Example Problem with Complete Solution

3C-4 : Enthalpy Change of N2 Using the IG Heat Capacity 5 pts
Nitrogen gas is heated from 450 K to 1120 K. Determine ΔU and ΔH in kJ/kg by:
a.) Integrating the Shomate Equation
b.) Treating the CP value as a constant, determined using the Shomate Equation at the average temperature, 800 K
c.) Treating the CP value as a constant, determined using the Shomate Equation at room temperature, 25oC
Read : The Shomate Equation will yield the most accurate estimate of the enthalpy change.  Assuming a constant value of Cp determined at the average temperature should yield a reasonable estimate of DH as well.  Using the Cp value at room temperature should not be very accurate.  We can compare this result to the value we get in part (a).
Diagram: A diagram is not necessary for this problem.
Given: T1 600 K T2 1000 K
Find: DH1-2 = ??? kJ/kg
a.) Nitrogen behaves as an ideal gas
b.) Nitrogen behaves as an ideal gas with a linear relationship between Cp and T.
This is equivalent to using a constant value of Cp that is equal to the average of Cp(T1) and Cp(T2).
c.) Nitrogen behaves as an ideal gas with a constant heat capacity.
Assumptions: The assumptions are part of the problem statement.
Equations / Data / Solve:
Let's begin by collecting the data we will need from the NIST Webbook : Temp (K) 298. - 6000.
A 26.092
MW 28.01 g/mole B 8.218801
C -1.976141
D 0.159274
E 0.044434
Part a.) The enthalpy change associated with a temperature change for an ideal gas can be determined from :
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
Plug in values for the temperatures and the constants to get : DH 12615 J/mol
Eqn 6
DH 450.4 kJ/kg
We can determine DU using the definition of enthalpy :
Eqn 7
For ideal gases, Eqn 7 becomes :
Eqn 8
We can then solve Eqn 8 for DU :
Eqn 9
R 8.314 J/mol-K DU 9289 J/mol
Eqn 10
DU 331.6 kJ/kg
Part b.) First we need to use the Shomate Equation, Eqns 2 & 3, to evaluate Cp(T1) and Cp(T2) :
t1 0.6 Cp(T1) 30.470 J/mol-K
t2 1 Cp(T2) 32.538 J/mol-K
Therefore, the average value of Cp is : Cp,avg 31.504 J/mol-K
When the heat capacity is a constant, Eqn 1 simplifies to :
Eqn 11
DH 12602 J/mol
DH 449.9 kJ/kg
DU 331.2 kJ/kg
This amounts to about 0.1% error relative to the result in part (a).
Part c.) We can use Eqns 2 & 3 to evaluate the heat capacity at 25oC or 298.15 K :
t1 0.29815 Cp(298.15K) 28.871 J/mol-K
DH 11548 J/mol
DH 412.3 kJ/kg
DU 293.6 kJ/kg
This amounts to almost 9% error relative to the result in part (a).
That is not acceptable.
Verify: No assumptions were made other than the ones in the problem statement.
Answers : a.) DU 332 kJ/kg DH 450 kJ/kg
b.) DU 331 kJ/kg DH 450 kJ/kg
c.) DU 294 kJ/kg DH 412 kJ/kg