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This topic covers Density -- Claimed by Ritvik Khanna, M08 Dr. Greco Spring 2016
Claimed by Deepthi Munagapati, Fall 2025


This topic covers density.


==The Main Idea==
== The Main Idea ==


Density can be simply defined as the relative compactness of a substance. The consistency of the matter at hand can be quantified by mass per unit volume. The density, also called the volumetric mass density, of a substance is denoted by the greek symbol ρ (the lowercase Greek letter rho) but sometimes it can also be represented by the Latin letter D. The density of materials varies, different materials will have different densities. For instance, osmium and iridium are different elements and therefore have different densities. Osmium has a density of 22.59 g/cm^3 while iridium has a density of 22.42 g/cm^3. Also the density of a material can be related to the materials buoyancy, packaging, and purity.  
Density describes how much mass is contained in a given volume. It is a measure of how “packed” matter is. Formally, density is usually written with the Greek letter <math>\rho</math> (rho), and in some engineering contexts you may also see the letter <math>D</math>.


===A Mathematical Model===
At a basic level:


Mathematically density is defined as mass divided by volume.  
* If two objects have the same volume, the one with larger mass has the higher density.
* If two objects have the same mass, the one that takes up less space has the higher density.


:<math> \rho = \frac{m}{V},</math>
Typical values at room temperature:


Density = Mass / Volume  
* Water: <math>\rho \approx 1.0~\text{g/cm}^3 = 1000~\text{kg/m}^3</math> 
* Aluminum: <math>\rho \approx 2.70~\text{g/cm}^3</math> 
* Copper: <math>\rho \approx 8.94~\text{g/cm}^3</math> 
* Ice: <math>\rho \approx 0.92~\text{g/cm}^3</math> 
* Air: <math>\rho \approx 0.0012~\text{g/cm}^3</math>  


In this equation ρ is the density, m is the mass, and V is the volume.The mass can be expressed in grams [g] or kilograms [kg], and the volume is measured in liters [L], cubic centimeters [cm^3], or milliliters [mL]. In this equation mass is divided by volume and in the US oil and gas industry sometimes density is referred to as weight per unit volume. This is technically incorrect as this quantity is actually called specific weight.  
These differences in density explain everyday phenomena like floating and sinking, as well as engineering choices such as why airplanes are not built out of very dense metals.


== A Mathematical Model ==


===A Computational Model===
For a uniform material, density is defined as mass divided by volume:


The density of a solution is the sum of the mass of the components of that solution. Mass concentrations of each component ρi in a solution sums to density of the solution. Therefore this can be modeled by the following equation.
:<math>\rho = \frac{m}{V}</math>


\rho = \sum_i \varrho_i \,
where


This expressed as a function of the densities of pure components of the mixture allows for the computation of the excess molar volumes through the following summation.  
* <math>\rho</math> is the density,
* <math>m</math> is the mass,
* <math>V</math> is the volume.


\rho = \sum_i  \rho_i \frac{V_i}{V}\,= \sum_i  \rho_i \varphi_i =\sum_i  \rho_i \frac{V_i}{\sum_i V_i + \sum_i V_i^{E}}
You can rearrange this expression depending on what quantity you want:


From this point by having the excess volumes and activity coefficients it is possible to determine the activity coefficients by using the equation below.
* To find mass from density and volume:
::<math>m = \rho V</math>
* To find volume from mass and density:
::<math>V = \dfrac{m}{\rho}</math>


\bar{V^E}_i= RT \frac{\partial (ln(\gamma_i))}{\partial P}
Common unit combinations:


==Examples==
* SI units: <math>\text{kg/m}^3</math>
* Chemistry and materials: <math>\text{g/cm}^3</math> or <math>\text{g/mL}</math>


It is essential to keep units consistent. For example:


===Simple===
* <math>1~\text{g/cm}^3 = 1000~\text{kg/m}^3</math>
* <math>1~\text{mL} = 1~\text{cm}^3</math>


What is the weight of the isopropyl alcohol that exactly fills a 200.0 mL container? The density of isopropyl alcohol is 0.786 g/mL.
=== Average vs Local Density ===


d=g/mL
For extended objects that may not be uniform, you can define:


g=(d)(mL)
* **Average density**:
::<math>\rho_{\text{avg}} = \dfrac{\text{total mass}}{\text{total volume}}.</math>


g=(0.789 g/mL)(200.0 mL) = 157.2 g
* **Local (or point) density**: for a nonuniform material the density can vary from point to point. In that case, density is defined in terms of differentials:
::<math>\rho = \frac{\mathrm{d}m}{\mathrm{d}V}</math>


...therefore the isopropyl alcohol weights <math>157.2</math> grams
The total mass of a nonuniform object is then


===Middling===
:<math>m = \int_V \rho(\vec{r})\,\mathrm{d}V,</math>


A copper ingot has a mass of 2.15 kg. If the copper is drawn into wire whose diameter is 2.27 mm, how many inches of copper wire can be obtained from the ingot?
where the integral is taken over the entire volume of the object and <math>\rho(\vec{r})</math> can depend on position.


a) Determine volume of copper:
Later in physics this same idea is used for charge density and current density in electricity and magnetism.


