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Chapter 12: Problem 2

What volume does \(300 \mathrm{~g}\) of mercury occupy? The density of mercury is \(13600 \mathrm{~kg} / \mathrm{m}^{3}\). From \(\rho=m / V\), $$ V=\frac{m}{\rho}=\frac{0.300 \mathrm{~kg}}{13600 \mathrm{~kg} / \mathrm{m}^{3}}=2.21 \times 10^{-5} \mathrm{~m}^{3}=22.1 \mathrm{~cm}^{3} $$

Short Answer

Expert verified
The volume occupied by 300 g of mercury is 22.1 cm鲁.

Step by step solution

01

Convert Mass to Kilograms

First, we need to ensure that the mass of mercury is in kilograms to match the units of the density. Given is 300 grams.Convert grams to kilograms:\[ 300 \text{ g} = 0.300 \text{ kg} \]
02

Calculate the Volume

Using the formula for density \( \rho = \frac{m}{V} \), we need to solve for volume \( V \):\[ V = \frac{m}{\rho} \]Substitute the values:\[ V = \frac{0.300 \text{ kg}}{13600 \text{ kg/m}^3} \]
03

Solve for Volume

Calculate the volume by dividing the mass by the density:\[ V = \frac{0.300}{13600} = 2.21 \times 10^{-5} \text{ m}^3 \]
04

Convert Volume to Cubic Centimeters

To express the volume in cubic centimeters, use the conversion factor: 1 m鲁 = 1,000,000 cm鲁.\[ V = 2.21 \times 10^{-5} \times 1,000,000 \text{ cm}^3 \]\[ V = 22.1 \text{ cm}^3 \]

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Key Concepts

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

Understanding Density
Density is a fundamental concept that relates mass and volume. It tells us how much mass is contained in a certain volume of a substance. The formula for density is \( \rho = \frac{m}{V} \), where \( \rho \) represents density, \( m \) is mass, and \( V \) is volume.

For instance, in our exercise, mercury has a density of \( 13600 \; \text{kg/m}^3 \). This means that every cubic meter of mercury has a mass of 13600 kilograms. Density can indicate how "heavy" a substance is for a given volume, and it plays a vital role in determining how substances will interact in physical contexts, such as whether an object will float or sink in water.
Importance of Unit Conversion
Unit conversion is crucial when calculating properties like volume in physics and chemistry because different measurements often use different units.

In our problem, the mass of mercury initially given is in grams (300 g), but the density is in kilograms per cubic meter (\( \text{kg/m}^3 \)). This discrepancy requires converting grams to kilograms to ensure consistent units: \( 300 \; \text{g} = 0.300 \; \text{kg} \).

When converting units, remember some key conversions:
  • 1 kilogram = 1000 grams
  • 1 meter = 100 centimeters
These conversions help maintain accuracy and enable you to process calculations correctly and effectively.
Mass and Volume Relationship
The relationship between mass and volume is directly tied to density. With the formula \( V = \frac{m}{\rho} \), you can find the volume of a substance if the mass and density are known.

This relationship implies that for a given mass, an increase in volume leads to a decrease in density, and vice versa. In our example, knowing mercury's mass (0.300 kg) and density (13600 \( \text{kg/m}^3 \)), we calculate its volume: \( V = \frac{0.300}{13600} = 2.21 \times 10^{-5} \text{ m}^3 \).

To make it more interpretable, we convert the volume into cubic centimeters: \( 2.21 \times 10^{-5} \; \text{m}^3 \times 1,000,000 = 22.1 \; \text{cm}^3 \). This way, we understand how much space 300 g of mercury occupies, showcasing the intimate connection between mass, volume, and density.

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Most popular questions from this chapter

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 \(15 \mathrm{MPa}\).

