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Diffusion Demonstration
March 29, 2021 By Emma Vanstone Leave a Comment
Imagine pulling a delicious cake out of an oven, the smell slowly spreads around the room and then through the house. This is diffusion! The lovely cake smelling particles move from where there are lots of them ( high concentration ) to where there are less of them ( low concentration ). Diffusion can be quite a slow process as the movement of particles is random. One very easy diffusion demonstration is to pour squash or food colouring into a glass of water and watch as the colour spreads through the glass.
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration until its concentration becomes equal throughout the available space.
This video shows diffusion in action .
Diffusion Demonstration with Squash and Water
Squash or juice in water is a great demonstration of diffusion. You can see the squash starts in one area and then starts to move from areas of high squash concentration to areas of low concentration. Eventually the squash spreads throughout the whole glass and the colour becomes paler and even throughout.
I used food colouring in the images below to make the process easier to see.
Diffusion using food colouring and water
What is diffusion?
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration.
Diffusion occurs in gases and liquids. Particles in gases and liquids move around randomly, often colliding with each other or whatever container they are in. When they collide they change direction which means eventually they spread out through the whole available space.
Examples of Diffusion
Diffusion in humans.
Oxygen diffuses from the alveoli of the lungs into red blood cells. This is because the concentration of oxygen in the alveoli is high and the concentration of oxygen in the red blood cells is low. Red blood cells have very thin cell walls which allows oxygen to diffuse easily in and out of them.
Diffusion in plants
Plants use carbon dioxide fro the air for photosynthesis. Carbon dioxide enters the leaves through small holes called stomata in the underside of the leaf. Spongy cells called mesophyll cells allow gases to diffuse easily in and out of the leaf. The stomata can open and close so the plant doesn’t lose too much water.
More diffusion demonstrations
Making a cup of tea is another great diffusion demonstration. This diffusion activity using different shaped tea bags is great fun.
Another example of movement of substances important for living things is osmosis .
Last Updated on November 3, 2021 by Emma Vanstone
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A Simple Demonstration of Diffusion
The molecular motion demonstrator (mmd) shows how the random motion of molecules in a gas causes two separate gases to diffuse..
WATCH THE VIDEO
Teachable Topics:
- molecular motion
- particle transport
- chemical potential
Two gases will mix into one another because of the random motion of molecules. Such mixing is called diffusion (not to be confused with convection, where large-scale currents mix gases together).
Two gases separated by a barrier with a hole will mix by passing through the hole. This can be represented with the MMD by using a metal barrier with a hole to partition MMD. Putting small blue beads on one side and large beads on the other represents having two different gases initially separated by a barrier. Eventually, the random "molecular motion" created by the MMD will cause some "molecules" to pass through the barrier and mix with the other molecules.
Figure 1: Molecules experiencing diffusion
This experiment begins with all small molecules on one side and all large molecules on the other side.
After some time, the molecules are mixed together, and both sides probably contain some large and some small molecules. There is nothing to keep like-sized molecules together, and the two sides of the barrier are identical.
At a given moment, then, one side may contain more molecules than the other, or may have a higher concentration of one kind of molecule than the other. On the average, though, the molecules are evenly distributed between the two sides, and the two kinds of molecules are evenly mixed with one another.
If more beads were used in the experiment, these kind of fluctuations wouldn't occur so drastically. In a real, macroscopic gas, where there are countless molecules, any fluctuations are so relatively small that they aren't noticed. This means that two gasses diffusing into one another will stay mixed, and will appear totally uniform in composition.
- Molecular Motion Demonstrator
- aluminum barrier with hole
- six blue beads
- six white beads
- Set up the Molecular Motion Demonstrator on a flat surface or on an overhead projector.
- Next, add the aluminum barrier to the frame.
- Place in six blue beads on one side of the barrier, and six white beads on the other side of the barrier.
- Turn on the MMD, and adjust the speed to a level that makes the beads diffuse through the barrier.
Random Demo
- Measuring the Speed of Light with Marshmallows
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1.3 diffusion.
Prior knowledge: You will need to have an understanding of how the states of matter differ, in terms of their particles.
Diffusion in liquids
When a drop of ink is placed into a beaker of water, the ink diffuses slowly throughout the whole volume of the water. It does this because both the water molecules and the ink particles are constantly moving, colliding with each other and randomly changing direction. It follows that the hotter the water is, the faster the particles will be moving around and the faster the ink will diffuse throughout the volume.
