Presentation on theme: "Properties of Gases & The gas Laws"— Presentation transcript:
1 Properties of Gases & The gas Laws Chapter 14 Section 1 & 2Properties of Gases & The gas Laws
2 Measurable Properties of Gases Pressure (P)Temperature (T)Volume (V)Moles (n)Unitsatm, Pa (SI unit), mmHg, torrK (SI unit), ºCL, mL, cm3, m3moles, molInstrumentbarometer, manometerthermometergraduated cylinderNAMeaningforce exerted over area by colliding gas particlesamount of average kinetic energy of particlesspace occupied by particles, not the volume of particles1 mol = 6.02×1023 particles
3 Properties of Gases Compressible Have mass Gas particles always in motionGas particles exert pressure when they run into a wallTake 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
4 Pressure vs. Volume (Boyle’s Law) Describe the picture.
5 Boyle’s Law Equations P1∙V1 = P2∙V2 “1” and “2” refer to two different sets of conditionMatch the unit for each variableThe volume is inversely proportional to the pressureThe volume increases (decreases) as the pressure decreases (increases)The temperature and amount of gas must remain unchanged for this law to work
6 Bolye’s Law (Pressure-Volume Relationship) hyperbola
7 ExampleA 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?
8 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?
9 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?
10 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?
11 Pressure-Volume Data (Boyle’s Law) Pressure(kPa)Volume (L)PV(kPa×L)1500.33450.12000.25050.02500.2003000.167
13 Charles’s Law (Temperature-Volume Relationship)
14 Charles’S Law1)2) The volume and temperature are directly proportional3) 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
15 Graph of Charles’s Law (Temperature-Volume Relationship)
16 ExampleA 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?
17 Practices1) 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?
18 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?
19 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?
20 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 unchangedWith this law, 2 variables out of 3 can change
21 Example520 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?
22 From Combined gas law to…. Boyle’s law: at constant temperatureCharles’ law: at constant pressureGay-Lussac’s law: at constant volume
25 Gay-Lussac’s Law (Temperature-Pressure Relationship) Temperature and pressure are directly proportional:Temperature MUST be in kelvinsWorks only if the volume and the amount of gas are kept constant
26 ExampleAn 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
27 Avogadro’s LawThe 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 does1 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
28 Combine Boyle, charles, & Avogadro’s law P1∙V1 = P2∙V2
29 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…
30 ExampleDetermine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 kPa.
31 Summary of gas laws Combine V, P, n, and T into one law. From the combined gas law, getBoyle’s lawCharles’ lawGay-Lussac’s lawAvogadro’s lawIdeal gas law
32 Kinetic Molecular Theory 1) Gas particles are in constant, rapid, and random motion2) 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 pressure4) Perfect elastic collisions between gas particles – no loss of energy during collisions but all transferred5) Average kinetic energy of gas particles is proportional to kelvin temperature6) Gas particles at the same temperature don’t have the same amount of kinetic energy (See the graph on the next slide)
34 Non-Ideal gas behavior Gas particles attract or repel each otherGas particles do have a volumeImperfect collision – energy lostNon-ideal behavior becomes more ideal at high temperature and low pressure
35 Summary of gas lawsBoyle’sCharles’sCombinedGay-Lussac’sAvogadro’s
36 Summary of gas laws P1∙V1 = P2∙V2 Boyle’s Charles’s Combined Gay-Lussac’sAvogadro’svolume-pressurevolume-temperaturevolume-pressure-temperaturepressure-temperaturevolume-moleV and P inversely proportionalV and T directly proportionalV and P inversely and V and T directly proportionalP and T directly proportionalV and n directly proportionalP1∙V1 = P2∙V2T and n constantP and n constantn constantn and V constantP and T constant
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