| 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. |
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| 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. |
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| Calculate the ΔU, in J/mole, of the mixture for the heating process. Assume the mixture is an ideal gas. |
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| 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. |
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| Diagram: |
The figure given in
the problem statement is adequate. Just include the initial and final
temperatures. |
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| Given: |
P1
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220 |
kPa |
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yNH3 = |
0.30 |
mol NO/mol |
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T1
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350 |
oC = |
623.15 |
K |
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yCH4 = |
0.50 |
mol CO/mol |
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T2
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735 |
oC = |
1008.15 |
K |
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yO2 = |
0.20 |
mol O2/mol |
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| Find: |
DU = |
??? |
J/mole |
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| Assumptions: |
1 - The initial state
and the final state are equilibrium states. |
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2 - There is no change in internal energy or enthalpy due to mixing of the gases. |
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3 - The pure components and the mixture behave as ideal gases. |
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| Equations
/ Data / Solve: |
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The
internal energy of an ideal
gas does not depend on pressure, only on temperature. |
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Therefore, the question
becomes, what is the change
in internal energy
from T1 = 400 oC, to T2 = 600 oC. |
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Eqn 1 |
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The Shomate
Equation for the ideal
gas heat capacity is : |
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Eqn 2 |
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where : |
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Eqn 3 |
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and : |
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Eqn 4 |
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Combining Eqns 1, 2 and 3
and integrating yields : |
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Eqn 5 |
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T in Kelvin ! |
Eqn 6 |
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Nitric
Oxide |
Carbon
Monoxide |
Oxygen |
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Heat
Capacity Constants from the NIST WebBook: |
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298. - 1200. |
298. - 1300. |
298. - 6000. |
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A |
23.83491 |
25.56759 |
29.659 |
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B |
12.58878 |
6.09613 |
6.137261 |
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R = |
8.314 |
J/mol K |
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C |
-1.139011 |
4.054656 |
-1.186521 |
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D |
-1.497459 |
-2.671301 |
0.09578 |
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E |
0.214194 |
0.131021 |
-0.219663 |
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Method #1: |
Calculate the
constants for the heat capacity polynomial for the gas mixture and then integrate to determine DH for
the mixture. |
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Mixture |
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25.86607 |
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8.05215 |
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1.44832 |
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-1.765732 |
DHmix = |
12528 |
J/mol |
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0.085836 |
DUmix = |
9327 |
J/mol |
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Method #2: |
Calculate DH and
then DU for EACH
gas and then compute the molar average DU and DH using the following equations: |
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Eqn 7 |
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Eqn 8 |
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NO |
CO |
O2 |
Mixture |
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DH = |
12633 |
12307 |
12923 |
12528 |
J/mol |
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DU = |
9433 |
9106 |
9722 |
9327 |
J/mol |
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| Verify: |
Assumptions 1 & 2 cannot be verified from the data given in the problem. |
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The ideal gas
assumption needs to be verified. |
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We need to determine
the specific volume and check if : |
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Eqn 9 |
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V1 |
23.55 |
L/mol |
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V2 |
38.10 |
L/mol |
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The ideal gas
assumption is valid because V > 20 L/mole For both the initial and final states. |
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| Answers : |
DUmix = |
9327 |
J/mol |
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