Cambridge IGCSE0610

Diffusion

Biology 0610 Chapter Notes

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DiffusionOsmosisActive transport
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1. The Principles of Diffusion

Diffusion is the fundamental way substances move over short distances in liquids and gases. Imagine spraying air freshener in a corner of a room. Initially, the scent particles are highly concentrated in that corner. These particles are in constant, random motion due to their own kinetic energy. They bump into each other and spread out, moving from the area of high concentration to areas of lower concentration. This continues until the particles are evenly distributed throughout the room. This process requires no energy from the surroundings, which is why it's called a passive process. The overall movement from high to low concentration is called the 'net movement'. When particles are evenly spread, a state of dynamic equilibrium is reached, where particles still move randomly, but there is no longer any net movement from one area to another.

Key term

Concentration Gradient: The difference in the concentration of a substance between two regions.

Examiner insight

Examiners reward clear, precise definitions. Always use the phrases 'net movement', 'from a high to low concentration', and 'down a concentration gradient' for full marks.

Common pitfall

Stating that particles stop moving at equilibrium. In reality, particles continue to move randomly, but there is no net change in concentration because the movement is equal in all directions.

Worked example 13 marks

A crystal of purple potassium permanganate is placed at the bottom of a beaker of still water. Describe and explain what will be observed over the next hour. [3]

  1. 1

    Step 1: Describe the observation. The purple colour will be seen to spread out from the crystal into the surrounding water. Over time, the entire beaker of water will become a uniform, pale purple colour.

  2. 2

    Step 2: Explain the process. This happens due to diffusion. The potassium permanganate particles move from an area of high concentration (around the crystal) to an area of low concentration (the rest of the water).

  3. 3

    Step 3: Explain the energy and movement. This movement is passive and happens because the particles have kinetic energy, causing them to move randomly down the concentration gradient until they are evenly distributed.

Recap

  • Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration.
  • This movement occurs down a concentration gradient.
  • It is a passive process, meaning it does not require metabolic energy from the cell.
  • Movement is due to the random kinetic energy of the particles.
  • Diffusion continues until a dynamic equilibrium is reached, where particles are evenly spread.

Quick check

  1. Define the term 'diffusion'.2 marks
  2. Is diffusion an active or passive process? Explain your answer.2 marks

2. Factors Affecting the Rate of Diffusion

The speed at which diffusion occurs is not constant. Several factors can influence how quickly particles spread out. A steeper concentration gradient (a bigger difference in concentration) will cause faster diffusion. Higher temperatures give particles more kinetic energy, making them move and collide more frequently, which speeds up diffusion. A larger surface area provides more space for particles to move across, increasing the overall rate. Finally, the diffusion distance is crucial; diffusion is very fast over short distances but extremely slow over long ones. This is why cell membranes are so thin.

Key term

Rate of Diffusion: The speed at which the net movement of particles occurs, often measured as the amount of substance moving across a surface per unit of time.

Examiner insight

Questions often ask you to apply your knowledge. Be prepared to explain which factor is most relevant in a given biological context, such as the thinness of alveoli walls relating to diffusion distance.

Common pitfall

Simply listing the factors without explaining *how* they affect the rate. For example, instead of 'temperature', write 'higher temperature increases kinetic energy, leading to a faster rate of diffusion'.

Worked example 14 marks

Explain why oxygen diffuses from the air in the alveoli into the blood faster during exercise than during rest. Refer to two factors in your answer. [4]

  1. 1

    Step 1: Identify the first factor. During exercise, breathing rate and depth increase. This brings more oxygen into the alveoli and removes carbon dioxide more quickly.

  2. 2

    Step 2: Explain the effect of the first factor. This action maintains a steeper concentration gradient for oxygen between the air in the alveoli (high O2) and the blood (low O2), causing a faster rate of diffusion.

  3. 3

    Step 3: Identify the second factor. During exercise, heart rate increases, causing blood to flow more quickly past the alveoli.

  4. 4

    Step 4: Explain the effect of the second factor. This rapid blood flow quickly carries away oxygenated blood and brings deoxygenated blood, which also helps to maintain a steep concentration gradient, further increasing the rate of diffusion.

Recap

  • A steeper concentration gradient increases the rate of diffusion.
  • A higher temperature increases the rate of diffusion by giving particles more kinetic energy.
  • A larger surface area increases the rate of diffusion by providing more area for exchange.
  • A shorter diffusion distance increases the rate of diffusion.

Quick check

  1. List three factors that can increase the rate of diffusion.3 marks

3. Diffusion in Biological Systems

Living organisms are critically dependent on diffusion for survival. Specialised exchange surfaces have evolved to maximise the rate of diffusion of essential substances. For example, in the lungs, millions of tiny air sacs called alveoli provide a massive surface area for gas exchange. Their walls are only one-cell thick, creating a very short diffusion distance for oxygen to enter the blood and carbon dioxide to leave. Constant breathing and blood flow ensure a steep concentration gradient is always maintained. Similarly, in the small intestine, digested food molecules like glucose diffuse into the bloodstream. The intestine is lined with millions of finger-like projections called villi, which massively increase the surface area for absorption. In plants, carbon dioxide for photosynthesis diffuses into the leaf through small pores called stomata.

Key term

Exchange Surface: A specialised region of an organism's body adapted for efficient transfer of substances, typically having a large surface area, thin walls, and a mechanism to maintain a concentration gradient.

Fun fact

The total surface area of the alveoli in an adult's lungs is approximately 70 square metres, which is about the size of one side of a volleyball court!

