Elementary Principles 173 - 6.14 (cont'd) Mass of air:...

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6- 7 6.14 (cont’d) Mass of air: mol H O 18.02 g 1 mol mol dry air 29.0 g 1 mol g 2 0038 0962 2858 .. . += Volume of air: 1 mol 22.4 L STP 273.2 32.2 K 1 mol 273.2K L bg b g + = 2504 . Density g 25.04 L gL == 1141 . Increase in increase in decrease in density Increase in more water (MW =18), less dry air (MW = 29) decrease in m decrease in density Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force on the ball must also be lower. Therefore, the statement is wrong. TV h r ⇒⇒ 6.15 a. hy P p rH O H O 50% 0.50 90 C 22 =⇒ = ° y HO 2 2 mm Hg 760.0 mm Hg mol H O / mol = × = 050 52576 0346 Dew Point: y p p T 0.346 760 mm Hg dp 2 =∗ = = di 262 9
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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