CH131_4A

CH131_4A - Homework assignment #4 Read 9.1 - 9.5, 9.7 - 9.8...

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Homework assignment #4 Read 9.1 - 9.5, 9.7 - 9.8 Problems Chap. 9 - 5, 9, 11, 19, 21, 25, 27, 29, 31, 33, 35, 37, 41, 43, 57, 59, 61
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Gaseous State (Chap. 9) Many familiar substances exist as gases at room temperature e.g. N 2 , O 2 , H 2 , CO 2 , CH 4 , NH 3 , CO, O 3 ,… At high enough T, all substances overcome intermolecular forces and become gases (vaporize) Gases => no definite volume, no definite shape; in contrast to solids and liquids Consequently, (1) gases are very compressible; (2) form homogeneous mixtures. Due to the fact that gas phase molecules are far apart (except at the highest pressures). They are mostly empty space!
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Three most important properties for describing gases: 1. Pressure ( P ) 2. Volume ( V ) 3. Temperature ( T ) V => space that gas occupies; gas fills any volume common units: Liters, ml =10 -3 L = cc cm 3
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Pressure - perhaps the property that is most unique to gases (vapor) Pr essure Force Area Gases could be exerting a force due to molecular collisions on any surface they are in contact with or due to gravitation (if I have alot). 1 m 2 Consider a column, 1 m 2 cross-section at the earth’s surface extending through the atmosphere Mass of air in this column = ~10,000 Kg Force = mass x gravitational constant due to weight of this air mass P = F A = 10 4 Kg ( ) 9.81 m /sec 2 ( ) 1 m 2 10 5 Kg m - 1 sec - 2 Define a pressure unit (pascal) Pa = Kg m -1 sec -2 P atm ~ 10 5 Pa 1 bar 1 atm = 1.01325 x 10 2 KPa atm unit of P
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Another way to picture atmospheric pressure… Hg 760 mm Hg Say we take a tube (closed at one end) and fill it w/ Hg (a very dense liquid) and invert it in a pool of Hg Some Hg comes out of the tube but you see a column of Hg in tube ~ 76 cm or 760 mm tall above the surface of the Hg pool (exact height Why? Atm P exerts force on liq. surface which is transmitted through the liq. & pushed column of liquid Hg up until: F mass Hg col = F atm Such a device is called a barometer ; height is a function of atmospheric P, and identity of liquid but independent of A (column area)
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To see this: when F mass Hg col = F atm mg = PA Rearrange P = mg/A Remember P = F/A so F atm = PA Previously, we defined density d = m/V; here V = hA (Hg in col.) where h = height of column, A = cross-sectional area of column
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CH131_4A - Homework assignment #4 Read 9.1 - 9.5, 9.7 - 9.8...

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