### Problem Explanation: Volume Change in Artificial Diamond Due to Pressure DifferenceArtificial diamonds can be created by using high-pressure, high-temperature presses. Suppose an artificial diamond with a volume of \(1.30 \times 10^{-6} \, \text{m}^3\) is formed under a pressure of \(4.70 \, \text{GPa}\). #### Objective:Find the change in its volume (\( \Delta V \) in \( \text{m}^3 \)) when it is released from the press and brought to atmospheric pressure. #### Given Data:- Initial volume, \( V = 1.30 \times 10^{-6} \, \text{m}^3 \)- Pressure under which it is formed, \( P = 4.70 \, \text{GPa} \)- Bulk modulus of diamond, \( B = 194 \, \text{GPa} \)- Atmospheric pressure is assumed to be negligible compared to the pressure under which the diamond was formed.#### Approach to Solution:1. **Understand Bulk Modulus**: - The bulk modulus \( B \) is a measure of a substance's resistance to uniform compression. - It is given by the formula: \[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \] Where \( \Delta P \) is the change in pressure and \( \frac{\Delta V}{V} \) is the relative change in volume.2. **Calculate the Change in Volume**: - Rearrange the bulk modulus formula to solve for \( \Delta V \): \[ \frac{\Delta V}{V} = -\frac{\Delta P}{B} \]3. **Substitute Given Values**: - \( \Delta P \approx 4.70 \, \text{GPa} \) (since atmospheric pressure is negligible) - \( B = 194 \, \text{GPa} \) - Calculate \( \frac{\Delta V}{V} \): \[ \frac{\Delta V}{V} = -\frac{4.70}{194} \] \[ \frac{\Delta V}{V} = -0.0242 \] - Thus, \[ \Delta V = V

Given data : pressure under which diamond is kept, P = 4.70 Gpa = 4.7 × 109 Pa Bulk modulus of…

Artificial diamonds can be made using high-pressure, high-temperature presses. Suppose an artificial diamond of volume 1.30 x 10-6 m3 is formed under a pressure of 4.70 GPa. Find the change in its volume (in m³) when it is released from the press and brought to atmospheric pressure. Take the diamond's bulk modulus to be B = 194 GPa. (Assume the system is at sea level.)

Artificial diamonds can be made using high-pressure, high-temperature presses. Suppose an artificial diamond of volume 1.30 x 10-6 m3 is formed under a pressure of 4.70 GPa. Find the change in its volume (in m³) when it is released from the press and brought to atmospheric pressure. Take the diamond's bulk modulus to be B = 194 GPa. (Assume the system is at sea level.)

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Chapter1: Units, Trigonometry. And Vectors

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Concept explainers

Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

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### Problem Explanation: Volume Change in Artificial Diamond Due to Pressure Difference

Artificial diamonds can be created by using high-pressure, high-temperature presses. Suppose an artificial diamond with a volume of \(1.30 \times 10^{-6} \, \text{m}^3\) is formed under a pressure of \(4.70 \, \text{GPa}\).

#### Objective:
Find the change in its volume (\( \Delta V \) in \( \text{m}^3 \)) when it is released from the press and brought to atmospheric pressure.

#### Given Data:
- Initial volume, \( V = 1.30 \times 10^{-6} \, \text{m}^3 \)
- Pressure under which it is formed, \( P = 4.70 \, \text{GPa} \)
- Bulk modulus of diamond, \( B = 194 \, \text{GPa} \)
- Atmospheric pressure is assumed to be negligible compared to the pressure under which the diamond was formed.

#### Approach to Solution:
1. **Understand Bulk Modulus**:
- The bulk modulus \( B \) is a measure of a substance's resistance to uniform compression.
- It is given by the formula:
\[
B = -\frac{\Delta P}{\frac{\Delta V}{V}}
\]
Where \( \Delta P \) is the change in pressure and \( \frac{\Delta V}{V} \) is the relative change in volume.

2. **Calculate the Change in Volume**:
- Rearrange the bulk modulus formula to solve for \( \Delta V \):
\[
\frac{\Delta V}{V} = -\frac{\Delta P}{B}
\]
3. **Substitute Given Values**:
- \( \Delta P \approx 4.70 \, \text{GPa} \) (since atmospheric pressure is negligible)
- \( B = 194 \, \text{GPa} \)
- Calculate \( \frac{\Delta V}{V} \):
\[
\frac{\Delta V}{V} = -\frac{4.70}{194}
\]
\[
\frac{\Delta V}{V} = -0.0242
\]
- Thus,
\[
\Delta V = V

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