Example Problem with Complete Solution

3D-1 : Calculating and Using the Heat Capacities of Ideal Gas Mixtures 4 pts
Three ideal gases, Nitric Oxide (NO), Carbon Monoxide (CO), and Oxygen (O2),  at 220 kPa and 350oC are held in a tank with three chambers, as shown below.
The dividers between the chambers are removed and the three gases are allowed to mix.The mixture contains 30 mole% NO, 50 mole% CO, and 20 mole% O2. The mixture is then heated to 735oC. 
Calculate the ΔU, in J/mole, of the mixture for the heating process. Assume the mixture is an ideal gas.
Read : The key to this problem is that enthalpy does not depend on pressure for an ideal gas.  So, the initial and final pressures are not relevant.  We want to determine the change in the internal energy, but only the constant pressure heat capacities are tabulated.  We can either use Cv = Cp - R and then integrate Cv with respect to T to get DU or we can integrate Cp with respect to T to get DH and then use the definiition of enthalpy to get DU.  The final aspect of the problem is that the system contains a mixture.  We can either use the mole fractions to determine the constants of the heat capacity polynomial for the mixture and then integrate Cp with respect to T one time, or we can integrate Cp for each chemical component with respect to T and sum the resulting DH values to get DH for the mixture.  Either way, once we have DH, we use the definition of enthalpy to determine DU.
Diagram: The figure given in the problem statement is adequate. Just include the initial and final temperatures.
Given: P1 = 220 kPa yNH3 =  0.30 mol NO/mol
T1 =  350 oC       = 623.15 K yCH4 =  0.50 mol CO/mol
T2 =  735 oC       = 1008.15 K yO2 =  0.20 mol O2/mol
Find: DU = ??? J/mole
Assumptions: 1 - The initial state and the final state are equilibrium states.
2 - There is no change in internal energy or enthalpy due to mixing of the gases.
3 - The pure components and the mixture behave as ideal gases.
Equations / Data / Solve:
The internal energy of an ideal gas does not depend on pressure, only on temperature.
Therefore, the question becomes, what is the change in internal energy from T1 = 400 oC, to T2 =  600 oC.
Eqn 1
The Shomate Equation for the ideal gas heat capacity is :
Eqn 2
where :
Eqn 3
and :
Eqn 4
Combining Eqns 1, 2 and 3 and integrating yields :
Eqn 5
T in Kelvin ! Eqn 6
Heat Capacity Constants from the NIST WebBook:   298. - 1200. 298. - 1300. 298. - 6000.
A 23.83491 25.56759 29.659
B 12.58878 6.09613 6.137261
R = 8.314 J/mol K C -1.139011 4.054656 -1.186521
D -1.497459 -2.671301 0.09578
E 0.214194 0.131021 -0.219663
Method #1: Calculate the constants for the heat capacity polynomial for the gas mixture and  then integrate to determine DH for the mixture.
-1.765732 DHmix = 12528 J/mol
0.085836 DUmix = 9327 J/mol
Method #2: Calculate DH and then DU for EACH gas and then compute the molar average DU and DH using the following equations:
Eqn 7
Eqn 8
  NO CO O2 Mixture  
DH = 12633 12307 12923 12528 J/mol
DU = 9433 9106 9722 9327 J/mol
Verify: Assumptions 1 & 2 cannot be verified from the data given in the problem.
The ideal gas assumption needs to be verified.
We need to determine the specific volume and check if :
Eqn 9
V1 23.55 L/mol V2 38.10 L/mol
The ideal gas assumption is valid because V > 20 L/mole For both the initial and final states.
Answers : DUmix = 9327 J/mol