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The surface-area-to-volume ratio also called the surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects. The surface-area-to-volume ratio is measured in units of inverse distance. A cube with sides of length a will have a surface area of 6a2 and a volume of a3. The surface to volume ratio for a cube is thus shown as $SA:V = \frac{6a^2}{a^3} = \frac{6}{a}$.

For a given shape, SA:V is inversely proportional to size. A cube 2 m on a side has a ratio of 3 m−1, half that of a cube 1 m on a side. On the converse, preserving SA:V as size increases requires changing to a less compact shape.

## Physical chemistry

In involving a solid material, the surface-area-to-volume is an important factor for the reactivity, that is, the rate at which the chemical reaction will proceed. Materials with large surface area to volume ratios (e.g., very small diameter, or very porous or otherwise not compact) react at much faster rates than monolithic materials, because more surface is available to react. Examples include grain dust; while grain is not typically flammable, grain dust is explosive. Finely ground salt dissolves much more quickly than coarse salt. It is same-case-applicable to a multiparticulate system or any system that has a surface coating, a very important parameter to be consider while performing coating for pharmaceutical solid oral-dosage form.

High surface-area-to-volume ratio provides a strong "driving force" to speed up thermodynamic processes that minimize thermodynamic free energy.

## Biology

The ratio between the surface area and volume of cells and organisms has an enormous impact on their biology. For example, many aquatic microorganisms have increased surface area to increase their drag in the water. This reduces their rate of sink and allows them to remain near the surface with less energy expenditure. Humans and other large animals cannot rely on diffusion for absorption and ejection of respiratory gases for their whole body; however, animals such as flatworms and leeches can, as they have more surface area per unit volume. For similar reasons, surface to volume ratio places a maximum limit on the size of a cell.

An increased surface area to volume ratio also means increased exposure to the environment. The many tentacles of jellyfish and anemones provide increased surface area for the acquisition of food. Greater surface area allows more of the surrounding water to be sifted for nutrients.

Individual organs in animals are often shaped by requirements of surface area to volume ratio. The numerous internal branchings of the lung increase the surface area through which oxygen is passed into the blood and carbon dioxide is released from the blood. The intestine has a finely wrinkled internal surface, increasing the area through which nutrients are absorbed by the body.

Smaller single celled organisms have a high surface area to volume ratio, which allows them to rely on oxygen and material diffusing into the cell (and wastes diffusing out) in order to survive. The higher the SA:Volume ratio they have, the more effective this process can be. Larger animals require specialized organs (lungs, kidneys, intestines, etc.) that effectively increase the surface area available for exchange processes, and a circulatory system to move material and heat energy between the surface and the core of the organism.

A wide and thin cell, such as a nerve cell, or one with membrane protrusions such as microvilli has a greater surface-area-to-volume ratio than a spheroidal one. Likewise a worm has proportionately more surface area than a rounder organism of the same mass does.

Increased surface area can also lead to biological problems. King Kong, the fictional giant gorilla, would have insufficient lung surface area to meet his oxygen needs, and could not survive. For small organisms with their high surface:volume ratio, friction and fluid dynamics (wind, water flow) are relatively much more important, and gravity much less important, than for large animals.

High surface-area-to-volume ratios also present problems of temperature control in unfavorable environments. More contact with the environment through the surface of a cell or an organ (relative to its volume) increases loss of water and dissolved substances for small organisms. Being large helps endotherms ("warm-blooded" animals) to maintain body temperatures different from their surroundings. This is a basis for Bergmann's rule, whereby in a group of closely related animal species, those at higher latitudes are larger than those the live nearer to the equator. But small organisms have a hard time keeping cooler than their surroundings, which is probably part of why small desert mammals are nocturnal, and some large mammals (hippos, elephants, rhinos, large ungulates) are well adapted to tropical environments.

Similar principles apply to plants, which is part of why succulent plants such as cacti, some euphorbs, and others are well-suited to hot dry conditions (there are additional, physiological reasons, including Crassulacean acid metabolism). Tiny leaves on some desert plants (creosote bush, acacias, palo verde, many shrubs) allows those leaves to shed heat by conduction and convection, reducing the need for cooling by evaporation of water (transpiration) (Nobel, 2009).

## Examples

Shape Length $a$ Area Volume SA/V ratio SA/V ratio for unit volume
Tetrahedron side $\sqrt{3} a^2$ $\frac{\sqrt{2}a^3}{12}$ $\frac{6\sqrt{6}}{a} \approx \frac{14.697}{a}$ 7.21
Cube side $6a^2$ $a^3$ $\frac{6}{a}$ 6
Octahedron side $2\sqrt{3}a^2$ $\frac{1}{3} \sqrt{2}a^3$ $\frac{3\sqrt{6}}{a} \approx \frac{7.348}{a}$ 5.72
Dodecahedron side $3\sqrt{25+10\sqrt{5}} a^2$ $\frac{1}{4} (15+7\sqrt{5}) a^3$ $\frac{12\sqrt{25+10\sqrt{5}}}{(15+7\sqrt{5})a} \approx \frac{2.694}{a}$ 5.31
Icosahedron side $5\sqrt{3}a^2$ $\frac{5}{12} (3+\sqrt5)a^3$ $\frac{12 \sqrt{3}}{(3+\sqrt{5})a} \approx \frac{3.970}{a}$ 5.148
Sphere radius $4\pi r^2$ $\frac{4\pi r^3}{3}$ $\frac{3}{r}$ 4.836
Example of Cubes of varying size
Side Area of Face Total Surface Area Volume of Cube Surface Area to Volume Ratio
1 m 1 m2 6 m2 1 m3 6.0 m−1 (or  m2•m−3)
2 m 4 m2 24 m2 8 m3 3.0 m−1
4 m 16 m2 96 m2 64 m3 1.5 m−1
6 m 36 m2 216 m2 216 m3 1.0 m−1
8 m 64 m2 384 m2 512 m3 0.75 m−1
12 m 144 m2 864 m2 1728 m3 0.5 m−1
20 m 400 m2 2400 m2 8000 m3 0.3 m−1

## References

• Nobel, P.S. (2009). Physicochemical and Environmental Plant Physiology. Oxford, UK: Academic Press.
• Schmidt-Nielsen, K (1984). Scaling: Why is Animal Size so Important?. New York, NY: Cambridge University Press.
• Vogel, S (1988). Life's Devices: The Physical World of Animals and Plants. Princeton, NJ: Princeton University Press.
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 2.1.6 Explain the importance of the surface area to volume ratio as a factor limiting cell sizeInternational Baccalaureate Biology Tutorial 2.1.6 Explain the importance of the surface area to volume ratio as a factor limiting cell size. Surface Area to Volume RatioA quick tutorial on Surface Area to Volume ratio, suitable for High School students. IB Biology 2.1.6 Surface area to volume ratio in cellshttp://ibguides.com/ Video covering topic 2.1.6 for IB biology sl and hl on the importance of surface area to volume ratio in cells. Surface Area to Volume RatioThis video tutorial will guide you through calculating a Surfae Area to Volume Ratio. Surface Area, Volume, Diffusion and OsmosisInitial discussion of why the surface area to volume ratio is important on the cellular level and a brief description of osmosis and diffusion. 2.1.6 Surface area-volume ratios limiting cell size Alka Seltzer Demo Surface Area to Volume Ratio (8th) Surface Area to Volume Ratio.mp4 Surface Area to Volume RatioWhy are cells so small? SBI3U / SBI4U Surface Area to Volume Ratio
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