Area of a Gold Leaf? | Calculate with Density

[bosburn93]

Homework Statement

Gold, which has a density of 19.32 g/cm^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, witha mass of 27.63g, is pressed into a leaf 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber?

Homework Equations

ρ=m/V and others?

The Attempt at a Solution

Not sure what to do. Got volume but how do i use that to get area? I also do not know what equations would apply to a gold leaf.

 

[Punkyc7]

isnt volume =area * thickness

 

[bosburn93]

isnt volume =area * thickness

Yes, if i assume that it is a cube or rectangular prism.
is that right?

 

[SteamKing]

In general, a prism does not have to be rectangular in cross section for the volume relation to hold, so long as all cross sections are similar. For the gold leaf, the thickness is so small, it would be reasonable to assume that the areas of the front and back of the leaf are equal, hence the leaf could be considered to be a very short prism.

 

[rwisz]

I would suggest using the equation for density, ρ=m/V, to solve for the volume of the gold leaf. Since the mass and density of the gold are given, we can rearrange the equation to solve for volume: V=m/ρ. Plugging in the given values, we get V= 27.63g/19.32 g/cm^3 = 1.43 cm^3.

To find the area of the leaf, we can use the formula for the volume of a cylinder, V=πr^2h, where r is the radius and h is the height. Since we are given the thickness of the leaf as 1.000 µm, we can convert this to cm by dividing by 10,000. So h= 1.000 µm/10,000 = 0.0001 cm.

Now we can rearrange the equation to solve for the area, A: A=V/(πr^2). Plugging in the values we already calculated, we get A= 1.43 cm^3/(π(0.00025 cm)^2) = 2.29 x 10^-3 cm^2. Therefore, the area of the gold leaf is approximately 2.29 x 10^-3 cm^2.

For part (b), to find the length of the cylindrical fiber, we can use the formula for the volume of a cylinder again, V=πr^2h. This time, we are given the radius as 2.500 µm, so r=2.500 µm/10,000 = 0.00025 cm. We can also rearrange the equation to solve for h: h=V/(πr^2). Plugging in the given values, we get h= 1.43 cm^3/(π(0.00025 cm)^2) = 1.83 cm. Therefore, the length of the cylindrical fiber is approximately 1.83 cm.