Learning undergraduate engineering thermodynamics might be less painful with a blog. I hope that students, faculty and interested observers will share their thoughts on the laws of thermodynamics, phase and chemical equilibrium and many related topics.
Please post here any questions you have about the final exam. It has been a pleasure working with you. Best of luck on the final exam !
Adios, Dr. B
32 comments:
Anonymous
said...
If a reaction was liquid -> gas + 1/2 gas, then would y1 = 0.5y2 ? This is like the situation in problem 13.23 in homework six just with a different stoichiometric rxn.
On the practice final that has solutions, the 2nd problem has a dH2 value that is way off from the number I got. I tried calculating it using a program I wrote on my calculator, doing the whole integral on the calculator, doing it by hand and using the intcp program in Excel (*R). I got the same number each time...dH =~-5400. Could Dr. B or someone else check this please?
cj: I just double checked the calculation of deltaH and I think I did it correctly. If I had to guess, I would say you forgot to divide by 2 and by 3 on the T^2 and T^3 terms in the integral. Don't know what to tell you here. Do your deltaA, deltaB terms, etc match ? Let me know if you figure what is wrong.
Like I said, I did the Cp integral using several different methods:intcp (Excel), intcp (program I wrote on my calculator), the built-in integral function on my calculator using the Cp equation, and by hand. They *all* came out to the same number. I'm pretty sure I divided by 2 and 3 where necessary...at the very least my calculator should have.
There are no delta A,B, and C for this problem, since there is only one chemical. I've been using the same values you gave, though.
I just used the excel macro to find the heat capacity and got -282.64, then multiplied by 8.314 to get -2350.23. I don't know if this helps . I usually have problems with dividing by R or not having the T's in K.
This is really bizarre...I used the macro as well, and got -647. Can you give me the exact thing you typed in so maybe I can figure out what I could possibly be doing wrong?
I was integrating from 298 instead of 203 K. That habit is possibly TOO ingrained now. Well, at least I know all my calculator programs are consisitent.
Thanks for confirming, Clare. I thought I was going crazy.
Although that is certainly not out of the question.
Ok, I know this is from the beginning of the quarter, but I am still confused about Maxwell's relations. I understand how to derive them but I don't understand what they "mean." Do you have any explanations that might help? This has been bothering me all quarter.
CJ & Claire: OK, I was confused. I was looking at problem #1. On problem #2, I integrated from -70C or 203.15K to -161.95C or 111.20K and I got -2350.2 J/mole. This includes multiplying by R because I put units on the result. Cp/R is dimensionless. CJ, you integrated from 298.15K to 11.2 K and got -5383.6 J/mole, right ? OK, I hope this one is resolved.
Loogel: What do Maxwell Eqns mean ? At the heart, the Maxwell Eqns are relationships between SLOPES on multidimensional surfaces. If you are standing on the side of a mountain and you walk due north and measure the rate at which your altitude changes, you are measuring something like the partial derivative of z (altitude) with respect to y (latitude) at constant x (longitude). The cool part about the Maxwell's are that the slope are all RELATED to each other. They are related because all of these abstract thermodynamic properties that we have made up (U,H,S,A,G) are all fxns of P, V & T ONLY (for a pure substance). The abstract properties are just complicated functions of the concrete, tangible properties, PVT. So, it makes sense that all of these slopes in the Maxwell's are related. It is just that the mountain(s) are multidimensional and so not as easy to picture as walking on Mt. Ranier.
I remember you saying before the second midterm that if Wilson was on the test, which it was, then the final would have either be Margules or van Laar. Is this true? I only ask because if there is no chance for another Wilson problem, then I won't bother making a program for it.
Not to bring up a sore subject or anything, but on that same problem #2 from the old final with solutions, how did you find T2 to solve for dH2? I don't see how to get it from Antoines equation.
I think I see what you are supposed to do. When you say that T2 is Tsat at 1 bar, this means that the fluid is at saturation Temp and Pressure under those conditions so you can use Antoine because P=Psat. Then you can put that pressure into the Antoine equation and solve for T2. I get the same answer as he does when I use P*=1bar=100kPa.
I also have a question on problem #2 of the practice final. Why did you not include a residual for the effluent conditions? Becuase for the entire process (excluding the condensation) shouldn't it be dH=(Hig2+Hr2)-(Hig1+Hr1)?
This would give us: dH=dHig+(Hr2-Hr1)
It seems to me that you only took into account Hr1. If that is the case why can we ignore Hr2?
erik: We ignore the Hr2 residual in problem 2 because it says in the problem statement that the methane at 1 bar is an ideal gas. Residual= Actual - Ideal. Since actual=ideal, the residual is zero. This is the same as what happened on problem 2 of our first test on with enthalpy rather than entropy. If you look at eqn 1 from that test you will see the how it cancels out.
claire: fugacity is for gases and activity is for liquids... generally.
