of 27 /27
R = 8.314 J/(mol.K) T = 323 K Tabel 12.4: Dietil-Keton/n-Hexan Margules Asli P/kPa x1 y1 90.15 0.000 0.000 0.0000 0.6929 0.0000 1.9995 91.78 0.063 0.049 0.9008 0.0328 1.4812 1.1117 0.0076 3.0396 88.01 0.248 0.131 0.4719 0.1206 1.1138 0.5788 0.1033 1.7839 81.67 0.372 0.182 0.3205 0.1655 0.9553 0.3395 0.2099 1.4042 78.89 0.443 0.215 0.2778 0.2097 0.9722 0.2382 0.2794 1.2689 76.82 0.508 0.248 0.2571 0.2642 1.0428 0.1652 0.3454 1.1796 73.39 0.561 0.268 0.1898 0.3056 0.9770 0.1181 0.3993 1.1254 66.45 0.640 0.316 0.1234 0.3368 0.8692 0.0660 0.4773 1.0682 62.95 0.702 0.368 0.1292 0.3927 0.9929 0.0380 0.5341 1.0387 57.70 0.763 0.412 0.0717 0.4624 0.9087 0.0195 0.5844 1.0197 50.16 0.834 0.490 0.0161 0.5362 0.7398 0.0070 0.6334 1.0070 45.70 0.874 0.570 0.0274 0.5481 0.8443 0.0032 0.6556 1.0032 29.00 1.000 1.000 0.0000 0.0000 1.3489 1.0000 persamaan a b A21 = A12 = ln g1 ln g2 GE/ x1.x2.R T ln g1 ln g2 g1 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000 1.6000 f(x) = − 0.656010018741054 x + 1.34894593987776 GE/x1x2RT Linear (GE/x1x2RT)

# Contoh perhitungan Margules

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Margules Equation

### Text of Contoh perhitungan Margules

R = 8.314 J/(mol.K)T = 323 K

Tabel 12.4: Dietil-Keton/n-Hexan Margules Asli

P/kPa x1 y190.15 0.000 0.000 0.0000 0.6929 0.0000 1.9995 1.000091.78 0.063 0.049 0.9008 0.0328 1.4812 1.1117 0.0076 3.0396 1.007788.01 0.248 0.131 0.4719 0.1206 1.1138 0.5788 0.1033 1.7839 1.108881.67 0.372 0.182 0.3205 0.1655 0.9553 0.3395 0.2099 1.4042 1.233678.89 0.443 0.215 0.2778 0.2097 0.9722 0.2382 0.2794 1.2689 1.322376.82 0.508 0.248 0.2571 0.2642 1.0428 0.1652 0.3454 1.1796 1.412573.39 0.561 0.268 0.1898 0.3056 0.9770 0.1181 0.3993 1.1254 1.490866.45 0.640 0.316 0.1234 0.3368 0.8692 0.0660 0.4773 1.0682 1.611762.95 0.702 0.368 0.1292 0.3927 0.9929 0.0380 0.5341 1.0387 1.706057.70 0.763 0.412 0.0717 0.4624 0.9087 0.0195 0.5844 1.0197 1.793950.16 0.834 0.490 0.0161 0.5362 0.7398 0.0070 0.6334 1.0070 1.884145.70 0.874 0.570 0.0274 0.5481 0.8443 0.0032 0.6556 1.0032 1.926229.00 1.000 1.000 0.0000 0.0000 1.3489 1.0000 3.8532

persamaan garis lurusa -0.656b 1.3489

A21 = 0.6929A12 = 1.3489

ln g1 ln g2GE/

x1.x2.RT ln g1 ln g2 g1 g2

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

1.4000

1.6000

f(x) = − 0.656010018741054 x + 1.34894593987776

GE/x1x2RTLinear (GE/x1x2RT)gamma1ln(gamma2)

