T and n fixed: drag the piston and watch the pressure
Controls
Choose a law
Readings
- Pressure, P
- 99.8kPa
- Pressure (atm)
- 0.985atm
- Volume, V
- 25.0L
- Moles, n
- 1.0mol
- Absolute temperature, T
- 300K
- Temperature in Celsius
- 26.85°C
- PV (stays constant)
- 2,494kPa·L
- PV/T
- 8.314kPa·L/K
- Hits on the piston per second
- 0.0/s
- PV/nRT (from counted hits)
- 1.00
How to use this simulation
- Boyle is selected first. Grab the piston handle with your mouse or finger and pull it down (or use the volume slider): the particles crowd together, hit the piston more often and the gauge needle climbs.
- Watch the red dot on the P–V graph slide along a curve (a hyperbola), and check the readout "PV (stays constant)": it does not move however far you push.
- Pick Charles and raise the temperature: the flame grows, the particles speed up, and the loaded piston rises. Follow the straight line backwards; the dashed part reaches zero volume at −273.15 °C.
- Pick Gay-Lussac and do the same: now the piston is locked, so the volume stays put and the pressure climbs in a straight line. Pick Avogadro and add moles: new particles arrive and the volume grows.
- Pick Ideal gas and move all three sliders. Whatever you do, "R = PV/nT" stays the same, and "PV/nRT", worked out by counting real particle hits, wobbles around 1.
Balloons, syringes and summer tyres
Leave a party balloon in a hot car and it may burst. Put it in the fridge and it goes a bit saggy. Nobody let air in or out, so why did the size change? Temperature: a warm gas expands, a cold gas contracts.
Put your thumb over the tip of an empty syringe and push the plunger. The first bit is easy, but the further you push, the harder it pushes back, as if the trapped air is fighting you. Let go and the plunger springs back out. Squeeze a gas into less space and its pressure goes up: that is Boyle's law in your hand.
On a hot afternoon, the air pressure in a car tyre is higher than it was on a cool morning, which is why car manuals tell you to check tyre pressure when the tyres are cold. The tyre's volume barely changes, but the temperature does, and the pressure follows. That is Gay-Lussac's law.
All three stories involve the same four quantities: pressure (P), volume (V), temperature (T) and the amount of gas in moles (n). Each gas law simply says: hold one or two of these fixed, and here is how the others change together. In the simulation above you can hold and change them yourself.
From zero: why does a gas push at all?
Picture a closed room with thousands of tiny balls flying around at random, bouncing off the walls. That is a gas. Its particles are far apart, move in straight lines in every direction, and keep hitting the walls of their container. Each hit gives the wall a tiny push. Add up trillions of pushes every second and you get a steady force on every square metre of wall. Force per unit area is pressure.
This picture is the kinetic theory of gases (also called the kinetic molecular theory). Its assumptions, in plain words: (1) a gas is made of a huge number of tiny particles whose own volume is negligible compared with the container; (2) they move randomly in straight lines; (3) they do not attract or repel each other; (4) their collisions with each other and the walls are perfectly elastic, so no energy is lost; (5) their average kinetic energy is proportional to the absolute temperature.
That last idea is the key one: temperature is a measure of the average kinetic energy of the particles. Heat a gas and the particles move faster. Faster particles do two things: they hit the walls more often, and each hit is harder. Both raise the pressure. In the simulation, watch "Hits on the piston per second" go up when you raise the temperature.
Not all particles move at the same speed. Some are slow, a few are very fast, and most are in between. This spread is called the Maxwell–Boltzmann distribution. The simulation colours the fast ones differently. Raise the temperature and the whole distribution shifts faster; the typical speed grows with the square root of the absolute temperature.
