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Gas Law Essential Questions

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Presentation on theme: "Gas Law Essential Questions"— Presentation transcript:

1 Gas Law Essential Questions
How do we know what is happening to atoms and molecules if we can not see them? How is what I am seeing related to what is happening on a molecular level? How do I represent or model the behavior or particles on a nanoscopic level?

2 Marshmallow in a Syringe
1. What will happen when we pull the syringe up? 2. The marshmallows expanded (inflated) 3. Temperature was constant (no heating or cooling) 4. When the syringe was pulled up the amount of space inside the syringe increased (increased volume) 5. There were less collisions of the gas particles with the walls of the container (decreased pressure) 6. Boyle’s Law involve PV changes at constant T 7. P & V are inversely proportional, this indirect relationship means as one variable increases the other decreases 8. P1V1 = P2V2 9.

3 Marshmallow in a Syringe

4 Marshmallow in a Syringe
V high V decrease

5 Marshmallow in a Syringe
BEFORE and AFTER T constant room temp V inc P dec More space for gas particles to move around means less collisions on the surface of the marshmallows, so gas inside them expands under the reduced pressure around them Syringe is pulled up

6 Hot Air Balloon

7 Hot Air Balloon 1. What will happen when we cover the flask with the balloon 2. The balloon went inside the flask 3. Pressure must be constant b/c V & T clearly changing 4. When flask is taken off the hot plate, the particles slow down. This means Temperature is decreasing. 5. The balloon is moving inside the flask so there is less space inside occupied by the gas. Volume decrease. 6. Charles’s Law involves V & T at constant P 7. V & T are directly proportional, this relationship means as one variable increases the other also increases 8. 9.

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9 Hot Air Balloon 10. BEFORE and AFTER T dec V dec P constant
Balloon is pushed inside by Pout P constant T dec V dec Flask is taken off hot plate and covered with a balloon As the particles move slower, the volume decreases to maintain constant pressure. The skin of the balloon is flexible and can change shape to adjust the amount of space in order to keep the # of collisions the same inside and out.

10 Candle in the Water

11 Candle in the Water 1. What will happen when we invert the candle with the flask 2. The water went inside the flask 3. P constant 4. T decreasing 5. V decreasing 6. Charles’ Law, V & T at constant P 7. Direct 8. 9.

12 Water pushed in by Pressure outside
Candle in Water BEFORE and AFTER Water pushed in by Pressure outside T dec V dec As the particles move slower, the volume decreases to maintain constant pressure. The water is pushed in by outside air pressure to keep the # of collisions the same inside and out. P constant Candle goes out P out

13 Cartesian Diver

14 Cartesian Diver 1. What will happen when we squeeze the bottle?
2. Water is pushed into the dropper and it sinks Temperature was constant (no heating or cooling) Average speed of the particles remained the same 4. When bottle was squeezed the pressure on the gas bubble inside the dropper increased increased (increased number of collisions) 5. The gas inside the dropper was taking up less space and there was more water inside the dropper 6. Boyle’s Law involve PV changes at constant T 7. P & V are inversely proportional, this indirect relationship means as one variable increases the other decreases 8. P1V1 = P2V2 9.

15 Cartesian Diver 10. BEFORE and AFTER P inc V dec TOP Bottle
T constant room temp P inc V dec As water is pushed in the gas shrinks making the overall (rubber, glass, water, air) density of the dropper greater than 1 g/mL. Therefore, it sinks! Bottle is squeezed BOTTOM Bottle

16 Soda Can

17 Soda Can 1. What will happen when we invert the can into the water 2. The can was crushed 3. V constant 4. T decreasing 5. P decreasing 6.Gay Lussac’s Law, P & T at constant V 7. Direct 8. 9.

18 Pressure inside is low, the can is crushed by outside air pressure
Can Crush BEFORE and AFTER Pressure inside is low, the can is crushed by outside air pressure T dec P dec V constant As the temperature decreases, the particles move slower and so the pressure inside the can drops. The pressure outside remains the same and nature is unable to correct itself, so the outside pressure crushes the can. P out = P in Can is inverted


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