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Temperature Scales

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barclay-jacobson
14 diapositives · 3.8K Views · Nov 13, 2014

Temperature Scales. Fahrenheit, Celsius & Kelvin. 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.

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Temperature Scales
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  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. Celsiusa 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. Kelvinanother 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? V1 V2 = T1 T2 T1 = 127°C + 273 = 400K V1 = 3.0 L T2 = 227°C + 273 = 500K V2 = ?

  11. = 2) Plug in the variables: 3.0L V2 400K 500K 3) Cross multiply and solve (500K)(3.0L) = V2(400K) V2 = 3.75 L

  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. P1 = P2 T1 T2

  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: P1= 2.0 atm P2 = T1 = 18°C + 273 T2 = 62°C + 273 = 291 K = 335 K ?

  14. Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P2: P1 = P2 T1 T2 P2 = P1 x T2 T1 P2 = 2.0 atm x 335 K = 2.30 atm 291 K