P · V / T = k
Enter any five values and leave one blank. It solves the moment you stop typing, shows every algebraic step, and plots both gas states on a live pressure–volume diagram. No submit button, no page reload.
The combined gas law calculator solves P₁V₁/T₁ = P₂V₂/T₂ for any one unknown among six gas state variables. Enter the five quantities you know: gas pressure, gas volume and gas temperature for the first gas state and the second gas state, then leave the sixth field blank. The tool converts every entry to SI units, isolates the unknown variable by algebraic rearrangement, and prints each substitution step below the answer.
Five features separate this combined gas law calculator from a hand calculation:
The combined gas law calculator answers 4 classes of problem, including chemistry homework on gas pressure, gas volume and gas temperature changes, laboratory correction of a measured gas volume to standard temperature and pressure (STP), cleanroom and pressure-vessel design where an overpressure has to hold across a temperature swing, and field checks on compressed-gas cylinders.
This page derives the combined gas law formula and all 6 rearrangements, extends the formula with moles, works three examples that find P₂, V₂ and T₂, compares the combined gas law against the ideal gas law, defines the 4 thermodynamic processes, states the first law of thermodynamics, traces the Carnot cycle, tabulates STP values, derives Boyle's law, Charles' law and Gay-Lussac's law from the combined form, and converts pressure and temperature units.
Definition
The combined gas law states that for a fixed amount of an ideal gas, pressure × volume ÷ absolute temperature is a constant: P·V/T = k. Because that constant does not change when the gas is compressed, expanded, heated or cooled, the value of PV/T before a change must equal its value after. Which is why the working form is P₁V₁/T₁ = P₂V₂/T₂.
It is called combined because it merges the three single-variable gas laws into one relationship: Boyle's law (pressure–volume at fixed temperature), Charles' law (volume–temperature at fixed pressure), and Gay-Lussac's law (pressure–temperature at fixed volume). Freeze any one of the three properties and the combined gas law collapses back into the corresponding single law.
Two conditions must hold. The gas must be a closed system, the same molecules before and after, so the number of moles n is unchanged, and the temperature must be absolute, measured in kelvin. If the amount of gas does change between states, use the extended form with moles instead.
Reference card
The primary form equates the two states directly. Every other version below is the same equation solved for a different unknown. Nothing new is being introduced, only rearranged.
alternate form · P · V / T = k (k is constant for a sealed sample of ideal gas)
| Solve for | Formula | Reads as |
|---|---|---|
| P₁ | P₁ = (P₂ · V₂ · T₁) / (V₁ · T₂) | Original pressure from the final state |
| V₁ | V₁ = (P₂ · V₂ · T₁) / (P₁ · T₂) | Original volume from the final state |
| T₁ | T₁ = (P₁ · V₁ · T₂) / (P₂ · V₂) | Original temperature, in kelvin |
| P₂ | P₂ = (P₁ · V₁ · T₂) / (V₂ · T₁) | Final pressure after the change |
| V₂ | V₂ = (P₁ · V₁ · T₂) / (P₂ · T₁) | Final volume after the change |
| T₂ | T₂ = (P₂ · V₂ · T₁) / (P₁ · V₁) | Final temperature, in kelvin |
| With moles | P₁V₁ / (n₁ · T₁) = P₂V₂ / (n₂ · T₂) | Use this when the amount of gas changes between the two states |
| Symbol | Quantity | SI unit | Also accepted |
|---|---|---|---|
| P₁, P₂ | Absolute pressure of state 1 and state 2 | Pa | kPa, MPa, bar, mbar, atm, mmHg, torr, psi, inHg |
| V₁, V₂ | Volume occupied by the gas | m³ | L, mL, dm³, cm³, ft³, in³, gal |
| T₁, T₂ | Absolute temperature | K | °C, °F, °R (converted to K internally) |
| n | Amount of gas. Constant unless the extended form is used | mol | mmol, kmol |
| k | The constant PV/T for the sample | J/K | equals n·R |
| R | Universal gas constant, also called the ideal gas constant | 8.314462618 J/(mol·K) | 0.08206 L·atm/(mol·K) |
Extended form
The standard combined gas law quietly assumes the amount of gas never changes. When it does (a cylinder is topped up, a valve vents, a reaction produces gas), the constant k itself changes and the ordinary form breaks. The fix is to divide both sides by the number of moles.
Use this when the container is sealed and no gas enters or leaves. n cancels out of both sides, so you never need to know how much gas you have.
