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Chapter 12: Problem 40

Determine the fractional change in volume as the pressure of the atmosphere \(\left(1 \times 10^{5} \mathrm{~Pa}\right)\) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is \(125 \mathrm{GPa}\)

Short Answer

Expert verified
The fractional change in volume is \(-8 \times 10^{-7}\).

Step by step solution

01

Understand the Problem

We need to find the fractional change in volume of a metal block when the atmospheric pressure is reduced from \(1 \times 10^5 \text{ Pa}\) to zero. The bulk modulus of the metal is given as \(125 \text{ GPa}\).
02

Define Bulk Modulus

Bulk modulus \(K\) is defined as \(K = -\frac{\Delta P}{\frac{\Delta V}{V_0}}\), where \(\Delta P\) is the change in pressure, \(\Delta V\) is the change in volume, and \(V_0\) is the initial volume.
03

Calculate Change in Pressure

The change in pressure \(\Delta P\) is the initial pressure minus the final pressure. Therefore, \(\Delta P = 1 \times 10^5 \text{ Pa} - 0 = 1 \times 10^5 \text{ Pa}\).
04

Solve for Fractional Change in Volume

Substitute \(\Delta P = 1 \times 10^5 \text{ Pa}\) and \(K = 125 \times 10^9 \text{ Pa}\) into the bulk modulus formula to get the fractional change in volume: \[ \frac{\Delta V}{V_0} = -\frac{\Delta P}{K} = -\frac{1 \times 10^5}{125 \times 10^9} \].
05

Simplify the Fraction

Perform the division: \(-\frac{1 \times 10^5}{125 \times 10^9} = -8 \times 10^{-7}\). Therefore, the fractional change in volume \(\frac{\Delta V}{V_0}\) is \(-8 \times 10^{-7}\).

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Key Concepts

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

Fractional Change in Volume
When a metal block is exposed to a change in external pressure, its volume is affected. The fractional change in volume is a measure of how much the volume of the block changes relative to its original volume. This concept is important because it tells us how susceptible a material is to expanding or contracting under pressure.
In mathematical terms, the fractional change in volume is given by \( \frac{\Delta V}{V_0} \), where \( \Delta V \) represents the change in volume and \( V_0 \) is the initial volume of the block.
  • It measures the relative change, making it a dimensionless quantity.
  • A negative value indicates a reduction in volume, while a positive value represents expansion.
Understanding this concept helps us in determining how materials will behave under different pressure conditions.
Pressure Change
Pressure change occurs when the external force applied per unit area of a surface changes. In the context of our metal block, the problem presents an initial atmospheric pressure of \(1 \times 10^5 \text{ Pa}\) which reduces to zero.
This difference in pressure is what we call a pressure change, denoted as \( \Delta P \).
  • Pressure change can lead to material deformation, as seen in this exercise.
  • The magnitude of pressure change directly impacts the degree of volume change in the block.
Calculating \( \Delta P \) is efficient as it simply involves subtracting the final pressure from the initial pressure, resulting in \( \Delta P = 1 \times 10^5 \text{ Pa} - 0 = 1 \times 10^5 \text{ Pa} \). This pressure change affects the block鈥檚 volume as it moves to a vacuum environment.
Volume Deformation
Volume deformation refers to the change in the shape or size of a material due to an external pressure force. In analyzing our metal block, we look at how the volume alters when exposed to a vacuum.
Volume deformation is closely tied to both the pressure change and the material's properties, like its bulk modulus.
  • The deformation is calculated using the formula from the bulk modulus \( K = -\frac{\Delta P}{\frac{\Delta V}{V_0}} \).
  • This allows us to express the fractional change in volume, which shows whether the block expands or contracts.
In this exercise, using the known bulk modulus, the fractional change in volume \( \frac{\Delta V}{V_0} \) calculates to \(-8 \times 10^{-7}\), indicating a slight contraction.
Metal Block
The metal block in this scenario serves as an example of how changes in pressure affect solid objects. Metals typically have specific mechanical properties such as the bulk modulus, which describes their response to pressure changes.
  • Metal blocks are often used in scientific problems due to their consistent and measurable properties.
  • The bulk modulus of a metal indicates how resistant it is to uniform compression.
  • In this problem, the block possesses a bulk modulus of \(125 \text{ GPa}\), a typical value for metals.
Studying how pressure impacts a metal block helps us understand similar principles in larger and more complex structures, as well as practical applications in engineering and materials science.

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Most popular questions from this chapter

Two parallel oppositely directed forces, each \(4000 \mathrm{~N}\), are applied tangentially to the upper and lower faces of a cubical metal block \(25 \mathrm{~cm}\) on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus for the metal is 80 GPa.

What is the density of the material in the nucleus of the hydrogen atom? The nucleus can be considered to be a sphere of radius \(1.2\) \(\times 10^{-15} \mathrm{~m}\) and its mass is \(1.67 \times 10^{-27} \mathrm{~kg} .\) The volume of a sphere is \((4 / 3) \pi r^{3}\).

A calibrated flask has a mass of \(30.0\) g when empty, \(81.0\) g when filled with water, and \(68.0 \mathrm{~g}\) when filled with an oil. Find the density of the oil. First find the volume of the flask from \(\rho=m / v\) using the water data: $$ V=\frac{m}{\rho}=\frac{(81.0-30.0) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=51.0 \times 10^{-6} \mathrm{~m}^{3} $$ Then, for the oil. $$ \rho_{\mathrm{oil}}=\frac{m_{\mathrm{cil}}}{V}=\frac{(68.0-30.0) \times 10^{-3} \mathrm{~kg}}{51.0 \times 10^{-6} \mathrm{~m}^{3}}=745 \mathrm{~kg} / \mathrm{m}^{3} $$

Air has a density of \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) under standard conditions. What is the mass of air in a room with dimensions \(10.0 \mathrm{~m} \times 8.00 \mathrm{~m} \times\) \(3.00 \mathrm{~m} ?\)

An electrolytic tin-plating process gives a tin coating that is \(7.50 \times\) \(10^{-5}\) thick. How large an area can be coated with \(0.500 \mathrm{~kg}\) of tin? The density of tin is \(7300 \mathrm{~kg} / \mathrm{m}^{3}\). The volume of \(0.500\) kg of tin is given by \(\rho=m / V\) to be $$ V=\frac{m}{\rho}=\frac{0.500 \mathrm{~kg}}{7300 \mathrm{~kg} / \mathrm{m}^{3}}=6.85 \times 10^{-5} \mathrm{~m}^{3} $$ The volume of a film with area \(A\) and thickness \(d\) is \(V=A d\). Solving for \(A\), we find $$ A=\frac{V}{d}=\frac{6.85 \times 10^{-5} \mathrm{~m}^{3}}{7.50 \times 10^{-7} \mathrm{~m}}=91.3 \mathrm{~m}^{2} $$ as the area that can be covered.
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