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Unformatted text preview: Problem 32 Pressure (P) = 14.7psia Density () = 0.103 lb/ft3 Therefore, the molecular weight of the compound is: M = 12n + 2n = 14n (1) Applying the ideal gas law, Therefore, Applying the values of , R, T and P, from equation (1) M = 14n = 42.102 Therefore, n = 3.0 Temperature(T) = 100 o F = 560 o R R = 10.73 psiaft 3 /lbmol o R The general formula for this compound is C n H 2n From the general formula for this compound ,C n H 2n , The compound is C 3 H 6 Propene RT M m nRT PV * = = P RT RT PV m M = = * 42.102 7 . 14 560 * 73 . 10 * 103 . = = M Problem 321 Component Mole Fractions Critical Pressure Critical Temperature Molecular w Methane 0.8700 666.4 579.8116.7 343.3 298.7 16.04 Ethane 0.0490 706.5 34.6 89.9 549.9 26.9 30.07 Propane 0.0360 616.0 22.2 206.1 666.1 24.0 44.10 iButane 0.0250 527.9 13.2 274.5 734.5 18.4 58.12 nButane 0.0200 550.6 11.0 305.6 765.6 15.3 58.12 1.0000 660.8 383.3 1.67 7.04 From Fig 37, Z 0.96 For P= 4650 Psia and T=(180+460) o R y i P ci , psia...
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 Spring '08
 Peters

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