Known Nitrogen and oxygen, initially separate at different temperatures and pressures, are allowed to mix without heat or work interactions with the surroundings until a final equilibrium state is attained.
Find Determine the final temperature of the mixture, in °R, the final pressure of the mixture, in atm, and the amount of entropy produced in the mixing process, in Btu/°R.
Schematic and Given Data:
Engineering Model
1. The system is taken to be the nitrogen and the oxygen together.
2. When separate, each of the gases behaves as an ideal gas.
3. The final mixture acts as an ideal gas and the Dalton model applies: Each mixture component occupies the total volume and exhibits the mixture temperature.
4. No heat or work interactions occur with the surroundings, and there are no changes in kinetic and potential energy.
Analysis
a. The final temperature of the mixture can be determined from an energy balance. With assumption 4, the closed system energy balance reduces to
\Delta U=Q^{\nearrow0}- W^{\nearrow0} \quad \text { or } \quad U_{2}-U_{1}=0
The initial internal energy of the system, U_{1}, equals the sum of the internal energies of the two gases when separate
U_{1}=n_{ N _{2}} \bar{u}_{ N _{2}}\left(T_{ N _{2}}\right)+n_{ O _{2}} \bar{u}_{ O _{2}}\left(T_{ O _{2}}\right)
where T_{ N _{2}}=460^{\circ} R is the initial temperature of the nitrogen and T_{ O _{2}}=540^{\circ} R is the initial temperature of the oxygen. The final internal energy of the system, U_{2}, equals the sum of the internal energies of the two gases evaluated at the final mixture temperature T_{2}
U_{2}=n_{ N _{2}} \bar{u}_{ N _{2}}\left(T_{2}\right)+n_{ O _{2}} \bar{u}_{ O _{2}}\left(T_{2}\right)
Collecting the last three equations
n_{ N _{2}}\left[\bar{u}_{ N _{2}}\left(T_{2}\right)-\bar{u}_{ N _{2}}\left(T_{ N _{2}}\right)\right]+n_{ O _{2}}\left[\bar{u}_{ O _{2}}\left(T_{2}\right)-\bar{u}_{ O _{2}}\left(T_{ O _{2}}\right)\right]=0
The temperature T_{2} can be determined using specific internal energy data from Table A-23E and an iterative procedure like that employed in part (a) of Example 12.4. However, since the specific heats of N _{2} \text { and } O _{2} vary little over the temperature interval from 460 to 540°R, the solution can be conducted accurately on the basis of constant specific heats. Hence, the foregoing equation becomes
n_{ N _{2}} \bar{c}_{v, N _{2}}\left(T_{2}-T_{ N _{2}}\right)+n_{ O _{2}} \bar{c}_{v, O _{2}}\left(T_{2}-T_{ O _{2}}\right)=0
Solving for T_{2}
T_{2}=\frac{n_{ N _{2}} \bar{c}_{v, N _{2}} T_{ N _{2}}+n_{ O _{2}} \bar{c}_{v, O _{2}} T_{ O _{2}}}{n_{ N _{2}} \bar{c}_{v, N _{2}}+n_{ O _{2}} \bar{c}_{v, O _{2}}}
Selecting c_{v} values for N _{2} \text { and } O _{2} from Table A-20E at the average of the initial temperatures of the gases, 500°R, and using the respective molecular weights to convert to a molar basis
\bar{c}_{v, N _{2}}=\left(28.01 \frac{ lb }{ lbmol }\right)\left(0.177 \frac{ Btu }{ lb \cdot{ }^{\circ} R }\right)=4.96 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }
\bar{c}_{v, O _{2}}=\left(32.0 \frac{ lb }{ lbmol }\right)\left(0.156 \frac{ Btu }{ lb \cdot{ }^{\circ} R }\right)=4.99 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }
Substituting values into the expression for T_{2}
(0.79 lbmol )\left(4.96 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }\right)\left(460^{\circ} R \right)
T_{2}=\frac{+(0.21 lbmol )\left(4.99 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }\right)\left(540^{\circ} R \right)}{(0.79 lbmol )\left(4.96 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }\right)}
+(0.21 lbmol )\left(4.99 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }\right)
= 477°R
b. The final mixture pressure p_{2} can be determined using the ideal gas equation of state, p_{2}=n \bar{R} T_{2} / V, where n is the total number of moles of mixture and V is the total volume occupied by the mixture. The volume V is the sum of the volumes of the two tanks, obtained with the ideal gas equation of state as follows:
V=\frac{n_{ N _{2}} \bar{R} T_{ N _{2}}}{p_{ N _{2}}}+\frac{n_{ O _{2}} \bar{R} T_{ O _{2}}}{p_{ O _{2}}}
where p_{ N _{2}}=2 atm is the initial pressure of the nitrogen and p_{ O _{2}}=1 atm is the initial pressure of the oxygen. Combining results and reducing
p_{2}=\frac{\left(n_{ N _{2}}+n_{ O _{2}}\right) T_{2}}{\left(\frac{n_{ N _{2}} T_{ N _{2}}}{p_{ N _{2}}}+\frac{n_{ O _{2}} T_{ O _{2}}}{p_{ O _{2}}}\right)}
Substituting values
p_{2}=\frac{(1.0 lbmol )\left(477^{\circ} R \right)}{\left[\frac{(0.79 lbmol )\left(460^{\circ} R \right)}{2 atm }+\frac{(0.21 lbmol )\left(540^{\circ} R \right)}{1 atm }\right]}
= 1.62 atm
c. Reducing the closed system form of the entropy balance
S_{2}-S_{1}=\int_{1}^{2}\left(\frac{\delta Q}{T}\right)_{ b }^{\nearrow0}+\sigma
where the entropy transfer term drops out for the adiabatic mixing process. The initial entropy of the system, S_{1}, is the sum of the entropies of the gases at the respective initial states
S_{1}=n_{ N _{2}} \bar{s}_{ N _{2}}\left(T_{ N _{2}}, p_{ N _{2}}\right)+n_{ O _{2}} \bar{s}_{ O _{2}}\left(T_{ O _{2}}, p_{ O _{2}}\right)
The final entropy of the system, S_{2} is the sum of the entropies of the individual components, each evaluated at the final mixture temperature and the partial pressure of the component in the mixture
S_{2}=n_{ N _{2}} \bar{s}_{ N _{2}}\left(T_{2}, y_{ N _{2}} p_{2}\right)+n_{ O _{2}} \bar{s}_{ O _{2}}\left(T_{2}, y_{ O _{2}} p_{2}\right)
Collecting the last three equations
\begin{aligned}\sigma=& n_{ N _{2}}\left[\bar{s}_{ N _{2}}\left(T_{2}, y_{ N _{2}} p_{2}\right)-\bar{s}_{ N _{2}}\left(T_{ N _{2}}, p_{ N _{2}}\right)\right] \\&+n_{ O _{2}}\left[\bar{s}_{ O _{2}}\left(T_{2}, y_{ O _{2}} p_{2}\right)-\bar{s}_{ O _{2}}\left(T_{ O _{2}}, p_{ O _{2}}\right)\right]\end{aligned}
Evaluating the change in specific entropy of each gas in terms of a constant specific heat \bar{c}_{p}, this becomes
\sigma=n_{ N _{2}}\left(\bar{c}_{p, N _{2}} \ln \frac{T_{2}}{T_{ N _{2}}}-\bar{R} \ln \frac{y_{ N _{2}} p_{2}}{p_{ N _{2}}}\right)
+n_{ O _{2}}\left(\bar{c}_{p, O _{2}} \ln \frac{T_{2}}{T_{ O _{2}}}-\bar{R} \ln \frac{y_{ O _{2}} p_{2}}{p_{ O _{2}}}\right)
The required values for \bar{c}_{p} can be found by adding \bar{R} \text { to the } \bar{c}_{v} values found previously (Eq. 3.45)
\bar{c}_{p}(T)=\bar{c}_{v}(T)+\bar{R} (ideal gas) (3.45)
\bar{c}_{p, N _{2}}=6.95 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }, \quad \bar{c}_{p, O _{2}}=6.98 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R }
Since the total number of moles of mixture n = 0.79 + 0.21 = 1.0, the mole fractions of the two gases are y_{ N _{2}}=0.79 \text { and } y_{ O _{2}}=0.21.
Substituting values into the expression for σ gives
\sigma=0.79 lbmol \left[6.95 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R } \ln \left(\frac{477^{\circ} R }{460^{\circ} R }\right)\right.
\left.-1.986 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R } \ln \left(\frac{(0.79)(1.62 atm )}{2 atm }\right)\right]
+0.21 lbmol \left[6.98 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R } \ln \left(\frac{477^{\circ} R }{540^{\circ} R }\right)\right.
\left.-1.986 \frac{ Btu }{ lbmol \cdot{ }^{\circ} R } \ln \left(\frac{(0.21)(1.62 atm )}{1 atm }\right)\right]
1 = 1.168 Btu/°R
1 Entropy is produced when different gases, initially at different temperatures and pressures, are allowed to mix.
Skills Developed
Ability to…
• analyze the adiabatic mixing of two ideal gases at constant total volume.
• apply energy and entropy balances to the mixing of two gases.
• apply ideal gas mixture principles assuming constant specific heats.
Quick Quiz
Determine the total volume of the final mixture, in f t ^{3}. Ans. 215 f t ^{3}.
