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The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m-1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
Its volume would be multiplied by the cube of 2 and become 8 m 3. The original cube (1 m sides) has a surface area to volume ratio of 6:1. The larger (2 m sides) cube has a surface area to volume ratio of (24/8) 3:1. As the dimensions increase, the volume will continue to grow faster than the surface area. Thus the square–cube law.
The resulting surface area to volume ratio is therefore 3/r. Thus, if a cell has a radius of 1 μm, the SA:V ratio is 3; whereas if the radius of the cell is instead 10 μm, then the SA:V ratio becomes 0.3. With a cell radius of 100, SA:V ratio is 0.03. Thus, the surface area falls off steeply with increasing volume.
The volume of such a mixture is slightly less than the sum of the volumes of the components. Thus, by the above definition, the term "40% alcohol by volume" refers to a mixture of 40 volume units of ethanol with enough water to make a final volume of 100 units, rather than a mixture of 40 units of ethanol with 60 units of water.
Specific surface area (SSA) is a property of solids defined as the total surface area (SA) of a material per unit mass, [1] (with units of m 2 /kg or m 2 /g). Alternatively, it may be defined as SA per solid or bulk volume [ 2 ] [ 3 ] (units of m 2 /m 3 or m −1 ).
Surface tension gives them their near-spherical shape, because a sphere has the smallest possible surface area to volume ratio. Formation of drops occurs when a mass of liquid is stretched. The animation (below) shows water adhering to the faucet gaining mass until it is stretched to a point where the surface tension can no longer keep the drop ...
As the vessels decrease in size, they increase their surface-area-to-volume ratio. This allows surface properties to play a significant role in the function of the vessel. Diffusion occurs through the walls of the vessels due to a concentration gradient, allowing the necessary exchange of ions, molecules, or blood cells.
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...