Calculate the surface area of the following:


Q. A cube of side length 25mm



This would be a fairly straightforward calculation, each side measures 25 mm therefore the surface area of one side is 25×25 = 625 mm². There are six sides to the cube therefore the total surface area would be 6×625 = 3750 mm²


Q. A cube of side length 50mm which has been cut into cubes of side length 1mm.


This is only a slight complication, our cube has been cut into 50×50×50 cubes, each of which will have a side of 1 mm. The surface area of one such cube is 6 mm² but we now have 50×50×50 of them = 125,000 so the total surface area is now 125,000×6 = 750,000 mm².


Q. A cube of side length 1 m which has been cut into cubes of side length 1 nm (we are now looking at a nano particle scale).


We are now going into some mind blowing mathematics, you need to go through this very carefully in order not to make mistakes. 


Step 1 - 1 nm is 1×10-9 m or one billionth of a metre. The cube has therefore been cut into (109)3 cubes = 1027 cubes each with a side length of 1 nm. 


Step 2 - The surface area of one of these cubes is 1 nm² x 6 = 6 nm² which it is probably best if we convert this into metres now.


Step 3 - The surface area of one of our small cubes in metres is therefore 0.000000001 x 0.000000001 x 6 = 6 x 10-18 m2 . 


Step 4 - Remember that we have 1027 of these small cubes, each of which has a surface area of 6 x 10-18 m, so the total surface area is 1027 x (6 x 10-18) m2  which equates to 6 x 109 m2 or 6 billion square metres (around 2,300 square miles).


Q. Calculate the surface area to volume ratio of a cube with sides equals 1 mm, then perform the same calculation with a cube of sides 50 mm and comment upon which would be best served for diffusion through the membrane.


If a cube has a side length of 1 mm it will have a total surface area of 6 mm², and a volume of 1 mm³. The surface area to volume ratio is therefore 6 to 1. In the case of a cube with 50 mm sides, the total surface area would be 6×50×50 = 1500 mm² and a volume of 125,000 mm³. This gives a surface area to volume ratio of 0.012 to 1 which shows that as the volume increases, the surface area to volume ratio and therefore the efficiency of this type of diffusion pathway, drops off dramatically.


Q. This question talks about diffusion.


A river running past an industrial estate is polluted with contaminants as a result of an industrial mistake. As a result of this pollution the concentration of dissolved oxygen in the water is reduced. What effect, if any would this have on the rate of diffusion of oxygen into the gills of fish that live in the water?


As this is a river we can assume that it will be flowing at a reasonable rate and cause the contaminants to disperse quickly, however the reduction of dissolved oxygen in the water would mean that the "concentration gradient" between the oxygen in the river and that present in the blood of the fish gills would reduce. A lower concentration gradient would mean that oxygen would diffuse at a slower rate from the water into the gills of the fish, this would put the fish in danger as they could potentially stop receiving the oxygen they need to survive.


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