<math>8.94</math> g/cm^3 = <math>2150</math> g/x
=== Density, Weight, and Specific Weight ===
x = <math>240.49217 </math> cm^3


Note: this is the volume of the wire.
Density uses **mass**, not weight. Weight is a force:


b) Determine h in the volume of a cylinder (the wire)
* Mass is measured in kilograms.
* Weight is measured in newtons and is given by <math>W = mg</math>.


<math>240.49217</math> cm^3 = <math>3.14159</math> * <math>0.1135</math>*2 *{h}
Sometimes in industry people informally talk about “weight per unit volume”. The correct term for that quantity is **specific weight**, which equals <math>\rho g</math>.


{h} = <math>5942.37</math> cm
== A Computational Model ==


Note: 0.1135 cm is the radius
In many physics and engineering problems you need to compute mass, volume, or density repeatedly, often for many objects or for small pieces of a larger system. This is where a computational model is useful.


c) Convert cm to inch:
A basic computational idea:


5942.37 cm divided by 2.54 cm/in = 2340 inches
* Store density <math>\rho</math> and volume <math>V</math> for each object.
* Compute mass with a simple relation:
::<math>m = \rho V</math>
 
For a continuously varying density <math>\rho(x)</math> along a one dimensional rod of length <math>L</math>:
 
* Divide the rod into many small pieces of width <math>\Delta x</math>.
* Approximate the mass of each piece as:
::<math>\Delta m_i \approx \rho(x_i) A \Delta x</math>
  where <math>A</math> is the cross sectional area.
* Sum over all the pieces:
::<math>m \approx \sum_i \rho(x_i) A \Delta x</math>
 
This numerical sum approximates the exact integral
 
:<math>m = \int_0^L \rho(x) A\,\mathrm{d}x.</math>
 
This same approach is used in GlowScript or VPython simulations where you might represent an object as many small elements and calculate total mass, center of mass, or gravitational forces by summing over these elements.
 
== Examples ==
 
=== Simple ===
 
What is the mass of isopropyl alcohol that exactly fills a <math>200.0~\text{mL}</math> container? The density of isopropyl alcohol is <math>0.786~\text{g/mL}</math>.
 
Start from
 
:<math>\rho = \dfrac{m}{V} \quad \Rightarrow \quad m = \rho V.</math>
 
Substitute:
 
:<math>m = \left(0.786~\dfrac{\text{g}}{\text{mL}}\right) \left(200.0~\text{mL}\right) = 157.2~\text{g}.</math>
 
So the alcohol has a mass of <math>1.572 \times 10^2~\text{g}</math>.
 
=== Middling ===
 
A copper ingot has a mass of <math>2.15~\text{kg}</math>. If the copper is drawn into a wire whose diameter is <math>2.27~\text{mm}</math>, how many inches of copper wire can be obtained? The density of copper is <math>8.94~\text{g/cm}^3</math>. Treat the wire as a cylinder.
 
Step (a): Find the volume of copper.
 
Convert the mass to grams:
 
:<math>m = 2.15~\text{kg} = 2150~\text{g}.</math>
 
Use <math>\rho = m/V</math>:
 
:<math>8.94~\dfrac{\text{g}}{\text{cm}^3} = \dfrac{2150~\text{g}}{V}</math>
 
So
 
:<math>V = \dfrac{2150~\text{g}}{8.94~\text{g/cm}^3} \approx 240.5~\text{cm}^3.</math>
 
Step (b): Relate this volume to the volume of a cylinder.
 
The diameter is <math>2.27~\text{mm} = 0.227~\text{cm}</math>, so the radius is <math>r = 0.1135~\text{cm}</math>.
 
Volume of a cylinder:
 
:<math>V = \pi r^2 h \quad \Rightarrow \quad h = \dfrac{V}{\pi r^2}.</math>
 
Substitute:
 
:<math>h = \dfrac{240.5~\text{cm}^3}{\pi (0.1135~\text{cm})^2} \approx 5.94 \times 10^3~\text{cm}.</math>
 
Step (c): Convert to inches.
 