Diffusion in gases
To demonstrate diffusion in gases, a long glass tube is set up with cotton wool soaked with hydrochloric acid at one end, and cotton wool soaked with ammonia at the other end.
The hydrogen chloride and the ammonia gases diffuse along the tube from either end, because the particles are constantly, randomly moving. Where they meet, they react forming a white ‘smoke ring’ of ammonium chloride.
ammonia + hydrogen chloride → ammonium chloride
The smoke ring is not formed in the middle of the tube, but nearer to the end with the hydrochloric acid. This tells us that the ammonia molecules travel further than the hydrogen chloride molecules in the same amount of time, in other words they diffuse more quickly. This is because they are lighter: ammonia has a mass of 17 units while hydrogen chloride has a mass of 36.5 units.
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Diffusion of Gases
Related Topics: More Lessons for IGCSE Chemistry Math Worksheets
A series of free IGCSE Chemistry Activities and Experiments (Cambridge IGCSE Chemistry).
This is the demonstration on the diffusion of gases in which ammonia and hydrogen chloride meet in a long tube. They react and produce a smoke ring.
- Using sticky tape, stick small pieces of cotton wool to the inside end of two rubber bungs.
- Add a few drops of concentrated hydrochloric acid to the cotton wool on one of the bungs and concentrated ammonia to the cotton wool on the other.
- Quickly push the bungs into opposite ends of the tube ensuring that they are placed in position at the same time.
- Observe for 10–15 minutes.
- Write a word equation and symbol equation for the reaction taking place that produces the smoke ring.
- Explain the position of the smoke ring?
- Explain why the reaction does not occur more quickly.
- hydrogen chloride + ammonia → ammonium chloride HCl(g) + NH 3 (g) → NH 4 Cl(s)
- The smoke ring is further from the ammonia than the hydrogen chloride. This means that the ammonia molecules move more quickly and travel further along the tube than the hydrogen chloride molecules. Ammonia molecules have a lower relative molecular mass (17) than hydrogen chloride (36.5) and so can move faster.
- There are air molecules in the tube which will get in the way of the ammonia and hydrogen chloride molecules and slow them down.
Diffusion and Brownian Motion What is the difference between Brownian Motion and Diffusion?
- In Brownian motion , a particle does not have a specific direction to travel. Instead, it will move in all directions. Brownian motion of a particle is governed by the other particles in the medium.
- Diffusion takes place according to a concentration or chemical potential gradient. The particles will travel from a high concentration to a low concentration.
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9.4 Effusion and Diffusion of Gases
Learning objectives.
By the end of this section, you will be able to:
- Define and explain effusion and diffusion
- State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 9.27 ). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 9.27 . The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
We are often interested in the rate of diffusion , the amount of gas passing through some area per unit time:
The diffusion rate depends on several factors: temperature; the mass of the atoms or molecules; the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving movement of gaseous species similar to diffusion is effusion , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum ( Figure 9.28 ). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones ( Figure 9.29 ). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion : The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles :
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
Example 9.20
Applying graham’s law to rates of effusion.
Hydrogen effuses four times as rapidly as oxygen.
Check Your Learning
Example 9.21, effusion time calculations.
and combine it with Graham’s law:
Noting that amount of A = amount of B , and solving for time for Ne :
and substitute values:
Finally, solve for the desired quantity:
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.
Example 9.22
Determining molar mass using graham’s law.
Plug in known data:
The gas could well be CH 4 , the only gas with this molar mass.
How Sciences Interconnect
Use of diffusion for nuclear energy applications: uranium enrichment.
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235 U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235 U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF 6 , the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235 UF 6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238 UF 6 molecules. The gas that has passed through the barrier is slightly enriched in 235 UF 6 and the residual gas is slightly depleted. The small difference in molecular weights between 235 UF 6 and 238 UF 6 only about 0.4% enrichment, is achieved in one diffuser ( Figure 9.30 ). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
The large scale separation of gaseous 235 UF 6 from 238 UF 6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10 –6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF 6 .
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.
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Top 5 Experiments on Diffusion (With Diagram)
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The following points highlight the top five experiments on diffusion. The experiments are: 1. Diffusion of S olid in Liquid 2. Diffusion of Liquid in Liquid 3. Diffusion of Gas in Gas 4. Comparative Rates of Diffusion of Different Solutes 5. Comparative rates of diffusion through different media.
