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BOYLE’S LAW. WHAT IS BOYLE’S LAW? Boyle’s Law is one of the laws in physics that concern the behaviour of gases. At constant temperature it relates pressure.

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Presentation on theme: "BOYLE’S LAW. WHAT IS BOYLE’S LAW? Boyle’s Law is one of the laws in physics that concern the behaviour of gases. At constant temperature it relates pressure."— Presentation transcript:

1 BOYLE’S LAW

2 WHAT IS BOYLE’S LAW? Boyle’s Law is one of the laws in physics that concern the behaviour of gases. At constant temperature it relates pressure and volume. When a gas is under pressure it takes up less space: - The higher the pressure, the smaller the volume - The law states that pressure is inversely proportional to the volume

3 BOYLE’S LAW EXAMPLE.

4 HOW CAN WE WRITE BOYLE’S LAW AS A FORMULA? This is more usually written as: Pressure = constant volume PV= k ( k = constant) P 1 V 1 =P 2 V 2

5 BELOW ARE SOME RESULTS OF AN EXPERIMENT Calculate pV (pressure x volume) for each set of results. What do you notice? Pressure p Volume V P x V 1.14044 1.726 2.220 2.617

6 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 P 1 x V 1 = P 2 x V 2 Let us look at the results again

7 HERE ARE THE RESULTS OF THE EXPERIMENT Did you notice that if p is doubled, V is halved? If p increases to 3 times as much, V decreases to a 1/3 rd. This means: Volume is inversely proportional to pressure, or V  1 p Pressure p Volume V P x V 1.14044 1.72644 2.22044 2.61744

8 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.

9 TODAY'S INVESTIGATION. - 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

10 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 cm 2. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

11 HOW WE WORK THIS OUT: We assume that the temperature is constant, so Boyle’s Law applies: Formula first:P 1 x V 1 = P 2 x V 2 Then numbers:= 3.0 x 2 = 1.0 x V 2 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.

12 HERE’S WHAT YOU SHOULD HAVE CALCULATED V2 = 3.0 x 2 1.0 therefore volume of bubbles = 6 cm 3 Note that P 1 and P 2 have the same unit, as will V 1 and V 2!!!

13 CHARLES’ LAW

14 In 1787, the French physicist Jacques Charles discovered that a gas’s volume is directly proportional to the temperature on the Kelvin scale if the pressure remains the same. Heating a gas makes it expand. Cooling a gas makes it contract. me

15 KINETIC MOLECULAR THEORY The kinetic-molecular theory states that gas particles move faster on average at higher temperatures, causing them to hit the walls of their container with more force. Repeated strong collisions cause the volume of a flexible container, such as a balloon, to increase. Gas volume decreases when the gas is cooled, because of the lower average kinetic energy of the gas particles at the lower temperature.

16 Charles’ Law- Graphically When the temperature scale is in kelvins, the graph shows a direct proportion between volume of a sample of gas and the temperature.

17 HOW CAN WE WRITE CHARLES’ LAW AS A FORMULA? This is more usually written as: Or, when all other conditions remain the same:

18 Solving Volume-Temperature Problems A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath at − 78.5°C. What is its volume, assuming the pressure remains constant?

19 HOW WE WORK THIS OUT: We assume that the pressure is constant, so Charles’ Law applies: Formula first:V 1 / T 1 = V 2 / T 2 Convert all Temperatures to Kelvin! 27C + 273.15 = 300.15 K -78.5 + 273.15 = 194.65 K Then numbers:= 665 mL / 300.15 k = V 2 / 194.65 K 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.

20 HERE’S WHAT YOU SHOULD HAVE CALCULATED V2 = 665 x 194.65 300.15 therefore volume of the balloon= 431.26 mL Note that T 1 and T 2 have the same unit, as will V 1 and V 2!!!

21 GAY-LUSSAC

22 GAY LUSSAC Pressure is the result of collisions of particles with the walls of the container and the average kinetic energy of particles is proportional to the sample’s average absolute temperature. Temperature and pressure have a directly proportional relationship. [If pressure is doubled, temperature is doubled.]

23 KINETIC MOLECULAR THEORY The kinetic-molecular theory states that gas particles move faster on average at higher temperatures, causing them to hit the walls of their container with more force. Repeated strong collisions cause the volume of a flexible container, such as a balloon, to increase. (Charles Law); HOWEVER, when volume is fixed (doesn’t move) the repeated strong collisions result in an increase in pressure.

24 Gay Lussac’s Law- Graphically When the temperature scale is in kelvins, the graph shows a direct proportion between pressure of a sample of gas and the temperature.

25 HOW CAN WE WRITE CHARLES’ LAW AS A FORMULA? This is more usually written as: Or, when all other conditions remain the same:

26 Solving Volume-Temperature Problems At 122C the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205C, assuming constant volume?

27 HOW WE WORK THIS OUT: We assume that the pressure is constant, so Charles’ Law applies: Formula first:P 1 / T 1 = P 2 / T 2 Convert all Temperatures to Kelvin! 122 C + 273.15 = 395.15 K 205 C+ 273.15 = 478.15 K Then numbers:= 1.07 atm / 395.15 k = P 2 / 478.15 K Now rearrange the numbers so that you have P2on one side, and the rest of the numbers on the other side of the ‘equals’ symbol.

28 HERE’S WHAT YOU SHOULD HAVE CALCULATED P2 = 1.07 x 478.15 395.15 therefore pressure =1.29 atm Note that T 1 and T 2 have the same unit, as will P 1 and P 2!!!


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