Use <math>1~\text{in} = 2.54~\text{cm}</math>:
 
:<math>h = \dfrac{5.94 \times 10^3~\text{cm}}{2.54~\text{cm/in}} \approx 2.34 \times 10^3~\text{in}.</math>
 
So you can draw about <math>2.3 \times 10^3</math> inches of wire.


=== Difficult ===
=== Difficult ===


Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter =<math>6.304 x 10^ {-3}</math> inches) can be produced from the copper that is in<math>5.88</math> pounds of covellite, an ore of copper that is <math>66%</math> copper by mass? (Hint: treat the wire as a cylinder. The density of copper is <math>8.94</math>g/ cm^3; one kg weighs <math>2.2046 </math> lb. The volume of a cylinder is πr2h.
Copper can be drawn into thin wires. How many meters of 34 gauge wire (diameter <math>6.304 \times 10^{-3}~\text{in}</math>) can be produced from the copper that is in <math>5.88~\text{lb}</math> of covellite, an ore of copper that is <math>66\%</math> copper by mass? Treat the wire as a cylinder. The density of copper is <math>8.94~\text{g/cm}^3</math>. One kilogram weighs <math>2.2046~\text{lb}</math>. The volume of a cylinder is <math>V_{\text{cyl}} = \pi r^2 h</math>.
 
Step (a): Find pounds of pure copper.
 
:<math>m_{\text{Cu, lb}} = 0.66 \times 5.88~\text{lb} = 3.8808~\text{lb}.</math>
 
Step (b): Convert to kilograms and grams.
 
:<math>m_{\text{Cu}} = \dfrac{3.8808~\text{lb}}{2.2046~\text{lb/kg}} \approx 1.760~\text{kg} \approx 1.76 \times 10^3~\text{g}.</math>
 
(You can keep extra digits in intermediate steps if you like.)
 
Step (c): Use density to find the volume of copper.
 
:<math>V = \dfrac{m}{\rho} = \dfrac{1.76 \times 10^3~\text{g}}{8.94~\text{g/cm}^3} \approx 197~\text{cm}^3.</math>
 
Step (d): Convert the wire diameter to radius in centimeters.
 
Diameter: <math>d = 6.304 \times 10^{-3}~\text{in}</math>.
 
Radius:
 
:<math>r = \frac{d}{2} = 3.152 \times 10^{-3}~\text{in}.</math>
 
Convert to centimeters:
 
:<math>r = (3.152 \times 10^{-3}~\text{in})(2.54~\text{cm/in}) \approx 8.00 \times 10^{-3}~\text{cm}.</math>
 
Step (e): Use the cylinder volume to find the length.
 
:<math>V = \pi r^2 h \quad \Rightarrow \quad h = \dfrac{V}{\pi r^2}.</math>
 
Substitute:
 
:<math>h = \dfrac{197~\text{cm}^3}{\pi (8.00 \times 10^{-3}~\text{cm})^2} \approx 9.80 \times 10^5~\text{cm}.</math>
 
Convert centimeters to meters:
 
:<math>h = 9.80 \times 10^5~\text{cm} \times \dfrac{1~\text{m}}{100~\text{cm}} \approx 9.80 \times 10^3~\text{m}.</math>
 
So you can draw on the order of <math>10^4~\text{m}</math> of thin copper wire.


== Common Mistakes ==


a) Determine pounds of pure copper in 5.88 lbs of covellite:
* **Mixing mass and weight** 
  Density uses mass, not weight. If you plug weight (in newtons) into the density formula, your units will be wrong.


5.88 lbs x 0.66 = 3.8808 lbs
* **Ignoring unit conversions** 
  Be very careful with units when moving between <math>\text{g/cm}^3</math>, <math>\text{kg/m}^3</math>, and milliliters. Convert everything into one consistent system before computing.


b) Convert pounds to kilograms:
* **Using the wrong volume** 
  For wires, rods, and spheres, you need the volume of the shape, not just its length or radius. Always write the appropriate geometric formula first.