A12 = 0.619261A21 = 1.112016

Margules dan Barker

P galat P P galat P90.1500 0.000E+00 1.1120 0.0000 3.0405 1.0000 90.1500 0.000E+0090.6708 1.209E-02 0.9218 0.0061 2.5138 1.0061 89.5821 2.395E-0288.0000 1.140E-04 0.4906 0.0837 1.6334 1.0873 85.4560 2.902E-0284.9861 4.060E-02 0.2940 0.1713 1.3418 1.1869 81.6700 5.334E-1182.7009 4.831E-02 0.2096 0.2293 1.2331 1.2577 78.9938 1.315E-0380.0298 4.178E-02 0.1480 0.2849 1.1595 1.3297 76.0581 9.918E-0377.3099 5.341E-02 0.1078 0.3311 1.1138 1.3924 73.2269 2.222E-0372.1315 8.550E-02 0.0624 0.3990 1.0644 1.4903 68.1203 2.514E-0266.9770 6.397E-02 0.0373 0.4499 1.0380 1.5682 63.2601 4.926E-0360.8918 5.532E-02 0.0202 0.4965 1.0204 1.6429 57.6815 3.214E-0452.5513 4.767E-02 0.0080 0.5445 1.0080 1.7238 50.1763 3.253E-0447.3074 3.517E-02 0.0040 0.5679 1.0040 1.7645 45.4903 4.588E-0329.0000 0.000E+00 0.0000 1.1120 1.0000 3.0405 29.0000 0.000E+00

sum = 4.839E-01 sum = 1.017E-01

persamaan garis lurus

(x1 = 1) (x1 = 0)

ln g1 ln g2 g1 g2

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

20.0000

30.0000

40.0000

50.0000

60.0000

70.0000

80.0000

90.0000

100.0000

percobaan

Margules Asli

Margules Barker

Cprata2

(J/mol) (J/mol) J/(mol.K)

CO -110525 -137169 3.507 29.1572

0 0 3.468 28.83295

-200660 -161960 5.547 46.11776

R, J/(mol.K) 8.314

Reaksi

CO -1

-2

1

-90135

-24791

K(298) = 22162.73

T, C 50 100 150 200T, K 323 373 423 473K(T) 1326.5658 14.7476 0.4750 0.0316

eps = 31.04162 22.81381 5.24693 0.45846Ky = 1325.2297 14.7476 0.4755 0.0317

galat = 1.33615 0.00002 0.00059 0.00011konversi CO = galat jelek 65.18% 14.99% 1.31%

1000/T 3.10 2.68 2.36 2.11ln(K,T) 7.19 2.69 -0.74 -3.45Perhatikan bahwa ln(K) vs 1/T merupakan garis lurus.Lihat Figure 13.2.

DHf,298 DGf,298 Cprata2/R

H2

CH3OH(g)

H2

CH3OH(g)

DH-R,298

DG-R,298

2.0 3.0 4.0

-4.00

-2.00

0.00

2.00

4.00

6.00

8.00

1/T

ln K

R,298298

GK exp

R.298

D

R,298

,298

HKT 1 1exp

K R T 298

D

B5
Herri Susanto: entalpi pembentukan dan energi bebas pembentukan senyawa beratom sama adalah NOL
C17
Herri Susanto: pers 13.11b ln(K) = -DdGr/(R.298)
C20
Herri Susanto: pers. 13.15 lnKT/lnK298= -dHr/R*(1/T-1/298) dHr bukan fungsi temp.

2.0 3.0 4.0

-4.00

-2.00

0.00

2.00

4.00

6.00

8.00

1/T

ln K

Cprata2

(J/mol) (J/mol) J/(mol.K)

CO -110525 -137169 3.507 29.1572

-393509 -394359 4.467 37.13864

0 0 3.468 28.83295

-200660 -161960 5.547 46.11776

-241818 -228572 4.038 33.57193

R, J/(mol.K) 8.314

Reaksi-1 Reaksi-2

CO -1 0

0 -1

-2 -3

1 1

0 1

-90135 -48969

-24791 3827

masuk K-298 22162.73 0.213384958

150 Temp , C 149.9839505 149.9839505

423 T, K 422.9839505 422.9839505K(T) 0.4754 0.0006

eps-tebak 5.6093 -1.2016

masuk keluar100.0 laju alir, mol/s 91.2

25% CO 21.27% mol atom C 40.0

15% 17.77% mol atom H 120.0

60% 57.45% mol atom O 55.0

0% 4.83%

0% -1.32% mol H2O 0.0100% 100%

Reaksi-1 Reaksi-2

Ky = 0.6887 -0.0189

galat = Ky - K(T) = 0.21328521 0.019527226

galat total = 0.232812436

-8262800.6139 entalpi, J/s -8262800.847Q-panas, J/s

-0.23281308636(keluar)

DHf,298 DGf,298 Cprata2/R

CO2

H2

CH3OH(g)

H2O(g)

CO2 CO + 2H2 → CH3OH (1)H2 CO2 + 3H2 → CH3OH + H2O (2)

CH3OH(g)

