Calculating density of a perovskite structure, a question, phase diagr

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SUMMARY

The discussion focuses on calculating the density of a perovskite structure, specifically addressing the lattice constant and the inclusion of titanium ions in mass calculations. The correct approach involves determining the lattice constant as the side length of the cube, rather than using the formula 2*(R1+R2+R3). Additionally, it is established that only 1/4 of the mass of titanium ions should be included in the density calculation due to their positioning within the structure.

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Junkwisch
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Homework Statement



See attachment q2 for the perovskite structure




The Attempt at a Solution


For the perovskite structure, I assume that the lattice constant is equal to 2*(R1+R2+R3) and calculate the density based on this. Am I correct? (The attachment q2ans is my attempt on the question)



For the phase diagram attachment, can anyone tell me what α, γ, ε represent?
i tried searching this on the internet, I assume that α is aluminium, ε is magnesium and γ is vapour phase. Is this correct?



Best Regards
Junkwisch
 

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  • phasediagram.png
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Junkwisch said:

Homework Statement



See attachment q2 for the perovskite structure




The Attempt at a Solution


For the perovskite structure, I assume that the lattice constant is equal to 2*(R1+R2+R3) and calculate the density based on this. Am I correct? (The attachment q2ans is my attempt on the question)

No. The titanium ion is small, it does not touch the other ions. See one face of the cube and figure out the side length so as the arrangement with one oxygen ion at the centre and barium ions at the corners can be established. The lattice constant is equal to the side length of the cube.

ehild
 
Last edited:
thank for your reply ehild

the attachment here is what I did by finding 'a' from a single face. I got a new value for volume which will result in higher density :D. Since Ti is so small, do I still need to include it in the mass calculation?


Junks
 
Sorry, I forgot to attach the pic


Junks
 

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Junkwisch said:
thank for your reply ehild

the attachment here is what I did by finding 'a' from a single face. I got a new value for volume which will result in higher density :D. Since Ti is so small, do I still need to include it in the mass calculation?Junks

You need to include it in the mass, but remember, only 1/4 th of the sides in the centre of the cube is filled with titanium ions.
The volume is correct now.

ehild
 
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ah, so only 1/4 of the mass of Ti is included, thank you ehild :D
 

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