Posted: 8/31/2004 11:00:45 AM EST
[Last Edit: 9/1/2004 12:37:31 PM EST by scottryan]
I need help with these two problems for my Metals 360 college class.....
Question 316 reads: AlphaMn has a cubic structure with side length a=0.8931 nm and a density of 7.47 g/cm^3. BetaMn has a different cubic structure with side length a=0.6326 nm and a density of 7.26 g/cm^3. The atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. Determine the percent volume change that would occur if AlphaMn transforms to BetaMn. When the problem says "cubic structure", that is on the atomic level with how the atoms are arranged. Question 350 reads: You would like to sort iron specimens, some of with are face centered cubic and others body centered cubic. Design an xray diffraction method by which this can be accomplished. thanks Scott 


USNA EE Final Exam: "Using this paperclip, this piece of string, the battery from your calculator, and your working knowledge of electrical engineering, construct a working model of the Aegis SPY1A radar." Sorry, dude, but I almost failed Materials, and I don't have my books handy (yes, I kept them) otherwise I'd be happy to help..... 

Yeshua u ackbar! <img src=/images/smilies/icon_smile_approve.gif border=0 align=middle>

thats tough, I'm a chemist and I think you need different help than I can give



UHHH 2+2 = 10 YEAH!
This is why I am POLS major, Izz stupidzz in mathzz! 


316 is just dimensional analysis. Mass is conserved, therefore 1gram A = 1 gram B, use the densities to calculate volume change.
It;'s been too long for me to know the xray solution off the top of my head 


yeah 316 is sort of a trick question in that they give you way too much info.



Volume of Alpha = 0.8931 ^ 3 = 0.7124
Volume of Beta = 0.6326 ^ 3 = 0.2532 Difference = 0.4592 % difference is 64.5% That said, I have no idea how Alpha would convert to Beta, or even if it can..... 

Yeshua u ackbar! <img src=/images/smilies/icon_smile_approve.gif border=0 align=middle>

Could you explain this further? I am not quite following how you would calculate the volume change from the densities? 


sorry i was wrong



ROKIT88's solution yields the mass change, that's why you don't understand.
Can't help with the second question, except with a smart ass answer. 


scottryan,
The volume would change by 2.9% (7.47/7.26 = 1.029). 


How? 


scottryan:
On 316, assume you have 1 gram of Alpha. Then Alpha converts to Beta, you still have 1 gram of material, but now in Beta form. You dont need the side lengths, its just the ratio of the densities. Kharn 


Man, with the Collective knowledge we have on this site, we could rule the world!



Well, if both Alpha and Beta have cubic structures, and the sides are given, then wouldn't the volume reduction be calculated as I did above, irrespective of density?
I don't think I made any arithmetic mistakes..... 

Yeshua u ackbar! <img src=/images/smilies/icon_smile_approve.gif border=0 align=middle>

You can determine the angle between the lattice points for fcc and bcc Fe by using the Bergers vector. By directing the xrays at a certain angle, you can cause xrays to refract at different angles, depending on which structure the sample is. 


Mary Ann +1 all the way. Unless of course you are making the WKRP calculation then its Bailey +1 all the way.


In most sports you only need one ball. In my sport you need both.
NRA,GOA,SAF. 
Your calculation relies on two unstated assumptions: (1) each bcc and fcc crystal is a cube and (2) that each bcc crystal has the same mass as a fcc crystal. The second assumption is not generally true. A single fcc crystal generally has more mass than a single bcc crystal due to a tighter "packing factor." Because mass is conserved, the number of crystals changes after the transition, which throws off your calculation. Xray diffraction question  I'm guessing, but I think that there are only a limited number of possible bragg angles for bcc and fcc crystals. If you know the bragg angle, you can tell by looking at a table whether it's bcc or fcc. Doesn't your textbook or class notes tell you something on the subject? 


The inverse of the density is the specific volume. Fractional volume change A > B: (1/DensityB  1/DensityA)/(1/DensityA) Multiply by 100 for % 


My classes are taught by foreigners. I had 8 problems in this assignment. I am lucky I could solve 6 of them on my own. 


+1, let's go for it. 


scottyryan, The exam question does not ask how, hence, in my answer I did not address the mechanism whereby the volumetric change occurs  neither should you. It asks only for the percent volumetric change. This, I gave. 


MikeMills has the right answer as does Weise. If you are still having trouble seeing it, here is a long form explanation that may help illustrate it for you:
Start with any known volume of A (1cc works nicely) Multiply by the density of A to eliminate the original dimension (volume) You are now left with grams of A grams A = grams B (no material is being added or removed from the system/sample) Now divide (or multiply by the inverse) by the density of B You are now left with the volume of B Vol.A subtracted from Vol.B = delta V divide delta V by Vol. A multiply by 100% voila Write it all out with the units, watch the units cancel. Get comfortable with the units and how to convert them backwards in your sleep because it sounds like you will be doing a lot of this in your class.Length, mass, heat, energy, time, velocity, and the coefficient of pie... 


.



Wow.....I'm just a CS major, but hanging around with a Metallurgical Engineering PHD student all the time musta rubbed off on me. I coulda solved the first one myself, and actually understood the answers for the 2nd one, even though I probably wouldn't have come up with them on my own.


C makes it easy to shoot yourself in the foot.
C++ makes it harder but if you do it blows away your whole leg. 
Thanks for helping me guys. If someone would like to explain this problem further, it would be appreciated. I need to know how these problems work, not just how to get the answer.



I don't think anyone has the correct answer on problem 316 yet. My teaching assistant told me I needed to find the number of atoms per unit cell and the mass does matter.
When the problem says "cubic structure", that is on the atomic level. 


I don't think the answer will be any different, but here is another method which combines all of the given info.
start with the density of A Divide by weight of Mn now you have moles/cc of A convert moles to Atoms convert cubic centimeters to cubic Angstroms You now have atoms per angstrom^3 multiply by the cubic structure volume (cube the given length) Now you have the Number of atoms in a unit cell Do the same for B Now that you have the number of atoms in each cube Take the cubic structure volume of B multiplied by the number of atoms in structure A and divide by the number of atoms in structure B (this gives you the volume of B that can be created from one cubic structure unit of A) Subtract this volume from the Cubic Structure of A and divide that number by the Volume of the cubic structure of A multiplied by 100% The answer should be the same. Find a nice cast iron frying pan to demonstrate its metallurgical properties on your TA. 


As far as I can tell it has somethign to do with 3.1415926



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