Key terms in the gas laws
Get these words exactly right and every definition question becomes easy.
| Term | Meaning | Units |
|---|---|---|
| Pressure (P) | Force per unit area that gas particles exert on the walls by hitting them | Pa, kPa, atm, mmHg, bar |
| Volume (V) | The space the gas fills; a gas always fills its whole container | L, mL, m³, dm³ |
| Absolute temperature (T) | Temperature on the Kelvin scale, T = t °C + 273.15 | K |
| Amount of gas (n) | Number of moles; 1 mol contains 6.022 × 10²³ particles | mol |
| Absolute zero | The temperature at which an ideal gas would have zero volume | 0 K = −273.15 °C |
| Ideal gas | A model gas that obeys the gas laws exactly at every pressure and temperature | — |
| Gas constant (R) | The universal constant in PV = nRT | 8.314 J mol⁻¹ K⁻¹ |
| Isothermal | At constant temperature; Boyle’s law describes an isothermal change | — |
| Isobaric / isochoric | At constant pressure (Charles) / at constant volume (Gay-Lussac) | — |
| STP | Standard temperature and pressure; see the molar volume section for the two definitions | — |
Boyle's law: squeeze it and the pressure rises
Statement: for a fixed amount of gas at constant temperature, the volume is inversely proportional to the pressure. Robert Boyle showed this in 1662 by trapping air in a J-shaped glass tube and pouring in mercury to squeeze it.
Inversely proportional means that when one doubles, the other halves. Triple the pressure and the volume drops to a third. So the product P × V stays the same. Try it in Boyle mode: drag the piston anywhere and "PV (stays constant)" does not change.
Why? Halve the volume and the same particles are crammed into half the space. Each one travels half as far between hits on the piston, so it hits twice as often. The temperature has not changed, so each hit is just as hard. Twice as many hits of the same strength: twice the pressure.
V ∝ 1/P (T, n fixed)
PV = kk depends on the temperature and the amount of gas
P₁V₁ = P₂V₂the version you use in problems
Graphs of Boyle's law
Plot P against V and you get a curve called a rectangular hyperbola; it never touches either axis. Each temperature gives its own curve, called an isotherm, and hotter isotherms sit further from the origin.
Plot P against 1/V and you get a straight line through the origin. Plot PV against P and you get a flat horizontal line, because PV is constant. Exams love asking you to match these three shapes.
Charles's law: heat it and it expands
Statement: for a fixed amount of gas at constant pressure, the volume is directly proportional to the absolute temperature. Jacques Charles noticed it around 1787; Joseph Gay-Lussac published careful measurements in 1802.
An older, Celsius-based version is still in many books: at constant pressure, a gas expands by 1/273 of its volume at 0 °C for every 1 °C rise, so Vₜ = V₀(1 + t/273).
In Charles mode a fixed load sits on the piston, so the pressure cannot change. Heat the gas and the faster particles push the piston up until the gas has spread out enough that the pressure inside balances the load again.
V ∝ T (P, n fixed)T must be in kelvin
V/T = k
V₁/T₁ = V₂/T₂the version you use in problems
The graph that discovered absolute zero
V against T in kelvin is a straight line through the origin. V against t in °C is also a straight line, but it does not pass through the origin; extend it backwards (extrapolate) and it hits zero volume at −273.15 °C. Lines drawn at different pressures all meet at exactly that point.
So at −273.15 °C an ideal gas would have no volume at all, and nothing can be colder, because a negative volume is meaningless. That temperature is absolute zero. Real gases turn into liquids long before this, which is why the simulation draws that part of the line dashed.
Gay-Lussac's law (the pressure law): locked volume, rising pressure
Statement: for a fixed amount of gas at constant volume, the pressure is directly proportional to the absolute temperature. It is also called the pressure law or Amontons's law.
In this mode the piston is pinned in place. Heat the gas and the particles hit the same walls more often and harder, so the gauge climbs. The P–t graph is a straight line that, extended backwards, also reaches zero at −273.15 °C.
P ∝ T (V, n fixed)
P₁/T₁ = P₂/T₂pressure cookers, tyres, aerosol cans
Avogadro's law: more gas, more room
Statement: at the same temperature and pressure, equal volumes of all gases contain equal numbers of particles. Put another way, at constant temperature and pressure the volume of a gas is proportional to the number of moles.
The surprising part is that the kind of gas does not matter. Hydrogen, oxygen or carbon dioxide: one mole of each takes up the same volume at the same T and P, because in an ideal gas the particles' own size is negligible and only their number counts. In Avogadro mode, raise n and watch new particles appear and the loaded piston rise.