H2O(g) CO : 25 – ε1 = yCO.N2

DH-R,298 H2 : 60 - 2 ε1 – ε2 = yN2.N2

DG-R,298 CO2 : 15 - ε2 = yCO2.N2

CH4OH : ε1 + ε2 = yCH3OH.N2

H2O : ε 2 = y H2O.N 2

100 - 2 ε1 –2 ε2 = N2

NERACA MASSA

CO2

H2

CH3OH(g)

H2O(g)

3

2

1 2

CH OH 1 21 2

CO. H 1 1 2

1 2 1 2

y 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

3 23

2

1 2 2

CH OH H O 1 2 1 22 3

CO. H1 1 2

1 2 1 2

y .y 100 2 2 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

D22
Herri Susanto: variasikan 120 dan 140 C

3 23

2

1 2 2

CH OH H O 1 2 1 22 3

CO. H1 1 2

1 2 1 2

y .y 100 2 2 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

40.0

120.0

55.0

-1.2

CO + 2H2 → CH3OH (1)

CO2 + 3H2 → CH3OH + H2O (2)

CO : 25 – ε1 = yCO.N2

H2 : 60 - 2 ε1 – ε2 = yN2.N2

CO2 : 15 - ε2 = yCO2.N2

CH4OH : ε1 + ε2 = yCH3OH.N2

H2O : ε 2 = y H2O.N 2

100 - 2 ε1 –2 ε2 = N2

3

2

1 2

CH OH 1 21 2

CO. H 1 1 2

1 2 1 2

y 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

3 23

2

1 2 2

CH OH H O 1 2 1 22 3

CO. H1 1 2

1 2 1 2

y .y 100 2 2 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

3 23

2

1 2 2

CH OH H O 1 2 1 22 3

CO. H1 1 2

1 2 1 2

y .y 100 2 2 100 2 2K

y y 25 60 2 3100 2 2 100 2 2

R = 8.314 J/(mol.K)T = 323 K

Problem 12.1: Metanol/air Margules Asli

P/kPa x1 y1 GE/RT19.953 0.0000 0.0000 0.0000 0.000039.223 0.1686 0.5714 0.4523 0.0133 0.0873 1.605342.984 0.2167 0.6268 0.3855 0.0260 0.1039 1.633248.852 0.3039 0.6943 0.2775 0.0725 0.1348 1.569152.784 0.3681 0.7345 0.2196 0.1057 0.1476 1.575756.652 0.4461 0.7742 0.1507 0.1462 0.1482 1.667060.614 0.5282 0.8085 0.0927 0.2095 0.1478 1.685963.998 0.6044 0.8383 0.0485 0.2708 0.1365 1.752367.924 0.6804 0.8733 0.0305 0.2998 0.1166 1.865670.229 0.7255 0.8922 0.0211 0.3237 0.1042 1.911872.832 0.7776 0.9141 0.0124 0.3435 0.0860 2.010284.562 1.0000 1.0000 0.0000 0.0000

persamaan garis lurusslop = 0.035

intersep = 1.6763

A21 = 0.5844A12 = 0.5966

ln g1 ln g2x1.x2/

(GE/RT)

0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

f(x) = 0.543154551486049 x + 1.41551811982626

grs lurus hasil pan-dangan HS

grs lurus hasil "trend-line Excel"

A12 = 0.619261A21 = 1.112016

Margules dan Barker

persamaan garis lurus

(x1 = 1) (x1 = 0)

0% 600.0 420.0 600.010% 582.0 419.4 600.020% 563.8 417.9 600.330% 545.5 415.7 601.140% 526.7 413.0 602.650% 507.5 410.0 605.060% 487.7 407.0 608.670% 467.1 404.3 613.780% 445.8 402.1 620.590% 423.4 400.6 629.2

100% 400.0 400.0 640.0Persamaan (A)

Persamaan (B)

Persamaan (C) Perhatikan bahwa: H = x1*h1 + x2 *h2bukan: H = x1*H1 + x2*H2

dH/dx = dilakukan secara analitik, Entalpi campuran bukan jumlah entalpi molar komponen murni,

karena H fungsi x diketahui. tetapi jumlah entalpi molar parsial komponen.

fraksi massa

entalpi campuran

H

entalpi molar

parsial h1

entalpi molar

parsial h2

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

400.0

450.0

500.0

550.0

600.0

650.0

700.0

entalpi campuranPolynomial (entalpi campuran)Hpar1Hpar2

Perhatikan bahwa: H = x1*h1 + x2 *h2

Entalpi campuran bukan jumlah entalpi molar komponen murni,

tetapi jumlah entalpi molar parsial komponen.