3.8808 lbs ÷ 2.6046 lbs kg¯1 = 1.489979 kg (I'm keeping a few guard digits)
* **Assuming density never changes** 
1.489979 kg = 1489.979 g
  For many materials density changes with temperature and phase. For example, liquid water and ice have different densities.


c) Determine volume this amount of copper occupies:
== Connectedness ==


8.94 g cm¯3 = 1489.979 g / x
=== Importance of Density ===
x = 166.664 cm3


Note: this is the volume of the cylinder.
Density is a fundamental property of matter. It connects microscopic structure to macroscopic behavior:


d) Convert the diameter in inches to a radius in centimeters:
* It helps determine whether objects float or sink in fluids.
* It is used to estimate mass when you can measure volume more easily.
* It appears in models of planets, stars, and galaxies where mass distribution matters.


dia = 6.304 x 10¯3 in; radius = 3.152 x 10¯3 in
Density is also frequently used when analyzing charge and mass distributions. In electricity and magnetism, charge density plays the same role that mass density plays in mechanics:
(3.152 x 10¯3 inch) (2.54 cm / inch) = 8.00 x 10¯3 cm


e) Determine h in the volume of a cylinder:
* Charge density influences electric fields and potentials.
* Mass density influences gravitational fields and motion under gravity.


166.664 cm3 = (3.14159) (8.00 x 10¯3 cm)2 h
=== How Density Changes ===
h = 8.29 x 105 cm = 5.63 x 103 m


Density can increase in either of two ways:


==Connectedness==
* Increasing mass while keeping volume fixed (for example, adding more material into the same container).
* Decreasing volume while keeping mass fixed (for example, compressing a gas or squeezing foam).


==Importance of Density==
It can decrease by:
:* Density is a very important physical characteristic of matter. Density can increase or decrease as the result of actions taken on an object. Overall the characteristics of density are very important in the the way the universe works and in our daily lives.


==How Density Changes==
* Removing mass from a fixed volume.
:* Density can increase either due to increasing mass or decreasing volume. For instance, if you have two objects that have the same mass and you compress one of the objects to a smaller size, the smaller sized object will have a higher density than the non compressed object. On the other hand, if you have two objets of equal volume but different mass, the object with the greater mass will have a high density. It is for this reason that a bowling ball has a higher density than a volleyball, even though both are similar in volume.  
* Allowing a compressed object to expand.


==Helium Balloons==
For example:
:* Helium balloons that we see at most birthday parties work in the way they do because of helium's density. Since the helium air inside the balloon is less dense than the air around the balloon, the balloon is able to float. This phenomenon is true for weather balloons and many other things.


==Ice and Icebergs==
* A bowling ball and a volleyball can have similar volumes, but the bowling ball has much greater mass, so its density is larger.
:* Icebergs often cause many problems and can be hazardous to ships. The reason that icebergs stay afloat is because the density of water changes as the temperature drops and ice floats. In addition, icebergs are made of frozen freshwater which has a lower density than the surrounding cold salty seawater which is more dense. Therefore icebergs remain afloat and pose quite the danger for ships.  
* If you compress a piece of foam into a smaller volume without losing material, its density increases.


==Density and Chemical Engineers==
=== Helium Balloons ===
:* It is very important for chemical engineers to understand the microscopic structure and macroscopic properties of matter. Chemical engineers and bio chemists also use the concept of density functional theory (DFT) in their work. DFT is a computational quantum mechanical modeling method used to investigate the electronic structure of particular atoms, molecules, and the condensed phases. Recently DFT advances have expanded on the knowledge of condensed fluids, polymer solutions, liquid crystal, and colloids all of which help with the fabrication of different types of materials.


Helium balloons float because the gas inside them is less dense than the surrounding air. According to Archimedes’ principle:


==History==
* The buoyant force on a balloon equals the weight of the displaced air.
* If the weight of the helium plus the balloon is less than the weight of the displaced air, the net force points upward.


:* Archimedes in 250 B.C. "discovered" density. He suspected that he was being cheated on by the metal craftsmen who had constructed his golden crown. Archimedes suspected that the craftsman was using a sizeable amount of silver with his gold crown, which is cheaper, and ultimately an insult to the king. Archimedes put the crown that the craftsmen had created as well as a crown of pure gold both into a bathtub full of water. He measured the amount of water that was in excess and confirmed his suspicions. He then proceeds to chant "Eureka! Eureka!" in the streets.  
Weather balloons and blimps use the same idea, just at different scales and with different gases.
 
=== Ice and Icebergs ===
 
Ice has a lower density than liquid water, which is unusual compared with many substances that get denser when they freeze. For water:
 
* The crystalline structure of ice creates more empty space.
* As a result, ice floats in liquid water.
 
Icebergs are made of frozen freshwater that is less dense than the cold, salty seawater that surrounds them. Because of this:
 
* Only a small fraction of an iceberg is visible above the surface.
* Most of the iceberg’s volume is below the waterline, which creates navigational hazards.
 
