Chem390_solution_1

# Chem390_solution_1 - 2 Chem390_solution_1.nb In[14]:=...

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Unformatted text preview: 2 Chem390_solution_1.nb In[14]:= constSI s. K  1(; RK300 Plot#PEthane #300, V', V, 0, 25, PlotRange  0, 10'; Exp300 ListPlot#data300 s. constSI'; RK350 Plot#PEthane #350, V', V, 0, 30, PlotRange  0, 10'; Exp350 ListPlot#data350 s. constSI'; Problem 1 : RK400 Plot#PEthane #400, V', V, 0, 35, PlotRange  0, 10'; Exp400 ListPlot#data400 s. constSI'; +1/ Let ' s first set up the RK equation. PRK #TB, VB' RT VB  Simplify\$ PRK #T, V' s bar s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 , V  V dm3 mol1 s. Chem 3900: Solution to Problem Set 1 In[13]:= PEthane #TB, VB' A Show#RK300, Exp300, RK350, Exp350, RK400, Exp400, AxesLabel  "V+L/", "P+bar/"' ; T V +V  B/ P+bar/ 10 As the volume V goes infinity +ideal gas limit/, the second term in the RK equation becomes negligible since it 8 is proportional to V  2 compared to V  1 for the first term. Also we can approximate 1 s +V  B/ with 1 s V . This makes the equation to be P R T s V , which is the IGES. As in the vdW equation of state, B corresponds to the molecular volume while A corresponds to the attractive intermolecular interaction strength. 3 6 Out[21]= 4 2 0 1 +2/ First evaluate the RK equation for ethane. ,V in units of dm mol , P in units of bar0 5 10 15 20 25 V+L/ +3/ Let ' s follow the procedure given in the problem. The values of critical parameters are the same as given in the textbook, Eq +16.14/ Then we plot the RK values together with the experimental values given in the notebook. In[51]:= d1 #TB, VB' V PRK #T, V'; dsolution Solve#d1 #T, V' m 0, T' 2 Out[52]= T  A2s3 - B4  V 4B V3  4 V2  2 +1/1s3 A2s3 - B4  V TC +1/2s3 A2s3 - B4  V  1s3 8 1 V +BV/ 4B V3  4 V2  16 V2 +BV/4  24 +BV/2 4B V3  4 V2  16 V2 +BV/4  24 +BV/2 R2s3 T s. dsolution##1'' 2 A2s3 - B4  Out[77]= 24 +BV/2 V  !, 8 1 V +BV/ 1s3 R2s3 2 In[77]:=  R2s3 T   T  16 V2 +BV/4 4B V3  4 V2  16 V2 +BV/4 R2s3  24 +BV/2  8 1 V +BV/ 1s3  !, 1s3 8 1 V +BV/ !! Chem390_solution_1.nb d2 #VB' PowerExpand#Simplify#PowerExpand#V d1 #T, V' s. T  TC s. B  1''' + Express in terms of V in units of B / 3 4 Chem390_solution_1.nb In[124]:= PR #TRB, VRB' Factor#FullSimplify#PowerExpand#PRK #TR TC , VR VC ' s PC ''' 3. ,0.333333  1.28244 VR  0.259921 TR3s2 VR  1. TR3s2 VR2 0 Out[124]= Out[74]= 2A 2s3 1s3 R +1  V/ +1  V/2 V4 +1  2 V/ 1  3 V2  4 3V In[72]:= Out[72]= In[76]:= +1  V/2 2 d2solution TR +0.259921  1. VR/ +0.259921  1. VR/ VR 2s3  V3 1  +1  V/2 +1  2 V/ V2 +1  V/2 +1  V/2 +1  2 V/ V2 +1  V/2 2s3 v -+1  V/ 7s3 V +1  2 V/  3 V5 1  1s3 +1  V/2 +1  2 V/ V2 +1  V/2 +6/ Plots of isotherms for Ethane.  