Determine the change in the specific internal energy of nitrogen (N2), in kJ/kg, as it is heated from 600 to 1500 K, using:
a.) The empirical specific heat equation (Shomate Equation) from the NIST Website.
b.) "The CoV value at the average temperature.
(Use the heat capacity polynomial to determine this CoV value.)
c.) The CoV value at room temperature, 25oC.
(Use the heat capacity polynomial to determine this CoV value.)
8 comments:
Why do the examples like this problem in 3c subtract the last term in the CP equation and NIST add the the last term?
Does anyone know what we're supposed to do for solving (b) and (c)? And are we really supposed to be using CoV for both of them or CoP? I'm confused because the title of the problem says DH but the problem description says internal energy instead of enthalpy. Help!
For part A, i keep getting the answer that is off by 1000. So instead of 1050 i get 1.050. The answer you get after u take the integral of Cp should be in J/mol so you divide by 1000 to get KJ/mol and multiply by 1000 to change 28.01 g/mol to kg/mol so 1000 in each case cancels eachother out. But if this is correct you just get 1.050 not 1050. What am I doing wrong?? anyone??
For your answer before any conversions you should have 29409.59 J/mol, right? So the conversion will be:
29410 J/mol x 1 KJ/1000 J = 29.41 KJ/mol
29.41 KJ/mol x 1mol/28g = 1.050 KJ/g
1.050 KJ/g x 1000g/kg = 1050 KJ/kg
Not sure what was going wrong, since you are correct that the 1000's cancel so it should just leave you with 29409.59 / 28.
Hope that helps.
Now... does anyone get (b) and (c)...
wondering:
I am not sure what you mean, but I will take a shot.
When you integrate the last term, E/T^2, it becomes -E/T.
I hope I figured out what you menat here. If not, blog again.
Desperate:
In class I pointed out the typo in this problem. Use Cp to determine deltaH.
In part (b), evaluate Cp at the avg temperature, 1050, and then treat it aas if it were a constant when you go to integrate it.
In part (c) you also treat Cp as a constant, but you evaluate it at 25 degC.
Anonymous & Desperate:
Desperate did this exactly correct and spelled it out very clearly here. Thank you, Desperate. I assume this was clear to you, Anonymous, since you did not blog again !
Desperate:
In parts (b) and (c) we treat Cp as a constant. When Cp is a constant, DH = Cp DT.
The VALUE of that constant Cp is determined by plugging a single T value into the Shomate polynomial for Cp. In part (b) that T value is the average T, 1050 K. In part (c), that T value is 198,15 K.
Hope that helps.
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