Both sides now equal the universal gas constant R = 8.3145 J/(mol·K), so the equality survives even when moles change. Switch the calculator's amount of gas changes toggle to use it.
Notice that if n₁ = n₂ the two moles terms cancel and the extended form reduces exactly to the standard one. It is a superset, not a different law. This is also the bridge to the ideal gas law: multiply either side by nT and you recover PV = nRT.
Four steps
The calculator has no submit button, it evaluates continuously as you type, so the working updates with every keystroke.
Pick a unit next to each field. Mix them freely, kPa with litres, °F with cubic feet. Everything is converted to SI before solving.
Fill the five values you were given, split across the initial state and the final state cards.
Tap a Solve for tab. That field locks, turns brass, and becomes the output slot.
The result appears with the full substitution underneath and both states plotted on the PV diagram. Copy the steps for your homework.
Worked solutions
Three complete solutions, one per unknown. Each shows the given values, the rearranged formula, the substitution, and the answer. The same sequence the calculator produces.
Choosing the right equation
Both describe ideal gases, but they answer different questions. The combined gas law compares two states of the same sample; the ideal gas law describes one state in absolute terms.
| Combined gas law | Ideal gas law | |
|---|---|---|
| Formula | P₁V₁/T₁ = P₂V₂/T₂ | PV = nRT |
| Variables needed | Five of the six state values. No n, no R. | Three of P, V, n, T, plus the constant R. |
| What it describes | A change between two states of one fixed sample. | A single state, on its own, in absolute terms. |
| When to use it | "The gas was at 2 atm and 300 K; now it is at 350 K in half the volume. What is the pressure?" | "I have 0.5 mol of gas in a 10 L flask at 298 K. What is the pressure?" |
| Requires moles? | No: n cancels, which is why it works when you don't know the amount of gas. | Yes: n is mandatory and must be in moles. |
| Limitations | Assumes a closed system and ideal behaviour. Fails at high pressure or near condensation. | Same ideal assumptions; also needs an accurate molar amount. Real gases near condensation or high pressure need a correction beyond this ideal form. |
| Dedicated tool | This page | Ideal gas law calculator (PV = nRT) |
Four special cases
A thermodynamic process is a path a gas takes from one state to another. Four of them are special because one property is held constant, which simplifies the combined gas law and the accompanying energy bookkeeping. Each diagram below plots pressure against volume, with the held-constant variable named underneath.
A rigid sealed container heated or cooled. Because the volume never moves, the gas does no work done by gas, every joule of heat goes straight into internal energy: ΔU = Q = Cv·n·ΔT.
P₁ / T₁ = P₂ / T₂A gas expanding against a fixed load, a piston under constant weight. The shaded rectangle is the work: W = p·ΔV. Heat supplied splits between work and internal energy, Q = Cp·n·ΔT.
V₁ / T₁ = V₂ / T₂A slow change in a heat bath. Internal energy is unchanged, so all heat absorbed converts to work: Q = W = n·R·T·ln(V₂/V₁). Real isothermal steps must be very slow.
P₁ · V₁ = P₂ · V₂Fast enough that no heat crosses the boundary, the heat of compression stays in the gas, which is why a bike pump warms up. Work comes entirely out of internal energy: W = −ΔU.
P₁·V₁γ = P₂·V₂γ, γ = Cp/CvEnergy bookkeeping
The combined gas law tells you where the gas ends up; the first law tells you what the change cost in energy. Internal energy rises by whatever heat goes in, minus whatever work the gas does on its surroundings: ΔU = Q − W.
Internal energy change depends only on temperature and the molecular complexity of the gas: ΔU = Cv · n · ΔT, where Cv is the molar heat capacity at constant volume. For ideal gases it takes three values, 3/2·R ≈ 12.47 J/(mol·K) for monatomic gases such as helium and argon, 5/2·R ≈ 20.79 J/(mol·K) for diatomic gases such as nitrogen and oxygen, and 3·R ≈ 24.94 J/(mol·K) for more complex molecules. At constant pressure use Cp = Cv + R. Real gases deviate slightly: nitrogen's measured Cv is about 20.81 J/(mol·K) rather than the theoretical 20.79.
Putting the processes together
Chain two isothermal processes and two adiabatic processes into a closed loop and you get the Carnot cycle, the most efficient heat engine the laws of thermodynamics permit. It is the same diagram style you just read, drawn as a single circuit.
