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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.

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Presentation on theme: "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."— Presentation transcript:

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” T K =  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 VTVT

4 Mathematically,

5 Using a glass syringe, a scientist draws exactly 25.5 cm 3 of dry oxygen at 20°C from a metal cylinder. She heats the syringe to 65 °C. What volume will the oxygen occupy? 1.What are you trying to determine? What volume will the oxygen occupy? V f or V 2

6 Step 2: Identify what is given: V i = 25.5 cm 3 T i = 20.0 °C T f = 65.0 °C

7 Step 3: Convert temperatures from °C to K T i = 20.0 °CT f = 65.0 °C = 20.0 °C = 65.0 °C = 293 K = 338 K

8 Step 4: Use equation to determine final volume: = = x 338 K = V f ViTiViTi VfTfVfTf V f V f = cm 3

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. 1.What are you trying to determine? What was the temperature (in °C) in the freezer. T f

10 Step 2: Identify what is given: V i = 2.50 L T i = 23.5 °C V f = 2.15 L Step 3: Convert temperature from °C to K T i = 23.5 °C = 297 K

11 Step 4: Use equation to determine final temperature: = 2.50 L (T f ) = 297K (2.15L) T f = K ViTiViTi VfTfVfTf T f C ross multiply Divide by 2.50L to isolate T f

12 Step 5: Convert the temperature back to °C T f = K – 273 = °C


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