Worked example 14 marks

Explain how the structure of a fish's gills is adapted for efficient gas exchange. [4]

  1. 1

    Step 1: Surface Area. The gills are made of many thin filaments, which are covered in lamellae. This structure creates a very large surface area for diffusion to occur.

  2. 2

    Step 2: Diffusion Distance. The walls of the lamellae and the capillaries within them are extremely thin, providing a very short diffusion distance for gases to travel between the water and the blood.

  3. 3

    Step 3: Concentration Gradient (Blood Supply). The gills have a rich blood supply that constantly flows, carrying oxygenated blood away and bringing deoxygenated blood. This maintains a steep concentration gradient for oxygen to diffuse into the blood.

  4. 4

    Step 4: Concentration Gradient (Ventilation). Water is constantly pumped over the gills. This ensures that water with a high oxygen concentration is always in contact with the gill surface, also helping to maintain the concentration gradient.

Recap

  • Alveoli in the lungs are adapted for gas exchange with a large surface area and thin walls.
  • Villi in the small intestine are adapted for absorbing nutrients via a large surface area.
  • Maintaining a steep concentration gradient is vital and is achieved by processes like breathing and blood flow.
  • Leaves in plants use diffusion for the intake of CO2 and release of O2 through stomata.
  • Efficient exchange surfaces share common features: large surface area, short diffusion path, and a steep concentration gradient.

Quick check

  1. State two features of a human alveolus that make it efficient at gas exchange.2 marks

4. Surface Area to Volume Ratio

The surface area to volume ratio (SA:Vol) is a critical concept that explains why cells are small and why large organisms need complex transport systems. As an object gets bigger, its volume (the space inside) increases much faster than its surface area (the area covering it). For a cube, volume increases by the cube of its side length (L³), while surface area only increases by the square (6L²). A single-celled organism has a large SA:Vol ratio, meaning it has enough surface area for diffusion to supply all its needs. However, a large multicellular organism has a small SA:Vol ratio. Its outer surface is not nearly large enough to supply oxygen and nutrients to its innermost cells by diffusion alone. This is why large organisms have developed specialised exchange surfaces (like lungs) and transport systems (like the circulatory system) to overcome this limitation.

Surface Area of a cube = 6 x (length of one side)²

Volume of a cube = (length of one side)³

Surface Area to Volume Ratio = Surface Area / Volume

Key term

Surface Area to Volume Ratio: A measure that compares the area of an object's surface to its internal volume; it decreases as an object's size increases.

Examiner insight

When asked to calculate a ratio, always show your full working for surface area and volume first. Marks are often awarded for these intermediate steps. Simplify the final ratio if possible (e.g., 6:1 is better than 600:100).

Worked example 15 marks

A cube-shaped cell has sides of 10 µm. A second cube-shaped cell has sides of 20 µm. Calculate the surface area to volume ratio for both cells and explain the significance of the results. [5]

  1. 1

    Step 1: Calculate for the 10 µm cell. Surface Area (SA) = 6 x (10 µm)² = 600 µm². Volume (V) = (10 µm)³ = 1000 µm³. SA:Vol Ratio = 600 / 1000 = 0.6.

  2. 2

    Step 2: Calculate for the 20 µm cell. Surface Area (SA) = 6 x (20 µm)² = 6 x 400 = 2400 µm². Volume (V) = (20 µm)³ = 8000 µm³. SA:Vol Ratio = 2400 / 8000 = 0.3.

  3. 3

    Step 3: Compare the ratios. The smaller cell has a larger SA:Vol ratio (0.6) compared to the bigger cell (0.3).

  4. 4

    Step 4: Explain the significance. The higher SA:Vol ratio in the smaller cell means that diffusion of substances into and out of the cell is more efficient relative to its volume. This is why cells are small and why large organisms cannot rely on diffusion alone.

Recap

  • As an organism or cell increases in size, its volume increases faster than its surface area.
  • This causes the surface area to volume ratio to decrease as size increases.
  • Small organisms have a large SA:Vol ratio and can rely on diffusion across their body surface.
  • Large organisms have a small SA:Vol ratio and require specialised exchange surfaces and transport systems.
  • Calculating the ratio involves finding the surface area and volume separately, then dividing SA by V.

Quick check

  1. What happens to the surface area to volume ratio as a cell gets larger?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define 'concentration gradient' and explain its importance for diffusion.2 marks
  2. A single-celled amoeba living in a pond gets all the oxygen it needs by diffusion across its cell membrane. Explain why a large animal like a whale cannot do this.4 marks
  3. Describe how the leaves of a plant are adapted for the efficient diffusion of carbon dioxide into the photosynthesising cells.4 marks
  4. A cube has a side length of 3 cm. Calculate its surface area, its volume, and its surface area to volume ratio. Show all your working.3 marks
  5. Explain how both breathing and blood circulation help to maximise the rate of gas exchange in the lungs.4 marks
  6. In an experiment, agar jelly containing an indicator was cut into cubes of different sizes. The cubes were placed in an acid. The time taken for the acid to diffuse to the centre of each cube was measured. Predict and explain the results for a 1 cm cube compared to a 3 cm cube.3 marks
  7. Why is diffusion referred to as a 'passive' process, and how does this differ from active transport?3 marks
  8. The small intestine is approximately 6 metres long and lined with villi. Explain the advantage of these two features for the absorption of digested food.3 marks
  9. A student suggests that putting a warm blanket on a person with a fever will help them cool down by increasing diffusion of heat away from the body. Evaluate this suggestion.2 marks
  10. Describe the journey of a molecule of carbon dioxide from a respiring muscle cell in the leg until it is breathed out from the lungs. Name the key transport and diffusion stages.5 marks

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