Analyzing an Ideal Gas Mixture Undergoing Compression
A mixture of 0.3 lb of carbon dioxide and 0.2 lb of nitrogen is compressed from p_{1}=1 atm , T_{1}=540^{\circ} R \text { to } p_{2}=3 atm in a polytropic process for which n = 1.25. Determine (a) the final temperature, in °R, (b) the work, in Btu, (c) the heat transfer, in Btu, (d) the change in entropy of the mixture, in Btu/°R.
Known A mixture of 0.3 lb of CO _{2} and 0.2 lb of N _{2} is compressed in a polytropic process for which n = 1.25. At the initial state, p_{1}=1 atm , T_{1}=540^{\circ} R. At the final state, p_{2}=3 \text { atm }.
Find Determine the final temperature, in °R, the work, in Btu, the heat transfer, in Btu, and the entropy change of the mixture in, Btu/°R.
Schematic and Given Data:
Engineering Model
1. As shown in the accompanying figure, the system is the mixture of CO _{2} \text { and } N _{2}. The mixture composition remains constant during the compression.
2. The Dalton model applies: Each mixture component behaves as if it were an ideal gas occupying the entire system volume at the mixture temperature. The overall mixture acts as an ideal gas.
3. The compression process is a polytropic process for which n = 1.25.
4. The changes in kinetic and potential energy between the initial and final states can be ignored.
Analysis
a. For an ideal gas, the temperatures and pressures at the end states of a polytropic process are related by Eq. 3.56:
\frac{T_{2}}{T_{1}}=\left(\frac{p_{2}}{p_{1}}\right)^{(n-1) / n}=\left(\frac{V_{1}}{V_{2}}\right)^{n-1} (ideal gas) (3.56)
T_{2}=T_{1}\left(\frac{p_{2}}{p_{1}}\right)^{(n-1) / n}
Inserting values
T_{2}=540\left(\frac{3}{1}\right)^{0.2}=673^{\circ} R
b. The work for the compression process is given by
W=\int_{1}^{2} p d V
Introducing p V^{n}= constant and performing the integration
W=\frac{p_{2} V_{2}-p_{1} V_{1}}{1-n}
With the ideal gas equation of state, this reduces to
W=\frac{m(\bar{R} / M)\left(T_{2}-T_{1}\right)}{1-n}
The mass of the mixture is m = 0.3 + 0.2 = 0.5 lb. The apparent molecular weight of the mixture can be calculated using
M = m/n, where n is the total number of moles of mixture. With Eq. 12.1, the numbers of moles of CO _{2} \text { and } N _{2} are, respectively,
n_{i}=\frac{m_{i}}{M_{i}} (12.1)
n_{ CO _{2}}=\frac{0.3}{44}=0.0068 lbmol , \quad n_{ N _{2}}=\frac{0.2}{28}=0.0071 lbmol
The total number of moles of mixture is then n = 0.0139 lbmol. The apparent molecular weight of the mixture is M = 0.5/0.0139 = 35.97.
Calculating the work
W=\frac{(0.5 lb )\left(\frac{1545 ft \cdot lbf }{35.97 lb \cdot{ }^{\circ} R }\right)\left(673^{\circ} R -540^{\circ} R \right)}{1-1.25}\left|\frac{1 Btu }{778 ft \cdot lbf }\right|
= −14.69 Btu
where the minus sign indicates that work is done on the mixture, as expected.
c. With assumption 4, the closed system energy balance can be placed in the form
Q=\Delta U+W
where ΔU is the change in internal energy of the mixture.
