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Gases and Temperature Changes

Gases and Temperature Changes. Kelvin Scale and Absolute Zero. x-intercept is –273  C For an “ideal” gas, –273  C is the point at which all molecular motion theoretically ceases 0 K is called “absolute zero” T K =  C + 273. CHARLES' LAW.

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Gases and Temperature Changes

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  1. Gases and Temperature Changes

  2. Kelvin Scale and Absolute Zero • x-intercept is –273C • For an “ideal” gas, • –273C is the point at which all molecular motion theoretically ceases • 0 K is called “absolute zero” • TK= C + 273

  3. 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

  4. Mathematically,

  5. 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

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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

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

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