What you need to know
- 2.1(i–ii)Gas-exchange surface properties and calculations using Fick’s lawRate of diffusion is proportional to (surface area × concentration difference) ÷ thickness of the exchange surface..
By the end, you should be able to…
- Explain why large active organisms need specialised exchange surfaces.
- Relate each surface feature to diffusionThe net movement of molecules from a region of higher concentration to a region of lower concentration. It is passive. rate.
- Calculate proportional changes using Fick’s law.
Build your understanding
Solve the size problem
A large organism has a lower external surface-area-to-volume ratio than a geometrically similar smaller organism. Many cells also lie far from the external surface. diffusionThe net movement of molecules from a region of higher concentration to a region of lower concentration. It is passive. across the body surface alone may not supply oxygen quickly enough for respirationThe breakdown of respiratory substrates such as glucose in cells to release energy, which is used to make ATP. or remove carbon dioxide at the required rate.
Specialised surfaces provide a large area within the organism, while circulation carries gases between the exchange surface and tissues. Keep the roles separate: ventilation moves the surrounding air or water, diffusion crosses the exchange barrier, and mass transportThe bulk flow of a fluid, such as blood or air, that carries substances over long distances. moves gases over longer distances.
The three main variables
Diffusion rate increases with exchange area and with the concentration difference across the surface. It decreases as the diffusion path becomes longer. Thin surfaces shorten that path; folding or branching increases area; ventilation and blood flow help maintain differences.
A moist surface helps gases dissolve before crossing cell surface membraneThe partially permeable membrane around a cell that controls which substances enter and leave., but excessive fluid can increase the diffusion distanceThe distance molecules have to travel. A shorter distance gives a faster rate of diffusion.. More thickness is not an advantage simply because the surface is wet.
Calculate with Fick’s law
For comparisons under otherwise identical conditions: diffusion rate is proportional to area × concentration difference ÷ diffusion distance. If the question supplies a diffusion coefficient D, an absolute amount-per-time rate can be calculated as rate = D × A × ΔC / d, using consistent units.
Without a coefficient or a reference rate, A × ΔC / d gives a proportional comparison, not automatically an absolute rate with valid units. For a given gas under the same conditions, the constant cancels when taking a ratio.
If area doubles while distance doubles and concentration difference is unchanged, the predicted rate is unchanged: 2/2 = 1. If distance halves with everything else unchanged, rate doubles. State the assumptions rather than assuming every physiologicalDescribes an adaptation of body processes, such as producing concentrated urine. factor remains fixed.
NOT: a thicker surface does not increase diffusion. Thickness is the divisor.
- rate = D × A × ΔC / d when D is supplied
- Without D, A × ΔC / d compares rates; it is not an absolute unit
- Doubling area and distance together leaves the rate unchanged
- A moist surface helps gases dissolve; extra fluid can increase distance
Two traps this diagram blocks
- Do not put thickness in the numerator.
- Do not claim a large area guarantees fast transfer if the barrier is thick or the gradient has collapsed.
From the diagram alone
Question
Area becomes three times larger and diffusion distance doubles. The concentration difference is unchanged. Calculate the new rate relative to the original. [2]
Show a model answer
Relative rate = 3 / 2 = 1.5. The rate is 1.5 times the original, a 50% increase, assuming the diffusion coefficient and other conditions stay the same.
| Change | Predicted rate |
|---|---|
| Area doubles | Twice as large |
| Concentration difference doubles | Twice as large |
| Diffusion distance doubles | Half as large |
Make your reasoning visible
Put thickness in the denominator
A thicker barrier reduces the predicted rate. Work with ratios when the question asks how many times faster or slower.
Convert units before calculating
A distance stated in micrometres must be converted if the other quantities and coefficient use metres. Include the units supplied by the model.
Key explanations
- Ventilation and blood flow help keep the concentration difference steep by bringing fresh medium and taking the exchanged substance away.
Avoid this misconception
The claim: Increasing the thickness of an exchange surface increases diffusionThe net movement of molecules from a region of higher concentration to a region of lower concentration. It is passive. because there is more tissue.
The correction: Extra thickness lengthens the diffusion pathway and reduces the rate when other factors stay the same.
Why can an amoeba breathe through its skin but you cannot?
Oxygen reaches cells by diffusion. That works over a fraction of a millimetre, but not across a whole body. Size changes the balance between surface and volume.
What the model shows
A growing cube with its surface area to volume ratio and oxygen supply, then an exchange barrier where you change area, concentration difference and thickness.
What to look for
How SA:V changes as the cube grows. What happens to the rate when the barrier gets thicker, and when two changes work against each other.
What it tells us
SA:V falls as organisms get larger, so they need specialised surfaces. Rate of diffusion ∝ area × concentration difference ÷ distance: 3 ÷ 2 = 1.5 and 0.5 × 3 ÷ 2 = 0.75.
Be careful what you claim
All values are illustrative. The barrier panel compares rates with the original; it does not give an absolute rate.
Try it in the model: drag the side slider from 1 mm to 10 mm and watch the oxygen meter.
Size, surfaces and Fick’s law
Grow a cube-shaped organism and watch its surface area to volume ratio fall, then change the area, concentration difference and thickness of an exchange barrier. All values are illustrative model values.
What you see
- Size: a cube-shaped organism, a cross-section showing how far oxygen diffuses in, a graph of SA:V against size and an oxygen supply-and-demand meter.
- Fick’s law: an exchange barrier with a high concentration on one side and low on the other. Molecules cross at a rate set by area, concentration difference and thickness, compared with the original on a graph.
