chemical engineering up 55%. 10:13 AM Problem 5.96 Kay\'s rulel | A gas consists

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chemical engineering

up 55%. 10:13 AM Problem 5.96 Kay's rulel | A gas consists of 40.0 mol% methane, 45.00 mol% ethane, and 15.0 mol% ethylene. A mass of 34.0 kilograms of this gas is to be compressed to a pressure of 200.0 bar at 900°C. Using Kay's rule, estimate the final volume of the gas. m3 Reference Materials Physical Property Tables (Reprinted with permission from Chemical Process Principles Charts, 2nd Edition,by O.A. Hougen, K. M Watson, and R. A. Ragatz, John Wiley& Sons, New York, 1960.) Generalized compressibility chart, low pressures (Based on L. C. Nelson and E. F Obert, Trans. ASME, 76, 1057(1954).) Generalized compressibility chart, medium pressures (Based on L. C. Nelson and E. F. Obert, Trans. ASME, 76, 1057(1954)) Generalized compressibility chart high pressures (Based on L. C. Nelson and E. F. Obert, Trans. ASME, 76, 1057(1954).) the tolerance is +/-4%

Explanation / Answer

For methane,   Tc= 190.6 K, Pc = 46.1 bar, y1=0.4

For ethane, Tc = 305 K, Pc= 49 bar, y2=0.45

For ethylene, Tc=282.5 K, Pc=50.6 bar, y3=0.15

Tc = 190.6*0.4 + 305*0.45 + 282.5*0.15 = 255.865 k

Pc=46.1*0.4 + 49*0.45 + 50.6*0.15 = 48.08 bar

Tr = T/Tc = 363/255.865 = 1.418

Pr = P/Pc = 200 bar/48.08 bar = 4.16

From generalized compressibility charts,

Z = 0.75 (approx)

Thus, molar volume = ZRT/P = 0.75 x 8.314 x 363/(20000) = 0.1132 L

Avg molar mass = 16*0.4 + 30*0.45 + 28*0.15 = 24.1 g/mol

Thus, moles in the given mass = 34 kg/24.1 g/mol = 1.4108 kmol

Volume = 1.4108 k mol x 0.1132 L/mol = 0.1597 kL = 0.1597 m3

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