Elementary Principles 158

# Elementary - 5.69 a V 50.0 mL 44.01 g = = 4401 mL mol n 5.00 g mol RT 82.06 mL atm 1000 K P= = = 186 atm mol K 440.1 mL mol V V= b For CO 2 Tc =

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5- 46 5.69 a. ± . . . . V= V n mL g g mol mL / mol == 50 0 500 44 01 4401 P= RT V 82.06 mL atm mol K 440.1 mL / mol atm ± = = 1000 K 186 b. For CO : T K, P atm 2c c = = 304 2 72 9 .. T T T K 304.2 K V VP RT mL mol atm 304.2 K mol K 82.06 mL atm r c r ideal c c = = 1000 32873 72 9 128 . ± . ideal rr Figure 5.4-3: V 1.28 and T 3.29 z=1.02 zRT 1.02 82.06 mL atm mol 1000 K P= 190 atm ˆ mol K 440.1 mL V c. a = mL atm / mol , b = mL / mol, m 22 3654 10 29 67 08263 1000 01077 6 . , ( ) . ×⋅ = = α K 82.06 440.1- 29.67 440 440.1+ 29.67 atm mL atm mol K mL mol mL atm mol mL mol 2 2 2 2 × = ch bg ej 1000 K 01077 3654 10 1 198 6 . 5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the presence of O 2 . b. Enough N 2 needs to be added to make x O 2 10 10 6 . Since the O 2 is so dilute at this condition, the properties of the gas will be that of N 2 . T K , P a t m , T cc r = 126 2 335 2 36 . nn PV RT atm 0.08206 L 298.2 K mol n m o l a i r 0.21 mol O mol air mol O n n m o l initial 1 Latm mol K
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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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