices 1) A sample of neon gas occupies a volume of 752 mL at 25 ºC. What volume will the gas occupy at 50 ºC if the pressure remains constant?
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2) A helium-filled balloon has a volume of 2. 75 L at 20 ºC
2) A helium-filled balloon has a volume of 2.75 L at 20 ºC. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature?
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3) A gas at 65 ºC occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure?
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Combined Gas Law (Volume, Temperature & Pressure)
Combine Boyle’s law and Charles’s law: For this law to work, the amount of gas must remain unchanged With this law, 2 variables out of 3 can change
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Example 520 mL of hydrogen gas at 750 mmHg and 25 °C is placed in a mL container and heated to 50 °C. What is the pressure of the gas in the container?
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From Combined gas law to….
Boyle’s law: at constant temperature Charles’ law: at constant pressure Gay-Lussac’s law: at constant volume
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Gay-Lussac’s Law
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Graph of Gay-Lussac’s Law
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Gay-Lussac’s Law (Temperature-Pressure Relationship)
Temperature and pressure are directly proportional: Temperature MUST be in kelvins Works only if the volume and the amount of gas are kept constant
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Example An aerosol can containing gas at 101 kPa and 22 ºC is heated to 55 ºC. Calculate the pressure in the heated can. Answer: P2=112kPa
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Avogadro’s Law The volume (V) of gas is directly proportional to the number of moles (n) of gas: The type of gas doesn’t affect the volume; only the # of moles of gas does 1 mol of ANY gas at 1 atm and 0˚C takes up 22.4 L volume. For this law to work, pressure and temperature must remain unchanged
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Combine Boyle, charles, & Avogadro’s law
P1∙V1 = P2∙V2
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Ideal Gas Law 1 mol of ANY gas at 1 atm and 0˚C takes up 22.4 L volume
Combine the two above information and get…
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Example Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 kPa.
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Summary of gas laws Combine V, P, n, and T into one law.
From the combined gas law, get Boyle’s law Charles’ law Gay-Lussac’s law Avogadro’s law Ideal gas law
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Kinetic Molecular Theory
1) Gas particles are in constant, rapid, and random motion 2) The distance between gas particles are much larger than the size of atoms *The size of gas particle is almost nothing. 3) Gas particles colliding with surface creates pressure 4) Perfect elastic collisions between gas particles – no loss of energy during collisions but all transferred 5) Average kinetic energy of gas particles is proportional to kelvin temperature 6) Gas particles at the same temperature don’t have the same amount of kinetic energy (See the graph on the next slide)
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Non-Ideal gas behavior
Gas particles attract or repel each other Gas particles do have a volume Imperfect collision – energy lost Non-ideal behavior becomes more ideal at high temperature and low pressure
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Summary of gas laws Boyle’s Charles’s Combined Gay-Lussac’s Avogadro’s
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Summary of gas laws P1∙V1 = P2∙V2 Boyle’s Charles’s Combined
Gay-Lussac’s Avogadro’s volume-pressure volume-temperature volume-pressure-temperature pressure-temperature volume-mole V and P inversely proportional V and T directly proportional V and P inversely and V and T directly proportional P and T directly proportional V and n directly proportional P1∙V1 = P2∙V2 T and n constant P and n constant n constant n and V constant P and T constant
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