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10.4 Buoyancy, volume, temperature, and pressure of gases  A hot-air balloon floats because the air inside is less dense than the air outside.  The balloon.

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Presentation on theme: "10.4 Buoyancy, volume, temperature, and pressure of gases  A hot-air balloon floats because the air inside is less dense than the air outside.  The balloon."— Presentation transcript:

1 10.4 Buoyancy, volume, temperature, and pressure of gases  A hot-air balloon floats because the air inside is less dense than the air outside.  The balloon example illustrates an important relationship, known as Charles’s law, discovered by Jacques Charles in 1787.

2 10.4 Charles’ Law  According to Charles’ law, the volume of a gas increases with increasing temperature.  Volume decreases with decreasing temperature.

3 10.4 Pressure-Temperature Relationship  The pressure of a gas is also affected by temperature changes.  If the mass and volume are kept constant, the pressure goes up when the temperature goes up, and the pressure goes down when the temperature goes down.

4 10.4 Pressure-Temperature Relationship  The mathematical relationship between the temperature and pressure of a gas at constant volume and mass was discovered by Joseph Gay- Lussac in 1802.

5 10.4 Converting to Kelvin  Any time you see a temperature in a formula in this section about gases, the temperature must be in Kelvins.

6 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.)

7  Looking for:  …final pressure in psi(P 2 )  Given  …P 1 = 300 psi, T 1 = 21  C, T 2 = 295  C  Relationships:  Convert temps using K =  C + 273  Charles’ Law: P 1 /T 1 = P 2 /T 2  Solution  Rearrange equation so P 2 = P 1 xT 2 / T 1  P 2 = 300 psi. x 568K / 294K = 580 psi. Solving Problems


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