= 1.7765 + 33.037 x 10-3 T (T: Kelvin) R Р (bar) T=340 K T=350 K T=360 K T=370 K T=380 K 0.10 0.99700 0.99719 0.99737 0.99753 0.99767 0.50 0.98745 0.98830 0.98907 0.98977 0.99040 2 0.95895 0.96206...


Calculate the enthalpy and entropy values of the saturated isobutane vapor at 350 K using the information in the table below. The vapor pressure at 350 K is 10 bar.


300 K and 1 bar (ideal gas reference state ) H0
ig
= 18,115 J/mol, S0
ig= 295.976 J/molK


= 1.7765 + 33.037 x 10-3 T<br>(T: Kelvin)<br>R<br>Р (bar)<br>T=340 K<br>T=350 K<br>T=360 K<br>T=370 K<br>T=380 K<br>0.10<br>0.99700<br>0.99719<br>0.99737<br>0.99753<br>0.99767<br>0.50<br>0.98745<br>0.98830<br>0.98907<br>0.98977<br>0.99040<br>2<br>0.95895<br>0.96206<br>0.96483<br>0.96730<br>0.96953<br>4<br>0.92422<br>0.93069<br>0.93635<br>0.94132<br>0.94574<br>6<br>0.88742<br>0.89816<br>0.90734<br>0.91529<br>0.92223<br>8<br>0.84575<br>0.86218<br>0.87586<br>0.88745<br>0.89743<br>10<br>0.79659<br>0.82117<br>0.84077<br>0.85695<br>0.87061<br>12<br>0.77310<br>0.80103<br>0.82315<br>0.84134<br>14<br>0.75506<br>0.78531<br>0.80923<br>15.41<br>0.71727<br>

Extracted text: = 1.7765 + 33.037 x 10-3 T (T: Kelvin) R Р (bar) T=340 K T=350 K T=360 K T=370 K T=380 K 0.10 0.99700 0.99719 0.99737 0.99753 0.99767 0.50 0.98745 0.98830 0.98907 0.98977 0.99040 2 0.95895 0.96206 0.96483 0.96730 0.96953 4 0.92422 0.93069 0.93635 0.94132 0.94574 6 0.88742 0.89816 0.90734 0.91529 0.92223 8 0.84575 0.86218 0.87586 0.88745 0.89743 10 0.79659 0.82117 0.84077 0.85695 0.87061 12 0.77310 0.80103 0.82315 0.84134 14 0.75506 0.78531 0.80923 15.41 0.71727

Jun 09, 2022
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