In[165]:= 2 3 ,1  V 0 1 TRindex 0.85, 0.9, 0.95, 1, 1.05, 1.1, 1.15, 1.2; Reducedplot Table#Plot#PR #TRindex##i'', VR', VR, 0.3, 3, PlotRange  1, 3.5, AxesLabel  "VsVC", "PsPC"', i, 1, 8'; Show#Reducedplot' NSolve#d2 #V' m 0, V' PsPC V  0.423661  0.283606 Ç, V  0.423661  0.283606 Ç, V  3.84732 VC 3 V B s. d2solution##3'' 2 3.84732 B Out[76]= Out[167]= In[81]:= TC 0.34504 A2s3 , Out[81]= In[85]:= 1 TC s. V ! VC 1 B2 1s3 0 1.0 R2s3 PC B5s3 0.000173718 meter3 mol , 261.907 K, 4.17847  106 J V  V dm3 mol1 s. constSI s. V  +1 s Vinv/ s. K  1( +P/ each. meter3 ! 1 0.1480 dm mol , 305.34 K, 48.714 bar ; 3 In[92]:= CriticalEt In[94]:= EstimateEt s CriticalEt s. constJSI Out[94]= VsVC Z#VinvB, TB' 6 1.  10 Out[177]= 1.17377, 0.857755, 0.857756 +5/ Law of Corresponding States. As can be seen below, the reduced pressure can be expressed as a function of the reduced temperature and volume. In[229]:=  4.5153  108 Vinv  6 1.  10 EstimateEt  3.0 Simplify\$PRK #T, V' V s +R T/ s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 , PowerExpand\$VC , TC , PC  s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 s. constJSI( Out[90]= 2.5 +7/ Compressibility. In[177]:= +4/ Critical parameters for Ethane is calculated as below. The difference between the two is 17.4 +V /, 14.2 +T/ and 14.2 In[90]:= 2.0 1 PowerExpand#PRK #TC , VC '' 0.0298944 A2s3 R1s3 Out[85]= 1.5 TcEt 0.000944285 Vinv T3s2 9  0. Vinv  2.03879  10  0.0000426373 Vinv2 T3s2 Vinv2 PowerExpand\$TC s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 s. constJSI(; Tindex TcEt, TcEt  200 K, TcEt  400 K, TcEt  600 K, TcEt  800 K, TcEt  1000 K s. K  1; Zplot Table#Plot#Z#Vinv, Tindex##i''', Vinv, 0, 20, PlotRange  0.2, 4, AxesLabel  "1sV +molsL/", "Z"', i, 1, 6' Show# Zplot' Chem390_solution_1.nb 4.0 5 6 Chem390_solution_1.nb Z 4.0 3.5 3.5 3.0 Out[231]=  3.0 2.5 , 2.5 2.0 1.5 1.0 5 10 15 20 1.0 1sV +molsL/ 5 Z 4.0 3.5 10 15 20 1sV +molsL/ Z 3.5 3.0 3.0 2.5 , 2.5 2.0 , 2.0 1.5 1.5 1.0 5 4.0 , 2.0 1.5 4.0 Z 10 15 20 1.0 1sV +molsL/ 5 Z 4.0 3.5 10 15 20 1sV +molsL/ Z 3.5 3.0 3.0 2.5 , 2.5 2.0 ! 2.0 1.5 1.5 1.0 5 10 15 20 1.0 1sV +molsL/ 5 4.0 10 15 20 5 10 15 20 1sV +molsL/ Z 3.5 3.0 Out[232]= 2.5 2.0 1.5 1.0 1sV +molsL/ Chem390_solution_1.nb 7 8 Chem390_solution_1.nb There is an upper limit to 1 s V , which is the zero of the equation below. The upper limit is 22.15 mol s L In[233]:= Out[233]= 4.0 3.5 3.0 NSolve\$106  2.038793409 109 Vinv2 m 0, Vinv( Out[296]= Vinv  22.1469, Vinv  22.1469 2.5  Out[254]= Out[255]= , 2.0 +8/ V irial equation. In[254]:= Z 1.5 Z#VinvB, TB' Simplify#PRK #T, V' V s +R T/ s. V  1 s Vinv' Series#Z#Vinv, T', Vinv, 0, 3' 1.0 1 1  