Split_00036 - 8.61(a i FG mIJ HV K Expt 1 = b4.4553 3.2551gkg = 0.600 kg bSGg 2.000 L liquid L b liquid = 0.600 g ii Expt 2 Mass of gas = 3.2571

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8- 36 8.61 (a) i) Expt 1 F H G I K J = =⇒ = m V SG liquid liquid kg 2.000 L kg L 4 4553 32551 0600 .. bg ii) Expt 2 Mass of gas =− = = 32571 32551 0 0020 2 0 . . kg kg g Moles of gas = = 2.000 L 273 K mm Hg 1 mol 363 K 760 mm Hg 22.4 liters STP mol 763 500 00232 . Molecular weight == 20 86 . g 0.0232 mol g mol b g iii) Expt. 1 2.000 liters 10 cm 0.600 g 1 mol 1 liter cm 86 g mol liquid 33 3 ⇒= = n 14 Energy balance : The data show that C v is independent of temperature QU n C T C Q nT v v = = ΔΔ Δ J 14 mols 2.4 K J mol K 284.2 K J 14 mols 2.4 K J mol K 331.2 K liquid b g b g 800 24 800 24 @ @ ⇒≡ C v liquid Jmo lK 24 Expt. 2 mol from ii vapor n . C a bT Q a bT dT a T T b TT a b v T T =+ ⇒ = + = − + L N M O Q P L N M O Q P L N M O Q P U V | | W | | = z 0 0232 0 0232 2 130 3669 3630 492 7 4900 4069 0 05052 21 2 2 1 2 22 1 2 .( ) . ( ) () . . ) . . ) . . J = 0.0232 a(366.9 -363.0) +
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.

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