Homework problem sets
Questions 7.1/7.2/7.7/7.15 and 9.3 specified problem 1 and 2 from J.R. Elliott, C.T. Lira, “Introductory Chemical
Engineering Thermodynamics”, 2nd ed.
Question 7.1
The compressibility factor chart provides a quick way to assess when the ideal gas law is valid. For the following
fluids, what is the minimum temperature in K where the fluid has a gas phase compressibility greater than 0.95
at 30 bar?
(a) Nitrogen
(b) Carbon dioxide
(c) Ethanol
On a generalized compressibility chart (e.g. Figure 7.4), we can read off the compressibility factors as a function
of the reduced pressure and the reduced temperature. We can find the properties of the various components
from the data on the inside backcover of the book:
Compound Tc (K) Pc (MPa)
Nitrogen 126.1 3.394 0.040
Carbon dioxide 304.2 7.382 0.228
Ethanol 516.4 6.384 0.637
The reduced pressure for these compounds at 30 bar can be determined
Compound Pr =30/Pc
Nitrogen 0.883913
Carbon dioxide 0.406394
Ethanol 0.469925
The overall compressibility factor can be found from
𝑍 = 𝑍 0 + 𝜔 ∙ 𝑍1
For nitrogen, at Pr=0.9 one need to go to reduced tempratures larger than 1.5 to obtain Z>0.95. Take T r = 2:
𝑍 = 0.95 + 0.044 ∙ 0.08 = 0.9532
Hence the temperature will need to be in excess of 252.2 K.
For carbon dioxide, the reduced pressure is only 0.41. Hence a reduced temperature larger than 1.1 (334 K)
should give the desired result .
Similarly, for ethanol with a reduced temperature larger than 1.2 (ca. 600 K)
,Question 7.2
A container having a volume of 40 liter contains one of the following fluids at the given initial conditions. After a
leak, the temperature and pressure are re-measured. For each option, determine the mass of fluid lost (in kg)
due to the leak, using
(a) Compressibility factor charts
(b) Peng-Robinson equation of state
For the following situations
Initial Final
Compound T, K p, bar T, K p, bar
Methane 300 100 300 50
Propane 300 50 300 0.9
n-Butane 300 50 300 10
Strategy:
𝑝∙𝑉
1. Calculate the number of moles initially and finally using 𝑛 = and using the molar weight
𝑍∙𝑅∙𝑇
to obtain the mass loss in kg
2. The compressibility factor can be obtained from Fig. 7.4 (knowing the reduced temperature
and pressure) or the Peng-Robinson equation of state.
3. Find critical properties of the compounds involved from the inside of the back-cover of the
textbook.
Compound M (g/mol) Tc (K) Pc (MPa)
Methane 16 190.6 4.604 0.011
Propane 44 369.8 4.249 0.152
n-butane 58 425.2 3.797 0.193
Compressibility chart
𝑇 𝑝
The reduced properties are given by 𝑇𝑟 = 𝑇 and 𝑝𝑟 = 𝑝
𝑐 𝑐
Initial Final
Compound Tr pr Tr pr
Methane 1.6 2.2 1.6 1.1
Propane 0.8 1.2 0.8 0.02
n-Butane 0.7 1.3 0.7 0.26
Going to Fig. 7.4 and reading off the graph:
Initial Final
Compound Z0 Z1 Z0 Z1
Methane 0.87 0.2 0.92 0.1
Propane 0.2 -0.05 1 0.0
n-Butane 0.2 -0.10 0.05 …..
