Chem 161-2010 Lecture 10

Chem 161-2010 Lecture 10 - CHEMISTRY 161-2010 Lecture 10...

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CHEMISTRY 161-2010 Lecture 10 ANNOUNCEMENTS E-MAIL ATTENDANCE EXAMS Chem 161-2010 Lecture 10 1
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PLAN FOR TODAY : CHAPTER 5 GASES (5.7-5.12): REAL GASES (Cont.) CHAPTER 6 (sections 6.1-6.4): • HEAT AND WORK • PROPERTIES OF ENTHALPY Chem 161-2010 Lecture 10 2
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TH EDITION CHEM 161-2008 RECITATION 6 TH WEEK CHAPTER 5 - GASES Real Gases ET: Discuss ideal vs. real gases in terms of going from H 2 O(g) to H 2 O(l). An ideal gas follows the ideal gas law, i.e., PV = nRT; therefore PV/nRT = 1. Real gas is at high pressure and/or low temperature, just before the ideal gas molecules are converted into liquid. Think of molecules like people. If the people move fast, then they pass each other with no attraction; but if the people are either pushed closer to each other, or slow down, then there will be an attraction. At high pressure: Molecules are pushed closer together so they attract each other, so the volume is lower (because one molecule needs less space than two), and the pressure is lower (because one molecule causes less pressure than two). At very high pressure or very high density: Unlike ideal gases in which the molecules don’t occupy significant space, in gases under very high pressure, the volume that the molecules occupy is now a large part of the total volume of the container, making less available volume for compression. Also, since the volume that the molecules occupy is so large, they bump into each other more frequently, so the distance they have to travel to the wall is now smaller (by as much as 2r), resulting in more frequent collisions with the container, i.e., greater pressure. (This can also be explained by the billiard ball effect.) Low temperature: Molecules go slowly and attract each other; when two molecules interact becoming one, the pressure and volume decrease (for the same reasons as at high pressure (above). 1.00 very high P 1.00----------------------- PV/nRT high P PV/nRT high T (behaves as ideal gas) low T Pressure Temperature 117 (mod). Calculate the pressure exerted by 1.00 mol CO 2 (g) when it is confined to a volume of 2.50 L at 200 and 1000 K by using (a) the ideal gas equation; (b) the van der Waals equation (page 204). Use data from Table 5.5. (c) Compare the two results, and comment on the reason(s) for the difference between them. Ideal gas equation: PV = nRT P = nRT/V @200K: P = (1.00mol x 0.08206Latmdeg -1 mol -1 x 298K)/2.50L = 6.56 atm @1000K: P = (1.00mol x 0.08206Latmdeg -1 mol -1 x 1000K)/2.50L = 32.82 atm van der Waals equation ( real gas equation): (P + n 2 a/V 2 ) x (V – nb) = nRT @200K: ((P + ((1.00 2 x 3.59)/(2.50 2 ))) x (2.50 – (1.00x0.0427))) = (1.00x0.08206x298); P = 6.10 atm @1000K: ((P + ((1.00 2 x 3.59)/(2.50 2 ))) x (2.50 – (1.00x0.0427))) = (1.00x0.08206x1000); P = 32.82 atm The van der Waals equation, which is for real gases, shows that at low temperature the internal pressure is lower than calculated for an ideal gas, while at high temperature the internal pressure is equivalent to that
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Chem 161-2010 Lecture 10 - CHEMISTRY 161-2010 Lecture 10...

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