1. Temperature Scales Fahrenheit, Celsius & Kelvin

2. Temperature  Is a measure of how hot or cold an object is compared to another object.  Indicates that heat flows from the object with a higher temperature to the object with a lower temperature.  Is measured using a thermometer.

3. Celsius a unit of temperature in the metric system Water freezes at 0° C. Water boils at 100° C. Normal body temperature is 37° C Room temperature is 21° C.

4. Kelvin another unit of temperature in the metric system. The lowest possible temperature is 0 Kelvin (-273 ° C). This is Absolute Zero. To convert from Celsius to Kelvin: K = C°+ 273

5. Temperature Scales  Fahrenheit, Celsius, and Kelvin reference points for the boiling and freezing points of water.

6. More About Gases Charles’ Law Gay-Lussac’s Law

7. Charles’ Law Jacques Charles determined the relationship between temperature and volume of a gas. He measured the volume of air at different temperatures, and observed a behavior pattern which led to his law. During his experiments pressure of the system and amount of gas were held constant.

8. Temperature The temperature of a gas is generally measured with a thermometer in Celsius. All calculations involving gases should be made after converting the Celsius to Kelvin temperature. Kelvin = C° + 273

9. Volume of balloon at room temperature Volume of balloon at 5°C

10. Charles’ Law Example: A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C? V 1 V 2 T 1 T 2 = T 1 = 127°C + 273 = 400K V 1 = 3.0 L T 2 = 227°C + 273 = 500K V 2 = ?

11. 2) Plug in the variables: 3) Cross multiply and solve (500K)(3.0L) = V 2 (400K) V 2 = 3.75 L = = 3.0L V 2 400K 500K

12. Gay-Lussac’s Law: P and T In Gay-Lussac’s Law the pressure exerted by a gas is directly related to the Kelvin temperature. Volume and the amount of gas are constant. P 1 = P 2 T 1 T 2

13. Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (Volume and the amount of gas are constant) 1. Set up the problem: P 1 = 2.0 atmP 2 = T 1 = 18°C + 273 T 2 = 62°C + 273 = 291 K = 335 K ?

14. Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P 2 : P 1 = P 2 T 1 T 2 P 2 = P 1 x T 2 T 1 P 2 = 2.0 atm x 335 K = 2.30 atm 291 K