(Note that the final conditions for n-butane is difficult to estimate in the intermediate pressure range for Tr=0.7
– this may indicate the presence of liquid in the vessel! The way to continue with n-butane is to assume that the
liquid phase hardly takes in any volume (its density is much higher or using the PR-EOS using the compressibility
of both the liquid and the vapour phase)
The overall compressibility factor can be found from
𝑍 = 𝑍 0 + 𝜔 ∙ 𝑍1
Compound Zinitial Zfinal
Methane 0.8722 0.9211
Propane 0.1924 1
n-Butane 0.1807
𝑝∙𝑉
Now solving for the number of moles using 𝑛 = 𝑍∙𝑅∙𝑇, e.g. for methane
, 𝑁⁄ 3
100∙𝑏𝑎𝑟∙105 ∙ 𝑚2 ∙40∙𝑙𝑖𝑡𝑒𝑟∙ 1∙𝑚
𝑏𝑎𝑟 1000∙𝑙𝑖𝑡𝑒𝑟
Initially: 𝑛𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑁∙𝑚 = 183.9 ∙ 𝑚𝑜𝑙
0.8722∙8.314∙ ∙300∙𝐾
𝑚𝑜𝑙∙𝐾
𝑁⁄
2 1∙𝑚3
50∙𝑏𝑎𝑟∙105 ∙ 𝑚 ∙40∙𝑙𝑖𝑡𝑒𝑟∙
𝑏𝑎𝑟 1000∙𝑙𝑖𝑡𝑒𝑟
Finally: 𝑛𝑓𝑖𝑛𝑎𝑙 = 𝑁∙𝑚 = 87.1 ∙ 𝑚𝑜𝑙
0.8722∙8.314∙ ∙300∙𝐾
𝑚𝑜𝑙∙𝐾
(a little less than half the initial amount – despite halving the pressure – due to the change in the compressibility
factor!)
Summarizing:
Compound ninitial, mol nfinal, mol n, mol m, kg
Methane 183.9 87.1 96.8 1.5
Propane 416.8 1.4 415.4 18.3
n-Butane 443.8
Using the Peng-Robinson equation of state:
𝑅∙𝑇 𝛼(𝑇)∙𝑎𝑐
𝑝 = 𝑉−𝑏 − 𝑉∙(𝑉+𝑏)+𝑏∙(𝑉−𝑏)
Can be re-written in terms of a cubic equation: 𝑍3 + 𝛼 ∙ 𝑍2 + 𝛽 ∙ 𝑍 + 𝛾 = 0
With 𝛼 = −1 + 𝐵; 𝛽 = A − 3B 2 − 2B; γ = −A ∙ B + B 2 + B 3
𝑎(𝑇)∙𝑝 𝑏∙𝑝
and 𝐴 = (𝑅𝑇)2 and 𝐵 = 𝑅𝑇
For methane
𝑎(𝑇) = 𝛼(𝑇) ∙ 𝑎𝑐
𝑁2 ∙𝑚2
𝑅 2 ∙𝑇𝑐 2 8.3142 ∙ ∙190.6∙𝐾 2 𝑃𝑎∙𝑚 6
𝑚𝑜𝑙2 ∙𝐾2
𝑎𝑐 = 0.457236 ∙ = 0.457236 ∙ 𝑁 = 0.2494 ∙
𝑝𝑐 4.604∙106 ∙ 2 𝑚𝑜𝑙2
𝑚
𝜅 = 0.37464 + 1.54226 ∙ 𝜔 − 0.26992 ∙ 𝜔 = 0.37464 + 1.54226 ∙ 0.011 − 0.26992 ∙ 0.0112 = 0.3916
2
2 2
𝑇 300
𝛼(𝑇) = [1 + 𝜅 ∙ (1 − √𝑇 )] = [1 + 0.3916 ∙ (1 − √190.16)] = 0.8208
𝑐
𝑃𝑎∙𝑚 6
𝑎(𝑇) = 𝛼(𝑇) ∙ 𝑎𝑐 = 0.8208 ∙ 0.2494 = 0.205 ∙
𝑚𝑜𝑙 2
𝑁∙𝑚
𝑅∙𝑇𝑐 8.314∙ ∙190.6∙𝐾 𝑚3
𝑏 = 0.077796 ∙ = 0.077796 ∙ 𝑚𝑜𝑙∙𝐾
𝑁 = 2.678 ∙ 10−5 ∙ 𝑚𝑜𝑙
𝑝𝑐 4.604∙106 ∙ 2
𝑚
Now the coefficients A and B can be calculated to solve the cubic equation of state (for each of the two
pressures):
At 1 bar at 200 bar
𝑁∙𝑚4 𝑁 𝑁∙𝑚4 𝑁
𝑎∙𝑝 0.205∙ ∙105 ∙ 2 𝑎∙𝑝 0.205∙ ∙200∙105 ∙ 2
𝑚𝑜𝑙2 𝑚 𝑚𝑜𝑙2 𝑚
𝐴 = (𝑅𝑇)2 = 𝑁∙𝑚 2 = 0.00345 𝐴 = (𝑅𝑇)2 = 𝑁∙𝑚 2 = 0.6892
