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Tags:
thermodynamics
gas law
behaviour of oxygen gas
several equations
Peng-Robinson equation
Peng-Robinson
experimental behaviour

Thermodynamics
Coursework
CHEMICAL ENGINEERING THERMODYNAMICS (3CET)
MOHAMMED AL-SHEREEM (BENG CHEMICAL ENGINEERING FT)
ID: 1599611
27/01/2019
Question 1
Plot the p-v-T diagram for oxygen, using an appropriate equation of state – identify critical regions
and places on the diagram.
Answer:
➢ Choosing and modelling an equation of state:
There are various equations of state that could model the behaviour of oxygen gas. With the
appropriate comparisons between the ideal gas law, Van der waals, Redlich-Kwong and the PengRobinson equations, it was deduced that the Peng-Robinson is the most suitable as it:
▪
▪
▪
▪
▪
Is well established within industries (Ashour, et al., 2011)
Is suitable for representing real gases such as the case for oxygen (Sandler, 2017)
Considers various essential physical properties such as vapour-liquid equilibria and
behaviours (Tsai & Chen, 1998).
𝑇 𝑃 𝑉
Includes reduced variables (𝑇𝑟 , 𝑃𝑟 , 𝑉𝑟 ) derived from dimensionless quantities (𝑇 , 𝑃 , 𝑉 ).
𝑐
𝑐
𝑐
Produces accurate results that are close to the real experimental behaviour (Dahm & Donald,
2014).
The Peng-Robinson equation is represented in (𝐸𝑞 1) below (Goodwin, et al., 2010):
𝑅𝑇
𝑎𝜑
𝑃=
−
(𝐸𝑞 1)
𝑉𝑚 − 𝑏 𝑉𝑚 (𝑉𝑚 + 𝑏) + 𝑏(𝑉𝑚 − 𝑏)
Where 𝑃 is pressure (𝑏𝑎𝑟), 𝑅 is the ideal gas constant (𝐽/𝑚𝑜𝑙 ∙ 𝐾), 𝑇 is the temperature (𝐾), 𝑉𝑚 is
the molar volume (𝑚3 /𝑚𝑜𝑙) and 𝑎, 𝑏 and 𝜑 are shown in (𝐸𝑞 1.1, 1.2 𝑎𝑛𝑑 1.3 ) below.
𝑎 = 0.45724
𝑅 2 𝑇𝑐 2
(𝐸𝑞 1.1)
𝑃𝑐
𝑏 = 0.07780
𝑅𝑇𝑐
(𝐸𝑞 1.2)
𝑃𝑐
𝜑 = [1 + (1 − 𝑇𝑟 0.5 )(0.37464 + 1.54226𝜔 − 0.26992𝜔2 )]2 (𝐸𝑞 1.3)
In which 𝑇𝑐 and 𝑃𝑐 are the critical temperature and pressure respectively, 𝑇𝑟 is the reduced
temperature and 𝜔 is the acentric factor.
➢ Plotting a pressure-volume-temperature (P-V-T) diagram
(𝐸𝑞 1) can be used to plot the (P-V-T) diagram for oxygen. Firstly, a set of thermodynamic
parameters must be defined for oxygen as shown in Table 1.
Table 1: Thermodynamic parameters for oxygen
Thermodynamic parameter
Critical Temperature
Critical Pressure
Ideal gas constant
Acentric factor
Corresponding value
154.55 (𝐾)
50.8 (𝑏𝑎𝑟)
83.14 (𝑐𝑚3 ∙ 𝑏𝑎𝑟/𝑚𝑜𝑙 ∙ 𝐾)
0.0226
Accordingly, by using MATLAB and (𝐸𝑞 1, 1.1, 1.2 𝑎𝑛𝑑 1.3 ), a set of isotherms where plotted at
temperatures below, at and above the critical temperature 𝑇𝑐 as shown in Figure 1. The MATLAB
script can be found in Appendix A.
The critical (inflection) point where:
𝑇𝑐 = 154.55 𝐾
𝑃𝑐 = 50.8 𝑏𝑎𝑟
𝑉𝑚 𝑐 = 74.7 𝑐𝑚3 /𝑚𝑜𝑙
Figure 1: The oxygen P-V-T diagram as drawn in MATLAB where the critical point is illustrated
As can been seen in Figure 1, the isotherms with 𝑇 > 𝑇𝑐 show an inverse relationship between
pressure and volume. Alternatively, for the isotherms where 𝑇 < 𝑇𝑐 , the relationship varies along
𝑑𝑃
the different molar volumes. In both cases 𝑑𝑉 ≠ 0, however at the critical point, it is noticeable that
𝑑𝑃
there is an inflection where 𝑑𝑉 = 0.
➢ Adding the oxygen saturation curve
In order to identify the different thermodynamic regions, the oxygen saturation curve has to be
added to the P-V-T diagram. By using the data for the range of pressures and corresponding
molar volumes as shown in Appendix A (NIST, 2018), the saturation curve can be added to the
original plot as shown in Figure 2 where the various regions are clearly shown.
The compressed liquid region
The critical point
The superheated vapour
region
The saturated vapour line
The saturated liquid line
Figure 2: The different thermodynamic regions for the oxygen saturation curve in the P-V-T diagram
➢ Adding the horizontal equilibrium lines