Experiment # 1
Diffusion of s olid in liquid:.
Experiment:
A beaker is almost filled with water. Some crystals of CuSO 4 or KMnO 4 are dropped carefully without disturbing water and is left as such for some time.
Observation:
The water is uniformly coloured, blue in case of CuSO 4 and pink in case of KMnO 4 .
The molecules of the chemicals diffuse gradually from higher concentration to lower concentration and are uniformly distributed after some time. Here, CuSO 4 or KMnO 4 diffuses independently of water and at the same time water diffuses independently of the chemicals.
Experiment # 2
Diffusion of liquid in liquid:.
Two test tubes are taken. To one 30 rim depth of chloroform and to the other 4 mm depth of water are added. Now to the first test tube 4 mm depth of water and to the other 30 mm depth of ether are added (both chloroform and ether form the upper layer).
Ether must be added carefully to avoid disturbance of water. The tubes are stoppered tightly with corks. The position of liquid layers in each test tube is marked and their thickness measured.
The tubes are set aside for some time and the thickness of the liquids in each test tube is recorded at different intervals.
The rate of diffusion of ether is faster than that of chloroform into water as indicated by their respective volumes.
The rate of diffusion is inversely proportional (approximately) to the square root of density of the substance. Substances having higher molecular weights show slower diffusion rates than those having lower molecular weights.
In the present experiment ether (C 2 H 5 -O-G 2 H 5 , J mol. wt. 74) diffuses faster into water than chloroform (CHCI 3 , mol. wt. 119.5). This ratio (74: 119-5) is known as diffusively or coefficient of diffusion.
Experiment # 3
Diffusion of gas in gas:.
One gas jar is filled with CO 2 (either by laboratory method: CaCO 3 + HCL, or by allowing living plant tissue to respire in a closed jar). Another jar is similarly filled with O 2 (either by laboratory method: MnO 2 + KClO 2 , or by allowing green plant tissue to photosynthesize in a dosed jar). The gases may be tested with glowing match stick.
The oxygen jar is then inverted over the mouth of the carbon dioxide jar and made air-tight with grease. It is then allowed to remain for some time. The jars are carefully removed and tested with glowing match stick.
The glowing match sticks flared up in both the jars.
The diffusion of CO 2 and O 2 takes place in both the jars until finally the concentrations are same in both of them making a mixture of CO 2 and O 2 . Hence the glowing match sticks flared up in both the jars.
Experiment # 4
Comparative rates of diffusion of different solutes:.
3.2gm of agar-agar is completely dissolved in 200 ml of boiling water and when partially cooled, 30 drops of methyl red solution and a little of 0.1 N NaOH are added to give an alkaline yellow colour. 3 test tubes are filled three-fourth full with agar mixture and allowed to set.
The agar is covered with 4 ml portion of the following solutions, stoppered tightly and kept in a cool place:
(a) 4 ml of 0-4% methylene blue,
(b) 4 ml of 0.05 N HCl, and (4.2 ml of 0.1ml HCL plus 2 ml of 0-4% methylene blue.
The diffusion of various solutes is recorded in millimeters after 4 hours. The top of the gel should be marked before the above solutions are added.
The rate of diffusion of HCL alone (tube b) is faster compared to the combination of methylene blue and HCl (tube c) and minimum in case of methylene blue alone (tube a).
Different substances like gases, liquids and solutes can diffuse simultaneously and independently at different rates in the same place without interfering each other.
HCL being gaseous in nature and of lower molecular weight can diffuse much faster than methylene blue which is a dye of higher molecular weight having an adsorptive property. Hence in combination, these; two substances diffuse more readily than methylene blue alone.
Experiment # 5
Comparative rates of diffusion through different media:.
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Diffusion of Gases: Introduction, Meaning, Graham’s Law
The smell of food greets us as we enter the kitchen. Similarly, if someone uses perfume, the aroma is carried around the room. This is due to the fact that the aroma of the meal or perfume mixes with the air. In this article, we will explore this process which is called diffusion of gases.
Before understanding the concept of diffusion of gases, let us see the meaning of diffusion in general.
What is Diffusion?
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. This process happens in gases and liquids faster than solids because their particles can move randomly.
The term diffusion comes from the Latin word diffundere, which means to spread out.
Increasing the temperature or pressure increases the rate of diffusion.