(8.314∙ ∙293.15∙𝐾) (8.314∙ ∙293.15∙𝐾)
𝑚𝑜𝑙∙𝐾 𝑚𝑜𝑙∙𝐾
𝑚3 𝑁 𝑚3 𝑁
𝑏∙𝑝 2.678∙10−5 ∙ ∙105 ∙ 2 𝑏∙𝑝 2.678∙10−5 ∙ ∙200∙105 ∙ 2
𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 𝑚
𝐵= 𝑅𝑇
= 𝑁∙𝑚 = 0.0011 𝐵= 𝑅𝑇
= 𝑁∙𝑚 = 0.2197
8.314∙ ∙293.15∙𝐾 8.314∙ ∙293.15∙𝐾
𝑚𝑜𝑙∙𝐾 𝑚𝑜𝑙∙𝐾
This means that the cubic equation to solve will be
𝑍 3 − 0.9989 ∙ 𝑍 2 + 0.001244 ∙ 𝑍 − 2.58 ∙ 10−6 = 0 𝑍 3 − 0.7803 ∙ 𝑍 2 + 0.10485 ∙ 𝑍 − 0.09254 = 0
Knowing Z, the number of moles initially and finally can be calculated and thus the mass loss. This can all be put
into a spreadsheet:
, Compound Methane Propane n-butane
Tc K 190.6 369.8 425.2
pc MPa 4.604 4.249 3.797
0.011 0.152 0.193
T 300 300 300 300 300 300
p 1.00E+07 5.00E+06 5.00E+06 9.00E+04 5.00E+06 1.00E+06
ac Pa m6/mol 2 0.249385 1.017201735 1.504894047
k 0.391572 0.602827288 0.66224193
a(T) 0.810563 0.810563 1.12331273 1.12331273 1.223187976 1.223188
6 2
a(T) Pa m /mol 0.202142 0.202142 1.142635657 1.142635657 1.840768303 1.840768
b m3/mol 2.68E-05 5.62922E-05 7.24303E-05
A 0.324934 0.162467 0.918364796 0.016530566 1.479471427 0.295894
B 0.1074 0.0537 0.1128 0.0020 0.1452 0.0290
alpha -0.8926 -0.9463 -0.8872 -0.9980 -0.8548 -0.8548 -0.9710 -0.9710
beta 0.075647 0.046467 0.654469833 0.012455729 1.125829533 1.1258 0.235285 0.2353
gamma -0.02212 -0.00568 -0.08946264 -2.9443E-05 -0.19067212 -0.1907 -0.00772 -0.0077
p 100 bar 50 bar 50 bar 0.9 bar 50 bar 1 bar
Z estimate 1 1 1 1 1 0.0001 1 0.001
Z estimate 0.87532 0.918133 0.63947081 0.985777416 0.552874786 0.169387 0.801603 0.0331
Z estimate 0.83734 0.902356 0.334404972 0.985358641 0.243593858 0.190678 0.682763 0.038633
Z estimate 0.833766 0.901785 0.164003795 0.985358283 0.190285912 0.19084 0.622448 0.038794
Z estimate 0.833735 0.901785 0.167556167 0.985358283 0.190840195 0.19084 0.602858 0.038794
Z estimate 0.833735 0.901785 0.167567362 0.985358283 0.190840291 0.19084 0.600686 0.038794
Z estimate 0.833735 0.901785 0.167567362 0.985358283 0.190840291 0.19084 0.600659 0.038794
n, mol 192.4 88.9 478.5 1.5
n, mol 103.4 477.1
m, kg 1.7 21.0
First it can be noted that the two different methods for methane and propane do not differ much.
