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Ch 11: Equilibrium & Elasticity
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 11: Equilibrium & ElasticityProblem 35
Chapter 11, Problem 35

A specimen of oil having an initial volume of 600 cm3 is subjected to a pressure increase of 3.6×106 Pa, and the volume is found to decrease by 0.45 cm3. What is the bulk modulus of the material and the compressibility?

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1
Understand the concept of bulk modulus, which is a measure of a material's resistance to uniform compression. It is defined as the ratio of the pressure increase to the relative decrease in volume.
Use the formula for bulk modulus (B): B = -ΔP / (ΔV/V₀), where ΔP is the change in pressure, ΔV is the change in volume, and V₀ is the initial volume.
Substitute the given values into the formula: ΔP = 3.6×10^6 Pa, ΔV = -0.45 cm³ (negative because the volume decreases), and V₀ = 600 cm³.
Calculate the bulk modulus using the substituted values. Remember to keep the units consistent, especially when dealing with volume changes.
To find the compressibility (κ), which is the reciprocal of the bulk modulus, use the formula: κ = 1/B. This will give you the material's compressibility in terms of Pa⁻¹.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bulk Modulus

Bulk modulus is a measure of a material's resistance to uniform compression. It is defined as the ratio of the pressure increase to the relative decrease in volume. Mathematically, it is expressed as B = -ΔP / (ΔV/V), where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.
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Compressibility

Compressibility is the reciprocal of the bulk modulus and indicates how much a material can be compressed under pressure. It is defined as the fractional change in volume per unit increase in pressure. A higher compressibility means the material is more easily compressed, and it is calculated as κ = 1/B.
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Pressure-Volume Relationship

The pressure-volume relationship describes how the volume of a material changes in response to applied pressure. In the context of bulk modulus, this relationship is crucial for determining how much a material compresses when subjected to external forces, allowing for calculations of bulk modulus and compressibility.
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Related Practice
Textbook Question

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. The bulk modulus of lead is one-fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

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Textbook Question

A brass wire is to withstand a tensile force of 350 N without breaking. What minimum diameter must the wire have?

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Textbook Question

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

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Textbook Question

In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16×108 Pa (about 1.15×103 atm). If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? (Normal atmospheric pressure is about 1.0×105 Pa. Assume that k for seawater is the same as the freshwater value given in Table 11.2.)

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Textbook Question

A square steel plate is 10.0 cm on a side and 0.500 cm thick. (a) Find the shear strain that results if a force of magnitude 9.0×105 N is applied to each of the four sides, parallel to the side. (b) Find the displacement x (in centimeters).

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