According to Figure 1, the isotherms in which 𝑇 > 𝑇𝑐 represent one root where those in which 𝑇 <
𝑇𝑐 represent real roots. Consequently, Maxwell constructs can be utilized for the isotherms below
the critical point in which horizontal equilibrium lines can be obtained. Another method of
constructing the horizontal lines is through the calculation of 𝑃𝑠𝑎𝑡 (saturation pressure) for the
isotherms with 𝑇 < 𝑇𝑐 . 𝑃𝑠𝑎𝑡 can be obtained from (𝐸𝑞 2) below (Green, 2008).
𝑃𝑆𝑎𝑡 = 𝑒𝑥𝑝
𝐶1+(
𝐶2
)+𝐶3 ln(𝑇)+𝐶4𝑇 𝐶5
𝑇
(𝐸𝑞 1.4)
In which the coefficients for oxygen are (Green, 2008):
𝐶1 = 51.245
𝐶2 = −1200.2
𝐶3 = −6.4361
𝐶4 = 2.8405 ∗ 10−2
𝐶5 = 1
Thus, the saturation pressures at 𝑇 = 130 𝐾 and 𝑇 = 145 𝐾 are 𝑃 = 17.54 𝑏𝑎𝑟 and 𝑃 = 34.56 𝑏𝑎𝑟
respectively. Accordingly, these values were added to the original plot, thus obtaining horizontal
equilibrium lines as shown in Figure 3. It is also noticeable that the horizontal lines show no change
in pressure during the phase changes.
Saturated liquid-vapour
region at 130 𝐾
(34.56 𝑏𝑎𝑟)
Quality region
Saturated liquid-vapour
region at 130 𝐾
(17.54 𝑏𝑎𝑟)
Figure 3: The P-V-T diagram with the horizontal equilibrium lines added at the saturation pressures
Question 2
Show the range of temperatures that can be achieved by burning acetylene under various conditions.
Adiabatic flame temperature calculation
Answer:
In the case where a fuel combusts without any heat loss (adiabatic), the final mixture of gases
reaches a maximum temperature called the adiabatic flame temperature (Strahle, 1993). The flame
temperature depends on various factors and conditions that will be assessed in this section.
➢ Adiabatic combustion at constant pressure
Under constant pressure, some of the energy is lost to expansion work, however no heat is lost to
the surroundings (Strahle, 1993).
At constant pressure, the first law of thermodynamics is as follows
∆𝑈 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑃∆𝑉 (𝐸𝑞 2)
Where 𝑈 is the internal energy (𝐽), 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 is the heat at constant pressure (𝐽) and 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 is the
work at constant pressure (𝐽). Further, enthalpy, by definition is
𝐻 = 𝑈 + 𝑃𝑉 (𝐸𝑞 2.1)
Where 𝐻 is the enthalpy (𝐽), thus, when there is a change in enthalpy at constant pressure, (𝐸𝑞 2.1)
becomes
∆𝐻 = ∆𝑈 + 𝑃∆𝑉 + 𝑉∆𝑃 = ∆𝑈 + 𝑃∆𝑉 (𝐸𝑞 2.1.1)
For an adiabatic process
𝑄𝐶𝑜𝑛𝑠𝑡𝑃 = 0 𝑎𝑛𝑑 ∆𝐻 = 0
At equilibrium, the energy balance becomes
∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 (𝑇, 𝑃) (𝐸𝑞2.2)
𝑇𝑓
∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 (𝐸𝑞2.2.1)
𝑇𝑖
Where 𝑛𝑖 is the number of moles of the 𝑖 𝑡ℎ species, 𝑇𝑖 is the initial temperature (𝐾), 𝑇𝑓 is the final
temperature (𝐾), 𝐶𝑃𝑖 is the specific heat capacity of the 𝑖 𝑡ℎ species (𝐽/𝑚𝑜𝑙 ∙ 𝐾) and 𝑄𝑉𝐿 is the total
latent heat required in the case of phase changes.
➢ Adiabatic combustion at constant volume
At constant volume, there is no shaft work (no expansion), thus the adiabatic flame temperature is
higher than that at constant pressure. From the first law of thermodynamics (Strahle, 1993).
𝑄𝐶𝑜𝑛𝑠𝑡𝑉 = ∆𝑈,
𝑎𝑛𝑑 𝑄𝐶𝑜𝑛𝑠𝑡𝑉 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑛𝑅𝑇 = ∆𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠−𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∆𝑛𝑅𝑇
For adiabatic conditions
∆𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠−𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∆𝑛𝑅𝑇 = 0
Thus, (𝐸𝑞2.2.1) becomes
𝑇𝑓
∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 − 𝑅(𝑛𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑇𝑖 − 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 𝑇𝑓 ) (𝐸𝑞 2.2.2)
