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Density of Regular Solids (Cube, Sphere, Cylinder, Cone, Prism)

How is the density of regular solids (such as a cube, sphere, cylinder, cone, and rectangular prism) calculated from their mass and dimensions?

Subject: General Chemistry Chapter: Matter Its Properties and Measurement Topic: Density of Regular Solids Answer included
density of regular solids density formula mass divided by volume volume of cube volume of sphere volume of cylinder volume of cone volume of prism
Accepted answer Answer included

Density of regular solids means determining how much mass is packed into a known geometric volume (cube, rectangular prism, cylinder, sphere, cone, etc.).

Core definition

Density is mass per unit volume: \[ \rho = \frac{m}{V}. \]

Common units: kg/m3, g/cm3. Consistency of units for \(m\) and \(V\) is essential.

Step-by-step method for any regular solid

1) Identify the solid and measure its dimensions. Examples: cube side \(a\), cylinder radius \(r\) and height \(h\), sphere radius \(r\), rectangular prism sides \(L,W,H\).

2) Compute the volume \(V\) using the correct geometric formula. (A reference table is provided below.)

3) Measure or obtain the mass \(m\). Use a balance/scale; keep units consistent (e.g., grams with cm3, kilograms with m3).

4) Compute density.

\[ \rho = \frac{m}{V}. \]

Volume formulas for common regular solids

Solid Typical dimensions Volume \(V\) Density expression \( \rho = \frac{m}{V} \)
Cube Side \(a\) \(V = a^3\) \(\rho = \frac{m}{a^3}\)
Rectangular prism \(L, W, H\) \(V = LWH\) \(\rho = \frac{m}{LWH}\)
Cylinder Radius \(r\), height \(h\) \(V = \pi r^2 h\) \(\rho = \frac{m}{\pi r^2 h}\)
Sphere Radius \(r\) \(V = \frac{4}{3}\pi r^3\) \(\rho = \frac{m}{\frac{4}{3}\pi r^3}\)
Cone Base radius \(r\), height \(h\) \(V = \frac{1}{3}\pi r^2 h\) \(\rho = \frac{m}{\frac{1}{3}\pi r^2 h}\)
Right circular pyramid (square base) Base side \(a\), height \(h\) \(V = \frac{1}{3}a^2 h\) \(\rho = \frac{m}{\frac{1}{3}a^2 h}\)
General prism Base area \(A_{\text{base}}\), height \(h\) \(V = A_{\text{base}} h\) \(\rho = \frac{m}{A_{\text{base}} h}\)

Worked example (cylinder): density from measurements

Consider a right circular cylinder with measured radius \(r = 2.50 \text{ cm}\), height \(h = 10.0 \text{ cm}\), and mass \(m = 420 \text{ g}\).

Compute the volume.

\[ V = \pi r^2 h = \pi (2.50\ \text{cm})^2 (10.0\ \text{cm}) = \pi \cdot 6.25\ \text{cm}^2 \cdot 10.0\ \text{cm} = 62.5\pi\ \text{cm}^3. \]

\[ V \approx 62.5 \cdot 3.1416\ \text{cm}^3 \approx 196.35\ \text{cm}^3. \]

Compute the density.

\[ \rho = \frac{m}{V} = \frac{420\ \text{g}}{196.35\ \text{cm}^3} \approx 2.14\ \text{g/cm}^3. \]

Unit check

Using grams and cubic centimeters produces g/cm3. Converting to kg/m3: \[ 2.14\ \text{g/cm}^3 = 2140\ \text{kg/m}^3. \]

Composite solids and average density

For an object made of multiple regular solids (or pieces) with total mass equal to the sum of the masses and total volume equal to the sum of the volumes, the average density is:

\[ \rho_{\text{avg}} = \frac{m_1 + m_2 + \cdots + m_n}{V_1 + V_2 + \cdots + V_n}. \]

This form is also used for an object with internal voids (subtract the void volume from the outer volume) as long as the mass corresponds to the material actually present.

Visualization: mass–volume relationship and slope as density

Density as the slope of a mass–volume graph

Volume, V Mass, m ΔV Δm slope = Δm/ΔV = ρ 0
The straight line shows the proportional relationship \(m = \rho V\) for a uniform material; on a mass–volume plot, the slope is density \( \rho = \frac{\Delta m}{\Delta V} \).

Common pitfalls (quick checks)

  • Mixed units: convert all lengths to the same unit before computing volume; keep mass units consistent with the desired density unit.
  • Radius vs diameter: for circles, \(r = \frac{d}{2}\) must be used in \( \pi r^2 \).
  • Hollow objects: use the material volume (outer volume minus inner void volume).
  • Significant figures: report density with precision consistent with the measurements of \(m\) and the dimensions.
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