Diffusion of Gases
The gases possess a remarkable property of spreading in the entire space available to them and of intermixing together even against gravity.
For example, if a bottle of ammonia is opened at one corner of a room, its smell can be felt at the other corner.
Let us understand with the help of an example how two gases intermix together even against gravity.
Suppose we have two gas jars, one containing colourless hydrogen gas and the other containing reddish-brown \({\text{N}}{{\text{O}}_2}.\) The two gases intermix when the jar containing hydrogen is inverted over the jar containing \({\text{N}}{{\text{O}}_2}\) and the cover in between the two jars is removed, and both jars look reddish-brown after some time.
\({\text{N}}{{\text{O}}_2}\) is heavier than hydrogen, yet it travels upwards while the lighter hydrogen gas travels downwards.
Thus, the intermixing of two gases even against gravity occurs due to the rapid movement of gas molecules and the presence of intermolecular spaces in a gas. Due to high velocity, the molecules of one gas penetrate into the intermolecular space of the other when brought in contact with each other. The velocities of the molecules are so high that they overcome the gravitational forces and mix up together even against gravity. The diffusion may be thus defined as-
The spontaneous intermixing of two or more gases even against gravity is called diffusion.
Examples of Diffusion of Gases in Daily Life
Diffusion of gases allows spreading the smell of various things quickly-
- The fragrance of perfume can be smelled from a distance.
- The smell of the burning incense stick can be smelled from a distance.
- The aroma of the food while being cooked.
Graham’s Law of Diffusion
Thomas Graham studied the diffusion of various gases and established a relationship between the rate of diffusion and the density of a gas. This relationship is known as Graham’s law of diffusion. This law can be stated as follows-
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.
Mathematically, at constant temperature and pressure-
\({\text{r}} \propto \frac{1}{{\sqrt {\text{d}} }}\)
where \({\text{d}}\) is the density of the gas and \({\text{r}}\) is the rate of diffusion.
If \({{\text{r}}_1}\) and \({{\text{r}}_2}\) are the rates of diffusion of two gases at the same temperature and pressure, and \({{\text{d}}_1},{{\text{d}}_2}\) are their densities respectively, then according to this law-
\({{\text{r}}_1} \propto \frac{1}{{\sqrt {{{\text{d}}_1}} }}\) \({{\text{r}}_2} \propto \frac{1}{{\sqrt {{{\text{d}}_2}} }}\) \(\therefore \frac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \sqrt {\frac{{{{\text{d}}_2}}}{{{{\text{d}}_1}}}} \)
Since molecular mass of a gas is twice its vapour density i.e., \({\text{M}} = 2 \times {\text{d}},\) we have-
\(\frac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \sqrt {\frac{{{{\text{d}}_2}}}{{{{\text{d}}_1}}}} = \sqrt {\frac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} \)
Where \({{\text{M}}_1}\) and \({{\text{M}}_1}\) are respectively the molecular masses of two gases under consideration. Thus, Graham’s law can also be stated as-
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
The rate of diffusion of a gas is defined as the ratio of the volume of the gas diffused to the time taken in diffusion, i.e.,
\({\text{r}} = \frac{{\text{V}}}{{\text{t}}}\)
where \({\text{V}}\) is the volume of the gas diffused and \({\text{r}}\) is the time taken for diffusion. If \({{\text{V}}_1}\) and \({{\text{V}}_2}\) are the volume of two gases that undergo diffusion in time \({{\text{t}}_1}\) and \({{\text{t}}_2}\) respectively then we have-
\({{\text{r}}_1} = \frac{{{{\text{V}}_1}}}{{{{\text{t}}_1}}}\) \({{\text{r}}_2} = \frac{{{{\text{V}}_2}}}{{{{\text{t}}_2}}}\) \(\therefore \frac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}} = \frac{{{{\text{V}}_1}/{{\text{t}}_1}}}{{~{{\text{V}}_2}/{{\text{t}}_2}}} = \sqrt {\frac{{{{\text{d}}_2}}}{{~{{\text{d}}_1}}}} = \sqrt {\frac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} \)
If \({{\text{V}}_1} = {{\text{V}}_2},\) we have-
\(\frac{{{{\text{t}}_2}}}{{{{\text{t}}_1}}} = \sqrt {\frac{{{{\text{d}}_2}}}{{~{{\text{d}}_1}}}} = \sqrt {\frac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} \)
Similarly, if \({{\text{t}}_1} = {{\text{t}}_2},\) then-
\(\frac{{{{\text{V}}_1}}}{{~{{\text{V}}_2}}} = \sqrt {\frac{{{{\text{d}}_2}}}{{~{{\text{d}}_1}}}} = \sqrt {\frac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} \)
It is a particular case of diffusion. When gas is allowed to escape through a fine orifice or a pinhole, the phenomenon is termed effusion.
