# Wednesday, Jan. 16 th : “A” Day Thursday, Jan. 17 th : “B” Day Agenda  Homework Questions/Collect  Quiz over sec. 12.1: “Characteristics of Gases” 

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• Wednesday, Jan. 16th: A DayThursday, Jan. 17th: B DayAgendaHomework Questions/CollectQuiz over sec. 12.1: Characteristics of GasesSec. 12.2: The Gas LawsBoyles Law, Charless Law, Gay-Lussacs Law, Avogadros Law, Combined Gas LawHomework: Sec. 12.2 review, pg. 432: #1-12Concept Review: The Gas LawsBe prepared for a quiz covering this section next time.

• HomeworkPg. 422: #1-12Questions/Problems on the homework?

Hand In

• Section 12.1 QuizCharacteristics of GasesYou should be OK to complete this on your own using your notes and your book.

Have Fun!

• Section 12.2: The Gas LawsIn this section, the relationship between the measurable properties of gases will be studied using the following variables:

P = pressure exerted by the gasT = temperature in KELVINS of the gasV = total volume occupied by the gasn= number of moles of the gas

• Pressure-VolumeIn 1662, English scientist Robert Boyle found that as the pressure on a gas is increased in a closed container, the volume of the gas decreases.

Pressure Volume

Pressure Volume

• Gas molecules in a car-engine cylinderPressure-Volume

• Boyles Law

P1V1 = P2V2

• Pressure is inversely related to volume at constant temperatureP V

• Sample Problem B, Pg. 425Solving Pressure-Volume ProblemsA given sample of gas occupies 523 mL at 1.00 atm. The pressure is increased to 1.97 atm, while the temperature remains the same. What is the new volume of the gas?Use Boyles Law: P1V1 = P2V2P1 = 1.00 atmV1 = 523 mLP2 = 1.97 atmPlug in variables and SOLVE for V2V2 = 265 mL(3 sig figs)

• Additional PracticeIf 650 mL of hydrogen is stored in a cylinder with a moveable piston at 225 kPa and the pressure is increased to 545 kPa at constant temperature, what is the new volume?Use Boyles Law: P1V1 = P2V2P1 = 225 kPaV1 = 650 mLP2 = 545 kPaPlug variables in and SOLVE for V2V2 = 268 mL(3 sig figs)

• Temperature-Volume Relationships

In 1787, French physicist Jacques Charles discovered that a gass volume is directly proportional to the temperature on the Kelvin scale if the pressure remains constant.

Temperature Volume

Temperature Volume

• Temperature-Volume

• Charless Law

V 1 = V 2 T1 T2

Remember: Temperature is in Kelvins!

• Volume is directly proportional to Kelvin temperature at constant pressureT V

• Sample Problem C, Pg. 428Solving Volume-Temperature ProblemsA balloon is inflated to 665 mL volume at 27C. It is immersed in a dry-ice bath at 78.5C. What is its volume, assuming the pressure remains constant?Use Charless Law: V 1 = V 2 T1 = T2V1 = 665 mLT1 = 27C + 273 = 300 KT2 = -78.5C + 273 = 194.5 KPlug in variables and solve for V2.V2 = 431 mL(3 sig figs)

• Additional PracticeIf the original temperature of a 62.2 L sample of a gas is 150C, what is the final temperature of the gas (in degrees C) if the new volume of gas is 24.4 L and the pressure remains constant?Use Charless Law: V 1 = V 2 T1 = T2V1 = 62.2 LT1 = 150C + 273 = 423 KV2 = 24.4 LPlug in variables and solve for T2. Then change to C.T2 = -107C (3 sig figs)

• Temperature-Pressure RelationshipsIn 1802, French scientist Joseph Gay-Lussac discovered that if the temperature of a gas is doubled in a closed container of fixed volume, the pressure will double as well.

Temperature Pressure

Temperature Pressure

• Gay-Lussacs Law

P1 = P2 T1 T2

Remember: Temperature is in Kelvins!

• Pressure is directly proportional to Kelvin temperature, at constant volume T P

• Sample Problem D, pg. 430Solving Pressure-Temperature ProblemsAn aerosol can containing gas at 101 kPa and 22C is heated to 55C. Calculate the pressure in the heated can.Use Gay-Lussacs Law: P1 = P2 T1 T2P1 = 101 kPaT1 = 22C + 273 = 295 KT2 = 55C + 273 = 328 KPlug in variables and solve for P2 112 kPa

• Additional PracticeThe pressure in a bottle of soda pop is 505 kPa at 20.0C. What is the new pressure if someone warms the sealed bottle to 65.0C?Use Gay-Lussacs Law: P1 = P2 T1 T2P1 = 505 kPaT1 = 20.0C + 273 = 293 KT2 = 65.0C + 273 = 338 KPlug in variables and solve for P2 583 kPa

• Volume-Molar RelationshipsIn 1811, Italian scientist Amadeo Avogadro proposed the idea that equal volumes of ALL gases, under the same conditions, have the same number of particles.

Quite the looker, eh?

• Avogadros LawGas volume is directly proportional to the number of moles of gas at the same temperature and pressure.V = kn or V1 = V2 n1 n2n = number of moles of gask = proportionality constant

• Volume-Molar Relationships

• Avogadros LawAt STP (0C and 1 atm pressure) the volume of 1 mole of ANY gas is 22.41 L.

The MASS of 22.41 L of a gas at STP will be equal to the gass molar mass.(molar mass = the mass, in grams, of 1 mole)

• Combined Gas LawThe combined gas law is a combination of Boyles law and Charles law.+P1V1 = P2V2 T1 T2

Remember: Temperature is in Kelvins!

• Combined Gas Law ExampleWhen 500 mL of O2 gas at 25C and 1.045 atm is cooled to -40C and the pressure is increased to 2.00 atm, what is the new volume of the gas?Use the combined gas law: P1V1 = P2V2 T1 T2P1 = 1.045 atmV1 = 500 mLT1 = 25C + 273 = 298 KP2 = 2.00 atmT2 = -40C + 273 = 233 K V2 = 204 mL

• Additional ExampleA gas at STP occupies 28 cm3 of space. If the pressure changes to 3.8 atm and the temperature increases to 203C, find the new volume.Use the combined gas law: P1V1 = P2V2 T1 T2

P1 = 1.00 atmV1 = 28.0 cm3T1 = 0C + 273 = 273 KP2 = 3.8 atmT2 = 203C + 273 = 476 K V2 = 12.8 cm3

• How do you know which gas law to use?Look at the data given in the problem. What do you know from the problem?Temperature, volume, pressure, etc.Choose the formula that uses the variables given and solve for the missing variable

• HomeworkSec. 12.2 review, pg. 432: #1-12Concept Review: The Gas Laws

Next Time:Quiz over Sec. 12.2Lab Write-upGas Laws and Drinking Straws Activity