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

AI Thread Summary
The discussion focuses on calculating the density of a perovskite structure using the lattice constant, which is derived from the arrangement of ions in the unit cell. The initial assumption about the lattice constant being equal to 2*(R1+R2+R3) is corrected, emphasizing that the side length of the cube should be determined based on the arrangement of ions. It is clarified that while titanium ions are small and occupy only a fraction of the cube, they must still be included in the mass calculation, specifically accounting for only 1/4 of their mass. The participants also seek clarification on the representation of phases in a phase diagram, with assumptions made about the elements involved. Overall, the conversation highlights the importance of accurate calculations and understanding of ionic arrangements in determining the properties of perovskite structures.
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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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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