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Is it possible to calculate the surface area of an irregularly shaped vessel, like a laboratory volumetric flask, by knowing its volumen?

By knowing how much liquid it can hold, one could use the formula for the surface area of a cylinder and find out what the total surface area for such a body is. For example r2*pi*h = V would yield for h = 1 cm and 1000 cm3 the value for r.

r would then be: 17.84 cm

The surface area for the cylinder containing 1000 cm3 would be:

2*rpih = 112.1 cm2 (mantle) + r2*pi= 1000 cm2 * 2 (2 circles, top and bottom)

Total surface area= 2000 cm2 + 112.1 cm2 = 2112.1 cm2

The question would be if this area is equivalent to the surface area of the irregularly shaped laboratory volumetric flask containing the same volume of 1000 cm3.

The only part remaining would be that the area of the free volume (not touching the walls of the flask = the small volume at the top of the flask) would have to be substracted from this total area.

Is this so?

Kind regards,

Ricardo

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  • $\begingroup$ You would need some detail for the geometry of your flask. If the container is irregular, It would be possible for a container to have a very high surface area compared to its volume. But without knowing the more about its shape that is the best that I can really say. $\endgroup$ – Doug M Apr 10 '18 at 0:50
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No, the surface area has no intrinsic connection to the volume, even for simple shapes.

Take the case of two right circular cylinders with open top.

$$\text{Volume: }V=\pi r^2h$$ $$\text{Surface area: }S=2\pi rh+\pi r^2=\pi r(2h+r)$$

Now let flask $1$ have dimensions $r=3$ and $h=4$. Then $V=36\pi$ and $S=33\pi$.

Next, let flask $2$ have dimensions $r=6$ and $h=1$. Then $V=36\pi$ but $S=48\pi$.

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