1. Overview
Diffusion is a fundamental physical process by which substances move into and out of cells. It is a passive process, meaning it requires no energy input from the cell’s metabolism, and it is essential for vital functions such as gas exchange and the absorption of nutrients.
Key Definitions
- Diffusion: The net movement of particles from a region of their higher concentration to a region of their lower concentration (down a concentration gradient), as a result of their random movement.
- Concentration Gradient: The difference in the concentration of a substance between two areas. The steeper the gradient (the larger the difference), the faster the rate of diffusion.
- Net Movement: The overall direction of movement of a substance (even if individual particles move in all directions, the majority move from high to low concentration).
- Kinetic Energy: The energy possessed by particles due to their motion; this provides the "power" for diffusion.
Core Content
The Process of Diffusion
Diffusion occurs because all molecules and ions are in constant, random motion.
- Particles possess kinetic energy, causing them to bounce off each other and spread out.
- While individual particles move randomly in all directions, the net movement will always be from where there are many particles to where there are fewer.
- Eventually, particles reach equilibrium, where they are spread evenly. (Note: Particles still move at equilibrium, but there is no further net change).
Diffusion in Living Organisms
Living cells are surrounded by a partially permeable cell membrane. Small molecules can diffuse directly through this membrane to enter or leave the cell.
Importance of Diffusion:
- Gas Exchange: Oxygen diffuses from the lungs into the blood, and Carbon Dioxide diffuses from the blood into the lungs. In plants, CO2 diffuses into leaves for photosynthesis.
- Absorption of Solutes: Dissolved food molecules (like glucose) diffuse from the small intestine into the bloodstream.
- Excretion: Waste products like urea diffuse out of cells into the blood for transport to the kidneys.
Factors Influencing the Rate of Diffusion
- Surface Area: A larger surface area (e.g., folded membranes like root hair cells or villi) provides more space for particles to pass through, increasing the rate.
- Temperature: Higher temperatures increase the kinetic energy of particles, making them move faster and thus diffuse faster.
- Concentration Gradient: A greater difference in concentration between two areas increases the "push" toward the lower concentration area.
- Distance: The shorter the distance particles have to travel (e.g., thin capillary walls), the faster diffusion occurs.
Extended Content (Extended Only)
There are no specific supplement objectives listed for topic 3.1 in the current syllabus requirements provided.
Key Equations
While there are no complex mathematical formulas for diffusion at this level, you must understand the Surface Area to Volume (SA:V) Ratio:
$$\text{Surface Area to Volume Ratio} = \frac{\text{Total Surface Area}}{\text{Total Volume}}$$
- Large organisms: Have a small SA:V ratio and need specialized exchange surfaces.
- Small organisms (e.g., bacteria): Have a large SA:V ratio, allowing diffusion to happen quickly enough to meet all their needs.
Common Mistakes to Avoid
- ❌ Wrong: Particles stop moving once they are evenly spread out.
- ✓ Right: Particles continue to move randomly, but there is no net movement in any one direction.
- ❌ Wrong: Diffusion requires energy from respiration (ATP).
- ✓ Right: Diffusion is a passive process; the energy comes from the intrinsic kinetic energy of the molecules.
- ❌ Wrong: Molecules move from low to high concentration.
- ✓ Right: Molecules move down the gradient, from high to low concentration.
Exam Tips
- Command Word - "Describe": If asked to describe diffusion, always mention "net movement," "high to low concentration," and "random movement."
- Command Word - "Explain": If asked to explain why a rate increased, link it to the factors (e.g., "The rate increased because higher temperature gave particles more kinetic energy").
- Contexts: Expect questions about the Alveoli (lungs) or Villi (intestines). In these cases, always mention that they are "one cell thick" to explain the factor of "short diffusion distance."
- Typical Values: Be prepared to compare rates. For example, if Surface Area doubles, the rate of diffusion generally increases significantly.
Exam-Style Questions
Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0610 Theory papers.
Exam-Style Question 1 — Short Answer [5 marks]
Question:
A student places a small crystal of potassium permanganate at the bottom of a beaker filled with distilled water. The beaker is left undisturbed for 24 hours.
(a) Define the term diffusion. [2]
(b) Describe what the student would observe in the beaker after 24 hours. [2]
(c) State one factor that would increase the rate of diffusion of the potassium permanganate in this experiment. [1]
Worked Solution:
(a)
- Diffusion is the net movement of particles $[particles \ from \ a \ region \ of \ their \ higher \ concentration \ to \ a \ region \ of \ their \ lower \ concentration]$ Definition of diffusion including the net movement and concentration gradient.
