Given information:
It is given that enthalpy and entropy are set equal to zero for the ideal gas state at 101.33 kPa and 273.15 K .
The vapor pressure of 1,3-butadiene at 380 K is 1919.4 kPa
For pure species 1,3-butadiene, the properties can be written down using Appendix B, Table B.1
ω=0.19, Tc=425.2 K, Pc=42.77 bar, ZC=0.267, Tn=268.7 K, VC=220.4 cm3mol
The molar volume for1,3-butadiene saturated vapor is calculated directly from equation,
V=ZRTP
For Z=Z0+ωZ1
Tr=T2TcTr=380 K425.2 K=0.894
Since vapor pressure is given as 1919.4 kPa and is final state pressure.
So,
Pr=PPcPr=1919.4 kPa×1 bar100 kPa42.77 bar=0.449
So, at above values of Tr and Pr, The values of Z0 and Z1 can be written from Appendix D
Tr=0.894 lies between reduced temperatures Tr=0.85 and Tr=0.90 and Pr=0.449 lies in between reduced pressures Pr=0.4 and Pr=0.6 .
At Tr=0.85 and Pr=0.4
Z0=0.0661, ( H R)RTC0=−4.309, ( S R)R0=−4.785
At Tr=0.85 and Pr=0.6
Z0=0.0983, ( H R)RTC0=−4.313, ( S R)R0=−4.418
At Tr=0.90 and Pr=0.4
Z0=0.78, ( H R)RTC0=−0.596, ( S R)R0=−0.463
At Tr=0.90 and Pr=0.6
Z0=0.1006, ( H R)RTC0=−4.074, ( S R)R0=−4.145
And
At Tr=0.85 and Pr=0.4
Z1=−0.0268, ( H R)RTC1=−4.753, ( S R)R1=−4.853
At Tr=0.85 and Pr=0.6
Z1=−0.0391, ( H R)RTC1=−4.754, ( S R)R1=−4.841
At Tr=0.90 and Pr=0.4
Z1=−0.1118, ( H R)RTC1=−0.751, ( S R)R1=−0.744
At Tr=0.90 and Pr=0.6
Z1=−0.0396, ( H R)RTC1=−4.254, ( S R)R1=−4.269
Applying linear interpolation of two independent variables, From linear double interpolation, if M is the function of two independent variable X and Y then the value of quantity M at two independent variables X and Y adjacent to the given values are represented as follows:
X1 X X2Y1 M1,1 M1,2Y M=? Y2 M2,1 M2,2
So,
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1Z0=[( 0.6−0.449 0.6−0.4)×0.0661+( 0.449−0.4 0.6−0.4)×0.0983] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×0.78+( 0.449−0.4 0.6−0.4)×0.1006] ×0.894-0.850.9−0.85Z0=0.549
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( H R )RTC0=[( 0.6−0.449 0.6−0.4)×−4.309+( 0.449−0.4 0.6−0.4)×−4.313] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×−0.596+( 0.449−0.4 0.6−0.4)×−4.074] ×0.894-0.850.9−0.85( H R )RTC0=−1.791
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( S R )R0=[( 0.6−0.449 0.6−0.4)×−4.785+( 0.449−0.4 0.6−0.4)×−4.418] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×−0.463+( 0.449−0.4 0.6−0.4)×−4.145] ×0.894-0.850.9−0.85( S R )R0=−1.823
And
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1Z1=[( 0.6−0.449 0.6−0.4)×−0.0268+( 0.449−0.4 0.6−0.4)×−0.0391] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×−0.1118+( 0.449−0.4 0.6−0.4)×−0.0396] ×0.894-0.850.9−0.85Z1=−0.086
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( H R )RTC1=[( 0.6−0.449 0.6−0.4)×−4.753+( 0.449−0.4 0.6−0.4)×−4.754] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×−0.751+( 0.449−0.4 0.6−0.4)×−4.254] ×0.894-0.850.9−0.85( H R )RTC1=−1.987
