Given information:
It is given that in the initial state, propane is assumed as an ideal gas. And at initial state
And final state pressure and temperature are given as, P1=1 bar, T1=308.15 K, P2=135 bar, T2=468.15 K
For pure species propane, the properties can be written down using Appendix B, Table B.1
ω=0.152, Tc=369.8 K, Pc=42.48 bar, M=44.1 gmmol .
Now For final state
Tr=T2TcTr=468.15 K369.8 K=1.266
And
Pr=PPcPr=135 bar42.48 bar=3.178
So, at above values of Tr and Pr, The values of Z0 and Z1 can be written from Appendix D
Tr=1.266 lies between reduced temperatures Tr=1.2 and Tr=1.3 and Pr=3.178 lies in between reduced pressures Pr=3.00 and Pr=5.00 .
At Tr=1.2 and Pr=3.00
Z0=0.5425, ( H R)RTC0=−2.801, ( S R)R0=−1.727
At Tr=1.3 and Pr=3.00
Z0=0.6344, ( H R)RTC0=−2.274, ( S R)R0=−1.299
At Tr=1.2 and Pr=5.00
Z0=0.7069, ( H R)RTC0=−3.166, ( S R)R0=−1.827
At Tr=1.3 and Pr=5.00
Z0=0.7358, ( H R)RTC0=−2.825, ( S R)R0=−1.554
And
At Tr=1.2 and Pr=3.00
Z1=0.1095, ( H R)RTC1=−0.934, ( S R)R1=−0.991
At Tr=1.3 and Pr=3.00
Z1=0.2079, ( H R)RTC1=−0.300, ( S R)R1=−0.481
At Tr=1.2 and Pr=5.00
Z1=−0.0141, ( H R)RTC1=−1.840, ( S R)R1=−1.767
At Tr=1.3 and Pr=5.00
Z1=0.0875, ( H R)RTC1=−1.066, ( S R)R1=−1.147
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=[( 5−3.178 5−3)×0.5425+( 3.178−3 5−3)×0.7069] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×0.6344+( 3.178−3 5−3)×0.7358] ×1.266-1.21.3−1.2Z0=0.614
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=[( 5−3.178 5−3)×−2.801+( 3.178−3 5−3)×−3.166] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×−2.274+( 3.178−3 5−3)×−2.825] ×1.266-1.21.3−1.2( H R )RTC0=−2.497
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=[( 5−3.178 5−3)×−1.727+( 3.178−3 5−3)×−1.827] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×−1.299+( 3.178−3 5−3)×−1.554] ×1.266-1.21.3−1.2( S R )R0=−1.462
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=[( 5−3.178 5−3)×0.1095+( 3.178−3 5−3)×−0.0141] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×0.2079+( 3.178−3 5−3)×0.0875] ×1.266-1.21.3−1.2Z1=0.164
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=[( 5−3.178 5−3)×−0.934+( 3.178−3 5−3)×−1.84] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×−0.300+( 3.178−3 5−3)×−1.066] ×1.266-1.21.3−1.2( H R )RTC1=−0.588
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=[( 5−3.178 5−3)×−0.991+( 3.178−3 5−3)×−1.767] ×1.3−1.2661.3−1.2 +[( 5−3.178 5−3)×−0.481+( 3.178−3 5−3)×−1.147] ×1.266-1.21.3−1.2( S R )R1=−0.717
Now from equation (1) at final state
Z=Z0+ωZ1Z=0.614+0.152×0.164Z=0.639
And
V=ZRTPV=0.639×83.14 cm 3barmol K×468.15 K135 barV=184.2 cm3mol
From equation (2) at final state,
HR=(HR)0+ω(HR)1
( HR )0RTC=−2.497⇒(HR)0=−2.497×8.314Jmol K×369.8 K(HR)0=−7677.069Jmol
And
( H R )RTC1=−0.588⇒(HR)1=−0.588×8.314Jmol K×369.8 K(HR)1=−1807.816Jmol
So,
HR=(HR)0+ω(HR)1HR=−7677.069Jmol+0.152×−1807.816JmolHR=−7951.86Jmol
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=468.15308.15=1.519
Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:
i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0
∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫T0TΔC∘PRdT=1.213×308.15×(1.519−1)+28.785×10−32308.152(1.5192−1) +−8.824×10−63×308.153(1.5193−1)+0×105308.15(1.519−11.519)∫T0TΔC∘PRdT=1765.1286K
ΔH=R∫T1T2 C P igRdT+H2RΔH=8.314 JK mol×1765.1286K−7951.86JmolΔH=6723.42 Jmol
( SR )0R=−1.462⇒(SR)0=−1.462×8.314Jmol K(SR)0=−12.15Jmol K
And
( S R )R1=−0.717(SR)1=8.314Jmol K×−0.717(SR)1=−5.96Jmol K
So,
SR=(SR)0+ω(SR)1SR=−12.15Jmol K+0.152×−5.96Jmol KSR=−13.06Jmol 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=468.15308.15=1.519
Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:
i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0
∫T1T2 C P igRdTT=[A+{BT0+(CT02+D τ 2 T 0 2 )( τ+12)}(τ−1lnτ)]×lnτ∫T1T2CP igRdTT= [1.213+{28.785×10−3×308.15+(−8.824×10−6×308.152+0× 10 5 1.519 2× 308.15 2)(1.519+12)}(1.519−1ln1.519)] ×ln1.519∫T1T2CPigRdTT=4.563
Hence,
ΔS=R∫T1T2 C P igRdTT−Rln P 2 P 1+S2RΔS=8.314Jmol K×4.563−8.314Jmol K×ln135 bar1 bar+(−13.06Jmol K)ΔS=−15.905 Jmol K