The Peng-Robinson equation of state for n-butane yields two solutions depending on the starting value for the
iteration. The starting value of Z=1 yields Z = 0.6 and the starting value of Z=0.001 yields Z = 0.039. The higher
value corresponds to the compressibility factor for the gas phase and the smaller value to the compressibility
factor for the liquid phase. For each phase, we can write:
𝑝∙𝑉 𝐿 𝑝∙𝑉 𝑉
𝑛𝐿 = 𝑍 𝐿∙𝑅∙𝑇 𝑛𝑉 = 𝑍 𝑉∙𝑅∙𝑇
The total number of moles in the vessel is given by:
𝑝 ∙ 𝑉𝐿 𝑝 ∙ 𝑉𝑉 𝑝 𝑉𝐿 𝑉𝑉
𝑛𝐿 + 𝑛𝑉 = 𝐿 + 𝑉 = ∙ ( 𝐿 + 𝑉)
𝑍 ∙𝑅∙𝑇 𝑍 ∙𝑅∙𝑇 𝑅∙𝑇 𝑍 𝑍
The total volume of the vessel is fixed:
𝑝 ∙ 𝑉𝐿 𝑝 ∙ 𝑉𝑉 𝑝 40 ∙ 𝑙𝑖𝑡𝑒𝑟𝑠 − 𝑉 𝑉 𝑉 𝑉
𝑛𝐿 + 𝑛𝑉 = 𝐿 + 𝑉 = ∙( + 𝑉)
𝑍 ∙𝑅∙𝑇 𝑍 ∙𝑅∙𝑇 𝑅∙𝑇 𝑍𝐿 𝑍
This equation has multiple solutions and cannot be solved without additional information on how the process
proceeded.
Questions 7.1/7.2/7.7/7.15 and 9.3 specified problem 1 and 2 from J.R. Elliott, C.T. Lira, “Introductory Chemical
Engineering Thermodynamics”, 2nd ed.
Question 7.1
The compressibility factor chart provides a quick way to assess when the ideal gas law is valid. For the following
fluids, what is the minimum temperature in K where the fluid has a gas phase compressibility greater than 0.95
at 30 bar?
(a) Nitrogen
(b) Carbon dioxide
(c) Ethanol
On a generalized compressibility chart (e.g. Figure 7.4), we can read off the compressibility factors as a function
of the reduced pressure and the reduced temperature. We can find the properties of the various components
from the data on the inside backcover of the book:
Compound Tc (K) Pc (MPa)
Nitrogen 126.1 3.394 0.040
Carbon dioxide 304.2 7.382 0.228
Ethanol 516.4 6.384 0.637
The reduced pressure for these compounds at 30 bar can be determined
Compound Pr =30/Pc
Nitrogen 0.883913
Carbon dioxide 0.406394
Ethanol 0.469925
The overall compressibility factor can be found from
𝑍 = 𝑍 0 + 𝜔 ∙ 𝑍1
For nitrogen, at Pr=0.9 one need to go to reduced tempratures larger than 1.5 to obtain Z>0.95. Take T r = 2:
𝑍 = 0.95 + 0.044 ∙ 0.08 = 0.9532
Hence the temperature will need to be in excess of 252.2 K.
For carbon dioxide, the reduced pressure is only 0.41. Hence a reduced temperature larger than 1.1 (334 K)
should give the desired result .