The change in internal energy of the mixture equals the sum of the internal energy changes of the components. With Eq. 12.30
U_{2}-U_{1}=\sum_{i=1}^{j} n_{i}\left[\bar{u}_{i}\left(T_{2}\right)-\bar{u}_{i}\left(T_{1}\right)\right] (12.30)
1 \Delta U=n_{ CO _{2}}\left[\bar{u}_{ CO _{2}}\left(T_{2}\right)-\bar{u}_{ CO _{2}}\left(T_{1}\right)\right]+n_{ N _{2}}\left[\bar{u}_{ N _{2}}\left(T_{2}\right)-\bar{u}_{ N _{2}}\left(T_{1}\right)\right]
This form is convenient because Table A-23E gives internal energy values for N _{2} \text { and } CO _{2}, respectively, on a molar basis. With values from this table
\begin{aligned}\Delta U &=(0.0068)(3954-2984)+(0.0071)(3340-2678) \\&=11.3 Btu\end{aligned}
Inserting values for ΔU and W into the expression for Q
Q=+11.3-14.69=-3.39 Btu
where the minus sign signifies a heat transfer from the system.
d. The change in entropy of the mixture equals the sum of the entropy changes of the components. With Eq. 12.34
S_{2}-S_{1}=\sum_{i=1}^{j} n_{i}\left[\bar{s}_{i}\left(T_{2}, p_{i 2}\right)-\bar{s}_{i}\left(T_{1}, p_{i 1}\right)\right] (12.34)
\Delta S=n_{ CO _{2}} \Delta \bar{s}_{ CO _{2}}+n_{ N _{2}} \Delta \bar{s}_{ N _{2}}
where \Delta \bar{s}_{ N _{2}} \text { and } \Delta \bar{s}_{ CO _{2}} are evaluated using Eq. 12.36 and values of \bar{s}^{\circ} \text { for } N _{2} \text { and } CO _{2} from Table A-23E. That is,
\Delta \bar{s}_{i}=\bar{s}_{i}^{\circ}\left(T_{2}\right)-\bar{s}_{i}^{\circ}\left(T_{1}\right)-\bar{R} \ln \frac{p_{2}}{p_{1}} (12.36)
\Delta S=0.0068\left(53.123-51.082-1.986 \ln \frac{3}{1}\right)
2 +0.0071\left(47.313-45.781-1.986 \ln \frac{3}{1}\right)
= −0.0056 Btu/°R
Entropy decreases in the process because entropy is transferred from the system accompanying heat transfer.
1 In view of the relatively small temperature change, the changes in the internal energy and entropy of the mixture can be evaluated alternatively using the constant specific heat relations, Eqs. 12.37 and 12.39, respectively. In these equations, \bar{c}_{v} \text { and } \bar{c}_{p} are specific heats for the mixture determined using Eqs. 12.23 and 12.24 together with appropriate specific heat values for the components chosen from Table A-20E.
\Delta \bar{u}=\bar{c}_{v}\left(T_{2}-T_{1}\right), \quad \Delta \bar{u}_{i}=\bar{c}_{v, i}\left(T_{2}-T_{1}\right) (12.37)
\Delta \bar{s}=\bar{c}_{p} \ln \frac{T_{2}}{T_{1}}-\bar{R} \ln \frac{p_{2}}{p_{1}}, \quad \Delta \bar{s}_{i}=\bar{c}_{p, i} \ln \frac{T_{2}}{T_{1}}-\bar{R} \ln \frac{p_{2}}{p_{1}} (12.39)
\bar{c}_{v}=\sum_{i=1}^{j} y_{i} \bar{c}_{v, i} (12.23)
\bar{c}_{p}=\sum_{i=1}^{j} y_{i} \bar{c}_{p, i} (12.24)
2 Since the composition remains constant, the ratio of partial pressures equals the ratio of mixture pressures, so Eq. 12.36 can be used to evaluate the component specific entropy changes required here.
Skills Developed
Ability to…
• analyze a polytropic process of a closed system for a mixture of ideal gases.
• apply ideal gas mixture principles.
• determine changes in internal energy and entropy for ideal gas mixtures using tabular data.
Quick Quiz
Recalling that polytropic processes are internally reversible, determine for the system the amount of entropy transfer accompanying heat transfer, in Btu/°R. Ans. −0.0056 Btu/°R.