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( S R )R1=[( 0.6−0.449 0.6−0.4)×−4.853+( 0.449−0.4 0.6−0.4)×−4.841] ×0.9−0.8940.9−0.85 +[( 0.6−0.449 0.6−0.4)×−0.744+( 0.449−0.4 0.6−0.4)×−4.269] ×0.894-0.850.9−0.85( S R )R1=−1.997
Now from equation (1) at final state
Z=Z0+ωZ1Z=0.549+0.19×−0.086Z=0.53
And
V=ZRTPV=0.53×8.314 l kPamol K×1000 cm 31 l×380 K1919.4 kPaVvap=872.38 cm3mol
From equation (2) at final state,
HR=(HR)0+ω(HR)1
( H R )RTC0=−1.791⇒(HR)0=−1.791×8.314Jmol K×425.2 K(HR)0=−6331.39Jmol
And
( H R )RTC1=−1.987⇒(HR)1=−1.987×8.314Jmol K×425.2 K(HR)1=−7024.27Jmol
So,
HR=(HR)0+ω(HR)1HR=−6331.39Jmol+0.19×−7024.27JmolHR=−7666.0013Jmol
ΔH=∫T1T2CPigdT+H2R−H1R
At final state H1R=0
ΔH=∫T1T2CPigdT+H2R
∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)
Where τ=TT0
τ=TT0=380273.15 =1.391
Values of above constants for 1,3-Butadiene in above equation are given in appendix C table C.1 and noted down below:
i A 103B 106 C 10−5D1,3−butadiene 2.734 26.786 −8.882 0
∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫T0TΔC∘PRdT=2.734×273.15×(1.391−1)+26.786×10−32273.152(1.3912−1) +−8.882×10−63×273.153(1.3913−1)+0×105273.15(1.391−11.391)∫T0TΔC∘PRdT=1124.13K
ΔH=R∫T1T2 C P igRdT+H2RΔH=8.314 JK mol×1124.13K−7666.0013JmolΔH=1680.026 Jmol=Hvap
( S R )R0=−1.823⇒(SR)0=−1.823×8.314Jmol K(SR)0=−15.16Jmol K
And
( S R )R1=−1.997(SR)1=8.314Jmol K×−1.997(SR)1=−16.6Jmol K
So,
SR=(SR)0+ω(SR)1SR=−15.16Jmol K+0.19×−16.6Jmol KSR=−18.31Jmol K
ΔS=∫T1T2CPigdTT−RlnP2P1+S2R−S1R
At final state S1R=0
ΔS=R∫T1T2CPigRdTT−RlnP2P1+S2R
∫T1T2CPigRdTT=[A+{BT0+(CT02+Dτ2T02)(τ+12)}(τ−1lnτ)]×lnτ
τ=TT0=380273.15 =1.391
Values of above constants for 1,3-Butadiene in above equation are given in appendix C table C.1 and noted down below:
i A 103B 106 C 10−5D1,3−butadiene 2.734 26.786 −8.882 0
∫T1T2 C P igRdTT=[A+{BT0+(CT02+D τ 2 T 0 2 )( τ+12)}(τ−1lnτ)]×lnτ∫T1T2CP igRdTT= [2.734+{26.786×10−3×273.15+(−8.882×10−6×273.152+0× 10 5 1.391 2× 273.15 2)(1.391+12)}(1.391−1ln1.391)] ×ln1.391∫T1T2CPigRdTT=3.453
Hence,
ΔS=R∫T1T2 C P igRdTT−Rln P 2 P 1+S2RΔS=8.314Jmol K×3.453−8.314Jmol K×ln1919.4 kPa101.33 kPa+(−18.31Jmol K)ΔS=Svap=−14.056 Jmol K
Now, for saturated liquid
Molar volume of saturated liquid can be found out using following correlation proposed by Rackett
Vsat=VCZC(1− T r)0.2857Vsat=220.4 cm3mol×0.267(1−0.894)0.2857Vliq=109.95 cm3mol
Now, for enthalpy and entropy of saturated liquid
ΔHnRTn=1.092(ln P C−1.013)0.93− T n T cΔHnRTn=1.092(ln42.77−1.013)0.93−268.7 K425.2 KΔHnRTn=10.05ΔHn=10.05×8.314 Jmol K×268.7 KΔHn=22448.8 Jmol
ΔH2ΔH1=( 1− T r2 1− T r1 )0.38ΔH222448.8 Jmol=( 1−0.894 1− 268.7 K 425.2 K )0.38ΔH2=13988.12 Jmol
ΔH2=Hvap−HliqHliq=Hvap−ΔH2Hliq=1680.026 Jmol−13988.12 JmolHliq=−12308.09 Jmol
ΔH2T=Svap−SliqSliq=Svap−ΔH2TSliq=−14.056 Jmol K−13988.12 Jmol380 KSliq=−50.87 Jmol K