Similarly, for ethanol with a reduced temperature larger than 1.2 (ca. 600 K)
,Question 7.2
A container having a volume of 40 liter contains one of the following fluids at the given initial conditions. After a
leak, the temperature and pressure are re-measured. For each option, determine the mass of fluid lost (in kg)
due to the leak, using
(a) Compressibility factor charts
(b) Peng-Robinson equation of state
For the following situations
Initial Final
Compound T, K p, bar T, K p, bar
Methane 300 100 300 50
Propane 300 50 300 0.9
n-Butane 300 50 300 10
Strategy:
𝑝∙𝑉
1. Calculate the number of moles initially and finally using 𝑛 = and using the molar weight
𝑍∙𝑅∙𝑇
to obtain the mass loss in kg
2. The compressibility factor can be obtained from Fig. 7.4 (knowing the reduced temperature
and pressure) or the Peng-Robinson equation of state.
3. Find critical properties of the compounds involved from the inside of the back-cover of the
textbook.
Compound M (g/mol) Tc (K) Pc (MPa)
Methane 16 190.6 4.604 0.011
Propane 44 369.8 4.249 0.152
n-butane 58 425.2 3.797 0.193
Compressibility chart
𝑇 𝑝
The reduced properties are given by 𝑇𝑟 = 𝑇 and 𝑝𝑟 = 𝑝
𝑐 𝑐
Initial Final
Compound Tr pr Tr pr
Methane 1.6 2.2 1.6 1.1
Propane 0.8 1.2 0.8 0.02
n-Butane 0.7 1.3 0.7 0.26
Going to Fig. 7.4 and reading off the graph:
Initial Final
Compound Z0 Z1 Z0 Z1
Methane 0.87 0.2 0.92 0.1
Propane 0.2 -0.05 1 0.0
n-Butane 0.2 -0.10 0.05 …..
(Note that the final conditions for n-butane is difficult to estimate in the intermediate pressure range for Tr=0.7
– this may indicate the presence of liquid in the vessel! The way to continue with n-butane is to assume that the
liquid phase hardly takes in any volume (its density is much higher or using the PR-EOS using the compressibility
of both the liquid and the vapour phase)
The overall compressibility factor can be found from
𝑍 = 𝑍 0 + 𝜔 ∙ 𝑍1
Compound Zinitial Zfinal
Methane 0.8722 0.9211
Propane 0.1924 1
n-Butane 0.1807
𝑝∙𝑉
Now solving for the number of moles using 𝑛 = 𝑍∙𝑅∙𝑇, e.g. for methane
, 𝑁⁄ 3
100∙𝑏𝑎𝑟∙105 ∙ 𝑚2 ∙40∙𝑙𝑖𝑡𝑒𝑟∙ 1∙𝑚
𝑏𝑎𝑟 1000∙𝑙𝑖𝑡𝑒𝑟
Initially: 𝑛𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑁∙𝑚 = 183.9 ∙ 𝑚𝑜𝑙
0.8722∙8.314∙ ∙300∙𝐾
𝑚𝑜𝑙∙𝐾
𝑁⁄
2 1∙𝑚3
50∙𝑏𝑎𝑟∙105 ∙ 𝑚 ∙40∙𝑙𝑖𝑡𝑒𝑟∙
𝑏𝑎𝑟 1000∙𝑙𝑖𝑡𝑒𝑟
Finally: 𝑛𝑓𝑖𝑛𝑎𝑙 = 𝑁∙𝑚 = 87.1 ∙ 𝑚𝑜𝑙
0.8722∙8.314∙ ∙300∙𝐾
𝑚𝑜𝑙∙𝐾
(a little less than half the initial amount – despite halving the pressure – due to the change in the compressibility
factor!)