Applications of Gaseous Diffusion
Diffusion is a remarkable property of gases and finds several applications, which are as follows-
i ) Separation of the constituents of a gaseous mixture
The rate of diffusion of a gas depends upon its density. If the constituents of a gaseous mixture have different densities, their rates of diffusion will be different, and this property can be utilised for their separation. If a mixture of two gases is subjected to diffusion, the gas with a higher density will diffuse less as compared to the lighter gas. Hence, the diffused mixture will be richer in the lighter gas. Thus by repeating the process, the two gases can be separated almost completely. This process of separation of gases is known as fractional diffusion and has been utilised in the separation of isotopes present as a gaseous mixture.
ii) Determination of molecular masses and densities
According to Graham’s law of diffusion,
Thus, by comparing the rate of diffusion of an unknown gas with that of a gas known density or known molecular mass, the density or molecular mass of the unknown gas can be determined with the above equation.
iii) Dispersal of poisonous and foul-smelling gases
Diffusion serves as a natural process for the dispersal of foul-smelling and poisonous gases into the atmospheric air. This dilutes their bad effects on our environment. It is because of this property that atmospheric air always remains homogeneous.
Diffusion of Gases in Liquids
This can be understood with the help of an example-
Gases like carbon dioxide and oxygen are essential for the survival of aquatic life. The oxygen and carbon dioxide present in the air diffuse into the water and dissolve in it. Thus, aquatic plants use dissolved carbon dioxide for preparing their food by photosynthesis, and aquatic animals use dissolved oxygen for respiration.
In this article, we studied that particles move from a higher concentration to a lower concentration. We studied in detail the Diffusion of gases and the various factors that can affect the rate of diffusion of the gas. Now we also know the applications of diffusion of gas and how the process of diffusion of gases in liquids is helpful.
Q.1. What is an example of gas diffusion? Ans: The burning of incense sticks is an example of the diffusion of a gas. We can smell the burning incense stick because it diffuses in the air and makes its way to our noses.
Q.2. What is the diffusion of gas? Ans: The spontaneous intermixing of two or more gases even against gravity is called diffusion.
Q.3. Can diffusion happen in gases? Ans: Yes, diffusion occurs in gas and liquids due to the random movement of their particles and the intermolecular space available.
Q.4. What are the three types of diffusion? Ans: The three types of diffusion are as follows- (i) Simple diffusion (ii) Osmosis (iii) Facilitated diffusion
Q.5. What tissue is specialized for the rapid diffusion of gases? Ans: Simple squamous epithelia tissue, which is present as the lining of alveoli, the lining of blood vessels is a specialised tissue that allows diffusion.
Q.6. What factors affect the rate of diffusion of gases? Ans: Several factors affect the rate of diffusion. Some of the factors are- Mass of the gases The density of the gases Concentration gradient Temperature
We hope this detailed article on the diffusion of gases will be helpful to you. If you have any questions regarding the article or, in general, about the diffusion of gases, please ping us through the comments section, and we will get back to you as soon as possible.
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Diffusion in Liquids
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Chemical Concept Demonstrated
- Diffusion of liquids
Demonstration
| 100 mL of three solutions are poured into a large filter funnel, being careful not to mix the layers. dissolved in chloroform (CHCl ). |
Observations
The top HCl layer diffuses into the middle layer and changes color indicating an acid. The bottom NH 3 layer is also diffusing into the middle layer and changes color indicating a base. Eventually once the HCl and the NH 3 begin to mix, the indicator turns the color of a neutral solution.
Explanation
Liquids diffuse from solutions of higher concentration to solutions of lower concentration. The indicator in the water showed how the HCl diffused downward into the water, but the NH 3 diffused upward. Both were going toward the solution with the lowest concentration. Then, they began to mix with each other in the same manner.
