Chapter 6 Part 3

Chapter 6 Part 3 - Extra Credit #2 Due Today None will be...

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• Extra Credit #2 Due Today – None will be accepted late without special permission • Chapter 6 Homework answers are posted • Read and Study Chapter 7 (Acids and Bases) • Homework – Chapter 7, 18 – 62 even Due Tuesday, November 20 th Another article for today will be posted. You will be notified by e-mail
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The Gas Laws Charles’s Law
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The Gas Laws Charles’s Law (1787) At constant pressure , the volume of a gas is directly proportional to its absolute temperature (in ºK) . V α T V = bT V/T = b V 1 /T 1 = V 2 /T 2 • However; for 100 °C to 200 °C change, the volume will NOT double •Why? • Relevance to car tires in summer vs. winter • See also article (to be posted) about filling tires with pure nitrogen (N 2 )
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The Gas Laws Charles’s Law • Draw contrast to situation with non-rigid container walls! • See video clips N 2 ( l )
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A balloon indoors, where the temperature is 27 ° C, has a volume of 2.00 L. What would its volume be (a) in a hot room where the temperature is 47 ° C, and (b) outdoors, where the temperature is –23 ºC? (Assume no change in pressure in either case.) EXAMPLE 6.15 Charles’s Law: Temperature-Volume Relationships Solution First, and most important, convert all temperatures to the Kelvin scale: T (K) = t ( ° C) + 273 The initial temperature ( T 1 ) in each case is (27 + 273) = 300 K and the final temperatures are (a) (47 + 273) = 320 K and (b) (–23 + 273) = 250 K. a. We start by separating the initial from the final condition. Initial Final Change t 1 = 27 ° C t 2 = 47 ° C T 1 = 300 K T 2 = 320 K V 1 = 2.00 L V 2 = ? Solving the equation = V 1 T 1 V 2 T 2
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EXAMPLE 6.15 Charles’s Law: Temperature-Volume Relationships continued for V 2 , we have V 2 = V 2 = = 2.13 L As expected, because the temperature increases, the volume must also increase.
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Chapter 6 Part 3 - Extra Credit #2 Due Today None will be...

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