=== Density and Chemical Engineers ===
 
For chemical engineers and materials scientists, density is not just a number; it connects structure to function. Some uses include:
 
* Monitoring concentration and purity of liquid mixtures by measuring density.
* Designing separation processes where density differences are used to separate phases.
* Modeling material behavior using density functional theory (DFT), a quantum mechanical method used to study the electronic structure of atoms, molecules, and condensed phases.
 
DFT and related approaches have helped expand understanding of condensed fluids, polymer solutions, liquid crystals, and colloids. Many of these are important in designing and fabricating new materials.
 
== History ==
 
A famous historical story about density involves Archimedes (around 250 BCE). The legend says that Archimedes was asked to determine whether a crown thought to be made of pure gold had been adulterated with a less valuable metal.
 
* He realized that if the crown had the same mass as a pure gold reference piece but displaced more water, it must have a lower density, and therefore contain another metal.
* By comparing water displacement, he could test the density without melting the crown.
 
Although the details of the story are probably simplified, the underlying method is exactly what we still use today: measure mass, measure volume (often by displacement), then compare densities.


== See also ==
== See also ==


===Videos===
=== Videos ===
 
These videos give additional intuition and worked examples on density.
 
:* https://www.youtube.com/watch?v=UggNVz87kys
:* https://www.youtube.com/watch?v=UggNVz87kys
:* https://www.youtube.com/watch?v=4tYXaCADxfE
:* https://www.youtube.com/watch?v=4tYXaCADxfE
Line 131: Line 298:
:* https://www.youtube.com/watch?v=5tVNGlG1GBc
:* https://www.youtube.com/watch?v=5tVNGlG1GBc


===Further reading===
=== Further reading ===
NOTE: This section contains interesting articles, stories, and real life applications revolving around density.  
 
NOTE: This section contains interesting articles, stories, and real life applications involving density.


:* http://www.propertiesofmatter.si.edu/titanic.html
:* http://www.propertiesofmatter.si.edu/titanic.html
Line 138: Line 306:
:* http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Density-and-Volume-Real-life applications.html
:* http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Density-and-Volume-Real-life applications.html


=== External links ===


===External links===
:* http://www.sciencedirect.com/science/article/pii/0016706189900402
:* http://www.sciencedirect.com/science/article/pii/0016706189900402
:* http://www.ssc.education.ed.ac.uk/BSL/chemistry/densityd.html
:* http://www.ssc.education.ed.ac.uk/BSL/chemistry/densityd.html


==References==
== References ==


https://en.wikipedia.org/wiki/Density
:* https://en.wikipedia.org/wiki/Density
http://dl6.globalstf.org/index.php/joc/article/view/980
:* http://dl6.globalstf.org/index.php/joc/article/view/980
http://www.chemteam.info/SigFigs/WS-Density-Probs1-10-Ans.html
:* http://www.chemteam.info/SigFigs/WS-Density-Probs1-10-Ans.html
http://www.ssc.education.ed.ac.uk/BSL/pictures/density.jpg
:* http://www.ssc.education.ed.ac.uk/BSL/pictures/density.jpg
http://www.softschools.com/examples/science/density_examples/9/
:* http://www.softschools.com/examples/science/density_examples/9/
http://www.ehow.com/about_5484217_importance-density.html
:* http://www.ehow.com/about_5484217_importance-density.html
http://onlinelibrary.wiley.com/doi/10.1002/aic.10713/abstract
:* http://onlinelibrary.wiley.com/doi/10.1002/aic.10713/abstract
http://archimedespalimpsest.org/about/history/archimedes.php
:* http://archimedespalimpsest.org/about/history/archimedes.php
Chabay & Sherwood: Matters and Interactions -- Modern Mechanics Volume 1, 4th Edition
:* Chabay & Sherwood: ''Matter and Interactions Modern Mechanics'', Volume 1, 4th Edition


[[Category:Properties of Matter]]
[[Category:Properties of Matter]]

Latest revision as of 13:54, 2 December 2025

Claimed by Deepthi Munagapati, Fall 2025

This topic covers density.

The Main Idea

Density describes how much mass is contained in a given volume. It is a measure of how “packed” matter is. Formally, density is usually written with the Greek letter [math]\displaystyle{ \rho }[/math] (rho), and in some engineering contexts you may also see the letter [math]\displaystyle{ D }[/math].

At a basic level:

  • If two objects have the same volume, the one with larger mass has the higher density.
  • If two objects have the same mass, the one that takes up less space has the higher density.