The shaded area inside the loop is the net work delivered per cycle. No engine operating between the same two temperatures can beat the Carnot efficiency η = 1 − T_cold / T_hot, with both temperatures in kelvin, the reason engineers push combustion temperatures up rather than exhaust temperatures down. Work through the adiabatic legs of this cycle in the Adiabatic Process calculator →
Lookup card
Textbook problems often say "at STP" and expect you to supply the missing pressure and temperature. Two conventions are in circulation. Check which one your course uses, because the molar volume differs by 1.3%.
| Convention · Notes | Temperature · Pressure · Molar volume |
|---|---|
| STP, classical The value most chemistry courses still use. | 273.15 K (0 °C) · 101.325 kPa (1 atm) · 22.414 L/mol |
| STP, IUPAC since 1982 Current IUPAC recommendation. | 273.15 K (0 °C) · 100 kPa (1 bar) · 22.711 L/mol |
| SATP, ambient Standard ambient temperature and pressure. | 298.15 K (25 °C) · 100 kPa (1 bar) · 24.790 L/mol |
| NTP, normal Common in ventilation and flow engineering. | 293.15 K (20 °C) · 101.325 kPa (1 atm) · 24.055 L/mol |
To use any of these in the calculator, enter the standard condition as state 1 and your unknown condition as state 2. More on standard conditions →
The three parent laws
Each parent law is the combined gas law with one variable frozen. Cancel the constant term from both sides of P₁V₁/T₁ = P₂V₂/T₂ and the corresponding law falls out immediately.
| Law · Constant variable | Formula · In one line |
|---|---|
| Boyle's law Temperature T | P₁ · V₁ = P₂ · V₂ Squeeze a gas at fixed temperature and pressure rises in exact inverse proportion. |
| Charles' law Pressure P | V₁ / T₁ = V₂ / T₂ Heat a gas under a fixed load and it expands in direct proportion to absolute temperature. |
| Gay-Lussac's law Volume V | P₁ / T₁ = P₂ / T₂ Heat a gas in a rigid vessel and pressure climbs in direct proportion to absolute temperature. |
A fourth relation, Avogadro's law (V₁/n₁ = V₂/n₂), brings in the amount of gas. Fold it into the combined gas law and you arrive at the full ideal gas law, PV = nRT, which is exactly what the extended moles form above expresses.
Conversion reference
The calculator converts automatically, but these are the factors it uses. Pressure units only need to match each other in a combined gas law problem. Temperature must always be absolute.
| Unit · Symbol · Typical use | Value in kPa · Value in Pa |
|---|---|
| Pascal Pa SI base unit | 0.001 · 1 |
| Kilopascal kPa Chemistry, meteorology | 1 · 1 000 |
| Bar bar Industry, IUPAC STP | 100 · 100 000 |
| Atmosphere atm Classical STP | 101.325 · 101 325 |
| Millimetre of mercury mmHg Blood pressure, vacuum | 0.133322 · 133.322 |
| Torr torr Vacuum systems | 0.133322 · 133.322 |
| Pound per square inch psi US engineering, tyres | 6.894757 · 6 894.757 |
| Inch of mercury inHg US barometric pressure | 3.386389 · 3 386.389 |
| From · Absolute? | Conversion to K · Example |
|---|---|
| Celsius No: must be converted | K = °C + 273.15 · 25 °C → 298.15 K |
| Fahrenheit No: must be converted | K = (°F + 459.67) × 5/9 · 77 °F → 298.15 K |
| Rankine Yes: absolute scale | K = °R × 5/9 · 536.67 °R → 298.15 K |
| Kelvin Yes: SI absolute scale | K = K · 298.15 K |
Volume units cancel the same way pressure units do, so litres on both sides is fine. See the pressure unit converter and Kelvin conversion calculator for standalone tools.
Measured readings reach the calculator from 4 instrument classes, and the instrument decides the unit you type in. A barometric pressure sensor with an RS485 Modbus RTU interface reports in hPa or mbar. A digital vacuum gauge with a 760 mmHg absolute range reports in mmHg or torr. A pressure data logger, such as a LEO Record (Ei) Pressure Data Logger, records in kPa, psi or bar depending on how it was configured. Wireless environmental sensors, such as a ProSens Environmental Monitor and Controller, that stream over MQTT into a SCADA or IoT platform (cleanroom overpressure monitoring and electrical panel temperature monitoring are two common installations) normally normalise to kPa and °C before logging.
Gauge instruments read zero at atmospheric pressure, so add 101.325 kPa (1 atm, 14.696 psi, 760 mmHg, 1.01325 bar) to a gauge reading before entering it as P₁ or P₂. The combined gas law needs absolute pressure, and a gauge reading of 0 kPa entered directly makes the equation divide by zero.