How to earn full marks:
- State that diffusion is the net movement of particles.
- State that particles move from a region of high concentration to a region of low concentration.
(b)
- The purple colour will have spread throughout the water. $[The \ colour \ will \ be \ more \ evenly \ distributed \ throughout \ the \ beaker]$ The potassium permanganate will have diffused through the water.
How to earn full marks:
- State that the purple colour will spread throughout the water.
- State that the colour will become more evenly distributed.
(c)
- Increasing the temperature of the water. $[Increase \ temperature]$ Increasing the temperature will increase the rate of diffusion.
How to earn full marks:
- State that increasing the temperature will increase the rate of diffusion.
Common Pitfall: Remember that diffusion is the net movement of particles. Particles are always moving randomly, but diffusion refers to the overall movement from high to low concentration. Also, make sure you know the factors that affect diffusion rate: temperature, concentration gradient, surface area, and diffusion distance.
Exam-Style Question 2 — Short Answer [6 marks]
Question:
A biologist investigates the diffusion of a dye into agar jelly. She cuts three cubes of agar jelly, each with a different surface area to volume ratio. She places each cube into a beaker containing the same concentration of dye solution. After 30 minutes, she measures the distance the dye has diffused into each cube.
(a) State two variables that the biologist should keep constant in this experiment. [2]
(b) Identify the independent variable in this experiment. [1]
(c) Explain why a higher surface area to volume ratio would result in a faster rate of diffusion into the agar jelly. [3]
Worked Solution:
(a)
The concentration of the dye solution. $[Concentration \ of \ dye]$ The concentration of the dye solution needs to be kept constant to ensure a fair test.
The temperature of the dye solution. $[Temperature \ of \ dye]$ The temperature of the dye solution needs to be kept constant to ensure a fair test.
How to earn full marks:
- State that the concentration of the dye solution must be kept constant.
- State that the temperature of the dye solution must be kept constant.
(b)
- The surface area to volume ratio of the agar jelly cubes. $[SA:V \ ratio]$ The surface area to volume ratio is the variable that is being changed.
How to earn full marks:
- State that the surface area to volume ratio is the independent variable.
(c)
A higher surface area to volume ratio means there is more surface area for diffusion to occur through. $[More \ surface \ area \ for \ diffusion]$ The rate of diffusion is proportional to the surface area.
A smaller volume means the dye has less distance to diffuse. $[Less \ distance \ to \ diffuse]$ The rate of diffusion is inversely proportional to the distance.
Therefore, the dye will diffuse into the agar jelly faster. $[Faster \ diffusion]$ Combining both points, the dye will diffuse faster in the agar jelly with a higher surface area to volume ratio.
How to earn full marks:
- State that a higher surface area to volume ratio means there is more surface area for diffusion to occur through.
- State that a smaller volume means the dye has less distance to diffuse.
- State that the dye will diffuse into the agar jelly faster.
Common Pitfall: Don't forget that a higher surface area to volume ratio means faster diffusion. Some students mistakenly think a smaller ratio is better. Also, be clear on the difference between independent, dependent, and controlled variables in experiments.
Exam-Style Question 3 — Extended Response [8 marks]
Question:
A student investigates the effect of temperature on the rate of diffusion of oxygen gas through a membrane. He sets up an experiment where oxygen is released on one side of a membrane and measures the time it takes for the oxygen to diffuse through the membrane to reach a sensor on the other side. He repeats the experiment at different temperatures. The results are shown in the table below.
| Temperature (°C) | Time for Oxygen to Diffuse (s) |
|---|---|
| 10 | 120 |
| 20 | 90 |
| 30 | 70 |
| 40 | 55 |
| 50 | 45 |
(a) Describe the trend shown in the data. [2]
(b) Explain why increasing the temperature increases the rate of diffusion. [4]
(c) Suggest one way to improve the reliability of this experiment. [2]
Worked Solution:
(a)
As the temperature increases, the time for oxygen to diffuse decreases. $[Inverse \ relationship]$ The relationship between temperature and time is inversely proportional.
The decrease in time is greater at lower temperatures than at higher temperatures. $[Rate \ of \ decrease \ slows \ down]$ The rate of decrease slows down as the temperature increases.
