= 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 340 K using the information in the table below. The vapor pressure at 340K is 8 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
SOLUTION.PDF
SOLUTION.PDF

Get Answer To This Question

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here
Have files?