B Vinv 1 B  5 A Vinv T3s2 +R  B R Vinv/ A R T3s2 2 Vinv  B  AB R T3s2 2 3 Vinv  B  A B2 R T3s2 4.0 3 Vinv  O#Vinv' 4 10 15 20 1sV +molsL/ Z 3.5 3.0 In[269]:= B2#TB' B A ; B3#TB' B2  R T3s2 TBoyle Solve#B2#T' m 0, T'; TBoyleEthane PowerExpand\$ AB ; B4#TB' R T3s2 B3  A B2 ; R T3s2 T s. TBoyle##1'' s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 s. constJSI( Out[271]= In[292]:= 2.5 , 2.0 1.5 1.0 759.063 K 5 Z1st#VinvB, TB' A Simplify%1  B  R T3s2 v V s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 , V  V dm3 mol1 s. constSI s. V  +1 s Vinv/ s. K  1); Zplot1st Table#Plot#Z1st#Vinv, Tindex##i''', Vinv, 0, 20, PlotRange  0.2, 4, AxesLabel  "1sV +molsL/", "Z"', i, 1, 6'; A AB Z2nd#VinvB, TB' Simplify% 1  B  v V  B2  v V2 s. B  0.045153 dm3 mol1 , R T3s2 R T3s2 A  77.486 dm6 atm mol2 K1s2 , V  V dm3 mol1 s. constSI s. V  +1 s Vinv/ s. K  1); Zplot2nd Table#Plot#Z2nd#Vinv, Tindex##i''', Vinv, 0, 20, PlotRange  0.2, 4, AxesLabel  "1sV +molsL/", "Z"', i, 1, 6'; Table#Show#Zplot##i'', Zplot1st##i'', Zplot2nd##i''', i, 1, 6' 4.0 10 15 20 1sV +molsL/ Z 3.5 3.0 2.5 , 2.0 1.5 1.0 5 10 15 20 1sV +molsL/ Chem390_solution_1.nb 4.0 9 10 Chem390_solution_1.nb In[352]:= Z Z#VinvB, TB' Simplify\$PRK #T, V' V s +R T/ s. B  0.045153 dm3 mol1 , A  77.486 dm6 atm mol2 K1s2 , V  V dm3 mol1 s. constSI s. V  +1 s Vinv/ s. K  1(; 3.5 Table#Table#PEthane #Tindex##i'', 1 s Vinv', Z#Vinv, Tindex##i''' s. K  1, Vinv, 0.1, 22, 0.1', i, 1, 6'; ListPlot#PZ, PlotRange  0, 3000, 0, 3, AxesLabel  "P +bar/", "Z"' PZ 3.0 2.5 , 2.0 3.0 1.5 Z 1.0 5 4.0 10 15 20 1sV +molsL/ 2.5 Z 2.0 3.5 Out[354]= 3.0 2.5 1.5 , 2.0 1.0 1.5 1.0 5 4.0 10 15 20 0.5 1sV +molsL/ 0.0 Z 3.5 In[363]:= 3.0 2.5 500 1000 1500 2000 2500 P +bar/ 3000 PZbelow Table#Table#PEthane #Tindex##1''  20 i, 1 s Vinv', Z#Vinv, Tindex##1''  20 i' s. K  1, Vinv, 0.1, 22, 0.1', i, 0, 6'; ListPlot#PZbelow, PlotRange  0, 200, 0, 1, AxesLabel  "P +bar/", "Z"' ! 2.0 0 1.0 Z 1.5 0.8 1.0 5 10 15 20 1sV +molsL/ 0.6 Out[364]= In each plot, the top curve is f rom the RK equation, the middle one f rom 2 nd V irial approximation, and the bottom f rom the f irst order V irial approximation. +9/ Plots of P, Z for Ethane. With increasing T, the Z becomes closer to 1 at large P. 0.4 0.2 0.0 0 50 100 150 200 P +bar/ As shown above, Z can have multiple values for a single P below a certain pressure. +Instability/ This indicates the phase transistion. ...
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## This note was uploaded on 05/04/2009 for the course CHEM 3900 taught by Professor Park during the Spring '08 term at Cornell University (Engineering School).

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