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!
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
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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