𝑇𝑖
Where 𝑛𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 and 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 is the number of moles of the reactants and products/inerts
respectively.
➢ Calculations
Before carrying out energy balances, the specific heat capacities must be determined. The heat
capacity varies dramatically as combustion results in a large range of temperatures, thus it is
necessary to obtain 𝐶𝑃 functions that could represent a more suitable heat capacity at the different
temperatures. Consequently, (𝐸𝑞 2.3) is utilised in which the coefficients A, B, C and D are specific
for the different gases (Smith, 2005).
𝐶𝑃𝑖 = 𝑅 ∗ [𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + 𝐷𝑇 −2 ] (𝐸𝑞 2.3)
Thus, by using the heats of formation, latent heats and sensible heats, different conditions were
examined for the resulting final adiabatic temperatures.
Combustion in pure oxygen
A complete combustion with pure of oxygen was examined in which exact stoichiometric amount
were considered. The reaction is as follows.
5
𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 𝐻2 𝑂(𝑙) (𝐸𝑞 2.4)
2
The energy balance is shown in Table 2
Table 2: Energy balance for the combustion in pure oxygen at constant pressure and constant volume
Constant
pressure
Constant
volume
∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙)
−1.3 × 106
𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙)
4.1 × 104
𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙)
1.2 × 106
∆𝑛𝑅𝑇
N/A
𝑇𝑓 (𝐾)
6179
−1.3 × 106
4.1 × 104
1.4 × 106
−1.6 × 105
6690
Where 𝑄𝐶𝑝 is the total sensible heat as per (𝐸𝑞 2.3) and the integral in (𝐸𝑞2.2.1). By examining
Table 2, it is clear to see that the temperature reached is unrealistic for both constant volume and
pressure. In a real system, the product gases will decompose thermally well below 6000K , thus this
is an invalid calculation as the kinetics must be considered for a realistic result for pure oxygen. Also,
due to the final temperature being different in the cases for constant volume and constant pressure,
𝑄𝐶𝑝 is different between them. This difference is present in all conditions. Also, due to the
∆𝑛𝑅𝑇 term in the constant volume calculation, the final temperature is higher than that of constant
pressure.
Combustion in air
Here, a complete combustion is modelled with exact stoichiometries, however the combustion now
occurs in air in which the compositions of oxygen and nitrogen gases are 21% and 79% respectively.
The resulting temperatures is shown in Table 3.
Table 3: Energy balance for the combustion in air at constant pressure and constant volume
Constant
pressure
Constant
volume
∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙)
−1.3 × 106
𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙)
4.1 × 104
𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙)
1.2 × 106
∆𝑛𝑅𝑇
N/A
𝑇𝑓 (𝐾)
2848
−1.3 × 106
4.1 × 104
1.6 × 106
−3.4 × 105
3408
The energy balance is consistent between Tables 2 and 3 as the reaction is exactly the same fro
constant pressure. The only difference is the final temperature due to the presence of significant
amount of nitrogen which contributed to the sensible heat. Alternatively, at constant volume, 𝑄𝐶𝑝
and ∆𝑛𝑅𝑇 are different. The temperatures calculated are much more realistic in this model.
Combustion in 50% excess air
In this case, an excess amount of air is included for the combustion. The resulting temperatures is
shown in Table 4.
Table 4: Energy balance for the combustion in excess air at constant pressure and constant volume
Constant
pressure
Constant
volume
∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙)
−1.3 × 106
𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙)
4.1 × 104
𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙)
1.2 × 106
∆𝑛𝑅𝑇
N/A
𝑇𝑓 (𝐾)
2294
−1.3 × 106
4.1 × 104
1.6 × 106
−3.9 × 105
2806
Again the energy balance is consistent in constant pressure, but different in constant volume. The
further reduction in temperature is due to the unreacted air (excess).
Incomplete combustion in air producing only carbon monoxide and water
An incomplete combustion is modelled for the combustion at stoichiometric amounts as shown in
(𝐸𝑞 2.5)
3
𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑙) (𝐸𝑞 2.5)
2
The energy balance for the combustion is presented in Table 5
Table 5: Energy balance for the incomplete combustion in air (only CO) at constant pressure and constant volume
Constant
pressure
Constant
volume
∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙)
−7.3 × 105
𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙)
4.1 × 104
𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙)
6.9 × 105
∆𝑛𝑅𝑇
N/A
𝑇𝑓 (𝐾)
2541
−7.3 × 105
4.1 × 104
9.1 × 105
−2.2 × 105
3118
It is clear to see that the ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 is lower since the reaction now is different. Accordingly the final
temperatures are now lower than for the combustion in air (Table 3) since carbon monoxide is
produced instead (which has a lower heat of formation).
Incomplete combustion in air producing carbon monoxide, soot (solids) and water
Finally, the incomplete combustion is modelled in which solid carbon (soot) is also produced, thus
altering the stoichiometries.
𝐶2 𝐻2 (𝑔) + 𝑂2 (𝑔) → 𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑙) + 𝐶(𝑠) (𝐸𝑞 2.6)
Table 6: Energy balance for the incomplete combustion in air (CO and C) at constant pressure and constant volume
Constant
pressure
Constant
volume
∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝐽/𝑚𝑜𝑙)
−6.2 × 105
𝑄𝑉𝐿 (𝐽/𝑚𝑜𝑙)
4.1 × 104
𝑄𝐶𝑝 (𝐽/𝑚𝑜𝑙)
5.8 × 105
∆𝑛𝑅𝑇
N/A
𝑇𝑓 (𝐾)
2754
−6.2 × 105
4.1 × 104
7.7 × 105
−2.2 × 105
3364
As seen in Table 6, ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 is even lower than in Table 5 since less overall chemical compounds
are produced (formation of solid carbon produces no energy). Also, it is noticeable that 𝑄𝐶𝑝 is lower
in this case as less gaseous products are present, thus requiring less sensible energy, and thus
leading to higher adiabatic flame temperatures at the end.
Question 3
The pressure of propane vs. density at 400 K can be fitted by the following expression:
𝑃 = 33.258𝜌 − 7.588𝜌2 + 1.0306𝜌3 − 0.0588𝜌4 − 0.0034𝜌5 + 0.0006𝜌6
Where the units of P and are bar and mol L-1 respectively.
Use the Van der Waals equation and the m to calculate the pressure for = 0 to 12.3 mol L-1, and
plot your results. Discuss the comparison of your results to the above expression.
Answer:
➢ Virial equation of state (From the question)
The virial equations of state are expressions that incorporate temperature, volume and pressure
that could illustrate the behaviour of gases during expansion/compression. They are constructed
from empirically from experimental data. Consequently, this is the most suitable mathematical
expression that would illustrate the behaviour of gaseous propane at 400K (Plischke, 2005).
In order to plot the pressure from the expression given in the question, the relationship between
molar density and molar volume must be defined as in (Eq 3) where:
𝜌=
1
𝑉𝑚
Where 𝑉𝑚 is the molar volume in 𝐿/𝑚𝑜𝑙. Accordingly a set of molar densities can be obtained from
the set of volumes given in the questions. This results in a range of pressures than can be plotted vs
volume.
➢ Van der waals equation of state
