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Chapter Thirteen: The Behavior of Gases  13.1 Gases, Pressure, and the Atmosphere  13.2 The Gas Laws.

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Presentation on theme: "Chapter Thirteen: The Behavior of Gases  13.1 Gases, Pressure, and the Atmosphere  13.2 The Gas Laws."— Presentation transcript:

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2 Chapter Thirteen: The Behavior of Gases  13.1 Gases, Pressure, and the Atmosphere  13.2 The Gas Laws

3 13.1 What’s in Earth’s atmosphere?  Nitrogen (N 2 ) gas makes up about 78 percent of Earth’s atmosphere.  Nitrogen is released into the air by volcanoes and decaying organisms and is a vital element for living things.

4 13.1 Comparing atmospheres  An atmosphere is a layer of gases surrounding a planet or other body in space.

5 13.1 Life changed Earth’s atmosphere  Over time, photosynthesis breaks down carbon dioxide, uses carbon to build the organism, and releases oxygen into the air.

6 13.1 Life changed Earth’s atmosphere  Earth stores carbon as calcium carbonate so it doesn’t return to the atmosphere.  “Fossil fuels” (oil, coal, and natural gas) are carbon from decaying plants and animals in the ground.

7 13.1 Atmospheric pressure  Atmospheric pressure is a measurement of the force of air molecules in the atmosphere at a given altitude.  Your ear drum is one way you can detect changes in pressure.

8 13.1 Pressure in the atmosphere  At sea level, the weight of the column of air above a person is about 9,800 newtons (2,200 pounds)!  This is equal to the weight of a small car.  Why aren’t we crushed by this pressure?

9 13.1 Measuring Pressure  A barometer is an instrument that measures atmospheric pressure.  Mercury barometers were common until we discovered their vapors were harmful.

10 13.1 Measuring Pressure  Today we use aneroid barometers.  They have an airtight cylinder made of thin metal.  The walls of the cylinder respond to changes in pressure.

11 13.1 Pressure in the atmosphere  The gas molecules closest to Earth’s surface are packed together very closely.  This means pressure is lower the higher up you go into the atmosphere.

12 13.1 Pressure changes with altitude

13 13.1 Units of pressure

14 13.2 Boyle’s Law  When you squeeze a fixed quantity of gas into a smaller volume the pressure goes up.  This rule is known as Boyle’s law.

15 13.2 Boyle’s Law

16 Solving Problems  A kit used to fix flat tires consists of an aerosol can containing compressed air and a patch to seal the hole in the tire.  Suppose 5 liters of air at atmospheric pressure (1 atm) is compressed into a 0.5 liter aerosol can. What is the pressure of the compressed air in the can?  Assume no change in temperature or mass.

17 1.Looking for:  …final pressure in atmospheres (P 2 ) 2.Given  …V 1 = 5 L, P 1 = 1 atm, V 2 =.5 L 3.Relationships:  Boyle’s Law: P 1 V 1 = P 2 V 2 4.Solution  Rearrange equation so P 2 = P 1 V 1 / V 2  P 2 = 1atm x 5.0 L/ 0.5 L = 10 atm. Solving Problems

18 13.2 Pressure and Temperature  The pressure of a gas is affected by temperature changes.  If the mass and volume are kept constant, the pressure goes up when the temperature goes up, and down when the temperature goes down.

19 13.2 Gay-Lussac’s Law

20 13.2 Pressure and Temperature  Any time we apply gas laws, the the temperature must be in Kelvins.  Only the Kelvin scale starts from absolute zero, when energy of particles is theoretically zero.

21 13.2 Charles’ Law  According to Charles’ law, the volume of a gas increases with increasing temperature.  Volume decreases with decreasing temperature.

22 13.2 Charles’ Law  A hot-air balloon floats because the air inside is less dense than the air outside.

23 Solving Problems  A can of hair spray has a pressure of 300 psi at room temperature 21°C.  The can is accidentally moved too close to a fire and its temperature increases to 295°C.  What is the final pressure in the can? (Round answer to nearest whole number.)

24 1.Looking for:  …final pressure in atmospheres (P 2 ) 2.Given  …P 1 = 300 atm, T 1 = 21  C, T 2 = 295  C 3.Relationships:  Convert temps using K =  C + 273  Charles’ Law: P 1 /T 1 = P 2 /T 2 4.Solution  Rearrange equation so P 2 = P 1 xT 2 / T 1  P 2 = 300 atm. x 568K / 294K = 580 atm. Solving Problems


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