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

400.0

450.0

500.0

550.0

600.0

650.0

700.0

entalpi campuranPolynomial (entalpi campuran)Hpar1Hpar2

metanol etanol metanol etanol metanol etanol metanolA 16.4948 16.1952 16.4948 16.1952 16.4948 16.1952 16.4948B 3593.39 3423.53 3593.39 3423.53 3593.39 3423.53 3593.39C -35.2259 -55.7152 -35.2259 -55.7152 -35.2259 -55.7152 -35.2259

T, C 71.8918 71.8918 63.42074T, K 344.8918 344.8918 335 335 300 300 336.4207ln(Psat) 4.890713 4.35631 4.507807 3.936996 2.923269 2.180697 4.56435Psat, kPa 133.0484 77.96889 90.72266 51.26438 18.60199 8.852477 96.00016x1, x2 40.0% 60.0% 40.0% 60.0% 40.0% 60.0% 40.0%P-total 100.0007 100.0007 67.04769 67.04769 12.75228 12.75228 71.12748y1= 0.53219 0.46781 0.541243 0.458757 0.583487 0.416513 0.539877

25.50457

350 350 substitusi menurut halaman 355-3565.079026 4.561809 alfa-tebak 160.6175 95.75659 Psat2 =

40.0% 60.0% Td2 =121.701 121.701 afla-hitung

0.527909 0.472091 selisih=243.4019

Psat2 =Td2 =

325 325 afla-hitung4.09414 3.481783 selisih=

59.98775 32.5176440.0% 60.0% Psat2 =

43.50568 43.50568 Td2 =0.551539 0.448461 afla-hitung87.01137 selisih=

Psat2 =337.5 337.5 Td2 =

4.606947 4.045751 afla-hitung100.1779 57.1541 selisih=

40.0% 60.0%74.36361 74.36361 Psat2 =0.538854 0.461146 Td2 =148.7272 afla-hitung

selisih=332 332

4.386634 3.80389280.36946 44.87551

40.0% 60.0%59.07309 59.073090.544204 0.455796118.1462

330 3304.304482 3.713539

74.03086 40.9986340.0% 60.0%

54.21152 54.211520.546237 0.453763

108.423

328 3284.221207 3.62185868.11568 37.40699

40.0% 60.0%49.69047 49.69047

0.54832 0.4516899.38094

etanol metanol etanol metanol etanol metanol etanol metanol etanol16.1952 16.4948 16.1952 16.4948 16.1952 16.4948 16.1952 16.4948 16.19523423.53 3593.39 3423.53 3593.39 3423.53 3593.39 3423.53 3593.39 3423.53

-55.7152 -35.2259 -55.7152 -35.2259 -55.7152 -35.2259 -55.7152 -35.2259 -55.7152

60 60336.4207 345 345 350 350 355 355 333 3333.999039 4.894766 4.360738 5.079026 4.561809 5.257523 4.756163 4.427296 3.8485854.54569 133.5888 78.3149 160.6175 95.75659 192.0054 116.2988 83.70481 46.92639

60.0% 40.0% 60.0% 40.0% 60.0% 40.0% 60.0% 40.0% 60.0%100.4245 100.4245 121.701 121.701 146.5814 146.5814 61.63776 61.63776

0.460123 0.532097 0.467903 0.527909 0.472091 0.523956 0.476044 0.543205 0.456795

substitusi menurut halaman 355-3561.759995 alfa-tebak 1.677352 alfa-tebak 1.650966 alfa-tebak 1.78374776.68723 Psat2 = 78.68187 Psat2 = 79.34074 Psat2 = 76.13254344.4875 Td2 = 345.1143 Td2 = 345.3185 Td2 = 344.31083.348582 afla-hitung 3.341161 afla-hitung 3.338754 afla-hitung 3.3506820.474406 selisih= 0.497973 selisih= 0.505514 selisih= 0.467647

51.56147335.13173.4653950.033708

50.34847334.58993.4725260.002054

50.27626334.55733.4729560.000124

50.27191 jawaban334.55533.4729827.47E-06 puas

nitrogen oksigen nitrogen oksigen air13.4477 13.6835 13.4477 13.6835 16.5362

658.22 780.26 658.22 780.26 3985.44-2.854 -4.1758 -2.854 -4.1758 -38.9974

-195.708 -182.8767 -165.0261 -165.0261 98.3140577.29199 90.12334 107.97388 107.9739 371.31414.605173 4.605172 7.1860869 6.166405 4.543301100.0002 100.0002 1320.9241 476.4701 94.0006