Contributors
- Dr. George Bodner ( Perdue University )
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Diffusion in liquids
In association with Nuffield Foundation
Demonstrate that diffusion takes place in liquids by allowing lead nitrate and potassium iodide to form lead iodide as they diffuse towards each other in this practical
In this experiment, students place colourless crystals of lead nitrate and potassium iodide at opposite sides of a Petri dish of deionised water. As these substances dissolve and diffuse towards each other, students can observe clouds of yellow lead iodide forming, demonstrating that diffusion has taken place.
This practical activity takes around 30 minutes.
- Eye protection
- White tile or piece of white paper
- Lead nitrate (TOXIC, DANGEROUS FOR THE ENVIRONMENT), 1 crystal
- Potassium iodide, 1 crystal
- Deionised water
Greener alternatives
To reduce the use of toxic chemicals in this experiment you can conduct the experiment in microscale, using drops of water on a laminated sheet, find full instructions and video here, and/or use a less toxic salt than lead nitrate, eg sodium carbonate and barium chloride. More information is available from CLEAPSS.
Health, safety and technical notes
- Read our standard health and safety guidance.
- Wear eye protection throughout.
- Lead nitrate, Pb(NO 3 ) 2 (s), (TOXIC, DANGEROUS FOR THE ENVIRONMENT) – see CLEAPSS Hazcard HC057a .
- Potassium iodide, KI(s) – see CLEAPSS Hazcard HC047b .
- Place a Petri dish on a white tile or piece of white paper. Fill it nearly to the top with deionised water.
- Using forceps, place a crystal of lead nitrate at one side of the petri dish and a crystal of potassium iodide at the other.
- Observe as the crystals begin to dissolve and a new compound is formed between them.
Source: Royal Society of Chemistry
As the crystals of potassium iodide and lead nitrate dissolve and diffuse, they will begin to form yellow lead iodide
Teaching notes
The lead nitrate and potassium iodide each dissolve and begin to diffuse through the water. When the lead ions and iodide ions meet they react to form solid yellow lead iodide which precipitates out of solution.
lead nitrate + potassium iodide → lead iodide + potassium nitrate
Pb(aq) + 2I – (aq) → PbI 2 (s)
The precipitate does not form exactly between the two crystals. This is because the lead ion is heavier and diffuses more slowly through the liquid than the iodide ion.
Another experiment – a teacher demonstration providing an example of a solid–solid reaction – involves the same reaction but in the solid state.
Additional information
This is a resource from the Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry. This collection of over 200 practical activities demonstrates a wide range of chemical concepts and processes. Each activity contains comprehensive information for teachers and technicians, including full technical notes and step-by-step procedures. Practical Chemistry activities accompany Practical Physics and Practical Biology .
The experiment is also part of the Royal Society of Chemistry’s Continuing Professional Development course: Chemistry for non-specialists .
© Nuffield Foundation and the Royal Society of Chemistry
- 11-14 years
- 14-16 years
- Practical experiments
- Physical chemistry
- Reactions and synthesis
Specification
- Precipitation is the reaction of two solutions to form an insoluble salt called a precipitate.
- Motion of particles in solids, liquids and gases.
- Diffusion (Graham's law not required).
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Diffusion of gases: ammonia and hydrogen chloride. Place concentrated ammonia solution on a pad in one end of a tube and concentrated hydrochloric acid on a pad at the other and watch as the two gases diffuse far enough to meet and form a ring of solid ammonium chloride. This demonstration is best performed in a fume cupboard.
We recommend using the latest version of Chrome, Firefox, Safari, or Edge. Mix two gases to explore diffusion! Experiment with concentration, temperature, mass, and radius and determine how these factors affect the rate of diffusion.
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. Diffusion occurs in gases and liquids. Particles in gases and liquids move around randomly, often colliding with each other or whatever container they are in. When they collide they change direction which means eventually they spread out ...
Two gases separated by a barrier with a hole will mix by passing through the hole. This can be represented with the MMD by using a metal barrier with a hole to partition MMD. ... Figure 1: Molecules experiencing diffusion. This experiment begins with all small molecules on one side and all large molecules on the other side. After some time, the ...
This experiment will demonstrate rates of diffusion, a property of gases investigated by Thomas Graham. In 1829, he proposed his law of diffusion which states that the rate of diffusion of a gas is inversely proportional to the square root of its density: However, since the ideal gas law indicates that the density of a gas and its molecular ...
In both experiments there is a gradation of colour change which depends on the rate of diffusion of the gas. It might be possible to compare the rate of diffusion of these two gases. Ammonia with a relative molecular mass of 17 should be faster than chlorine with a relative molecular mass of 71.
