Thermodynamics calculators and simulators
Gas law, calorimetry, latent heat and Gibbs free energy calculators, plus heat engine, kinetic theory and states of matter simulators, with kelvin handled.
15 tools
Thermodynamics is where units stop being bookkeeping and start being the physics. Absolute temperature is not a convention you can opt out of: the gas laws are proportional in kelvin, and a Celsius substitution does not merely shift the answer, it produces a meaningless one.
Every tool here converts to kelvin, pascals and cubic metres internally, so you can enter atmospheres and millilitres and degrees Celsius together without the mixture being an error.
Most of these calculators answer one of two questions: what a gas does when its conditions change, and how much energy moved when something changed temperature. Both answers carry a restriction that is easy to walk past. The gas laws assume molecules ignore each other entirely, and q = mcΔT assumes nothing changes phase along the way, which is why latent heat gets calculators of its own.
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Visualisers and practice
Temperature and temperature difference are different quantities
A temperature of 20 °C is 293.15 K. A temperature rise of 20 °C is a rise of 20 K, not 293.15 K. The two need different conversions, and treating a difference as an absolute value is one of the most persistent errors in calorimetry.
The tools here keep them as separate quantities for exactly that reason: a field asking for ΔT will never apply the 273.15 offset, and a field asking for T always will.
The same distinction demolishes a common intuition. Going from 20 °C to 40 °C is not doubling the temperature in any thermodynamic sense. In kelvin it is 293.15 to 313.15, an increase of under 7 percent, which is why the pressure in a sealed vessel rises by about 7 percent rather than doubling.
Absolute pressure, and which R to use
The pressure in the gas law is absolute, measured from vacuum, not gauge pressure measured from the surrounding air. A tyre gauge reading 32 psi describes a gas at about 46.7 psi absolute, because atmospheric pressure is around 14.7 psi. Feeding the gauge figure into PV = nRT understates the amount of gas by roughly a third, and nothing about the answer looks unusual.
The other half of the same problem is the gas constant, since R comes in as many versions as there are pressure units. It is 8.314 J/(mol·K) in SI, and equivalently 8.314 L·kPa/(mol·K), because a litre kilopascal is exactly a joule. In litre atmospheres it is 0.082057 L·atm/(mol·K). An answer wrong by a factor of about a hundred is nearly always an R that does not match the units it was given.
Where the ideal gas law stops being true
PV = nRT assumes molecules occupy no volume and do not attract one another. Both hold well at modest pressure and well above the boiling point, where the error is typically under one percent. Both fail near condensation and at high pressure, where molecular volume becomes a significant fraction of the container and attraction pulls the pressure below the ideal prediction.
The size of the failure has a name. The compressibility factor Z is PV divided by nRT, exactly 1 for an ideal gas. Approaching condensation, attraction dominates and Z falls below 1, so a real gas occupies less volume than predicted. At very high pressure the finite size of the molecules takes over and Z rises above 1. When Z is far from 1, reach for a real-gas equation such as van der Waals.
There is also no single set of standard conditions. Molar volume is 22.414 L/mol at 0 °C and 1 atm, 24.465 L/mol at 25 °C and 1 atm, 22.711 L/mol against the IUPAC standard pressure of 1 bar, and 24.790 L/mol at 25 °C and 1 bar. The widely memorised 22.4 belongs to conditions almost no laboratory work is carried out at.
Calorimetry: the vessel counts, and phase changes do not fit
q = mcΔT describes heating a single substance that stays in one phase. Specific heat capacity is what varies between materials, and the range is wide: water is 4.184 J/(g·K), aluminium 0.897 and copper 0.385, so the same energy warms a gram of copper roughly eleven times as much as a gram of water.
The formula stops at a phase boundary. Latent heat is not in it, and the amounts are large: melting a gram of ice absorbs about 334 J at constant temperature, and boiling a gram of water about 2257 J. Heating that gram from 0 °C to 100 °C takes only 418 J, so boiling it off costs more than five times as much as heating it the whole way there. A ΔT run straight through 0 °C or 100 °C will be badly low. The heating curve calculator adds the latent heat at each boundary the range crosses, and the latent heat calculator covers a phase change on its own.
Two more things explain most disagreements in a real result. The calorimeter itself absorbs heat, so accounting only for the water understates the energy released, which is why careful work determines the vessel’s heat capacity first. And the word calorie is ambiguous: the thermochemical calorie is exactly 4.184 J while the dietary Calorie is a kilocalorie, a factor of a thousand apart.
Common questions
Which value of the gas constant R should I use?
Whichever one matches the units already in your calculation. It is 8.314 J/(mol·K) with pressure in pascals and volume in cubic metres, and the identical 8.314 works with kilopascals and litres because a litre kilopascal is exactly a joule. For pressure in atmospheres and volume in litres, use 0.082057 L·atm/(mol·K). Temperature is in kelvin regardless. An answer out by a factor of around a hundred almost always means R and the pressure unit disagree.
Why is my calorimetry result lower than the accepted value?
Usually because heat went somewhere the calculation did not account for. The calorimeter itself absorbs energy, so measuring only the temperature rise of the water understates the total, which is why the vessel’s heat capacity should be determined separately. Heat also escapes to the room during the experiment, and reading the temperature before the contents have equilibrated records a peak the whole system never reached. If the range crosses a melting or boiling point, q = mcΔT misses the latent heat entirely, and that error is large: boiling a gram of water takes about 2257 J against the 418 J needed to heat it from 0 °C to 100 °C.