Victor Wall

2022-01-07

Gold, which has a density of $19.32\frac{g}{c}{m}^{3}$, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber.
(a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of $1.000\mu m$ thickness, what is the area of the leaf?
b) If, instead, the gold is drawn out into a cylindrical fiber of radius $2.500\mu m$, what is the length of the fiber?

### Answer & Explanation

Matthew Rodriguez

a) Using the provided information about the density of gold, the sample size, thickness, and the following
equations and conversion factors, find the area of the gold leaf:
$V=ell\cdot w\cdot h=A\cdot h$
$m=p\cdot V$
$1\mu ={10}^{-6}m$
$Gol{d}_{p}=19.32\frac{g}{c}{m}^{3}$
First, find the volume of the sample and then find the area of the sample.
$V=\frac{m}{p}=\frac{27.6g}{19.32g/c{m}^{3}}\cdot \left(\begin{array}{c}\frac{0.01m}{1cm}\end{array}{\right)}^{3}$
$\approx 1.429×{10}^{-6}{m}^{3}$
$V=A\cdot h⇒A=\frac{V}{h}=\frac{1.429×{10}^{-6}{m}^{3}}{1x{10}^{-6}m}\approx 1.429{m}^{2}$
b) Using the provided information from part a), the radius of the cylinder, and the following equation for the
volume of a cylinder, find the length of the fiber
$V=\pi {r}^{2}h⇒h=\frac{V}{\pi {r}^{2}}$
$h=\frac{1.429×{10}^{-6}{m}^{3}}{\pi \cdot {\left(2.5×{10}^{-6}m\right)}^{2}}\approx 72778m$

Ethan Sanders

From the equation(1 - 8), the density (p) of the gold is defined as the mass of the gold (m) per its volume (V):
$p=\frac{m}{V}=19.32\frac{g}{c}{m}^{3}\left(1\right)$
(a) The volume of the thin leaf is found to be;
${V}_{\ell eaf}=\left(area\right)\left(thickness\right)={A}_{\ell eaf}×{t}_{\ell eaf}\left(2\right)$
From equation (1), the volume of the thin leaf will be;
${V}_{\ell eaf}=\frac{{m}_{\ell eaf}}{{p}_{\ell eaf}}=\frac{27.63g}{19.32g/c{m}^{3}}=1.43c{m}^{3}$
Now, by converting the volume to SI units, we get
${V}_{Leaf}=\left(1.43c{m}^{3}\right)\left(\begin{array}{c}\frac{1m}{100cm}\end{array}{\right)}^{3}=1.43×{10}^{-6}{m}^{3}$
The thickness of the thin leaf: Therefore, from the equation (2), we can find the area of the thin leaf as following;
${A}_{elleaf}=\frac{{V}_{Leaf}}{{t}_{Leaf}}=\frac{1.43×{10}^{{}_{6}}{m}^{3}}{1×{10}^{-6}m}1.43{m}^{2}$
(b) The volume of cylindrical fiber is equal to the volume of the thin leaf (same gold)
${V}_{Fiber}={V}_{Leaf}$
As known, the volume of the cylindrical fiber is;
${V}_{Fiber}=\left(cross-sectionalarea\right)\left(length\right)=A\cdot {\ell }_{Fiber}\left(3\right)$
This cross-section area is circular and given by: $A=\pi {r}^{2},wherer=2.5\mu m=2.5×{10}^{-6}m$ Thus from equation (3), we have
${\ell }_{Fiber}=\frac{{V}_{Fiber}}{\pi {r}^{2}}=\frac{1.43×{10}^{-6}{m}^{3}}{\left(3.14\right)\left(2.5×{10}^{-6}m{\right)}^{2}}=7.286×{10}^{4}m$
$\therefore el{l}_{Fiber}=7.286×{10}^{4}m\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}72.866km$

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