Procedure. Soak a ball of cotton wool in aqueous ammonia and insert a few centimetres into the glass tube. Rinse the tweezers in water, and dispose of excess ammonia. Close the end with the rubber bung. At the same time, insert a cotton wool ball soaked in hydrochloric acid at the other end. Rinse the tweezers again in water to avoid corrosion.
Wear splash-proof goggles and work in a running fume cupboard. Place the vials into 100 cm 3 beakers to catch drips and reduce the risk of tipping. Add 4 cm 3 of concentrated hydrochloric acid to one sample vial and 4 cm 3 of concentrated ammonia to the other. When ready for the demonstration, take them to the bench and leave them at least one ...
Diffusion - PhET Interactive Simulations
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To demonstrate diffusion in gases, a long glass tube is set up with cotton wool soaked with hydrochloric acid at one end, and cotton wool soaked with ammonia at the other end. The hydrogen chloride and the ammonia gases diffuse along the tube from either end, because the particles are constantly, randomly moving. Where they meet, they react ...
A series of free IGCSE Chemistry Activities and Experiments (Cambridge IGCSE Chemistry). This is the demonstration on the diffusion of gases in which ammonia and hydrogen chloride meet in a long tube. They react and produce a smoke ring. Method. Using sticky tape, stick small pieces of cotton wool to the inside end of two rubber bungs.
Figure 9.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion. If a mixture of gases is placed in a container with porous walls, the gases effuse through ...
The diffusion coefficient, D, will be proportional to the distance by the following formula x1/2 = sqrt (Dt) (1) where x1/2 is the distance at which the concentration is 1/2 of the initial concentration of the diffusant tracked, and t is the time at temperature or the time of diffusion. Experiment: 1. Diffusion of two gases.
The rate of diffusion of HCl gas is faster in case of tube 2 containing filter paper strip and slower in case of tube 1 containing agar-gel. Inference: The rate of diffusion of gases through a medium is inversely related to the density of the medium. Hence HCL diffuses faster in gaseous medium (tube 2) than in semisolid medium (tube 1). N.B.
Diffusion of gases on a microscale - teacher notes. This experiment looks at the spread of ammonia and chlorine gases as a result of their interaction with copper(II) sulfate and potassium iodide/starch solutions. The experiment is done using drops of solutions placed on a clear plastic sheet with the top of a well-plate as a lid.
r = V t. where V is the volume of the gas diffused and r is the time taken for diffusion. If V 1 and V 2 are the volume of two gases that undergo diffusion in time t 1 and t 2 respectively then we have-. r 1 = V 1 t 1. r 2 = V 2 t 2. ∴ r 1 r 2 = V 1 / t 1 V 2 / t 2 = d 2 d 1 = M 2 M 1. If V 1 = V 2, we have-.
Explanation. Liquids diffuse from solutions of higher concentration to solutions of lower concentration. The indicator in the water showed how the HCl diffused downward into the water, but the NH 3 diffused upward. Both were going toward the solution with the lowest concentration. Then, they began to mix with each other in the same manner.
Diffusion occurs in liquids and gases when their particles close particle A single piece of matter from an element or a compound, which is too small to be seen. Particles can be atoms, molecules ...
It's worth noting that in this experiment, gas diffusion rates are being utilized to confirm Graham's law, even though Graham's law is intended for effusion into vacuum. Q3. There would be a difference in the experimental results and actual rates of diffusion if this experiment was performed in an open atmosphere.
Seal with sellotape and remove to a fume cupboard. The demonstration can be performed along with a similar set-up using bromine to show that gases diffuse at different rates. To fill a 1 litre gas jar, use no more than 2 ml of liquid bromine. Adjust the volume of bromine liquid to the capacity of the gas jar that is available.
In the procedure the progress of diffusion is determined by measuring the change in conductivity over time. ... Diffusion of gases: A physical chemistry experiment. Thomas A. Gover ; Cite this: J. Chem. Educ. 1967, 44, 7, 409. Publication Date (Print): July 1, 1967. Publication History.
Procedure. Place a Petri dish on a white tile or piece of white paper. Fill it nearly to the top with deionised water. Using forceps, place a crystal of lead nitrate at one side of the petri dish and a crystal of potassium iodide at the other. Observe as the crystals begin to dissolve and a new compound is formed between them. Show Fullscreen.