This is the first practical equation of state in which compound specific coefficients are added to the
ideal gas equation in order to model a real gas behaviour. The van der Waals equation is shown in
(𝐸𝑞 3) (Dahm & Donald, 2014).
𝑃=
𝑅𝑇
𝑎
−
(𝐸𝑞 3)
𝑉𝑚 − 𝑏 𝑉𝑚 2
𝐿
Where 𝑎 = 8.779 (𝐿2 ∙ 𝑏𝑎𝑟/𝑚𝑜𝑙 2 ), 𝑏 = 0.08445 (𝑚𝑜𝑙) and 𝑅 = 8.314 ∗ 10−3 (𝐿 ∙ 𝐵𝑎𝑟/𝐾 ∙ 𝑚𝑜𝑙).
𝑎
Moreover, 𝑉
𝑚
2
accounts for the attractive forces and 𝑏 accounts for the fact that molecules have a
finite size (Israelachvili, 2011).
➢ Redlich-Kwong equation of state
The Redlich-Kwong equation of state was a successful improvement on the van der Waals equation.
The improvements were possible due to the empirical findings of the temperature dependency on
the attraction parameter with little attention on modifying the parameter b (Tuckerman & Adewumi,
2016). The equation is represented in (𝐸𝑞 3.1) while the expressions for the coefficients 𝑎 and 𝑏
isare shown in (𝐸𝑞 3.1.1) and (𝐸𝑞 3.1.2) respectively.
𝑃=
𝑅𝑇
𝑎
−
(𝐸𝑞 3.1)
𝑉𝑚 − 𝑏 √𝑇𝑉𝑚 (𝑉𝑚 + 𝑏)
𝑎 = 0.42780
𝑅 2 𝑇𝑐 2.5
(𝐸𝑞 3.1.1)
𝑃𝑐
𝑏 = 0.086640
𝑅𝑇𝑐
(𝐸𝑞 3.1.2)
𝑃𝑐
➢ Plotting the equations of state
Figure 4 shows the equations of state plotted for pressure vs molar volume. Similarly, having stated
that the virial model is the most accurate of the three, it is possible to it as a reference point, and
thus Figure 5 compares the behaviour of van der Waals and Redlich-Kwong and the virial equation of
state.
Figure 4: A pressure vs volume lot for the three equations of state
Figure 5: A comparison between the virial equation of state to the van der Waals and Redlich-Kwong
𝑚𝑜𝑙
),
𝐿
According to Figure 4, all three models show very similar results up to about 5 (
which is also
evident in Figure 5. This is due to the fact that as pressure increases at constant temperature, the
compressibility factor 𝑍 =
𝑃𝑉
𝑛𝑅𝑇
increases. The compressibility factor ultimately explains how ideal a
real gas is, thus at 𝑍 = 1, a real gas could be modelled as ideal, otherwise it is necessary to consider
the gas as a real gas (Rao, 2010). Accordingly, due to the increasing compressibility factor, the van
𝑚𝑜𝑙
).
𝐿
der Waals and Redlich-Kwong models deviate from the Virial at around 5 (
Moreover, it is
noticeable in Figure 4 that the van der Waals model deviates dramatically from both the RedlichKwong and Virial models. This is more clearly illustrated in Figure 5 in which the deviation grows
exponentially between the van der Waals and Virial models, whereas the Redlich-Kwong shows a
maximum deviation of about 58 (𝑏𝑎𝑟). The increased accuracy of the Redlich-Kwong equation is
due to the modification on the temperature reliance of the attractive forces. Therefore, in the case
where the virial equation is not available, the use of the Redlich-Kwong equation acceptable to some
extent. Finally, there are more complex, yet more accurate models such as the Soave modification of
Redlich-Kwong, or even the different forms of the Peng-Robinson equation that could produce more
suitable results for real gases (Goodwin, et al., 2010).
Question 4
Discuss the use of electrochemical equilibrium measurements in thermodynamics, including relevant
derivations and results and illustrative examples where appropriate.
Answer:
➢ Electrochemical processes
Electrochemical processes occur when electrons flow from one chemical substance to another, with
the driving force being the oxidation-reduction reactions (redox). Since all redox reactions occur with
both oxidation and reduction, it is common to describe a redox reaction as two half reactions. For
example, a typical electrochemical reaction is the redox reaction between Zinc and Bromine as
shown in (𝐸𝑞 4.8) (Br ...