Summarizing:
Compound ninitial, mol nfinal, mol n, mol m, kg
Methane 183.9 87.1 96.8 1.5
Propane 416.8 1.4 415.4 18.3
n-Butane 443.8
Using the Peng-Robinson equation of state:
𝑅∙𝑇 𝛼(𝑇)∙𝑎𝑐
𝑝 = 𝑉−𝑏 − 𝑉∙(𝑉+𝑏)+𝑏∙(𝑉−𝑏)
Can be re-written in terms of a cubic equation: 𝑍3 + 𝛼 ∙ 𝑍2 + 𝛽 ∙ 𝑍 + 𝛾 = 0
With 𝛼 = −1 + 𝐵; 𝛽 = A − 3B 2 − 2B; γ = −A ∙ B + B 2 + B 3
𝑎(𝑇)∙𝑝 𝑏∙𝑝
and 𝐴 = (𝑅𝑇)2 and 𝐵 = 𝑅𝑇
For methane
𝑎(𝑇) = 𝛼(𝑇) ∙ 𝑎𝑐
𝑁2 ∙𝑚2
𝑅 2 ∙𝑇𝑐 2 8.3142 ∙ ∙190.6∙𝐾 2 𝑃𝑎∙𝑚 6
𝑚𝑜𝑙2 ∙𝐾2
𝑎𝑐 = 0.457236 ∙ = 0.457236 ∙ 𝑁 = 0.2494 ∙
𝑝𝑐 4.604∙106 ∙ 2 𝑚𝑜𝑙2
𝑚
𝜅 = 0.37464 + 1.54226 ∙ 𝜔 − 0.26992 ∙ 𝜔 = 0.37464 + 1.54226 ∙ 0.011 − 0.26992 ∙ 0.0112 = 0.3916
2
2 2
𝑇 300
𝛼(𝑇) = [1 + 𝜅 ∙ (1 − √𝑇 )] = [1 + 0.3916 ∙ (1 − √190.16)] = 0.8208
𝑐
𝑃𝑎∙𝑚 6
𝑎(𝑇) = 𝛼(𝑇) ∙ 𝑎𝑐 = 0.8208 ∙ 0.2494 = 0.205 ∙
𝑚𝑜𝑙 2
𝑁∙𝑚
𝑅∙𝑇𝑐 8.314∙ ∙190.6∙𝐾 𝑚3
𝑏 = 0.077796 ∙ = 0.077796 ∙ 𝑚𝑜𝑙∙𝐾
𝑁 = 2.678 ∙ 10−5 ∙ 𝑚𝑜𝑙
𝑝𝑐 4.604∙106 ∙ 2
𝑚
Now the coefficients A and B can be calculated to solve the cubic equation of state (for each of the two
pressures):
At 1 bar at 200 bar
𝑁∙𝑚4 𝑁 𝑁∙𝑚4 𝑁
𝑎∙𝑝 0.205∙ ∙105 ∙ 2 𝑎∙𝑝 0.205∙ ∙200∙105 ∙ 2
𝑚𝑜𝑙2 𝑚 𝑚𝑜𝑙2 𝑚
𝐴 = (𝑅𝑇)2 = 𝑁∙𝑚 2 = 0.00345 𝐴 = (𝑅𝑇)2 = 𝑁∙𝑚 2 = 0.6892
(8.314∙ ∙293.15∙𝐾) (8.314∙ ∙293.15∙𝐾)
𝑚𝑜𝑙∙𝐾 𝑚𝑜𝑙∙𝐾
𝑚3 𝑁 𝑚3 𝑁
𝑏∙𝑝 2.678∙10−5 ∙ ∙105 ∙ 2 𝑏∙𝑝 2.678∙10−5 ∙ ∙200∙105 ∙ 2
𝑚𝑜𝑙 𝑚 𝑚𝑜𝑙 𝑚
𝐵= 𝑅𝑇
= 𝑁∙𝑚 = 0.0011 𝐵= 𝑅𝑇
= 𝑁∙𝑚 = 0.2197
8.314∙ ∙293.15∙𝐾 8.314∙ ∙293.15∙𝐾
𝑚𝑜𝑙∙𝐾 𝑚𝑜𝑙∙𝐾
This means that the cubic equation to solve will be
𝑍 3 − 0.9989 ∙ 𝑍 2 + 0.001244 ∙ 𝑍 − 2.58 ∙ 10−6 = 0 𝑍 3 − 0.7803 ∙ 𝑍 2 + 0.10485 ∙ 𝑍 − 0.09254 = 0
Knowing Z, the number of moles initially and finally can be calculated and thus the mass loss. This can all be put
into a spreadsheet:
, Compound Methane Propane n-butane
Tc K 190.6 369.8 425.2
pc MPa 4.604 4.249 3.797
0.011 0.152 0.193
T 300 300 300 300 300 300
p 1.00E+07 5.00E+06 5.00E+06 9.00E+04 5.00E+06 1.00E+06
ac Pa m6/mol 2 0.249385 1.017201735 1.504894047
k 0.391572 0.602827288 0.66224193
a(T) 0.810563 0.810563 1.12331273 1.12331273 1.223187976 1.223188
6 2
a(T) Pa m /mol 0.202142 0.202142 1.142635657 1.142635657 1.840768303 1.840768