V ∝ n (P, T fixed)
V₁/n₁ = V₂/n₂
Putting them together: the combined gas law and PV = nRT
Now join the laws. Boyle gives V ∝ 1/P, Charles gives V ∝ T and Avogadro gives V ∝ n. When all three change at once, V ∝ nT/P.
Replace the proportional sign with a constant, call it R, and you get V = nRT/P, or PV = nRT. That is the ideal gas law (ideal gas equation), and R is the molar gas constant. R is the same for every gas, which is why it is also called the universal gas constant.
For a fixed amount of gas (n constant), PV/T = nR is constant, so P₁V₁/T₁ = P₂V₂/T₂. That is the combined gas law, the one to reach for when pressure, volume and temperature all change in the same problem. Watch the "PV/T" readout in the simulation: it only moves when n does.
V ∝ nT/Pall three laws at once
PV = nRTthe ideal gas law
P₁V₁/T₁ = P₂V₂/T₂the combined gas law (n fixed)
PV = (m/M)RTwhen you are given a mass m and molar mass M
The gas constant R in different units
R is one physical quantity; only the number changes with the units. Every value below is converted from the SI value (1 atm = 101.325 kPa = 760 mmHg, 1 bar = 100 kPa, 1 cal = 4.184 J). R's real unit is energy per mole per kelvin, because pressure × volume is work.
| Value of R | Units | Use it with |
|---|---|---|
| 8.314 | J mol⁻¹ K⁻¹ | Pa, m³ (SI) |
| 8.314 | kPa L mol⁻¹ K⁻¹ | kPa, L |
| 0.08314 | L bar mol⁻¹ K⁻¹ | bar, L |
| 0.0821 | L atm mol⁻¹ K⁻¹ | atm, L |
| 62.36 | L mmHg mol⁻¹ K⁻¹ | mmHg, L |
| 1.987 | cal mol⁻¹ K⁻¹ | cal |
Absolute zero and the Kelvin scale
Temperatures in gas-law problems must be in kelvin. The Kelvin scale starts at absolute zero, −273.15 °C, and a kelvin is the same size as a Celsius degree, so converting is just adding: T (K) = t (°C) + 273.15. Most textbook problems round this to 273.
Why not Celsius? Going from 0 °C to 10 °C has not made anything "infinitely hotter". A gas expands in proportion to its kelvin temperature: 273 K to 283 K is only about 4% more. Put Celsius values into a ratio and you get zeros and negatives that wreck the answer.
At absolute zero the particles would have the least possible kinetic energy. Laboratories have cooled atoms to within a billionth of a kelvin of absolute zero, but reaching exactly 0 K is impossible.
| Where | Celsius | Kelvin |
|---|---|---|
| Absolute zero | -273.15 °C | 0.00 K |
| A very cold gas | -100.00 °C | 173.15 K |
| Ice melts | 0.00 °C | 273.15 K |
| Room temperature | 25.00 °C | 298.15 K |
| Body temperature | 37.00 °C | 310.15 K |
| Water boils | 100.00 °C | 373.15 K |
Molar volume at STP: where does 22.4 L come from?
The classic STP is 0 °C (273 K) and 1 atm. There, one mole of any ideal gas fills V = RT/P = 0.0821 × 273 ÷ 1 = 22.41 L, the famous 22.4 L molar volume (with exact constants it is 22.41 L).
Careful: since 1982 IUPAC has defined standard pressure as 1 bar (100 kPa), not 1 atm. At 273.15 K and 1 bar the molar volume is 22.71 L. Both numbers are right; they answer different definitions. Use whatever your question or syllabus states, and if it says nothing, most school syllabi still expect 22.4 L.
Dalton's law of partial pressures (briefly)
In a mixture of gases that do not react, the total pressure is the sum of the partial pressures of each gas. A partial pressure is the pressure that gas would exert if it had the container to itself.
Kinetic theory makes this obvious: each particle hits the wall on its own, and the wall does not care whether it is nitrogen or oxygen. So each gas's partial pressure equals its mole fraction times the total pressure.