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Running Head: THERMODYNAMICS REPORT

Thermodynamics Report

Student’s Name

Institution Affiliation

1

THERMODYNAMICS REPORT

2

Thermodynamics Report

Q1

The model demonstrating the behaviour of oxygen gas can be illustrated by several equations.

These include the Peng-Robinson equation, Van der Waals, Ideal gas law and the Redlichkwong. Out of the four equations, the Peng-Robinson was found to be the most useful because it

is appropriate in representing real gases such as oxygen (Sandler, 2017). As opposed to the other

methods, the equation is also well established within manufacturing industries (Ashour et,al

2011. According to Tsai & Chen, the method also covers several vital physical characteristics,

hich include vapour-liquid equilibriums (1998). As such, it illustrates reduced variables, which

are a result of deriving dimensionless quantities. For instance, (𝑇𝑟 , 𝑃𝑟 , 𝑉𝑟) has been derived from

𝑇

𝑃

𝑉

𝑐

𝑐

𝑐

(𝑇 , 𝑃 , 𝑉 ). Therefore, Dam & Donald assert that the method guarantees precise results similar to

the actual experimental behaviour (2014).

According to Goodwin, et al, The Peng-Robinson equation can be illustrated as follows (2010)

𝑃=

𝑅𝑇

𝑎𝜑

−

(𝐸𝑞 1)

𝑉𝑚 − 𝑏 𝑉𝑚 (𝑉𝑚 + 𝑏) + 𝑏(𝑉𝑚 − 𝑏)

In this case, P represents Pressure,

R is the constant representing the ideal gas

T stands for temperature

Vm stands for molar volume

a represents 0.45724

b stands for 0.07780

𝑅2 𝑇𝑐 2

𝑅𝑇𝑐

𝑃𝑐

𝑃𝑐

(𝐸𝑞 )

𝜑 = [1 + (1 − 𝑇𝑟 0.5 )(0.37464 + 1.54226𝜔 − 0.26992𝜔2 )]2

THERMODYNAMICS REPORT

3

Here Tc represents the critical temperature, while P c stands for Pressure. Tr and 𝜔 represesent the

reduced temperature and the acentric factor respectively.

The above equations were then used in plotting the P-V-T Diagram after the identiiction of the

thermodynamic properties of oxygen. Additionally, a MATLAB Code was established to help in

the plotting of isotherms sets for various molar volumes and temperature ranges. The values for

the temperature range was at the critical temperature, above it or below it (𝑇 ≤ Tc ≤ 𝑇).

The following is the data representation in the form of a table

Thermodynamic properties of

Values

Oxygen

The

The gas constant (R)

83.14 cm3 bar/mol K

Critical temperature (Tc )

154.60 K

Critical Pressure ( Pc )

50.80 bar

Acentric factor (𝜔)

0.02220

graphical representation of

Oxygen Isotherms on a P-V

diagram ranging on the

temperatures of 110, 140, 154, 180, and 210 are demonstrated as follows.

THERMODYNAMICS REPORT

4

Oxygen isotherm at 154 K is shown as follows with the plotting of equilibrium lines against the

line of saturation and isotherms of oxygen at the critical point.

The complete P-V-T diagram for the five oxygen isotherms at varied temperatures can be shown

as follows.

THERMODYNAMICS REPORT

5

Material

C1

C2

C3

C4 (1E-02)

C5

Oxygen

51.2450

-1200.20

-6.43610

2.84050

1.0

Q2

According to Strahle (1993), in scenarios where fuel burns without any loss of heat, the resulting

mixture of gases gets to the optimum temperature known as adiabatic flame temperature.

However, the flam temperature varies, depending on several factors or conditions. The

following are some of the points for consideration.

•

Adiabatic combustion when the pressure is constant

Strahle further argues that when the pressure is constant, a certain amount of energy is lost to

expansion processes, but there is no heat loss to the surroundings. Therefore, the first law of

thermodynamics applies in the following scenario

∆𝑈 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑃∆𝑉 (𝐸𝑞 2)

Here, 𝑈 represents internal energy

𝑄𝐶𝑜𝑛𝑠𝑡𝑃 Represents the heat at constant pressure

THERMODYNAMICS REPORT

6

𝑊𝐶𝑜𝑛𝑠𝑡𝑃 Stands for work at constant pressure

Enthalpy (H) is equal to 𝑈 + 𝑃𝑉

Where, 𝐻 refers to the enthalpy. Therefore, any change in the enthalpy when the pressure is

constant results to the following

∆𝐻 = ∆𝑈 + 𝑃∆𝑉 + 𝑉∆𝑃 = ∆𝑈 + 𝑃∆𝑉

The representation of the adiabatic process is

𝑄𝐶𝑜𝑛𝑠𝑡𝑃 = 0 𝑎𝑛𝑑 ∆𝐻 = 0

However, at equilibrium, the balance of energy is

∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 (𝑇, 𝑃) (𝐸𝑞2.2)

𝑇𝑓

∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 (𝐸𝑞2.2.1)

𝑇𝑖

In this case, 𝑛𝑖 stands for moles number of 𝑖 𝑡ℎ species,

𝑇𝑖 Stands for initial temperature 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 (𝐾)

𝑇𝑓 Illustrates final temperature

THERMODYNAMICS REPORT

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