Notes on Chapter Five

Notes on Chapter Five - 5.9 P = 1 atm 10 cm H 2 O 1m 1 atm...

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5- 4 5.9 P =1 atm + 10 cm H O m 10 cm atm 10.333 m H O atm 2 2 2 1 1 101 . T = 25 C = 298.2 K , V = 2.0 m 5 min m min = 400 L min 3 3 D ± . 0 40 ± ± m = n mol / min MW g / mol b g b g a. ± ± . m = PV RT MW = 1.01 atm 0.08206 298.2 K g min L atm mol K L min g mol 400 28 02 458 b. ± . min m = 400 K 298.2 K mol 22.4 L STP g L min g mol 273 1 28 02 458 b g 5.10 Assume ideal gas behavior: u m s V m s A m nRT P D 4 u u nR nR T T P P D D 3 2 2 2 1 2 1 1 2 1 2 2 2 F H G I K J ± ± ± ± d i d i S ± ² ± ² ± ² ± ² 2 2 2 1 1 2 1 2 2 1 2 2 60.0 m 333.2K 1.80 1.013 bar 7.50 cm T P D u u 165 m sec T P D sec 300.2K 1.53 1.013 bar 5.00 cm ³ ³ 5.11 Assume ideal gas behavior: n PV RT atm 5 L 300 K mol L atm mol K ³ 100 100 0 08206 0 406 . . . . b g MW g 0.406 mol g mol Oxygen 130 32 0 . . 5.12 Assume ideal gas behavior: Say m t mass of tank, n mol g of gas in tank N : 37.289 g m n g mol CO : 37.440 g m n mol n mol m g 2 t g 2 t g g t ³ ³ U V | W | 28 02 44.1 g 0 009391 37 0256 . . . b g b g unknown: MW g mol g mol Helium ´ 37 062 37 0256 0 009391 39 . . . . b g 5.13 a. ± . V cm STP min V liters 273K mm Hg cm t min 296.2K mm Hg 1 L V t std 3 3 3 763 10 760 9253 b g ' ' ' ' I I ± .

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