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Chapter 12: Problem 35

The compressibility of water is \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}\). Find the decrease in volume of \(100 \mathrm{~mL}\) of water when subjected to a pressure of \(15 \mathrm{MPa}\).

Short Answer

Expert verified
The decrease in volume is 7.5 mL.

Step by step solution

01

Understanding Given Values

We are given the compressibility of water, which is a measure of the fractional change in volume per unit change in pressure. The problem states that the compressibility is \( 5.0 \times 10^{-10} \, \mathrm{m}^2 / \mathrm{N} \). This can also be expressed in terms of pressure in pascals, where \( 1 \, \mathrm{MPa} = 10^6 \, \mathrm{Pa} \). We need to find the volume decrease for an initial volume of \( 100 \, \mathrm{mL} \), which is \( 100 \times 10^{-3} \, \mathrm{L} \) or \( 100 \times 10^{-6} \, \mathrm{m}^3 \). The pressure applied is \( 15 \, \mathrm{MPa} = 15 \times 10^6 \, \mathrm{Pa} \).
02

Using the Compressibility Formula

The formula for volume change due to compressibility is given by: \[ \Delta V = - \beta V \Delta P \] where \( \Delta V \) is the change in volume, \( \beta \) is the compressibility, \( V \) is the original volume, and \( \Delta P \) is the change in pressure.
03

Substituting Values

Substitute the given values into the formula: \[ \Delta V = - (5.0 \times 10^{-10} \, \mathrm{m}^2/\mathrm{N}) \times (100 \times 10^{-6} \, \mathrm{m}^3) \times (15 \times 10^6 \, \mathrm{Pa}) \] This simplifies to: \[ \Delta V = - (5.0 \times 10^{-10} \times 100 \times 10^{-6} \times 15 \times 10^6) \]
04

Calculating the Decrease in Volume

Calculate the expression: \[ \Delta V = - (5.0 \times 10^{-10} \times 100 \times 10^{-6} \times 15 \times 10^6) \] \[ \Delta V = -7.5 \times 10^{-6} \, \mathrm{m}^3 \]
05

Converting Decrease to Milliliters

Convert \( \Delta V \) from cubic meters to milliliters: \[ \Delta V = -7.5 \times 10^{-6} \, \mathrm{m}^3 = -7.5 \times 10^{-3} \, \mathrm{L} = -7.5 \, \mathrm{mL} \] The negative sign indicates a decrease in volume.

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Key Concepts

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

Volume Change
In the context of compressibility, volume change is a measure of how much the volume of a substance decreases when pressure is applied. When a material like water is subjected to increased pressure, its particles are forced closer together, which results in a reduction of its overall volume. This concept is crucial when learning about fluid dynamics and is relevant across various scientific and engineering applications.

In our exercise, we start with a given amount of water, specifically 100 mL, and aim to find out how much volume decreases when a given pressure is applied. The change in volume is often represented by the symbol \( \Delta V \), where \( \Delta \) denotes 'change' and \( V \) stands for volume. In the given solution, the volume changes by \(-7.5 \ \mathrm{mL}\), indicating a reduction in the water volume under the specified conditions.
Pressure in Pascals
Understanding pressure, specifically in pascals (Pa), is vital for solving problems related to compressibility. Pressure is defined as the force applied per unit area, and it鈥檚 measured in units of pascals in the International System of Units (SI). One pascal is equivalent to one newton per square meter.

In our exercise, pressure is applied in megapascals (MPa), a larger unit where \(1 \, \mathrm{MPa} = 10^6 \, \mathrm{Pa}\). This means the applied pressure of 15 MPa is equivalent to \(15 \times 10^6 \, \mathrm{Pa}\). The exercise requires you to convert and understand these units because they directly affect how you calculate the volume change when pressure is applied.
Compressibility Formula
The compressibility formula is the key to calculating how a substance's volume changes when pressure is applied. Compressibility \(\beta\) is a measure of the fractional change in volume per unit change in pressure, expressed in \(\mathrm{m}^2/\mathrm{N}\).

The formula used to calculate the volume change \(\Delta V\) is:
  • \[\Delta V = - \beta V \Delta P\]
Here, \(\Delta V\) represents the change in volume, \(\beta\) is the compressibility of the material, \(V\) denotes the initial volume, and \(\Delta P\) is the change in pressure applied. The negative sign indicates that volume decreases when pressure increases. By substituting the known values into the formula, you can calculate how much the water's volume decreases when it's subjected to the specified pressure, an important skill in fluid mechanics and physics.

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Most popular questions from this chapter

The specific gravity of cast iron is \(7.20\). Find its density and the mass of \(60.0 \mathrm{~cm}^{3}\) of it. Make use of $$ \text { sp } \mathrm{gr}=\frac{\text { Density of substance }}{\text { Density of water }} \quad \text { and } \quad \rho=\frac{m}{V} $$ From the first equation, Density of iron \(=(\) sp gr \()\) (Density of water) \(=(7.20)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=7200 \mathrm{~kg} / \mathrm{m}^{3}\) and so \(\quad\) Mass of \(60.0 \mathrm{~cm}^{3}=\rho V=\left(7200 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(60.0 \times 10^{-6} \mathrm{~m}_{3}\right)=0.432 \mathrm{~kg}\)

A metal wire \(75.0 \mathrm{~cm}\) long and \(0.130 \mathrm{~cm}\) in diameter stretches \(0.0350 \mathrm{~cm}\) when a load of \(8.00 \mathrm{~kg}\) is hung on its end. Find the stress, the strain, and the Young's modulus for the material of the wire. $$ \begin{array}{l} \sigma=\frac{F}{A}=\frac{(8.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{\pi\left(6.50 \times 10^{-4} \mathrm{~m}\right)^{2}}=5.91 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=5.91 \times 10^{7} \mathrm{~Pa} \\ \varepsilon=\frac{\Delta L}{L_{0}}=\frac{0.0350 \mathrm{~cm}}{75.0 \mathrm{~cm}}=4.67 \times 10^{-4} \\ Y=\frac{\sigma}{\varepsilon}=\frac{5.91 \times 10^{7} \mathrm{~Pa}}{4.67 \times 10^{-4}}=1.27 \times 10^{11} \mathrm{~Pa}=127 \mathrm{GPa} \end{array} $$

A \(60-\mathrm{kg}\) motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a crosssectional area of \(15 \mathrm{~cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa} .(a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

A horizontal rectangular platform is suspended by four identical wires, one at each of its corners. The wires are \(3.0 \mathrm{~m}\) long and have a diameter of \(2.0 \mathrm{~mm}\). Young's modulus for the material of the wires is \(180 \mathrm{GPa}\). How far will the platform drop (due to elongation of the wires) if a \(50-\mathrm{kg}\) load is placed at the center of the platform?

Compute the volume change of a solid copper cube, \(40 \mathrm{~mm}\) on each edge, when subjected to a pressure of \(20 \mathrm{MPa}\). The bulk modulus for copper is \(125 \mathrm{GPa}\).
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