How to earn full marks:
- State that as the temperature increases, the time for oxygen to diffuse decreases.
- State that the decrease in time is greater at lower temperatures than at higher temperatures.
(b)
At higher temperatures, the oxygen molecules have more kinetic energy. $[Higher \ kinetic \ energy]$ Higher temperature means more kinetic energy.
The oxygen molecules move faster. $[Faster \ movement]$ Molecules with more kinetic energy move faster.
This increases the rate of collision with the membrane. $[More \ collisions]$ Faster molecules collide more frequently.
The oxygen molecules diffuse through the membrane faster. $[Faster \ diffusion]$ More frequent collisions and faster movement results in faster diffusion.
How to earn full marks:
- State that at higher temperatures, the oxygen molecules have more kinetic energy.
- State that the oxygen molecules move faster.
- State that this increases the rate of collision with the membrane.
- State that the oxygen molecules diffuse through the membrane faster.
(c)
Repeat the experiment multiple times at each temperature and calculate the mean time. $[Repeat \ and \ calculate \ mean]$ Repeating the experiment and calculating the mean will reduce the effect of random errors.
Use a more precise timer to measure the time for oxygen to diffuse. $[Use \ a \ more \ precise \ timer]$ A more precise timer will reduce the effect of random errors.
How to earn full marks:
- State that the experiment should be repeated multiple times at each temperature and the mean time should be calculated.
- State that a more precise timer should be used to measure the time for oxygen to diffuse.
Common Pitfall: When describing trends, be specific! Don't just say "there's a relationship." Say how the variables relate (positive, negative, inverse, etc.). Also, remember that reliability is improved by repeating experiments and calculating means.
Exam-Style Question 4 — Extended Response [9 marks]
Question:
A farmer wants to investigate how different factors affect the rate of absorption of a fertiliser solution by the roots of his tomato plants. He sets up four different experiments, each varying one factor.
Experiment 1: Different concentrations of fertiliser solution (1%, 2%, 3%). Experiment 2: Different temperatures of the fertiliser solution (10°C, 20°C, 30°C). Experiment 3: Different surface areas of the roots (by trimming the roots). Experiment 4: One set of plants is treated with a respiration inhibitor, and one set is not.
(a) Define the term concentration gradient. [2]
(b) For Experiments 1, 2 and 3, explain how the factor being varied affects the rate of absorption of the fertiliser solution by diffusion. [6]
(c) Explain why the respiration inhibitor in Experiment 4 would affect the absorption of some nutrients by the plant roots. [1]
Worked Solution:
(a)
Concentration gradient is the difference in concentration of a substance. $[Difference \ in \ concentration]$ Concentration gradient is the difference in concentration of a substance.
Between two regions. $[Between \ two \ regions]$ The difference in concentration must be between two regions.
How to earn full marks:
- State that the concentration gradient is the difference in concentration of a substance.
- State that the difference in concentration must be between two regions.
(b)
Experiment 1: A higher concentration gradient (3% vs. 1%) means a faster rate of diffusion because there is a greater difference in concentration between the fertiliser solution and the inside of the root cells. $[Higher \ concentration \ gradient \ = \ faster \ diffusion]$ A higher concentration gradient results in faster diffusion.
Experiment 2: A higher temperature (30°C vs. 10°C) means the water and nutrient particles have more kinetic energy, move faster, and therefore diffuse faster. $[Higher \ temperature \ = \ faster \ diffusion]$ Higher temperature results in faster diffusion.
Experiment 3: A larger surface area of the roots means there is more area for diffusion to occur across the root cell membranes. This allows for a faster rate of absorption. $[Larger \ surface \ area \ = \ faster \ diffusion]$ Larger surface area results in faster diffusion.
How to earn full marks:
- State that a higher concentration gradient results in faster diffusion.
- State that higher temperature results in faster diffusion.
- State that a larger surface area results in faster diffusion.
(c)
- Respiration provides energy for active transport, which is needed to absorb some nutrients. The inhibitor will reduce the rate of active transport. $[Respiration \ provides \ energy \ for \ active \ transport]$ Respiration provides energy for active transport.
How to earn full marks:
- State that respiration provides energy for active transport.
Common Pitfall: Make sure you understand the difference between diffusion and active transport. Diffusion doesn't require energy, while active transport does. Also, remember that respiration is the process that provides the energy for active transport.