The Combined Gas Law: when all three variables change
The Combined Gas Law unifies Boyle's, Charles's and Gay-Lussac's Laws into a single equation: P₁V₁/T₁ = P₂V₂/T₂. It describes a fixed amount of gas whose pressure, volume and temperature can all change between an initial state and a final state. When one variable is held constant, the combined law reduces to the relevant individual law.
The equation states that the quantity PV/T is conserved for a fixed sample of gas. This is a direct consequence of the Ideal Gas Law: since PV = nRT and n and R are constant, PV/T = nR is also constant. The Combined Gas Law is simply the Ideal Gas Law written as a ratio between two states of the same gas sample, which makes it useful when the amount of gas does not change.
Deriving the individual laws
If temperature is constant, T₁ = T₂, and P₁V₁/T₁ = P₂V₂/T₂ simplifies to P₁V₁ = P₂V₂: Boyle's Law. If pressure is constant, P₁ = P₂, and the equation simplifies to V₁/T₁ = V₂/T₂: Charles's Law. If volume is constant, V₁ = V₂, and it simplifies to P₁/T₁ = P₂/T₂: Gay-Lussac's Law.
This is why the Combined Gas Law is more general: it handles cases where none of the three variables stays constant, which is the typical situation in real systems. A gas heated in a flexible container changes volume and pressure simultaneously. The combined law accounts for both.
Using the six-variable calculator
This calculator has six inputs: P₁, V₁, T₁ (initial state) and P₂, V₂, T₂ (final state). Select which variable to solve for. Enter five known values with their units. The calculator handles three pressure units, four volume units, and three temperature units, converting everything to a consistent basis and applying the formula.
Worked examples
Example 1: Compressing and heating. A gas at 1.00 atm, 298 K and 2.00 L is compressed to 0.500 L and heated to 400 K. Final pressure? P₂ = P₁V₁T₂/(T₁V₂) = (1.00 × 2.00 × 400) / (298 × 0.500) = 5.37 atm.
Example 2: Tire pressure with temperature change. A car tire at 2.20 atm (gauge: 32 psi) and 20 °C warms to 50 °C. Assuming constant volume, P₂ = P₁T₂/T₁ = 2.20 × 323.15/293.15 = 2.43 atm. The gauge reading would rise to about 35.6 psi.
Example 3: Scuba tank cooling. A scuba tank filled to 200 atm at 30 °C is cooled to 10 °C in cold water. At constant volume, P₂ = 200 × 283.15/303.15 = 187 atm. The pressure drops by about 6.5 percent. Divers account for this when planning gas consumption.
Why the Combined Gas Law is more useful than the individual laws
In real systems, it is unusual for only one variable to change while the others stay constant. A gas being compressed in an engine cylinder changes temperature as well as pressure and volume. The Combined Gas Law handles this without requiring you to decide which individual law applies. It is the most practical form of the gas laws for engineering calculations because it makes no assumptions about which variables are held constant.
Consider a diesel engine at the moment of compression. The piston moves from bottom dead center to top dead center, reducing the volume by a factor of about 18. The pressure rises from roughly 1 atm to about 50 atm, and the temperature climbs from perhaps 350 K to 900 K. These three quantities change simultaneously. The Combined Gas Law correctly relates the initial and final states: P₁V₁/T₁ = P₂V₂/T₂. Any single-variable law would give the wrong answer.
The Combined Gas Law in weather and aviation
Weather balloons demonstrate the Combined Gas Law dramatically. Launched at ground level (1 atm, 15 °C), a typical weather balloon ascends to about 30 km, where the pressure is roughly 0.01 atm and the temperature is approximately -50 °C. Using the Combined Gas Law with P₁ = 1 atm, T₁ = 288 K, P₂ = 0.01 atm, T₂ = 223 K: V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = V₁ × 100 × 0.774 = 77.4 × V₁. The balloon expands to about 77 times its launch volume before bursting.
Aircraft altimeters rely on the relationship between pressure, volume, and temperature. The static port on an aircraft measures ambient pressure, and the altimeter converts that pressure to an altitude using a standard atmosphere model. Changes in temperature affect the pressure-altitude relationship, which is why altimeter settings must be adjusted for local conditions.