P(total) = p₁ + p₂ + p₃ + …
p₁ = x₁ × P(total)x₁ = n₁ / n(total), the mole fraction
Ideal gases vs real gases
An ideal gas is a model. Real gas particles do take up some space, and they do attract each other slightly. At low pressure and high temperature the particles are far apart and fast, both effects are tiny, and real gases behave almost ideally.
At high pressure the particles are crowded, so their own volume matters. At low temperature they move slowly enough for attractions to pull them together, and eventually the gas liquefies. Under these conditions PV/nRT drifts away from 1. The van der Waals equation corrects for both effects and is the usual next step in advanced courses.
Experiments to try in the simulation
Grab a notebook. Predict first, then run it and compare.
Experiment 1: halve the volume (Boyle)
It starts at V = 25 L, T = 300 K, n = 1 mol and P = 99.77 kPa. Set the volume to 12.5 L. Predict the pressure. Answer: 199.5 kPa, exactly double, and the hits per second roughly double too.
Experiment 2: double the kelvins (Charles)
In Charles mode, raise T from 300 K to 600 K. The volume goes from 25 L to 50.0 L, exactly double. In Celsius, though, that was 26.85 °C to 326.85 °C, about 12.2 times. Proof that the law needs kelvin.
Experiment 3: find absolute zero
In Charles or Gay-Lussac mode, follow the green line back along its dashed part to where it meets the temperature axis: the red dot at −273.15 °C. Change the moles or the pressure and look again. The slope changes, the meeting point does not.
Experiment 4: measure pressure by counting hits
In Ideal gas mode, watch "PV/nRT (from counted hits)". The simulation really does count every particle striking the piston and adds up the push. It wobbles around 1 because there are only a few dozen particles. A real mole has 6 × 10²³ of them, so real pressure looks perfectly steady.
Solved problems
In every solution: write what is given, write the formula, then substitute. Temperatures go into kelvin first (T = t + 273, as most textbooks round it).
Problem 1: the simulation's starting state
Given n = 1 mol, T = 300 K, V = 25 L and R = 8.314 kPa L mol⁻¹ K⁻¹, find P.
P = nRT/V = 1 × 8.314 × 300 ÷ 25 = 2,494.2 ÷ 25 = 99.77 kPa = 0.985 atm. Open the simulation and the readings panel shows exactly this.
Problem 2: squeezing a syringe (Boyle's law)
A sealed syringe holds 60 mL of air at 1 atm. The plunger is pushed in to 20 mL at constant temperature. What is the new pressure?
P₁V₁ = P₂V₂, so P₂ = P₁V₁/V₂ = 1 × 60 ÷ 20 = 3 atm. A third of the volume, 3 times the pressure.
Problem 3: a balloon in the sun (Charles's law)
A balloon holds 2.0 L at 27 °C. At constant pressure it warms to 87 °C. Find the new volume.
T₁ = 27 + 273 = 300 K and T₂ = 87 + 273 = 360 K. V₂ = V₁T₂/T₁ = 2.0 × 360 ÷ 300 = 2.4 L. Dividing the Celsius values instead would give a wildly wrong answer.
Problem 4: tyre pressure on a hot road (Gay-Lussac's law)
A tyre is at 200 kPa at 27 °C. By the afternoon the air inside reaches 57 °C. Assuming the volume is constant, find the pressure.
P₂ = P₁T₂/T₁ = 200 × 330 ÷ 300 = 220 kPa, a rise of 20 kPa. That is why you check pressure on cold tyres.
Problem 5: pumping more gas in (Avogadro's law)
A balloon holds 0.20 mol of gas in 5.0 L. Another 0.10 mol is added at the same T and P. Find the new volume.
n₂ = 0.20 + 0.10 = 0.30 mol, so V₂ = V₁n₂/n₁ = 5.0 × 0.30 ÷ 0.20 = 7.5 L.
Problem 6: correcting to STP (combined gas law)
A gas occupies 380 mL at 27 °C and 800 mmHg. What is its volume at STP (273 K, 760 mmHg)?