Chipmunk 4:28: You algebraically manipulate the Antoine Eqn until only T is on the left side of the eqn. Then, you plug numbers into the right side to evaluate T. Maybe I misunderstood your question ?
Erik: EXCELLENT question. I assumed that since the heat of vaporization was given, and most likely measured, that the final state based on this measurement was a REAL fluid (that is the kind you find when experimental measurements are involved). Therefore, the residual was "included" in the latent heat of vaporization. I further assumed that sat'd propane vapor at 1 bar is an ideal gas. So, I did not compute a residual here either. My goal in this problem was to limit the work to one residual calculation b/c it is a test problem.
Claire, For the purpose of tomorrow's test, use activity when you are given some method to determine activity coefficients. Use fugacity if you are given things like critical properties and the accentric factor. Often, either will work.
In practice use activity for liquids in all cases and use fugacity for gases if they cannot be considered ideal (ideal if molar volume > 20 L/mol or 5 L/mol for diatomic and noble gases).
Eric: I thought you were worried about the liquid ! Treat as an IG because I said so, but this is shakey. T2 is very low and the vapor phase is unlikely to be ideal. The error may be significant in practice !
On the 2004 final, problem 2: There seems to be a conflict of species between the problem and solution, is it methane or propane? I’m concerned that maybe both species are meant to be in the problem; if that’s the case, then I’m REALLY confused about that one. Seems to me it should be just one species.
On our second exam, problem 3: Why do we use the vapor pressure instead of the given pressure? I’m still confused about that. Will you take off points for using 298 instead of 298.15? (old habits die hard)
It is methane. The reference to propane is a typo. Cut and paste is the best way to make lots of errors.
Use P* because the PF is used to correct PHIsat or fsat. In order to get PHIsat or fsat you must use P*. If you use P, you get SRKs best estimate of f. But this is not good. Use the PF and PHIsat.
Please try to use 298.15. It makes my life easier But to answer your question: no.
32 comments:
If a reaction was liquid -> gas + 1/2 gas, then would y1 = 0.5y2 ? This is like the situation in problem 13.23 in homework six just with a different stoichiometric rxn.
Claire,
Yes.
Should we expect any essay (thought related) questions?
On the practice final that has solutions, the 2nd problem has a dH2 value that is way off from the number I got. I tried calculating it using a program I wrote on my calculator, doing the whole integral on the calculator, doing it by hand and using the intcp program in Excel (*R). I got the same number each time...dH =~-5400. Could Dr. B or someone else check this please?
ahayles:
No. There will be no "Reading Questions," short answer or essay questions on the final.
cj:
I just double checked the calculation of deltaH and I think I did it correctly. If I had to guess, I would say you forgot to divide by 2 and by 3 on the T^2 and T^3 terms in the integral.
Don't know what to tell you here. Do your deltaA, deltaB terms, etc match ? Let me know if you figure what is wrong.
Like I said, I did the Cp integral using several different methods:intcp (Excel), intcp (program I wrote on my calculator), the built-in integral function on my calculator using the Cp equation, and by hand. They *all* came out to the same number. I'm pretty sure I divided by 2 and 3 where necessary...at the very least my calculator should have.
There are no delta A,B, and C for this problem, since there is only one chemical. I've been using the same values you gave, though.
CJ
I just used the excel macro to find the heat capacity and got -282.64, then multiplied by 8.314 to get -2350.23. I don't know if this helps . I usually have problems with dividing by R or not having the T's in K.
This is really bizarre...I used the macro as well, and got -647. Can you give me the exact thing you typed in so maybe I can figure out what I could possibly be doing wrong?
Oh wait. Ahahaha, I'm an idiot.
I was integrating from 298 instead of 203 K. That habit is possibly TOO ingrained now. Well, at least I know all my calculator programs are consisitent.
Thanks for confirming, Clare. I thought I was going crazy.
Although that is certainly not out of the question.
Ok, I know this is from the beginning of the quarter, but I am still confused about Maxwell's relations. I understand how to derive them but I don't understand what they "mean." Do you have any explanations that might help? This has been bothering me all quarter.
CJ & Claire:
OK, I was confused. I was looking at problem #1.
On problem #2, I integrated from -70C or 203.15K to -161.95C or 111.20K and I got -2350.2 J/mole. This includes multiplying by R because I put units on the result. Cp/R is dimensionless.
CJ, you integrated from 298.15K to 11.2 K and got -5383.6 J/mole, right ? OK, I hope this one is resolved.
Loogel:
What do Maxwell Eqns mean ?