More worked examples
Example 4: Weather balloon. A weather balloon contains 2.00 m³ of helium at 1.00 atm and 20 °C at launch. At 25 km altitude, the pressure is 0.025 atm and temperature -50 °C. Volume? V₂ = 2.00 × (1.00/0.025) × (223.15/293.15) = 60.9 m³.
Example 5: Internal combustion. In a cylinder, gas at 1.00 atm and 300 K with volume 0.500 L is compressed to 0.030 L and 900 K. Pressure? P₂ = 1.00 × (0.500/0.030) × (900/300) = 50 atm.
Example 6: Gas storage. A tank holds gas at 150 atm, 25 °C. If half the gas is released and the tank cools to 5 °C, what is the new pressure? P₂ = 150 × (278.15/298.15) = 140 atm. Note the volume is constant and the amount change doesn't apply here since we're measuring the remaining gas.
Frequently asked questions
In real systems, it is unusual for only one variable to change while the others stay constant. A gas being compressed in an engine cylinder changes temperature as well as pressure and volume. The Combined Gas Law handles this without requiring you to decide which individual law applies. It is the most practical form of the gas laws for engineering calculations because it makes no assumptions about which variables are held constant.
Consider a diesel engine at the moment of compression. The piston moves from bottom dead center to top dead center, reducing the volume by a factor of about 18. The pressure rises from roughly 1 atm to about 50 atm, and the temperature climbs from perhaps 350 K to 900 K. These three quantities change simultaneously. The Combined Gas Law correctly relates the initial and final states: P₁V₁/T₁ = P₂V₂/T₂. Any single-variable law would give the wrong answer.
Weather balloons demonstrate the Combined Gas Law dramatically. Launched at ground level (1 atm, 15 °C), a typical weather balloon ascends to about 30 km, where the pressure is roughly 0.01 atm and the temperature is approximately -50 °C. Using the Combined Gas Law with P₁ = 1 atm, T₁ = 288 K, P₂ = 0.01 atm, T₂ = 223 K: V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = V₁ × 100 × 0.774 = 77.4 × V₁. The balloon expands to about 77 times its launch volume before bursting.
Aircraft altimeters rely on the relationship between pressure, volume, and temperature. The static port on an aircraft measures ambient pressure, and the altimeter converts that pressure to an altitude using a standard atmosphere model. Changes in temperature affect the pressure-altitude relationship, which is why altimeter settings must be adjusted for local conditions.
Example 4: Weather balloon. A weather balloon contains 2.00 m³ of helium at 1.00 atm and 20 °C at launch. At 25 km altitude, the pressure is 0.025 atm and temperature -50 °C. Volume? V₂ = 2.00 × (1.00/0.025) × (223.15/293.15) = 60.9 m³.
Example 5: Internal combustion. In a cylinder, gas at 1.00 atm and 300 K with volume 0.500 L is compressed to 0.030 L and 900 K. Pressure? P₂ = 1.00 × (0.500/0.030) × (900/300) = 50 atm.
Example 6: Gas storage. A tank holds gas at 150 atm, 25 °C. If half the gas is released and the tank cools to 5 °C, what is the new pressure? P₂ = 150 × (278.15/298.15) = 140 atm. Note the volume is constant and the amount change doesn't apply here since we're measuring the remaining gas.
What is the Combined Gas Law? P₁V₁/T₁ = P₂V₂/T₂. It relates the pressure, volume, and temperature of a fixed amount of gas between two states.
How does it relate to the Ideal Gas Law? For a fixed amount of gas, PV/T = nR is constant. Setting the initial and final values equal gives the combined law.
When should I use the Combined vs Ideal Gas Law? Use the Combined Gas Law when the amount of gas stays the same and you only need initial and final states. Use the Ideal Gas Law when you need to find the amount of gas or when you have only one state.
Can I use Celsius in the Combined Gas Law? No. All temperatures must be in Kelvin. This calculator converts Celsius and Fahrenheit automatically.