40.0% 60.0% 38.3% 61.7%100.0002 100.0002 799.9996 799.9996 1 atm

40.0% 60.0% 79.0% 21.0% komps. Udara

Ptotal 800 kPay1 79% [N2]galat 0%

g/mL fraksi mol mL/g mL/mol v1 v20 0.998 0.0% 1.002 18.032 53.760 18.032

10% 0.982 4.2% 1.018 19.521 52.608 18.03220% 0.969 8.9% 1.032 21.159 53.392 18.08230% 0.954 14.4% 1.048 23.087 54.956 18.00540% 0.935 20.7% 1.069 25.442 56.080 17.74350% 0.914 28.1% 1.094 28.315 56.751 17.45060% 0.891 37.0% 1.122 31.820 57.205 17.18770% 0.868 47.7% 1.153 36.147 57.583 16.92180% 0.843 61.0% 1.186 41.597 57.935 16.57590% 0.818 77.9% 1.223 48.666 58.277 16.025

100% 0.789 100.0% 1.267 58.277 58.277 14.822

fraksi massa

0.0% 10.0% 20.0% 30.0% 40.0% 50.0% 60.0% 70.0% 80.0% 90.0% 100.0%0.000

10.000

20.000

30.000

40.000

50.000

60.000

70.000

etanol

Polynomial (etanol)

v-etanol

v-air

0.0% 10.0% 20.0% 30.0% 40.0% 50.0% 60.0% 70.0% 80.0% 90.0% 100.0%0.000

10.000

20.000

30.000

40.000

50.000

60.000

70.000

etanol

Polynomial (etanol)

v-etanol

v-air

Ujian Modul-2; Soal No. 5Perhitungan Koefisien FugasitasSenyawa H2O

Pc, bar 220Tc, K 647

0.345V-cair, 25C, dm3/kg 1.003V-cair, 25C, m3/mol 0.000018054R, J/mol/K 8.314

SOAL A Koefisien Fugasitas cairan jenuh pd titik didih normal P, kPa 100Titik didih normal; T, C 100Pr = P/Pc 0.0045454545Tr = T/Tc 0.5765069552B0 = -0.935650421B1 = -1.599398474

-0.0117277060.9883407955

SOAL B Koefisien Fugasitas cairan pada 25C dan 100 kPaP, kPa 100 --> Pa 100000P-sat pada 25C, kPa 3.166 --> Pa 3166T, C 25

a)P, kPa 3.166 --> Pa 3166T, C 25Pr 0.0001439091Tr 0.4605873261B0 = -1.375858861B1 = -4.323909874

-0.0008959750.9991044261

b)f = 3165.3974141

0.0316539741

omega, w

lnf =f =

Mencari f-sat

lnf =f =Mencari f (25C; 100 kPa)

f = f/P

lnφ=PrTr

(B0+ω .B1 )

B0=0,083−0,422

Tr1,6

B1=0,139−0,172

Tr4,2

f=φsat . Psat . exp(V l .(P−Psat )R .T )

document.xlsNIM : ...................................

Nama : .................................

TK2211: Bahan kuliah dan Kuis, Senin 30 Maret 2009

Table 12.1 Kesetimbangan fasa uap-cair MEK-Toluen (50 C)HASIL PERCOBAAN Raoult

P, kPa x1 y1 x1.Psat y1 hampiran1 12.30 0.0000 0.0000 0 0 0.00%2 15.51 0.0895 0.2716 3.230055 0.208256 6.33%3 18.61 0.1981 0.4565 7.149429 0.384171 7.23%4 21.63 0.3193 0.5934 11.52354 0.532757 6.06%5 24.01 0.4232 0.6815 15.27329 0.636122 4.54%6 25.92 0.5119 0.7440 18.47447 0.71275 3.13%7 27.96 0.6096 0.8050 22.00046 0.786855 1.81%8 30.12 0.7135 0.8639 25.75022 0.854921 0.90%9 31.75 0.7934 0.9048 28.63381 0.901852 0.29%

10 34.15 0.9102 0.9590 32.84912 0.961907 0.29%11 36.09 1.0000 1.0000 36.09 1 0.00%

total 30.59%SOAL

A Gambarlah kurva P-x,y jika sistem dianggap mengikuti Hukum RaoultB Berapakah galat (penghampiran) total semua titik pengamatan

jika sistem dianggap mengikuti Hukum Raoult,

C Jika sistem mengikuti persamaan "modified Raoult Law", tentukan koefisien aktifitas masing-masing komponen.