A \(15-\mathrm{kg}\) ball of radius \(4.0 \mathrm{~cm}\) is suspended from a point \(2.94 \mathrm{~m}\) above the floor by an iron wire of unstretched length \(2.85 \mathrm{~m}\). The diameter of the wire is \(0.090 \mathrm{~cm}\), and its Young's modulus is \(180 \mathrm{GPa}\). If the ball is set swinging so that its center passes through the lowest point at \(5.0 \mathrm{~m} / \mathrm{s}\), by how much does the bottom of the ball clear the floor? Discuss any approximations that you make. Call the tension in the wire \(F_{T}\) when the ball is swinging through the lowest point. Since \(F_{T}\) must supply the centripetal force as well as balance the weight, $$ F_{T}=m g+\frac{m v^{2}}{r}=m\left(9.81+\frac{25}{r}\right) $$ all in proper SI units. This is complicated, because \(r\) is the distance from the pivot to the center of the ball when the wire is stretched, and so it is \(r_{0}+\Delta r\), where \(r_{0}\), the unstretched length of the pendulum, is $$ r_{0}=2.85 \mathrm{~m}+0.040 \mathrm{~m}=2.89 \mathrm{~m} $$ and where \(\Delta r\) is as yet unknown. However, the unstretched distance from the pivot to the bottom of the ball is \(2.85 \mathrm{~m}+0.080 \mathrm{~m}=2.93 \mathrm{~m}\), and so the maximum possible value for \(\Delta r\) is $$ 2.94 \mathrm{~m}-2.93 \mathrm{~m}=0.01 \mathrm{~m} $$ We will therefore incur no more than a \(1 / 3\) percent error in \(r\) by using \(r=r_{0}=2.89 \mathrm{~m}\). This gives \(F_{T}=277 \mathrm{~N}\). Under this tension, the wire stretches by $$ \Delta L=\frac{F L_{0}}{A Y}=\frac{(277 \mathrm{~N})(2.85 \mathrm{~m})}{\pi\left(4.5 \times 10^{-4} \mathrm{~m}\right)^{2}\left(1.80 \times 10^{11} \mathrm{~Pa}\right)}=6.9 \times 10^{-3} \mathrm{~m} $$ Hence, the ball misses by $$ 2.94 \mathrm{~m}-(2.85+0.0069+0.080) \mathrm{m}=0.0031 \mathrm{~m}=3.1 \mathrm{~mm} $$ To check the approximation we have made, we could use \(r=2.90 \mathrm{~m}\), its maximum possible value. Then \(\Delta L=6.9 \mathrm{~mm}\), showing that the approximation has caused a negligible error.

Atmospheric pressure is about \(1.01 \times 10^{5} \mathrm{~Pa}\). How large a force does the atmosphere exert on a \(2.0-\mathrm{cm}^{2}\) area on the top of your head? Because \(P=F / A\), where \(F\) is perpendicular to \(A\), we have \(F=P A\). Assuming that \(2.0 \mathrm{~cm}^{2}\) of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), $$ F=P A=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-4} \mathrm{~m}^{2}\right)=20 \mathrm{~N} $$

A box-shaped piece of gelatin dessert has a top area of \(15 \mathrm{~cm}^{2}\) and a height of \(3.0 \mathrm{~cm}\). When a shearing force of \(0.50 \mathrm{~N}\) is applied to the upper surface, the upper surface displaces \(4.0 \mathrm{~mm}\) relative to the bottom surface. What are the shearing stress, the shearing strain, and the shear modulus for the gelatin? $$ \begin{array}{l} \sigma_{s}=\frac{\text { Tangential force }}{\text { Area of face }}=\frac{0.50 \mathrm{~N}}{15 \times 10^{-4} \mathrm{~m}^{2}}=0.33 \mathrm{kPa} \\ \varepsilon_{s}-\frac{\text { Displacement }}{\text { Height }}-\frac{0.40 \mathrm{~cm}}{3.0 \mathrm{~cm}}-0.13 \\ S=\frac{0.33 \mathrm{kPa}}{0.13}=2.5 \mathrm{kPa} \end{array} $$

An electrolytic tin-plating process gives a tin coating that is \(7.50 \times 10^{-5} \mathrm{~cm}\) 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 \(0.500 \mathrm{~kg}\) 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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