15 answers
The combined gas law formula is P₁V₁/T₁ = P₂V₂/T₂, equivalent to saying P·V/T = k for a fixed amount of ideal gas. P is absolute pressure, V is volume, and T is absolute temperature in kelvin. For example, a gas at 200 kPa, 4 L, and 300 K gives P·V/T = 2.67, so any later state of that same gas must also satisfy P·V/T = 2.67.
Rearrange to P₂ = (P₁ · V₁ · T₂) / (V₂ · T₁). Convert both temperatures to kelvin first, use the same pressure unit for P₁ and P₂, and the same volume unit for V₁ and V₂. Select the P₂ tab in the calculator and it does the substitution for you.
V₂ = (P₁ · V₁ · T₂) / (P₂ · T₁). For example, with P₁ = 101.325 kPa, V₁ = 2.50 L, T₁ = 300 K, P₂ = 250 kPa and T₂ = 450 K: V₂ = (101.325 × 2.50 × 450) / (250 × 300) = 1.52 L.
T₂ = (P₂ · V₂ · T₁) / (P₁ · V₁). T₁ must already be in kelvin and the answer comes out in kelvin, subtract 273.15 afterwards if you need Celsius. For example, with P₁ = 100 kPa, V₁ = 2 L, T₁ = 300 K, P₂ = 150 kPa, and V₂ = 1.5 L: T₂ = (150 × 1.5 × 300) / (100 × 2) = 337.5 K.
It means that for a sealed sample of ideal gas, pressure × volume ÷ absolute temperature keeps the same numerical value however the gas is squeezed, expanded, heated or cooled. Because k is unchanged before and after, P₁V₁/T₁ must equal P₂V₂/T₂. Physically, k = n·R.
Isochoric (constant volume, W = 0), isobaric (constant pressure, W = p·ΔV), isothermal (constant temperature, ΔU = 0 so Q = W), and adiabatic (no heat exchange, Q = 0 so W = −ΔU). Each holds one of pressure, volume, or temperature fixed, except adiabatic, which holds heat exchange at zero instead. Each is diagrammed above with its own formula and worked description.
A change in which pressure stays constant while volume and temperature change, so V₁/T₁ = V₂/T₂. Work done by the gas is W = p·ΔV, the rectangle under the line on a PV diagram, and the heat absorbed is Q = ΔU + W = Cp·n·ΔT.
The final temperature is 375 K. In an isochoric process volume is constant, so pressure and temperature change together as p₁/T₁ = p₂/T₂. Rearranging: T₂ = (T₁ × p₂) / p₁ = (300 × 125) / 100 = 375 K.
The law is a ratio relationship, so the temperature scale must have a true zero. Celsius and Fahrenheit have arbitrary zero points, which makes ratios meaningless. 20 °C is not twice as hot as 10 °C, but 400 K genuinely is twice 200 K. Convert with K = °C + 273.15. The calculator does this internally, so you may enter °C or °F safely.
Yes, gas pressure and gas temperature are directly proportional in the combined gas law, but only when gas volume is held constant. If V₁ = V₂ the law reduces to P₁/T₁ = P₂/T₂, which is Gay-Lussac's law and describes an isochoric process. The proportionality breaks the moment volume changes as well.
Pressure: Pa, kPa, MPa, bar, mbar, atm, mmHg, torr, psi, inHg. Volume: m³, L, mL, dm³, cm³, ft³, in³, US gallons. Temperature: K, °C, °F, °R. Amount: mol, mmol, kmol. Every field converts to SI before solving, so units may be mixed between fields without error.
The combined gas law compares two states of the same fixed sample and needs neither the amount of gas nor the gas constant R. The ideal gas law, PV = nRT, describes a single state absolutely and requires the number of moles. Use the combined gas law when a gas changes state and you do not know how much gas you have. See the full comparison.
It is the merger of all three. Hold temperature constant → Boyle's law, P₁V₁ = P₂V₂. Hold pressure constant → Charles' law, V₁/T₁ = V₂/T₂. Hold volume constant → Gay-Lussac's law, P₁/T₁ = P₂/T₂. The derivation table shows each step.
Yes. The standard form assumes a closed system in which the number of moles does not change, nothing leaks in or out. If the amount of gas does change, switch to the extended form P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂) using the calculator's moles toggle.
P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂). Both sides equal the universal gas constant R = 8.314462618 J/(mol·K), which is why the equality holds even when the amount of gas differs between the two states. Use this form whenever gas is added, removed, or produced between the two states, such as topping up a cylinder. See the side-by-side comparison.