Answer:
The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Explanation:
From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.
Let this distance = AC
the two adjacent sides = AB and BC
AB = a = 2R
BC = a = 2R
Using Pythagoras theorem
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2a²
AC = a√2
But a = 2R
AC = 2R√2
Therefore, the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Answer:
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Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
a true b false thos is the solution