Gas Pressure Calculator (Combined Gas Law)
Final Pressure (atm)
—
Final Pressure (kPa)
—
Final Pressure (psi)
—
How the Combined Gas Law Works
The combined gas law is a fundamental equation in chemistry and physics that describes how the pressure, volume, and temperature of a fixed amount of gas are interrelated. It states that the ratio of the product of pressure and volume to temperature remains constant for an enclosed gas sample. According to the International Union of Pure and Applied Chemistry (IUPAC), this law is a unification of three individual gas laws discovered in the 17th and 18th centuries.
The combined gas law merges Boyle's law (pressure inversely proportional to volume at constant temperature, discovered 1662), Charles's law (volume proportional to temperature at constant pressure, discovered 1787), and Gay-Lussac's law (pressure proportional to temperature at constant volume, discovered 1808). This calculator uses the combined gas law to find the final pressure when you change temperature and volume, which is useful in chemistry labs, ideal gas calculations, meteorology, HVAC engineering, and scuba diving applications.
The Combined Gas Law Formula
The combined gas law is expressed as:
(P1 x V1) / T1 = (P2 x V2) / T2
- P1, P2 — Initial and final pressure (in atm, kPa, psi, or any consistent unit)
- V1, V2 — Initial and final volume (in liters or any consistent unit)
- T1, T2 — Initial and final temperature (must be in Kelvin; K = C + 273.15)
Solving for final pressure: P2 = (P1 x V1 x T2) / (T1 x V2)
Worked example: A gas occupies 10 L at 1 atm and 273.15 K (0 C). It is heated to 373.15 K (100 C) while the volume stays at 10 L. P2 = (1 x 10 x 373.15) / (273.15 x 10) = 373.15 / 273.15 = 1.366 atm (138.4 kPa or 20.08 psi).
Key Terms You Should Know
- Kelvin (K) — The absolute temperature scale used in gas law calculations. Zero Kelvin (-273.15 C) is absolute zero, where molecular motion theoretically ceases. Gas laws require Kelvin because ratios of Celsius values can produce meaningless results when values cross zero.
- Atmosphere (atm) — A unit of pressure equal to 101.325 kPa or 14.696 psi. One atmosphere is the approximate air pressure at sea level.
- Ideal Gas — A theoretical gas model where molecules have zero volume and no intermolecular forces. Real gases approximate ideal behavior at high temperatures and low pressures (far from their condensation point).
- STP (Standard Temperature and Pressure) — Defined by IUPAC as 273.15 K (0 C) and exactly 100 kPa (approximately 1 atm). At STP, one mole of ideal gas occupies 22.414 liters.
- Boyle's Law — At constant temperature, pressure and volume are inversely proportional: P1V1 = P2V2. Doubling the pressure halves the volume.
- Charles's Law — At constant pressure, volume is directly proportional to temperature: V1/T1 = V2/T2. Heating a gas expands it.
Gas Laws Comparison
Each gas law holds one variable constant while relating the other two. The combined gas law encompasses all three in a single equation.
| Law | Formula | Constant | Relationship |
|---|---|---|---|
| Boyle's Law (1662) | P1V1 = P2V2 | Temperature | P inversely proportional to V |
| Charles's Law (1787) | V1/T1 = V2/T2 | Pressure | V proportional to T |
| Gay-Lussac's Law (1808) | P1/T1 = P2/T2 | Volume | P proportional to T |
| Combined Gas Law | (P1V1)/T1 = (P2V2)/T2 | Amount of gas | Relates all three variables |
| Ideal Gas Law | PV = nRT | None (includes moles) | Complete state equation |
Practical Examples
Example 1: Scuba diving. A diver's tank holds 12 L of air at 200 atm and 293 K (20 C) on the surface. At 30 meters depth, the ambient pressure is about 4 atm and the water temperature is 283 K (10 C). Using the combined gas law: V2 = (P1 x V1 x T2) / (T1 x P2) = (200 x 12 x 283) / (293 x 4) = 679,200 / 1,172 = 579.5 liters of air available at that depth. Use our Boiling Point Calculator to explore how pressure affects boiling points underwater.
Example 2: Weather balloon. A weather balloon is filled with 500 L of helium at 1 atm and 300 K at ground level. At 10 km altitude, pressure drops to approximately 0.26 atm and temperature to 223 K. V2 = (1 x 500 x 223) / (300 x 0.26) = 111,500 / 78 = 1,429.5 L. The balloon expands nearly 3 times its original volume.
Example 3: Car tire in winter. A tire is inflated to 2.2 atm (gauge) at 300 K (27 C) in summer. In winter, the temperature drops to 260 K (-13 C) and the volume decreases slightly from 12.5 L to 12.3 L. Absolute pressure: P2 = (3.2 x 12.5 x 260) / (300 x 12.3) = 10,400 / 3,690 = 2.818 atm absolute = 1.818 atm gauge. That is roughly a 12% pressure drop, explaining why tire pressure warning lights come on in cold weather. Use our Force Calculator for related physics problems.
