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Charles's Law

Charles's Law. The relationship between temperature and volume. How Volume Varies With Temperature. If we place a balloon in liquid nitrogen it shrinks:. So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon). Is 20  C twice as hot as 10  C?

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Charles's Law

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  1. Charles's Law The relationship between temperature and volume

  2. How Volume Varies With Temperature If we place a balloon in liquid nitrogen it shrinks: So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon).

  3. Is 20C twice as hot as 10C? Is 20 kg twice as heavy as 10 kg? Temperature scales No. 68F (20C) is not double 50F (10C) Yes. 44 lb (20 kg) is double 22 lb (10 kg) What’s the difference? • Weights (kg or lb) have a minimum value of 0. • But the smallest temperature is not 0C. • We saw that doubling P yields half the V. • An experiment with a fixed volume of gas in a cylinder will reveal the relationship of V vs. T…

  4. CHARLES' LAW If a gas has a constant mass and is held at a constant pressure then the volume divided by the temperature (in kelvins) is a constant value. V = kT or = k V T

  5. Mathematically,

  6. Graphically

  7. Using a glass syringe, a scientist draws exactly 25.5 cm3 of dry oxygen at 20°C from a metal cylinder. She heats the syringe to 65 °C. What volume will the oxygen occupy? • What are you trying to determine? What volume will the oxygen occupy? Vf or V2

  8. Step2: Identify what is given: Vi = 25.5 cm3 Ti = 20.0 °C Tf = 65.0 °C

  9. Step 3: Convert temperatures from °C to K Ti = 20.0 °C Tf = 65.0 °C = 20.0 °C + 273 = 65.0 °C + 273 = 293 K = 338 K

  10. Step 4: Use equation to determine final volume: = = x 338 K = Vf Vi Ti Vf Tf Vf 338 25.5 293 25.5 293 Vf = 29.42 cm3

  11. A balloon is filled with 2.50 L of dry helium at 23.5°C. the balloon is placed in a freezer overnight. The resulting volume is found to be 2.15L. What was the temperature (in °C) in the freezer. • What are you trying to determine? What was the temperature (in °C) in the freezer. Tf

  12. Step2: Identify what is given: Vi = 2.50 L Ti = 23.5 °C Vf = 2.15 L Step 3: Convert temperature from °C to K Ti = 23.5 °C + 273 = 297 K

  13. Step 4: Use equation to determine final temperature: = = 2.50 L (Tf) = 297K (2.15L) Tf = 255.42K Vi Ti Vf Tf 2.50 297 2.15 Tf Cross multiply Divide by 2.50L to isolate Tf

  14. Step 5: Convert the temperature back to °C Tf = 255.42K – 273 = -17.6 °C

  15. V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K Using Charles’ law: V1/T1 = V2/T2 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L 3. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 4. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2,1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L

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