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A specimen of oil having an initial volume of 600 \(\mathrm{cm}^{3}\) is subjected to a pressure increase of \(3.6 \times 10^{5} \mathrm{Pa},\) and the volume is found to decrease by 0.45 \(\mathrm{cm}^{3} .\) What is the bulk modulus of the material? The compressibility?

Short Answer

Expert verified
The bulk modulus is \(4.8 \times 10^{8} \mathrm{Pa}\) and the compressibility is \(2.08 \times 10^{-9} \mathrm{Pa}^{-1}\).

Step by step solution

01

Understand the Bulk Modulus Formula

The bulk modulus of a material is given by the formula \( B = -\frac{\Delta P}{\frac{\Delta V}{V}} \), where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume. The negative sign indicates compression.
02

Substitute Values into Bulk Modulus Formula

Given \( \Delta P = 3.6 \times 10^{5} \mathrm{Pa} \), \( \Delta V = -0.45 \mathrm{cm}^3\) (the volume decreases, hence it's negative), and \( V = 600 \mathrm{cm}^3\). Substitute these values into the formula: \[B = -\frac{3.6 \times 10^{5} \mathrm{Pa}}{\frac{-0.45 \mathrm{cm}^3}{600 \mathrm{cm}^3}}.\]
03

Calculate the Bulk Modulus

Calculate the fraction: \(-\frac{0.45 \mathrm{cm}^3}{600 \mathrm{cm}^3} = -0.00075\). Then compute the bulk modulus: \[B = -\frac{3.6 \times 10^{5} \mathrm{Pa}}{-0.00075} = 4.8 \times 10^{8} \mathrm{Pa}.\]
04

Understand the Compressibility Formula

Compressibility \( \beta \) is the reciprocal of the bulk modulus, given by \( \beta = \frac{1}{B} \).
05

Calculate the Compressibility

Using the bulk modulus found in Step 3, compressibility is \( \beta = \frac{1}{4.8 \times 10^{8} \mathrm{Pa}}.\) Upon calculation, \( \beta \approx 2.08 \times 10^{-9} \mathrm{Pa}^{-1}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Compressibility
Compressibility is a measure of how much a material can be compressed or reduced in volume under pressure. This property is particularly important in fluids, where understanding how substances respond to pressure changes is crucial. - It is denoted by the symbol \( \beta \). - Mathematically, compressibility is defined as the reciprocal of the bulk modulus. Thus, you have \( \beta = \frac{1}{B} \), where \( B \) stands for the bulk modulus.- A high compressibility means that a material decreases significantly in volume when pressure is applied.In the context of this exercise, oil's compressibility tells us how susceptible it is to volume changes under different pressure conditions. Given its low compressibility, oil is quite resistant to volume changes, maintaining its volume even under substantial pressure increases.
Pressure Change
Pressure change is a critical concept in understanding how fluids behave when subjected to external forces. It refers to the difference in pressure exerted on a fluid, either increasing or decreasing based on external conditions.- Represented mathematically as \( \Delta P \).- In practical situations, pressure changes can originate from environmental shifts or mechanical forces applied.In this specific problem, the oil specimen undergoes a pressure increase of \( 3.6 \times 10^5 \mathrm{Pa} \). This increase in pressure results in a decrease in the oil's volume, showcasing the impact of external pressure on fluid volumes. The ability to calculate how volume responds to pressure shifts is critical to numerous scientific and engineering applications, helping to predict and control fluid behavior.
Volume Change
Volume change in fluids indicates how the space occupied by a substance varies when it is subjected to external pressure. It is defined as the difference in the initial volume compared to the volume after pressure is applied.- Often represented with \( \Delta V \).- This concept is crucial when determining the bulk modulus, given its formula \( B = -\frac{\Delta P}{\frac{\Delta V}{V}} \).In the given problem, the oil's volume decreases by \( 0.45 \mathrm{cm}^3 \) from an initial volume of \( 600 \mathrm{cm}^3 \) under a specific pressure increase. This volume reduction is used to calculate the bulk modulus of the material, which reveals the oil's resistance to pressure-induced volume changes. Understanding volume change is vital in fields like fluid mechanics and engineering, where precise control of fluid properties is necessary.
Elasticity in Fluids
Elasticity in fluids, also known as fluid elasticity, describes the ability of fluids to return to their original volume after the removal of an external force. While elasticity is typically associated with solids, in fluids, it is quantified through parameters like bulk modulus. - Fluid elasticity is tightly linked to the fluid's compressibility and bulk modulus. - It indicates how well a fluid can recover its initial volume after being compressed. The exercise illustrates the concept through oil, where the bulk modulus calculation shows the degree of elasticity by measuring its resistance to volume reduction under pressure. In engineering and physics, comprehending fluid elasticity helps in designing systems where fluid resilience to pressure changes is critical. It ensures safety and efficiency in applications like hydraulic systems and packaging of compressible fluids.

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Most popular questions from this chapter

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) 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.

A \(240-\mathrm{kg}, 50.0 \mathrm{cm}\) -long uniform bar has a small \(1.10-\mathrm{kg}\) mass glued to its left end and a small \(2.20-\mathrm{kg}\) mass glued to the other cad. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

A diving board 3.00 \(\mathrm{m}\) long is supported at a point 1.00 \(\mathrm{m}\) from the end, and a diver weighing 500 \(\mathrm{N}\) stands at the free end \((\text { (Fig. } 11.24) .\) The diving board is of uniform cross section and weighs 280 \(\mathrm{N}\) . Find (a) the force at the support point and (b) the force at the left-hand end.

A uniform, \(7.5-\mathrm{m}\) -long beam weighing 9000 \(\mathrm{N}\) is hinged to a wall and supported by a thin cable attached 1.5 \(\mathrm{m}\) from the free end of the beam. The cable runs between the beam and the wall and makes a \(40^{\circ}\) angle with the beam. What is the tension in the cable when the beam is at an angle of \(30^{\circ}\) above the horizontal?

A metal wire 3.50 \(\mathrm{m}\) long and 0.70 \(\mathrm{mm}\) in diameter was given the following test. A load weighing 20 \(\mathrm{N}\) was originally hung from the wire to keep it taut. The position of the lower end of the wire was read on a scale as load was added. $$ \begin{array}{cc}{\text { Added Load (N) }} & {\text { Scale Reading (cm) }} \\\ \hline 0 & {3.02} \\ {10} & {3.07} \\ {20} & {3.12} \\ {30} & {3.17} \\\ {40} & {3.22} \\ {50} & {3.27} \\ {60} & {3.32} \\ {70} & {4.27}\end{array} $$ (a) Graph these values, plotting the increase in length horizontally and the added load vertically. \((b)\) Calculate the value of Young's modulus. (c) The proportional limit occurred at a scale reading of 3.34 \(\mathrm{cm} .\) What was the stress at this point?

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