Typical values at room temperature:

  • Water: [math]\displaystyle{ \rho \approx 1.0~\text{g/cm}^3 = 1000~\text{kg/m}^3 }[/math]
  • Aluminum: [math]\displaystyle{ \rho \approx 2.70~\text{g/cm}^3 }[/math]
  • Copper: [math]\displaystyle{ \rho \approx 8.94~\text{g/cm}^3 }[/math]
  • Ice: [math]\displaystyle{ \rho \approx 0.92~\text{g/cm}^3 }[/math]
  • Air: [math]\displaystyle{ \rho \approx 0.0012~\text{g/cm}^3 }[/math]

These differences in density explain everyday phenomena like floating and sinking, as well as engineering choices such as why airplanes are not built out of very dense metals.

A Mathematical Model

For a uniform material, density is defined as mass divided by volume:

[math]\displaystyle{ \rho = \frac{m}{V} }[/math]

where

  • [math]\displaystyle{ \rho }[/math] is the density,
  • [math]\displaystyle{ m }[/math] is the mass,
  • [math]\displaystyle{ V }[/math] is the volume.

You can rearrange this expression depending on what quantity you want:

  • To find mass from density and volume:
[math]\displaystyle{ m = \rho V }[/math]
  • To find volume from mass and density:
[math]\displaystyle{ V = \dfrac{m}{\rho} }[/math]

Common unit combinations:

  • SI units: [math]\displaystyle{ \text{kg/m}^3 }[/math]
  • Chemistry and materials: [math]\displaystyle{ \text{g/cm}^3 }[/math] or [math]\displaystyle{ \text{g/mL} }[/math]

It is essential to keep units consistent. For example:

  • [math]\displaystyle{ 1~\text{g/cm}^3 = 1000~\text{kg/m}^3 }[/math]
  • [math]\displaystyle{ 1~\text{mL} = 1~\text{cm}^3 }[/math]

Average vs Local Density

For extended objects that may not be uniform, you can define:

  • **Average density**:
[math]\displaystyle{ \rho_{\text{avg}} = \dfrac{\text{total mass}}{\text{total volume}}. }[/math]
  • **Local (or point) density**: for a nonuniform material the density can vary from point to point. In that case, density is defined in terms of differentials:
[math]\displaystyle{ \rho = \frac{\mathrm{d}m}{\mathrm{d}V} }[/math]

The total mass of a nonuniform object is then

[math]\displaystyle{ m = \int_V \rho(\vec{r})\,\mathrm{d}V, }[/math]

where the integral is taken over the entire volume of the object and [math]\displaystyle{ \rho(\vec{r}) }[/math] can depend on position.

Later in physics this same idea is used for charge density and current density in electricity and magnetism.

Density, Weight, and Specific Weight

Density uses **mass**, not weight. Weight is a force:

  • Mass is measured in kilograms.
  • Weight is measured in newtons and is given by [math]\displaystyle{ W = mg }[/math].

Sometimes in industry people informally talk about “weight per unit volume”. The correct term for that quantity is **specific weight**, which equals [math]\displaystyle{ \rho g }[/math].

A Computational Model

In many physics and engineering problems you need to compute mass, volume, or density repeatedly, often for many objects or for small pieces of a larger system. This is where a computational model is useful.

A basic computational idea:

  • Store density [math]\displaystyle{ \rho }[/math] and volume [math]\displaystyle{ V }[/math] for each object.
  • Compute mass with a simple relation:
[math]\displaystyle{ m = \rho V }[/math]

For a continuously varying density [math]\displaystyle{ \rho(x) }[/math] along a one dimensional rod of length [math]\displaystyle{ L }[/math]:

  • Divide the rod into many small pieces of width [math]\displaystyle{ \Delta x }[/math].
  • Approximate the mass of each piece as:
[math]\displaystyle{ \Delta m_i \approx \rho(x_i) A \Delta x }[/math]
 where [math]\displaystyle{ A }[/math] is the cross sectional area.
  • Sum over all the pieces:
[math]\displaystyle{ m \approx \sum_i \rho(x_i) A \Delta x }[/math]

This numerical sum approximates the exact integral

[math]\displaystyle{ m = \int_0^L \rho(x) A\,\mathrm{d}x. }[/math]

This same approach is used in GlowScript or VPython simulations where you might represent an object as many small elements and calculate total mass, center of mass, or gravitational forces by summing over these elements.

Examples

Simple

What is the mass of isopropyl alcohol that exactly fills a [math]\displaystyle{ 200.0~\text{mL} }[/math] container? The density of isopropyl alcohol is [math]\displaystyle{ 0.786~\text{g/mL} }[/math].