V₂ = P₁V₁T₂/(T₁P₂) = 800 × 380 × 273 ÷ (300 × 760) = 364.0 mL.
Problem 7: oxygen in a cylinder (PV = nRT)
A 10 L cylinder holds oxygen at 27 °C and 5 atm. How many moles, and how many grams, of O₂ are inside?
n = PV/RT = (5 × 10) ÷ (0.0821 × 300) = 50 ÷ 24.63 = 2.030 mol. Mass = n × M = 2.030 × 32 = 65.0 g.
Problem 8: proving the molar volume
Find the volume of 1 mol of an ideal gas at 273 K and 1 atm.
V = nRT/P = 1 × 0.0821 × 273 ÷ 1 = 22.41 L ≈ 22.4 L. With IUPAC's 1 bar and 273.15 K: V = 8.314 × 273.15 ÷ 100 = 22.71 L.
Problem 9: partial pressures in a mixture (Dalton's law)
A vessel holds 2 mol N₂ and 1 mol O₂ at a total pressure of 150 kPa. Find each partial pressure.
Mole fractions: x(N₂) = 0.667, x(O₂) = 0.333. p(N₂) = 0.667 × 150 = 100 kPa and p(O₂) = 50 kPa, which add back to 150 kPa.
Problem 10: why aerosol cans say "do not incinerate"
An aerosol can is at 3 atm at 27 °C. It ends up in a fire at 327 °C. What is the pressure now?
P₂ = P₁T₂/T₁ = 3 × 600 ÷ 300 = 6 atm. The kelvin temperature doubled, so the pressure doubled, which is more than many cans can hold.
Problem 11: a diver’s air bubble
In seawater the pressure rises by about 1 atm for every 10 m of depth. A 1.0 L bubble is released at 20 m (about 3 atm in total). What is its volume at the surface (1 atm), at constant temperature?
V₂ = P₁V₁/P₂ = 3 × 1.0 ÷ 1 = 3.0 L. That is why divers are trained never to hold their breath while ascending: the air in their lungs expands the same way.
Common mistakes
Avoid these and gas-law questions become free marks.
- Using Celsius. Charles, Gay-Lussac and PV = nRT all need kelvin.
- Dropping the conditions from a law: "at constant temperature" and "for a fixed amount of gas" are part of the statement.
- Using an R that does not match the units: 0.0821 for atm and litres, 8.314 for pascals and cubic metres.
- Drawing Boyle's P–V graph as a straight line. It is a curve; the straight line is P against 1/V.
- Mixing mL and L inside PV = nRT. In P₁V₁ = P₂V₂ the units only need to match on both sides.
- Saying particles get bigger when heated. They do not; they move faster and spread further apart.
- Using 22.4 L at any temperature. It is the molar volume at 0 °C and 1 atm only.
Gas laws in real life
The gas laws are not just exam material; they explain half the things that hiss, pop or float around you.
- Filling a syringe: pulling the plunger enlarges the space, the pressure inside drops (Boyle), and the higher outside pressure pushes the liquid in.
- Bicycle pump: pushing the handle shrinks the air, raises its pressure, and once it beats the tyre pressure the air flows in.
- Hot-air balloon: heated air expands (Charles), so the same volume holds less air; the lighter air inside lifts the balloon.
- Pressure cooker: trapped steam heats up and its pressure rises, which raises the boiling point of water above 100 °C and cooks food faster.
- Aerosol cans: the label says do not expose to temperatures above 50 °C or incinerate, because heat raises the pressure (Gay-Lussac) until the can can burst.
- Tyres in summer: pressure climbs on hot days, so fill tyres to the recommended value when they are cold.
- Scuba diving: deep water compresses the air in a diver’s lungs; on the way up it expands, so divers rise slowly and keep breathing.
- Breathing: your diaphragm enlarges your chest, the pressure in your lungs falls below the air outside, and air rushes in.
Exam corner
Gas laws turn up in middle-school and high-school chemistry and physics everywhere, from GCSE and IGCSE to AP Chemistry and the IB. The pattern of questions is remarkably stable: state a law with its conditions, sketch or identify a graph, convert Celsius to kelvin, and run a one- or two-step calculation with P₁V₁ = P₂V₂, V₁/T₁ = V₂/T₂, the combined gas law or PV = nRT.
