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Chapter 9: Problem 145

For a simple cubic array, solve for the volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the array.

Short Answer

Expert verified
The volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the simple cubic array is: \[V_{interior} = r^3(8 - \frac{32}{3}\pi)\]

Step by step solution

01

Express the Length of the Cube in terms of the Sphere's Radius

The cubic array contains eight identical spheres, which means there are two spheres along each edge of the cube. Since we need to find the volume of the interior space, we need to express the length of the cube in terms of the radius of the spheres. Let the radius of the sphere be r, then the length of the cube is 2r.
02

Calculate the Volume of the Cube

Now that we know the length of the cube, we can calculate its volume. The volume of the cube (V_cube) is given by the formula: \[V_{cube} = l^3\] Where l is the length of the cube. In our case, l = 2r, so: \[V_{cube} = (2r)^3 = 8r^3\]
03

Calculate the Volume of a Single Sphere

The cubic array contains eight identical spheres. The volume (V_sphere) of a single sphere is given by the formula: \[V_{sphere} = \frac{4}{3}\pi r^3\]
04

Calculate the Total Volume of all Spheres in the Array

Now that we have the volume of a single sphere, we can calculate the total volume of all spheres in the array (V_all_spheres). Since there are eight identical spheres in the array: \[V_{all\_spheres} = 8 \cdot V_{sphere} = 8 \cdot \frac{4}{3}\pi r^3 = \frac{32}{3}\pi r^3\]
05

Calculate the Volume of the Interior Space (Cubic Hole)

Finally, we can find the volume of the interior space (cubic hole, V_interior) by subtracting the total volume of all spheres from the volume of the cube: \[V_{interior} = V_{cube} - V_{all\_spheres} = 8r^3 - \frac{32}{3}\pi r^3 = r^3(8 - \frac{32}{3}\pi)\] The volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the simple cubic array is: \[V_{interior} = r^3(8 - \frac{32}{3}\pi)\]

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Most popular questions from this chapter

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69 ). Given that the density of cesium chloride is \(3.97 \mathrm{g} / \mathrm{cm}^{3},\) and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm},\) and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

Cake mixes and other packaged foods that require cooking often contain special directions for use at high elevations. Typically these directions indicate that the food should be cooked longer above 5000 ft. Explain why it takes longer to cook something at higher elevations.

A \(20.0-\mathrm{g}\) sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with \(100.0 \mathrm{g}\) water at \(80.0^{\circ} \mathrm{C}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 2.03 and \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C},\) respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{kJ} / \mathrm{mol}\).

Is it possible for the dispersion forces in a particular substance to be stronger than the hydrogen bonding forces in another substance? Explain your answer.

The compounds \(\mathrm{Na}_{2} \mathrm{O},\) CdS, and \(\mathrm{ZrI}_{4}\) all can be described as cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes is occupied for each case?
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