Elementary Principles 172

# Elementary Principles 172 - 6.13(cont'd hm = ha = 0.0281 =...

This preview shows page 1. Sign up to view the full content.

6- 6 6.13 (cont’d) 2 0.0281 0.0289 mol H O mol dry air 1 0.0281 m h == 22 2 2 0.0289 mol H O 18.02 g H O mol dry air 0.0180 g H O g dry air mol dry air mol H O 29.0 g dry air a h () [] 0.0289 100% 100% 86.5% 24.559 759.5 24.559 25.56 C 25.56 C m p h h pP p = × = ⎡⎤ ∗°− ∗° ⎣⎦ 6.14 Basis I : 1 mol humid air @ 70 F (21.1 C), 1 atm, °° = h r 50% hy P p rH O H O 50% 0.50 21.1 C =⇒ = ° bg y HO 2 2 mm Hg 760.0 mm Hg mol H O mol = × = 050 18 765 0012 .. . Mass of air: mol H O 18.02 g 1 mol mol dry air 29.0 g 1 mol g 2 0 012 0 988 2887 . += Volume of air: 1 mol 22.4 L STP 273.2 21.1 K 1 mol 273.2K L b g + = 2413 . Density of air g 24.13 L gL 1196 . Basis II 1 mol humid air @ 70 F (21.1 C), 1 atm, : = h r 80% P p O H O 80% 0.80 21.1 C = °
This is the end of the preview. Sign up to access the rest of the document.

## 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.

Ask a homework question - tutors are online