Tips for Solving Gas Law Problems
- Always convert to Kelvin. This is the number one source of errors in gas law problems. Add 273.15 to any Celsius temperature before plugging into the formula. A temperature of 25 C = 298.15 K.
- Use consistent units. Pressure must be in the same units on both sides (both in atm, both in kPa, etc.). Volume units must also match. Only temperature must specifically be in Kelvin.
- Identify what is constant. If the problem states "at constant temperature," you can simplify to Boyle's law. If "at constant pressure," use Charles's law. Use the full combined law only when both temperature and volume (or pressure) change.
- Watch for gauge vs. absolute pressure. Tire pressure gauges show gauge pressure (above atmospheric). For gas law calculations, convert to absolute pressure by adding 1 atm (14.7 psi) to the gauge reading.
- Check your answer's reasonableness. If you heat a gas at constant volume, pressure must increase. If you compress a gas at constant temperature, pressure must increase. If your answer contradicts these intuitions, check your calculation.
Pressure Unit Conversions
Gas pressure is measured in various units depending on the field. Here are the standard conversion factors used in scientific and engineering applications:
| Unit | Equivalent to 1 atm | Common Use |
|---|---|---|
| atm (atmosphere) | 1 | Chemistry, standard reference |
| kPa (kilopascal) | 101.325 | SI unit, engineering |
| psi (pounds/sq inch) | 14.696 | US engineering, tires |
| mmHg (torr) | 760 | Medicine, barometers |
| bar | 1.01325 | Meteorology, scuba diving |
Frequently Asked Questions
What is the combined gas law?
The combined gas law is the equation (P1 x V1) / T1 = (P2 x V2) / T2, which relates the initial and final states of pressure, volume, and temperature for a fixed amount of gas. It was formed by combining three earlier laws: Boyle's law (1662), Charles's law (1787), and Gay-Lussac's law (1808). The equation allows you to solve for any one unknown variable when you know the other five. For example, if you know the initial pressure, volume, and temperature plus the final volume and temperature, you can calculate the final pressure. The law assumes the gas behaves ideally and that the amount of gas remains constant (no gas escapes or is added).
Why must temperature be in Kelvin for gas law calculations?
Gas laws require an absolute temperature scale because they describe proportional relationships. The Kelvin scale starts at absolute zero (-273.15 C), where molecular motion theoretically ceases. Using Celsius creates problems because 0 C is not zero energy -- it is an arbitrary reference point. For example, if a gas at 10 C is heated to 20 C, it has not doubled in thermal energy (which the Celsius ratio 20/10 = 2 would incorrectly suggest). In Kelvin, the same change is 283 K to 293 K, a ratio of 1.035, correctly reflecting only a 3.5% increase in thermal energy. Using Celsius or Fahrenheit in gas law formulas produces mathematically nonsensical results.
What is an ideal gas and when does the approximation break down?
An ideal gas is a theoretical model where gas molecules have zero volume and exert no attractive or repulsive forces on each other. Real gases approximate ideal behavior at high temperatures (well above their boiling point) and low pressures (below about 10 atm for most gases). The approximation breaks down near a gas's condensation point, at very high pressures (above 100 atm), or at very low temperatures. For real-world corrections, scientists use the van der Waals equation, which adds terms for molecular volume and intermolecular attractions. For most chemistry classroom problems and everyday engineering applications (tire pressure, HVAC systems), the ideal gas approximation is accurate within 1-5%.
What is standard temperature and pressure (STP)?
STP is a reference condition defined by IUPAC as 273.15 K (0 C) and exactly 100 kPa (approximately 0.987 atm). An older definition used 1 atm (101.325 kPa), which is still common in many textbooks. At STP, one mole of an ideal gas occupies 22.414 liters (under the 1 atm definition) or 22.711 liters (under the 100 kPa definition). STP provides a standard reference for comparing gas volumes across experiments. Always check which STP definition your textbook or course uses, as the difference affects calculations by about 1.3%.
How does altitude affect gas pressure?
Atmospheric pressure decreases approximately exponentially with altitude. At sea level, pressure is about 1 atm (101.3 kPa). At 1,000 meters elevation, it drops to roughly 0.89 atm. At 5,500 meters (the approximate altitude of the highest permanent settlements), it is about 0.5 atm -- half of sea level. At 10,000 meters (typical cruising altitude for commercial aircraft), it drops to 0.26 atm. This pressure decrease is why aircraft cabins are pressurized and why water boils at lower temperatures at altitude. The combined gas law helps predict how sealed gas containers (aerosol cans, sealed food packages) expand when transported to higher altitudes.
How do I convert between pressure units?
The key conversions are: 1 atm = 101.325 kPa = 14.696 psi = 760 mmHg = 1.01325 bar. To convert between units, multiply by the appropriate ratio. For example, to convert 2.5 atm to kPa: 2.5 x 101.325 = 253.3 kPa. To convert 35 psi to atm: 35 / 14.696 = 2.38 atm. When using the combined gas law, ensure pressure is in the same units on both sides of the equation. This calculator outputs results in atm, kPa, and psi simultaneously so you can use whichever unit your application requires. Our Density Calculator can help with related gas density problems.