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The Gas Laws ISN pg. 35 Chapter 3 Section 2 Pg. 75-81.

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Presentation on theme: "The Gas Laws ISN pg. 35 Chapter 3 Section 2 Pg. 75-81."— Presentation transcript:

1 The Gas Laws ISN pg. 35 Chapter 3 Section 2 Pg. 75-81

2 Pressure Pressure is the result of a force distributed over an area. – Pressure = Force/area Force is measured in newtons (N) Area is measured in square meters (m 2 ) Pressure is N/m 2, the SI unit is pascal (Pa) – 1 kPa = 1000 Pa

3 Pressure Collisions between particles of a gas and the walls of the container cause the pressure in a closed container of gas. – More frequent collisions = greater pressure – Speed and mass of particles also affect pressure

4 Factors that Affect Gas Pressure Factors that affect the pressure of an enclosed gas are its temperature, its volume, and the number of its particles

5 Factors that Affect Gas Pressure Temperature – Raising the temperature of a gas will increase its pressure if the volume of the gas and the number of particles are constant. Ex: Tires on highway- temp. rises -> kinetic energy increases-> particles move faster and collide more-> hit with greater force-> increase in pressure

6 Factors that Affect Gas Pressure Volume – Reducing the volume of a gas increases its pressure if the temperature of the gas and the number of particles are constant. Ex: your lungs and diaphragm when you breathe

7 Factors that Affect Gas Pressure Number of Particles – Increasing the number of particles will increase the pressure of a gas if the temperature and the volume are constant. Ex: inflating tires

8 Boyle’s Law Relationship between pressure and volume of a gas Boyle’s Law states that the volume of a gas is inversely proportional to its pressure if the temperature and the number of particles are constant. P 1 V 1 = P 2 V 2

9 Boyle’s Law Example: To make an air horn, 1.50 L of air at 101 kPa are compressed into a can with a volume of 0.462 L. Assuming a constant temperature, what is the pressure on the compressed air? – Given: P 1 = 101 kPa, V 1 = 1.50 L, V 2 = 0.462 L – Unknown: P 2 = ? P 1 V 1 = P 2 V 2 P 1 V 1 P 2 = V 2

10 Charles’s Law Charles’s Law states that the volume of a gas is directly proportional to its temperature in kelvins if the pressure and the number of particles of the gas are constant. Temperatures must be expressed in kelvins

11 Charles’s Law Example: The volume of a gas is 7.5 L at a temperature of 435 K. What will the volume of the gas be if the temperature is lowered to 300 K? – Given: V 1 = 7.5 L T 1 = 435 K T 2 = 300 K – Unknown: V 2 =?

12 Boyle’s Law GraphCharles’s Law Graph See pg. 78 for example of each graph

13 Combined Gas Laws The combined gas law describes the relationship among the temperature, volume, and pressure of a gas when the number of particles is constant.

14 Combined Gas Laws Example: A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 20°C and the pressure was 1.00 atm. If the child were to let go of the balloon and it rose into the sky where the pressure is 0.667 atm and the temperature is -10°C, what would the new volume of the balloon be? – Given: V 1 = 1.80 L T 1 = 273+20= 293K P 1 = 1.0 atm P 2 = 0.667 atm T 2 = 273+(-10)= 263K – Unknown: V 2 = ?


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