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Margules Equation
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
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
galat adiabatik -6.5002886E-07
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