Use your textbooks to complete the chart below on page 122 in your ISN

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Presentation transcript:

Use your textbooks to complete the chart below on page 122 in your ISN Boyle’s Law Charles’s Law Gay-Lussac’s Laws Definition Formula Explanation

Quick Write p124: What happens when volume is increased inside of a balloon?

Class Norms 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) Class Norms

Parking Lot Burning Issues Questions Comments Ideas to Share

Boyle’s Law

What is Boyle’s Law? Boyle’s Law is one of the laws in physics that concern the behavior of gases When a gas is under pressure it takes up less space The higher the pressure, the smaller the volume Boyles Law tells us about the relationship between the volume of a gas and its pressure at a constant temperature The law states that pressure is inversely proportional to the volume

How can we write Boyle’s Law as a formula? Pressure is inversely proportional to the volume and can be written as: Pressure a 1/volume P=pressure in N/m2 V=volume in dm3 (litres) k=constant This is more usually written as: Pressure = constant volume PV=k P1V1=P2V2

How can we investigate Boyle’s Law? When investigating Boyles law a given volume of gas is sucked into a cylinder and the end is sealed The temperature of the gas is kept constant Using several equal weights we can apply increasing pressure to the gas We can calculate the pressure by dividing the force applied by the area of the top of the cylinder The volume will be shown on the scale on the cylinder

Boyle’s Law Apparatus

Below are some results of an experiment Pressure p Volume V P x V 1.1 40 44 1.7 26   2.2 20 2.6 17 Calculate pV (pressure x volume) for each set of results. What do you notice?

What these experimental results show The pressure x volume for each set of results remains constant This is called Boyle’s Law For a fixed mass of gas, at constant temperature, pV = constant or P1 x V1 = P2 x V2 Let us look at the results again

Here are the results of the experiment Pressure p Volume V P x V 1.1 40 44 1.7 26 2.2 20 2.6 17 Did you notice that if p is doubled, V is halved? If p increases to 3 times as much, V decreases to a 1/3rd . This means: Volume is inversely proportional to pressure, or V  1 p

What sort of graphs would this data give? If we plot volume directly against pressure we would get a downwards curve showing that volume gets smaller as the pressure gets larger, and vice versa.

Another way of plotting the data Curved lines are hard to recognize, so we plot the volume against the reciprocal of pressure (i.e. 1/p) This time the points lie close to a straight line through the origin. This means volume is directly proportional to 1/pressure or volume is inversely proportional to pressure

This leads us back to Boyle’s Law Boyle’s Law: for a fixed mass of gas kept at constant temperature the volume of the gas is inversely proportional to its pressure.

Problem: A deep sea diver is working at a depth where the pressure is 3.0 atmospheres. He is breathing out air bubbles. The volume of each air bubble is 2 cm2. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

How we work this out: We assume that the temperature is constant, so Boyle’s Law applies: Formula first: P1 x V1 = P2 x V2 Then numbers:= 1.0 x 2 = 3.0 x V2 Now rearrange the numbers so that you have V2 on one side, and the rest of the numbers on the other side of the ‘equals’ symbol.

Here’s what you should have calculated V2 = 3.0 x 2 1.0 therefore volume of bubbles = 6 cm3 Note that P1 and P2 have the same unit, as will V1 and V2

Boyle’s Law Worksheet: ISN p 123 Abbreviations atm – atmosphere mm Hg - millimeters of mercury torr - another name for mm Hg Pa - Pascal (kPa = kilo Pascal) K - Kelvin °C - degrees Celsius Conversions K = °C + 273 1 cm3 (cubic centimeter) = 1 mL (milliliter) 1 dm3 (cubic decimeter) = 1 L (liter) = 1000 mL Standard Conditions 0.00 °C = 273 K 1.00 atm = 760.0 mm Hg = 101.325 kPa = 101,325 Pa Problems: A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg? 2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm? 3. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm? 4. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L? 5. 500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard pressure? 6. Convert 350.0 mL at 740.0 mm of Hg to its new volume at standard pressure. 7. Convert 338 L at 63.0 atm to its new volume at standard pressure.