b m3/mol 2.68E-05 5.62922E-05 7.24303E-05
A 0.324934 0.162467 0.918364796 0.016530566 1.479471427 0.295894
B 0.1074 0.0537 0.1128 0.0020 0.1452 0.0290
alpha -0.8926 -0.9463 -0.8872 -0.9980 -0.8548 -0.8548 -0.9710 -0.9710
beta 0.075647 0.046467 0.654469833 0.012455729 1.125829533 1.1258 0.235285 0.2353
gamma -0.02212 -0.00568 -0.08946264 -2.9443E-05 -0.19067212 -0.1907 -0.00772 -0.0077
p 100 bar 50 bar 50 bar 0.9 bar 50 bar 1 bar
Z estimate 1 1 1 1 1 0.0001 1 0.001
Z estimate 0.87532 0.918133 0.63947081 0.985777416 0.552874786 0.169387 0.801603 0.0331
Z estimate 0.83734 0.902356 0.334404972 0.985358641 0.243593858 0.190678 0.682763 0.038633
Z estimate 0.833766 0.901785 0.164003795 0.985358283 0.190285912 0.19084 0.622448 0.038794
Z estimate 0.833735 0.901785 0.167556167 0.985358283 0.190840195 0.19084 0.602858 0.038794
Z estimate 0.833735 0.901785 0.167567362 0.985358283 0.190840291 0.19084 0.600686 0.038794
Z estimate 0.833735 0.901785 0.167567362 0.985358283 0.190840291 0.19084 0.600659 0.038794
n, mol 192.4 88.9 478.5 1.5
n, mol 103.4 477.1
m, kg 1.7 21.0
First it can be noted that the two different methods for methane and propane do not differ much.
The Peng-Robinson equation of state for n-butane yields two solutions depending on the starting value for the
iteration. The starting value of Z=1 yields Z = 0.6 and the starting value of Z=0.001 yields Z = 0.039. The higher
value corresponds to the compressibility factor for the gas phase and the smaller value to the compressibility
factor for the liquid phase. For each phase, we can write:
𝑝∙𝑉 𝐿 𝑝∙𝑉 𝑉
𝑛𝐿 = 𝑍 𝐿∙𝑅∙𝑇 𝑛𝑉 = 𝑍 𝑉∙𝑅∙𝑇
The total number of moles in the vessel is given by:
𝑝 ∙ 𝑉𝐿 𝑝 ∙ 𝑉𝑉 𝑝 𝑉𝐿 𝑉𝑉
𝑛𝐿 + 𝑛𝑉 = 𝐿 + 𝑉 = ∙ ( 𝐿 + 𝑉)
𝑍 ∙𝑅∙𝑇 𝑍 ∙𝑅∙𝑇 𝑅∙𝑇 𝑍 𝑍
The total volume of the vessel is fixed:
𝑝 ∙ 𝑉𝐿 𝑝 ∙ 𝑉𝑉 𝑝 40 ∙ 𝑙𝑖𝑡𝑒𝑟𝑠 − 𝑉 𝑉 𝑉 𝑉
𝑛𝐿 + 𝑛𝑉 = 𝐿 + 𝑉 = ∙( + 𝑉)
𝑍 ∙𝑅∙𝑇 𝑍 ∙𝑅∙𝑇 𝑅∙𝑇 𝑍𝐿 𝑍
This equation has multiple solutions and cannot be solved without additional information on how the process
proceeded.