At the heart, the Maxwell Eqns are relationships between SLOPES on multidimensional surfaces. If you are standing on the side of a mountain and you walk due north and measure the rate at which your altitude changes, you are measuring something like the partial derivative of z (altitude) with respect to y (latitude) at constant x (longitude). The cool part about the Maxwell's are that the slope are all RELATED to each other. They are related because all of these abstract thermodynamic properties that we have made up (U,H,S,A,G) are all fxns of P, V & T ONLY (for a pure substance). The abstract properties are just complicated functions of the concrete, tangible properties, PVT. So, it makes sense that all of these slopes in the Maxwell's are related. It is just that the mountain(s) are multidimensional and so not as easy to picture as walking on Mt. Ranier.
I hope this is helpful.
I remember you saying before the second midterm that if Wilson was on the test, which it was, then the final would have either be Margules or van Laar. Is this true? I only ask because if there is no chance for another Wilson problem, then I won't bother making a program for it.
Steve:
I suggest you study Wilson, but it is more likely that you will see Margules or van Laar on the final.
Not to bring up a sore subject or anything, but on that same problem #2 from the old final with solutions, how did you find T2 to solve for dH2? I don't see how to get it from Antoines equation.
Chipmunk:
I think I see what you are supposed to do. When you say that T2 is Tsat at 1 bar, this means that the fluid is at saturation Temp and Pressure under those conditions so you can use Antoine because P=Psat. Then you can put that pressure into the Antoine equation and solve for T2. I get the same answer as he does when I use P*=1bar=100kPa.
Thanks danny, much appreciated
I also have a question on problem #2 of the practice final. Why did you not include a residual for the effluent conditions? Becuase for the entire process (excluding the condensation) shouldn't it be dH=(Hig2+Hr2)-(Hig1+Hr1)?
This would give us: dH=dHig+(Hr2-Hr1)
It seems to me that you only took into account Hr1. If that is the case why can we ignore Hr2?
When do we use fugacity vs. activity?
erik:
We ignore the Hr2 residual in problem 2 because it says in the problem statement that the methane at 1 bar is an ideal gas. Residual= Actual - Ideal. Since actual=ideal, the residual is zero. This is the same as what happened on problem 2 of our first test on with enthalpy rather than entropy. If you look at eqn 1 from that test you will see the how it cancels out.
claire:
fugacity is for gases and activity is for liquids... generally.
Thanks Danny. That clears everything up. I missed that it said we could treat it as an IG.
Chipmunk 4:28:
You algebraically manipulate the Antoine Eqn until only T is on the left side of the eqn. Then, you plug numbers into the right side to evaluate T. Maybe I misunderstood your question ?
Danny:
Yes, exactly. Just be sure to convert P from 1 bar to 100 kPa because the vapor or saturation pressure in this form of Antoine is in kPa.
Erik:
EXCELLENT question.
I assumed that since the heat of vaporization was given, and most likely measured, that the final state based on this measurement was a REAL fluid (that is the kind you find when experimental measurements are involved). Therefore, the residual was "included" in the latent heat of vaporization.
I further assumed that sat'd propane vapor at 1 bar is an ideal gas. So, I did not compute a residual here either.
My goal in this problem was to limit the work to one residual calculation b/c it is a test problem.
Claire,
For the purpose of tomorrow's test, use activity when you are given some method to determine activity coefficients. Use fugacity if you are given things like critical properties and the accentric factor. Often, either will work.
In practice use activity for liquids in all cases and use fugacity for gases if they cannot be considered ideal (ideal if molar volume > 20 L/mol or 5 L/mol for diatomic and noble gases).
Eric:
I thought you were worried about the liquid !
Treat as an IG because I said so, but this is shakey. T2 is very low and the vapor phase is unlikely to be ideal. The error may be significant in practice !
Here are 3 questions I got by eamil tonight:
On the 2004 final, problem 2: There seems to be a conflict of species between the problem and solution, is it methane or propane? I’m concerned that maybe both species are meant to be in the problem; if that’s the case, then I’m REALLY confused about that one. Seems to me it should be just one species.
On our second exam, problem 3: Why do we use the vapor pressure instead of the given pressure? I’m still confused about that.
Will you take off points for using 298 instead of 298.15? (old habits die hard)
And here is my reply:
It is methane. The reference to propane is a typo. Cut and paste is the best way to make lots of errors.
Use P* because the PF is used to correct PHIsat or fsat. In order to get PHIsat or fsat you must use P*. If you use P, you get SRKs best estimate of f. But this is not good. Use the PF and PHIsat.
Please try to use 298.15. It makes my life easier
But to answer your question: no.
Best of luck tomorrow !
Gnite,
Dr. B
Dr. Baratuci,
I thought your first name was William.
Bill is your middle name, right?
Wouldn't that make him Bill Bill Baratuci?
Hi Jerry,
Bill is a short form of William. My middle name is Brian.
Enjoy the break,
Dr. B
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