Start from

[math]\displaystyle{ \rho = \dfrac{m}{V} \quad \Rightarrow \quad m = \rho V. }[/math]

Substitute:

[math]\displaystyle{ m = \left(0.786~\dfrac{\text{g}}{\text{mL}}\right) \left(200.0~\text{mL}\right) = 157.2~\text{g}. }[/math]

So the alcohol has a mass of [math]\displaystyle{ 1.572 \times 10^2~\text{g} }[/math].

Middling

A copper ingot has a mass of [math]\displaystyle{ 2.15~\text{kg} }[/math]. If the copper is drawn into a wire whose diameter is [math]\displaystyle{ 2.27~\text{mm} }[/math], how many inches of copper wire can be obtained? The density of copper is [math]\displaystyle{ 8.94~\text{g/cm}^3 }[/math]. Treat the wire as a cylinder.

Step (a): Find the volume of copper.

Convert the mass to grams:

[math]\displaystyle{ m = 2.15~\text{kg} = 2150~\text{g}. }[/math]

Use [math]\displaystyle{ \rho = m/V }[/math]:

[math]\displaystyle{ 8.94~\dfrac{\text{g}}{\text{cm}^3} = \dfrac{2150~\text{g}}{V} }[/math]

So

[math]\displaystyle{ V = \dfrac{2150~\text{g}}{8.94~\text{g/cm}^3} \approx 240.5~\text{cm}^3. }[/math]

Step (b): Relate this volume to the volume of a cylinder.

The diameter is [math]\displaystyle{ 2.27~\text{mm} = 0.227~\text{cm} }[/math], so the radius is [math]\displaystyle{ r = 0.1135~\text{cm} }[/math].

Volume of a cylinder:

[math]\displaystyle{ V = \pi r^2 h \quad \Rightarrow \quad h = \dfrac{V}{\pi r^2}. }[/math]

Substitute:

[math]\displaystyle{ h = \dfrac{240.5~\text{cm}^3}{\pi (0.1135~\text{cm})^2} \approx 5.94 \times 10^3~\text{cm}. }[/math]

Step (c): Convert to inches.

Use [math]\displaystyle{ 1~\text{in} = 2.54~\text{cm} }[/math]:

[math]\displaystyle{ h = \dfrac{5.94 \times 10^3~\text{cm}}{2.54~\text{cm/in}} \approx 2.34 \times 10^3~\text{in}. }[/math]

So you can draw about [math]\displaystyle{ 2.3 \times 10^3 }[/math] inches of wire.

Difficult

Copper can be drawn into thin wires. How many meters of 34 gauge wire (diameter [math]\displaystyle{ 6.304 \times 10^{-3}~\text{in} }[/math]) can be produced from the copper that is in [math]\displaystyle{ 5.88~\text{lb} }[/math] of covellite, an ore of copper that is [math]\displaystyle{ 66\% }[/math] copper by mass? Treat the wire as a cylinder. The density of copper is [math]\displaystyle{ 8.94~\text{g/cm}^3 }[/math]. One kilogram weighs [math]\displaystyle{ 2.2046~\text{lb} }[/math]. The volume of a cylinder is [math]\displaystyle{ V_{\text{cyl}} = \pi r^2 h }[/math].

Step (a): Find pounds of pure copper.

[math]\displaystyle{ m_{\text{Cu, lb}} = 0.66 \times 5.88~\text{lb} = 3.8808~\text{lb}. }[/math]

Step (b): Convert to kilograms and grams.

[math]\displaystyle{ m_{\text{Cu}} = \dfrac{3.8808~\text{lb}}{2.2046~\text{lb/kg}} \approx 1.760~\text{kg} \approx 1.76 \times 10^3~\text{g}. }[/math]

(You can keep extra digits in intermediate steps if you like.)

Step (c): Use density to find the volume of copper.

[math]\displaystyle{ V = \dfrac{m}{\rho} = \dfrac{1.76 \times 10^3~\text{g}}{8.94~\text{g/cm}^3} \approx 197~\text{cm}^3. }[/math]

Step (d): Convert the wire diameter to radius in centimeters.

Diameter: [math]\displaystyle{ d = 6.304 \times 10^{-3}~\text{in} }[/math].

Radius:

[math]\displaystyle{ r = \frac{d}{2} = 3.152 \times 10^{-3}~\text{in}. }[/math]

Convert to centimeters:

[math]\displaystyle{ r = (3.152 \times 10^{-3}~\text{in})(2.54~\text{cm/in}) \approx 8.00 \times 10^{-3}~\text{cm}. }[/math]

Step (e): Use the cylinder volume to find the length.