(bahan kuliah hari ini)

0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

P-x

P-y

P-y Raoult

x dan y, fraksi mol

P-t

ota

l, k

Pa

document.xlsNIM : ...................................

Nama : .................................

TK2211: Bahan kuliah dan Kuis, Senin 30 Maret 2009

Table 12.1 Kesetimbangan fasa uap-cair MEK-Toluen (50 C)HASIL PERCOBAAN

P, kPa x1 y11 12.30 0.0000 0.00002 15.51 0.0895 0.27163 18.61 0.1981 0.45654 21.63 0.3193 0.59345 24.01 0.4232 0.68156 25.92 0.5119 0.74407 27.96 0.6096 0.80508 30.12 0.7135 0.86399 31.75 0.7934 0.9048

10 34.15 0.9102 0.959011 36.09 1.0000 1.0000

SOALA Gambarlah kurva P-x,y jika sistem dianggap mengikuti Hukum RaoultB Berapakah galat (penghampiran) total semua titik pengamatan

jika sistem dianggap mengikuti Hukum Raoult,

C Jika sistem mengikuti persamaan "modified Raoult Law", tentukan koefisien aktifitas masing-masing komponen.

(bahan kuliah hari ini)

0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

P-x

P-y

P-y Raoult

x dan y, fraksi mol

P-t

ota

l, k

Pa

R, J/(mol.K) 8.314 Antoine EquationPc, Bar 220.55 A = 16.3872

Tc, K 647.1 B = 3885.7Omega = 0.253 C = 230.17

T, C 100P, Pa = 100000 Psat, kPa 101.3332

T, K = 337Pr = 0.00453411925Tr = 0.52078504095

Persamaan Keadaan van der WaalsalphaTr = 1

tau = 0epsilon = 0

OMEGA = 0.125PSY = 0.4219

Zc = 0.3750a(T); (3.45) = 0.5537

b; (3.46) = 3.049E-05

beta (3.50) = 0.0011q (3.52) = 6.4806

fasa gas fasa cairZ tebak = 0.9940 1.343E-03Z (3.52) = 0.9940 1.344E-03

tools = 4.196E-07 4.937E-04I, (6.65b) = 0.00109485788koef.fug. = 0.99 1.430E+03

fugasitas, Pa = 99404 142960554fugasitas, bar = 0.994 1429.606

NIM Nama 1 2 3 4 5 6 5Peb13007056 Daniel 6 6 10 6 10 4 7.0013007058 Thomas 6 6 10 6 10 4 7.0013007060 Anthony 8 8 10 10 10 4 8.3313007061 Harris 8 6 4 6 10 4 6.3313007063 Barry 8 8 10 6 10 4 7.6713007064 Richard 8 8 10 6 10 4 7.6713007067 Maria 8 8 10 6 10 4 7.6713007068 Dave 8 8 10 6 10 4 7.6713007069 Siti 8 8 10 6 10 4 7.6713007070 Rani 8 6 10 6 10 4 7.3313007071 Herdadi 6 8 10 6 10 4 7.3313007072 Hubert 8 6 10 6 10 4 7.3313007073 Lasa 8 8 10 6 10 0 7.0013007074 Riko 6 6 10 6 10 4 7.0013007075 Laras 8 8 10 6 10 4 7.6713007076 Alvin 8 8 10 6 10 4 7.6713007077 Arlavinda 8 8 10 6 10 4 7.6713007078 Rizal 8 6 10 6 10 4 7.3313007082 Herdinas 6 6 10 6 10 4 7.0013007086 Mohammad 6 6 10 6 10 4 7.0013007087 Rian 8 6 10 6 10 4 7.3313007090 Sarmedi 8 8 10 10 10 4 8.3313007091 Vipassi 6 6 10 6 10 4 7.0013007095 Frebigia 8 6 10 6 10 4 7.3313007096 Rikardo 8 8 10 6 10 4 7.6713007098 Iffan 8 8 10 6 10 4 7.6713007100 Sabrina 8 8 10 5 10 4 7.5013007101 Joseph 6 6 10 6 10 4 7.0013007105 Rusnang 8 8 10 6 10 4 7.6713007106 Phelia 8 8 10 6 10 4 7.67

rata-rata 7.5 7.1 9.8 6.2 10.0 3.9 7.4

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