Three habits earn marks: convert every temperature to kelvin before anything else, write the formula before substituting, and choose R to match your pressure and volume units. In multiple-choice questions, check which quantity is held constant before picking a law.
A sample structured question
A 10 L cylinder holds oxygen at 27 °C and 5 atm.
(a) Define absolute zero. (b) Explain, using kinetic theory, why the pressure of a gas rises when it is compressed at constant temperature. (c) Calculate the mass of oxygen in the cylinder. (d) A student says doubling the Celsius temperature will double the pressure. Evaluate this claim.
(c) 65.0 g (see Problem 7). (d) 27 °C to 54 °C is 300 K to 327 K, only about 9% more, so the pressure rises by about 9%, not 100%. Doubling the Celsius value is not doubling the absolute temperature.
One-screen revision summary
Read this list and the R table the night before the exam and you are covered.
- Gas pressure = force per unit area from countless particle hits; temperature = average kinetic energy of the particles.
- Boyle: T, n fixed → PV = constant, P₁V₁ = P₂V₂; the P–V graph is a hyperbola.
- Charles: P, n fixed → V/T = constant; the V–t line extrapolates to −273.15 °C.
- Gay-Lussac: V, n fixed → P/T = constant. Avogadro: P, T fixed → V/n = constant.
- Combined gas law P₁V₁/T₁ = P₂V₂/T₂; ideal gas law PV = nRT.
- R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 L atm mol⁻¹ K⁻¹ = 0.08314 L bar mol⁻¹ K⁻¹.
- T (K) = t (°C) + 273.15; molar volume 22.4 L at 0 °C and 1 atm, 22.71 L at 0 °C and 1 bar.
- Total pressure = sum of partial pressures; real gases behave ideally at low pressure and high temperature.
Frequently asked questions
What are the gas laws?
Four relationships between pressure, volume, temperature and amount of a gas: Boyle's law (PV constant at fixed T), Charles's law (V/T constant at fixed P), Gay-Lussac's law (P/T constant at fixed V) and Avogadro's law (V/n constant at fixed P and T). Combined, they give PV = nRT.
What is Boyle's law?
For a fixed amount of gas at constant temperature, volume is inversely proportional to pressure: PV = constant, or P₁V₁ = P₂V₂.
What is Charles's law?
For a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature: V/T = constant, or V₁/T₁ = V₂/T₂, with T in kelvin.
What is Gay-Lussac's law?
For a fixed amount of gas at constant volume, pressure is directly proportional to absolute temperature: P/T = constant, or P₁/T₁ = P₂/T₂.
What is the ideal gas law?
PV = nRT, where P is pressure, V volume, n moles, T absolute temperature and R the molar gas constant. It combines Boyle’s, Charles’s and Avogadro’s laws.
What is the value of R?
R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 L atm mol⁻¹ K⁻¹ = 0.08314 L bar mol⁻¹ K⁻¹ = 1.987 cal mol⁻¹ K⁻¹. It is one constant written in different units.
Why must temperature be in kelvin in the gas laws?
Volume and pressure are proportional to absolute temperature, which starts at absolute zero. Celsius has an arbitrary zero, so ratios of Celsius values are meaningless and can even be negative.
What is absolute zero?
0 K, or −273.15 °C: the temperature at which an ideal gas would have zero volume and its particles the least possible energy. It can be approached but never reached.
What is the combined gas law?
P₁V₁/T₁ = P₂V₂/T₂ for a fixed amount of gas. Use it when pressure, volume and temperature all change in the same problem.
Is the molar volume 22.4 L or 22.7 L?
Both, under different definitions: 22.41 L (≈ 22.4 L) at 0 °C and 1 atm, and 22.71 L at 0 °C and 1 bar, IUPAC's standard pressure since 1982. Use the one your question specifies.
When does a real gas behave like an ideal gas?
At low pressure and high temperature, when the particles are far apart and moving fast, so their own volume and their attractions for each other are negligible.
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