How to use it
- Under Part, choose Size or Fick’s law.
- Size: move the side slider and turn the exchange surface on or off.
- Fick’s law: change area, concentration difference and distance as multiples of the original, or press a preset from the lesson.
What it shows
- Why large active organisms need specialised exchange surfaces: SA:V falls as size increases (2.1(i)).
- Fick’s law: rate ∝ area × concentration difference ÷ diffusion distance, as ratios to the original (2.1(ii)).
Limits of the model
- The organism is a cube and every number is an illustrative model value, including the oxygen penetration depth.
- The Fick panel compares relative rates. The diffusion coefficient cancels in a ratio; an absolute rate would need D and consistent units.
- Real exchange surfaces also depend on ventilation, blood flow and the solubility of the gas, which the barrier panel holds fixed.
Guided tour · step 1 of 11
- In Size, just below these steps, set Side of cube to 2 mm or less.
- Read SA:V and the oxygen meter.
A small organism has a large surface area for its volume, and no cell is far from the surface, so diffusion across the body surface is enough.
Size
Questions
Check your understanding
Try each question before opening the suggested answer.
Practice question
Retrieval and explanation · suggested answer
Area triples and diffusion distance doubles, with concentration difference unchanged. Calculate the rate relative to the original.
Show suggested answer
The rate becomes 3/2 = 1.5 times the original, a 50% increase.
Practice question
Retrieval and explanation · suggested answer
Explain how blood flow helps maintain oxygen diffusion into capillaries at a gas-exchange surface.
Show suggested answer
Blood carries oxygen away from the exchange surface and brings blood with lower oxygen levels, helping maintain the difference across the barrier.
Take the idea further
Apply the same specification knowledge to an unfamiliar problem. Any extra assumptions are supplied in the question.
Stretch question
An illustrative model has half the original area, twice the original distance and three times the concentration difference. Predict the relative rate and explain why improving one variable need not restore function.
Show a hint
Multiply the changes in the numerator and divide by the change in distance.
Explore the reasoning
Relative rate = 0.5 × 3 / 2 = 0.75. It is 75% of the original. The increased concentration difference does not fully compensate for lost area and increased distance under these assumptions.
Recall this lesson
Lesson recall questions
Question 1 of 5: List the three properties of a gas exchange surface that determine the rate of diffusion in living organisms.
Gas exchange surfaces and Fick’s law
Practice question
Retrieval and explanation · suggested answer
List the three properties of a gas exchange surface that determine the rate of diffusion in living organisms.
Lesson: 2A 5: The Need for Gas Exchange Surfaces
Try this from memory first.
Put your biology to work
Explore where these ideas appear in the world. Enrichment, not extra required recall.
An artificial lung still needs gradients
2A · The Need for Gas Exchange SurfacesECMO circulates blood through an artificial lung to add oxygen and remove carbon dioxide before returning blood to the body. Designing an exchange surface requires control of flow…Specification 2.1(i) · 2.1(ii)Read & use the scienceClose article
Why this matters
ECMO circulates blood through an artificial lung to add oxygen and remove carbon dioxide before returning blood to the body. Designing an exchange surface requires control of flow and diffusion conditions. The clinical overview describes the device; applying Fick’s law is the student’s explanatory model. A large exchange area and a thin barrier help, but blood and gas flow must also sustain the relevant differences across that barrier. This gives a biomedical engineer several variables to investigate together. A successful device cannot be judged by area alone, just as a lung cannot be judged only by its size. It must exchange gases effectively while operating as part of a circulation.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 1, pp. 15–18. The application extends beyond required recall.
- 2.1(i) know the properties of gas exchange surfaces in living organisms (large surface area to volume ratio, thickness of surface and difference in concentration)
- 2.1(ii) understand how the rate of diffusion is dependent on these properties and can be calculated using Fick’s Law of Diffusion
Use the science
Why would a large membrane area alone not guarantee rapid gas transfer?
Explore a possible response
Transfer also depends on the barrier’s permeability and thickness, the gas difference across it and the flows maintaining that difference. A large area cannot compensate for every other limitation.
Explore the source
A lung interface on a microchip
2A · The Need for Gas Exchange SurfacesResearchers built a small device with human lung and blood-vessel cells on opposite sides of a flexible membrane. They could expose one side to air and the other…Specification 2.1(i) · 2.1(ii)Read & use the scienceClose article
Why this matters
Researchers built a small device with human lung and blood-vessel cells on opposite sides of a flexible membrane. They could expose one side to air and the other to flowing liquid, and apply stretching movements. The aim was to model selected properties of the lung interface for experiments, not to make an entire replacement lung. The design makes familiar variables experimental choices: barrier thickness, movement and exposure can be controlled and measured. A bioengineer asks which features must be reproduced for the scientific question being investigated, and which missing features limit what the model can tell us.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 1, pp. 15–18. The application extends beyond required recall.
- 2.1(i) know the properties of gas exchange surfaces in living organisms (large surface area to volume ratio, thickness of surface and difference in concentration)
- 2.1(ii) understand how the rate of diffusion is dependent on these properties and can be calculated using Fick’s Law of Diffusion
Use the science
If a model’s exchange barrier were much thicker than the real lung barrier, what effect would you predict on diffusion, assuming other factors were held constant?
Explore a possible response
A longer diffusion distance should reduce transfer rate under Fick’s law. A model may still answer some questions, but absolute rates need careful interpretation.
Explore the source
- Reconstituting Organ-Level Lung Functions on a ChipHuh et al., Science (2010)