quick write p126: what happens to a plastic bottle placed in the freezer overnight?
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Quick Write p126:
What happens to a plastic bottle placed in the freezer overnight?
Honor time limits
Actively participate (Have S.W.A.G.)
Listen respectfully to your colleagues
Place cell phones on vibrate or silent mode
Participants may write burning questions on a sticky note and place on the parking lot
BE PRESENT (Developing H.O.T.S. for Science)
Burning Issues
Questions
Comments
Ideas to Share
Charles’ LawCharles’ LawThe Temperature-Volume
Relationship
Charles’ LawCharles’ Law• French chemist Jacques CharlesJacques Charles
discovered that the volume of a gas at constant pressure changes with temperature.
• As the temperature of the gas increases, so does its volume, and as its temperature decreases, so does its volume.
C h a r l e s ’ L a wC h a r l e s ’ L a w
V= Volumek= Charles’ Law constant of ProportionalityT= Temperature in Kelvin
ExplanationExplanation• Raising the temperature of a
gas causes the gas to fill a greater volume as long as pressure remains constant.
• Gases expand at a constant rate as temperature increases, and the rate of expansion is similar for all gases.
ExampleExample• If the temperature of a given
amount of gas is doubled, for example, its volume will also double (as long as pressure remains unchanged).
2V = 22V = 2kkTT
Charles’ LawCharles’ LawCharles’ Law can be modified to a convenient form by solving for k.
Charles’ LawCharles’ LawIn a sample with volume V1 & temperature T1, changing either volume or temperature converts these variables to V2 & T2.
Demonstration Demonstration of of
Charles’ LawCharles’ Law
Charles’s law states that when a gas is kept at constant pressure, the volume of the gas will change with temperature.
In this experiment, balloons keep a small amount of gas (air) at an approximately constant pressure.
As the balloons are dipped into a beaker of liquid nitrogen (-196°C; -320°F), the air inside them quickly cools. The volume of the air inside the balloons decreases as the temperature of the balloons decreases.
As the balloons are dipped into a beaker of liquid nitrogen (-196°C; -320°F), the air inside them quickly cools. The volume of the air inside the balloons decreases as the temperature of the balloons decreases.
As the balloons are dipped into a beaker of liquid nitrogen (-196°C; -320°F), the air inside them quickly cools. The volume of the air inside the balloons decreases as the temperature of the balloons decreases.
As the balloons are dipped into a beaker of liquid nitrogen (-196°C; -320°F), the air inside them quickly cools. The volume of the air inside the balloons decreases as the temperature of the balloons decreases.
Relationship of Relationship of Boyle’s Law and Boyle’s Law and
Charles’ LawCharles’ Law
Temperature in kelvins
Pressuree in kilograms per square centimeter
Practical Practical ApplicationsApplications
Hot AIR BalloonHot AIR BalloonThe hot air that gives the hot-air
balloon its name is commonly created by a propane gas burner that sends
powerful jets of flame into the colorful rip-stop nylon envelope. Once the
balloon is aloft, its height is maintained by opening and closing the blast valve,
which controls the flow of the gas to the burner.
Charles’ Law Charles’ Law CalculationsCalculations
1. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00 L.2. 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?
3. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?
4. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?
5. Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?
6. A gas occupies 1.00 L at standard temperature. What is the volume at 333.0 °C?
7. At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
8. At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0 °C?
Abbreviationsatm – atmospheremm Hg - millimeters of mercurytorr - another name for mm HgPa - Pascal (kPa = kilo Pascal)K - Kelvin°C - degrees Celsius
ConversionsK = °C + 2731 cm3 (cubic centimeter) = 1 mL (milliliter)1 dm3 (cubic decimeter) = 1 L (liter) = 1000 mLStandard Conditions0.00 °C = 273 K1.00 atm = 760.0 mm Hg = 101.325 kPa = 101,325 Pa
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