Chapter Thirteen: The Behavior of Gases

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Presentation transcript:

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

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.

13.2 Boyle’s Law

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.

Looking for: Given Relationships: Solution Solving Problems …final pressure in atmospheres (P2) Given …V1 = 5 L , P1= 1 atm, V2 = .5 L Relationships: Boyle’s Law: P1V1 = P2V2 Solution Rearrange equation so P2 = P1V1 / V2 P2 = 1atm x 5.0 L/ 0.5 L = 10 atm.

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.

13.2 Gay-Lussac’s Law

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

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

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

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.)

Solving Problems Looking for: Given Relationships: Solution …final pressure in atmospheres (P2) Given …P1= 300 atm, T1 = 21 C , T2 = 295 C Relationships: Convert temps using K = C + 273 Charles’ Law: P1/T1 = P2/T2 Solution Rearrange equation so P2 = P1xT2 / T1 P2 = 300 atm. x 568K / 294K = 580 atm.