[math]\displaystyle{ V = \pi r^2 h \quad \Rightarrow \quad h = \dfrac{V}{\pi r^2}. }[/math]

Substitute:

[math]\displaystyle{ h = \dfrac{197~\text{cm}^3}{\pi (8.00 \times 10^{-3}~\text{cm})^2} \approx 9.80 \times 10^5~\text{cm}. }[/math]

Convert centimeters to meters:

[math]\displaystyle{ h = 9.80 \times 10^5~\text{cm} \times \dfrac{1~\text{m}}{100~\text{cm}} \approx 9.80 \times 10^3~\text{m}. }[/math]

So you can draw on the order of [math]\displaystyle{ 10^4~\text{m} }[/math] of thin copper wire.

Common Mistakes

  • **Mixing mass and weight**
 Density uses mass, not weight. If you plug weight (in newtons) into the density formula, your units will be wrong.
  • **Ignoring unit conversions**
 Be very careful with units when moving between [math]\displaystyle{ \text{g/cm}^3 }[/math], [math]\displaystyle{ \text{kg/m}^3 }[/math], and milliliters. Convert everything into one consistent system before computing.
  • **Using the wrong volume**
 For wires, rods, and spheres, you need the volume of the shape, not just its length or radius. Always write the appropriate geometric formula first.
  • **Assuming density never changes**
 For many materials density changes with temperature and phase. For example, liquid water and ice have different densities.

Connectedness

Importance of Density

Density is a fundamental property of matter. It connects microscopic structure to macroscopic behavior:

  • It helps determine whether objects float or sink in fluids.
  • It is used to estimate mass when you can measure volume more easily.
  • It appears in models of planets, stars, and galaxies where mass distribution matters.

Density is also frequently used when analyzing charge and mass distributions. In electricity and magnetism, charge density plays the same role that mass density plays in mechanics:

  • Charge density influences electric fields and potentials.
  • Mass density influences gravitational fields and motion under gravity.

How Density Changes

Density can increase in either of two ways:

  • Increasing mass while keeping volume fixed (for example, adding more material into the same container).
  • Decreasing volume while keeping mass fixed (for example, compressing a gas or squeezing foam).

It can decrease by:

  • Removing mass from a fixed volume.
  • Allowing a compressed object to expand.

For example:

  • A bowling ball and a volleyball can have similar volumes, but the bowling ball has much greater mass, so its density is larger.
  • If you compress a piece of foam into a smaller volume without losing material, its density increases.

Helium Balloons

Helium balloons float because the gas inside them is less dense than the surrounding air. According to Archimedes’ principle:

  • The buoyant force on a balloon equals the weight of the displaced air.
  • If the weight of the helium plus the balloon is less than the weight of the displaced air, the net force points upward.

Weather balloons and blimps use the same idea, just at different scales and with different gases.

Ice and Icebergs

Ice has a lower density than liquid water, which is unusual compared with many substances that get denser when they freeze. For water:

  • The crystalline structure of ice creates more empty space.
  • As a result, ice floats in liquid water.

Icebergs are made of frozen freshwater that is less dense than the cold, salty seawater that surrounds them. Because of this:

  • Only a small fraction of an iceberg is visible above the surface.
  • Most of the iceberg’s volume is below the waterline, which creates navigational hazards.

Density and Chemical Engineers

For chemical engineers and materials scientists, density is not just a number; it connects structure to function. Some uses include:

  • Monitoring concentration and purity of liquid mixtures by measuring density.
  • Designing separation processes where density differences are used to separate phases.
  • Modeling material behavior using density functional theory (DFT), a quantum mechanical method used to study the electronic structure of atoms, molecules, and condensed phases.

DFT and related approaches have helped expand understanding of condensed fluids, polymer solutions, liquid crystals, and colloids. Many of these are important in designing and fabricating new materials.

History

A famous historical story about density involves Archimedes (around 250 BCE). The legend says that Archimedes was asked to determine whether a crown thought to be made of pure gold had been adulterated with a less valuable metal.

  • He realized that if the crown had the same mass as a pure gold reference piece but displaced more water, it must have a lower density, and therefore contain another metal.
  • By comparing water displacement, he could test the density without melting the crown.

Although the details of the story are probably simplified, the underlying method is exactly what we still use today: measure mass, measure volume (often by displacement), then compare densities.

See also

Videos

These videos give additional intuition and worked examples on density.

Further reading

NOTE: This section contains interesting articles, stories, and real life applications involving density.

External links

References