Chapter3, Lesson C - Calculating and Using the Heat Capacities of Ideal Gas Mixtures

3C-5 :	Calculating and Using the Heat Capacities of Ideal Gas Mixtures								4 pts

Three ideal gases, Ammonia (NH3), Methane (CH4), and Oxygen (O2), are allowed to mix at 200 kPa and 400°C. The mixture contains 30 mole% NH3, 50 mole% CH4, and 20 mole% O2. The mixture is heated to 600°C. Calculate the change in the molar internal energy of the mixture for the heating process. You can assume the mixture is an ideal gas.

Read :	The key to this problem is that enthalpy and entropy do 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 ΔU or we can integrate Cp with respect to T to get ΔH and then use the definiition of enthalpy to get ΔU. 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 ΔH values to get ΔH for the mixture. Either way, once we have ΔH, we use the definition of enthalpy to determin ΔU.








Given:	P1 =	200	kPa			yNH3 =	0.30	mol NH3/mol
	T1 =	400	oC =	673.15	K	yCH4 =	0.50	mol CH4/mol
	T2 =	600	oC =	873.15	K	yO2 =	0.20	mol O2/mol

Find:	ΔU =	???	J/mole

Assumption:		- The initial state and the final state are equilibrium states.

Solution :	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
			T in Kelvin !		Eqn 2

	Heat Capacity Constants from the NIST Webbook:				Ammonia	Methane	Oxygen
	Heat Capacity Constants from the NIST Webbook:				298. - 1400.	298. - 1300.	298. - 6000.
				A	19.99563	-0.703029	29.659
				B	49.77119	108.4773	6.137261
	R =	8.314	J/mol K	C	-15.37599	-42.52157	-1.186521
				D	1.921168	5.862788	0.09578
				E	0.189174	0.678565	-0.219663

	Method #1:		Calculate the constant for the heat capacity polynomial for the gas mixture and then integrate to determine ΔH for the mixture.
	Method #1:
						Mixture
						11.57897
						70.39746
						-26.11089
						3.526900	ΔHmix =	10514	J/mol
Answer :						0.352102	ΔUmix =	8851	J/mol

	Method #2:		Calculate ΔH and then ΔU for EACH gas and then compute the molar average ΔU and ΔH using the following equations:



		NH3	CH4	O2	Mixture
	ΔH =	10092	12303	6672	10514	J/mol
Answer :	ΔU =	8429	10640	5010	8851	J/mol

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