3B-2 : | Internal Energy of Superheated Ammonia Vapor | 2 pts |
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Superheated ammonia vapor is stored in two rigid tanks, as shown below. Can you determine, by observation and reasoning alone, which has the higher molar internal energy, A or B? | ||||||||||||||
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Calculate the difference in molar internal energy between the two tanks using data from the NIST WebBook. | ||||||||||||||
Read : | Because the ammonia vapor is superheated, it has 2 degrees of freedom. In this case both the T and P must be specified to completely determine the state. Because the state is completely determined, we can use the given T and P values to look up properties like U and H in the Superheated Tables Ammonia. | |||||||||||||
Diagram: | Given in the problem statement. | |||||||||||||
Given: | PA = | 1.55 | atm | PB = | 1.55 | atm | ||||||||
TA = | 23 | oC | TB = | 4 | oC | |||||||||
Find: | DU = UA - UB = ??? | kJ/mol | ||||||||||||
Assumptions: | None. | |||||||||||||
Equations / Data / Solve: | ||||||||||||||
The internal energy of a substance is the sum of the kinetic energies stored in the vibrational, rotational, and translational motion of the molecules. Tank A has more energy by virtue of its higher temperature. Therefore, it must have the higher intern | ||||||||||||||
We must look up the isobaric properties of superheated ammonia in the NIST WebBook. Use the ASHRAE convention. A portion of the thermodynamic table used in this problem is given below. | ||||||||||||||
T (°C) |
P (atm) |
U (kJ/mol) |
Phase | |||||||||||
2 | 1.55 | 22.849 | vapor | |||||||||||
3 | 1.55 | 22.878 | vapor | |||||||||||
4 | 1.55 | 22.908 | vapor | |||||||||||
5 | 1.55 | 22.937 | vapor | |||||||||||
21 | 1.55 | 23.401 | vapor | |||||||||||
22 | 1.55 | 23.430 | vapor | |||||||||||
23 | 1.55 | 23.459 | vapor | |||||||||||
24 | 1.55 | 23.487 | vapor | |||||||||||
The internal energies at the two given temperatures are: | ||||||||||||||
T = 23oC | T = 4oC | |||||||||||||
UA | 23.459 | KJ/mol | UB | 22.908 | KJ/mol | |||||||||
As we predicted, the internal energy of the water vapor in Tank A is greater than in Tank B. | ||||||||||||||
The U of Tank A is greater by: | ||||||||||||||
DU = UA - UB = | 0.551 | KJ/mol | ||||||||||||
Verify: | No assumptions to verify this time. | |||||||||||||
Answers : | DU | 0.551 | KJ/mol |