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Question: State Boyles laws You have 5 minutes

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Agenda: Discuss Ch. 14 sec. 1-2 Go through how to do the math HW: Worksheet with these equations

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I can… Apply the 3 gas laws to problems involving pressure, volume and temperature of a gas. Pressure v. Volume Temperature v. Volume Pressure v. Temperature

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Warning All RED text slides are embedded notes

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The theory explains the behavior of particles that make up gas, liquids and solids- particles are in constant motion. What is the kinetic theory?

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1. Gas particles do not attract or repel each other. Briefly summarize each assumption about gas behaviors (pg. 419-420) 2. Gas particles are much smaller than the distances between them. 3. Gas particles are in constant, random motion 4. No kinetic energy is lost when gas particles collide witheach other or container. 5. All gases have the same average kinetic energy at a given temperature.

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1. Number of gas particles. What are the four factors is the basis of the kinetic theory? 2. Temperature 3. Pressure 4. Volume

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Pressure and volume are the two factors affected. Assuming that the number of particles and temperature is constant, what happens to the gas in a plastic balloon if you squeeze it? Provide the two factors that are affected? What is the relationship between the two? As one increases the other decreases (inverse relationship)

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Its states that the volume of a given amount of gas held at a constant temperature varies inversely with the pressure. State Boyles Law

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P 1 V 1 = P 2 V 2 What is Boyles Law’s Formula

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1. The volume of a gas at 99.0 kPa is 300 ml. If the pressure is increased to 188 kPa, what will be the new volume? P 1 = 99.0 kPa V 1 = 300 mL P 2 = 188 kPa V 2 = X (99.0 kPa) (300 mL) = (188 kPa) (X) 29700 mL = X 188 158 ml = X

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2. The pressure of a sample of helium in a 1.99 –L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00 L container? P 1 = 0.988 atm V 1 = 1.99 L P 2 = X V 2 = 2.00 L (0.988 atm) (1.99 L) = (X) (2.00 L) 1.966 atm = X 2.00 0.983 atm = X

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3. Air trapped in a cylinder fitted with a piston occupies 145.7 ml at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? P 1 = 1.08 atm V 1 = 145.7 mL P 2 = 1.43 atm V 2 = X (1.08 atm) (145.7 mL) = (1.43 atm) (X) 15735.6 mL = X 1.43 1.10 x10 2 mL = X

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4. If it takes 0.5000L of oxygen gas kept in a cylinder under pressure to fill an evacuated 4.00 L reaction vessel in which the pressure is 0.980 atm, what was the initial pressure of the gas in the cylinder? P 1 = X atm V 1 = 0.5000 L P 2 = 0.980 atm V 2 = 4.00 L (X) (0.5000 L) = (0.980 atm) (4.00L) X= 3.92 atm 0.5000 7.84 atm = X

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5. A sample of neon gas occupies 0.220 L at 0.860 kPa. What will be its volume at 29.2 kPa pressure? P 1 = 0.860 kPa V 1 = 0.220 L P 2 = 29.2 kPa V 2 = X L (0.860 kPa) (0.220 L) = (29.2 kPa) (X L) 0.1892 L = X 29.2 6.00 x10 -3 L= X

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Its states that the volume of a given mass of a gas is directly proportional to its kelvin temperature at constant pressure. State Charles Law

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V 1 V 2 T 1 T 2 What is Charles Law’s Formula =

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-273.15 ˚C After graphing temperature verses volume, and extrapolating temperature verses volume to zero volume, what is the temperature that would match this volume?

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-273.15 ˚C or 0 Kelvin, is absolute zero What is the temperature of absolute zero? ___________ Define what IS the theory behind absolute zero? The kinetic energy is 0, no motion!

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6. A gas at 89 C ˚ occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? T 1 = 89 + 273 = 362 K V 1 = 0.67 L T 2 = X V 2 = 1.12 L 405.44 = (X) (0.67 L) 362K 0.67 L 605 – 273 = 332 ˚C X 1.12 L = 405.44 K = X 0.67

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7. The Celsius temperature of a 3.00 L sample of gas is lowered from 80.0 ˚C to 30.0˚C. What will be the resulting volume of gas? T 1 = 80 + 273 = 353 K V 1 = 3.00 L T 2 = 30 + 273 = 303 K V 2 = X L 909 = (X) (353 K) 353 K 3.00 L 2.58 L 303 K X = 909 L = X 353

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8. What is the volume of the air in a balloon that occupies 0.620 L at 25˚C if the temperature is lowered to 0.00˚C? T 1 = 25 + 273 = 298 K V 1 = 0.620 L T 2 = 0.0 + 273 = 273 K V 2 = X L 169 = (X) (298 K) 298 K 0.620 L 0.568 L 273 K X = 169 L = X 298

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Its states that the pressure of a given mass of gas varies directly with the kelvin temperature when the volume remains constant. State Gay-Lussac’s Law

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What is Gay-Lussac’s Formula P 1 P 2 T 1 T 2 =

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9. A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0 ˚C. If the pressure in the container is increased to 201 kPa, what is the new temperature- in Celsius? P 1 = 125 kPa T 1 = 30 + 273 = 303K P 2 = 201 kPa T 2 = X 60903 = (X) (125 kPa) 125 kPa 303 K 487 – 273 = 214 ˚C 201 kPa X = 60903 K = X 125

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10. The pressure in an automobile tire is 1.88 atm at 25 ˚C. What will be the pressure if the temperature warms up to 37.0 ˚C? P 1 = 1.88 atm T 1 = 25 + 273 = 298K P 2 = X T 2 = 37 + 273 = 310K 583 = (X) (298K) 1.88 atm 298 K 1.96 atm or 2.0 atm X 310 K = 583 atm = X 298

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11. Helium gas in a 2.00 L cylinder is under 1.12 atm pressure. At 36.5 ˚C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas cylinder? P 1 = 1.12 atm T 1 = X P 2 = 2.56 atm T 2 = 36.5 + 273 = 309.5K 347 = (X) (2.56 atm) 1.12 atm X K 135 K- 273 = -138 ˚C 2.56 atm 309.5 K = 347 K = X 2.56

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12. A gas sample has a pressure of 30.7 kPa at 0.00˚C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa? P 1 = 30.7 kPa T 1 = 0.0 + 273 = 273 K P 2 = 28.4 kPa T 2 = X 7753.2 = (X) (30.7 kPa) 30.7 kPa 273 K 253 K- 273 = -20 ˚C 28.4 kPa X = 7753 K = X 30.7

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13. A ridged plastic container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0˚C. How much more pressure will the gas exert if the temperature is raised to 44.6˚C? P 1 = 660 torr T 1 = 22 + 273 = 295K P 2 = X T 2 = 44.6 + 273 = 317.6K 209616 = (X) (295 torr) 660 torr 295 K 711 torr- 660 = 51 torr X torr 317.6 K = 209616 K = X 295

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BOYLES CHARLES & GAY-LUSSAC Worksheet

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