Relationship between SA:V and rate of diffusion Watch

Catchetat
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This experiment is done towards cubic agar jelly. I don't really understand what the results should look like because I think mines are done wrong. A bigger volume agar jelly would have a lower SA:V, so would the rate of diffusion be the same for all jelly of different sizes or would it be slower for a jelly with a bigger size? Thanks.
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justinliu
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the larger the surface area, the greater the rate of diffusion. i think surely that with a larger cube of jelly, the surface is larger so therefore there is a greater rate of diffusion with a larger cube of jelly but im not too sure so i would get someone elses opinion before you write anything.
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afoneleri
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the bigger the jelly size, the slower the rate of diffusion. When you increase surface area and volume of a cube, the volume increases four fold as the surface area doubles. This means that there is more volume for each unit of surface area.

Think about animals - very small animals like earthworms gain nutrients and water through diffusions. Bigger animals need a transport system (guts and blood stream) to ensure things get everywhere.
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Peel
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(Original post by afoneleri)
the bigger the jelly size, the slower the rate of diffusion. When you increase surface area and volume of a cube, the volume increases four fold as the surface area doubles. This means that there is more volume for each unit of surface area.

Think about animals - very small animals like earthworms gain nutrients and water through diffusions. Bigger animals need a transport system (guts and blood stream) to ensure things get everywhere.
Thats exactly right.

The key ratio to remeber is really the SA:Volume ratio which you mentioned. Larger organisms have a larger suface area, but (usually) an even larger volume, meaning their SA : Volume ratio is lower than that of smaller organisms. You also have to keep in mind that as you increase the volume, the diffusion distance increases considerably.

Hope this helps!
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Catchetat
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You know for Fick's law.. that rate of diffusion is directly proportional to Concentration gradient x Surface area / distance

What does the distance bit mean?

And are you sure that "When you increase surface area and volume of a cube, the volume increases four fold as the surface area doubles."?
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Peel
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I haven't come across Fick's Law before oddly, but I think the distance would be the distance the diffusion takes place over, which I think equates to half the width of the object / organism in question.

There is an explanation for the SA and volume increases using cubes, but I've forgotten it.
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silent ninja
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I havent come across Fick's law but from the equation you've typed up, diffusion is inverseley proportional to distance. So the thicker the diffusing distance, the longer it takes. For example, alveoli are only 1 cell thick= very short distance, therefore diffusion rate is fast, or similarly look at the ileum.
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afoneleri
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(Original post by silent ninja)
I havent come across Fick's law but from the equation you've typed up, diffusion is inverseley proportional to distance. So the thicker the diffusing distance, the longer it takes. For example, alveoli are only 1 cell thick= very short distance, therefore diffusion rate is fast, or similarly look at the ileum.
That's right. The distance in Fick's law is the distance across which a substance has to diffuse. So going back to the cube example, if you imagine a cube as being hollow, to fill the space in the middle, a gas must diffuse through the cube wall - the thicker the wall, the longer it takes to get across (which goes back to surface area and volume).

For the cube: If you imagine you have a cube that has sides of 1cm each, the surface area of that cube is (1 cm x 1 cm)x6 = 6cm, as the cube has six walls. The volume is 1^3 = 1 cubic centimetre. So the SA:V ratio is 6:1

Increase the sides of the cube to 2cm each, the values are 24 cm squared and 8 cubic centimetres. The ratio is now 3:1

Keep incresing the size and you keep changing the ratio in the same way. If we jump forward and say the sides are now 10cm each, the values have become 600 centimetres squared and 1000 cubic centimetres. The ratio is now 3:5.

So you see, the volume is increasing much more rapidly that the surface area, hence the change in the ratio.

Hope this makes sense!

P.S. IB student and medicine? Good on ya'
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bright star
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we did this experiment for our practical exam, it's totally ridiculous!!!
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Catchetat
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hmm how is it ridiculous?
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bright star
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it's a really bad experiment.... how do you go about getting results with any practical application? it more or less tells you that it takes longer for liquid to move across a big thing than a small thing. plus there's a lot of logistical issues, like cutting the jelly pieces to the correct size for one.....
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- skyhigh -
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(Original post by bright star)
it's a really bad experiment.... how do you go about getting results with any practical application? it more or less tells you that it takes longer for liquid to move across a big thing than a small thing. plus there's a lot of logistical issues, like cutting the jelly pieces to the correct size for one.....
Wow. I have never come across an experiment for Fick's Law. Do you have to test the rate of difusion across jelly pieces with different lengths? What is the range of the lengths?
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sportygal2010
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usually they like it to be in basic sizes like
1cm, 2cm, 3cm
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Jimmy hard nails
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Lol
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1233kb
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what does SA:V ratio have to do with diffusion and transport systems and how does this link to life on earth?
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carrotstar
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(Original post by 1233kb)
what does SA:V ratio have to do with diffusion and transport systems and how does this link to life on earth?
Asking the big questions here... think it might need its own (newer) thread?
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1233kb
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(Original post by carrotstar)
Asking the big questions here... think it might need its own (newer) thread?
yeah maybe
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