Gas Law Calculator
Calculate gas properties using various gas laws
Ideal Gas Law: PV = nRT
Calculate any one variable given the other three
Boyle's Law: P₁V₁ = P₂V₂
Temperature and moles constant
Initial State
Final State
Selected law inputs will appear here
Result
Calculation Steps
Gas Properties
Standard Conditions (STP)
• Temperature: 273.15 K (0°C)
• Pressure: 1 atm (101.325 kPa)
• Molar Volume: 22.414 L/mol
• R = 0.08206 L·atm/(mol·K)
Gas Laws Overview
Ideal Gas Law
PV = nRT
Relates pressure, volume, temperature, and amount of gas. Valid for ideal gases at low pressure and high temperature.
R = 0.08206 L·atm/(mol·K) = 8.314 J/(mol·K) = 62.36 L·mmHg/(mol·K)
Boyle's Law
P₁V₁ = P₂V₂ (constant T and n)
At constant temperature, pressure and volume are inversely proportional. Doubling pressure halves volume.
Charles's Law
V₁/T₁ = V₂/T₂ (constant P and n)
At constant pressure, volume and temperature are directly proportional. Temperature must be in Kelvin.
Gay-Lussac's Law
P₁/T₁ = P₂/T₂ (constant V and n)
At constant volume, pressure and temperature are directly proportional.
Combined Gas Law
(P₁V₁)/T₁ = (P₂V₂)/T₂ (constant n)
Combines Boyle's, Charles's, and Gay-Lussac's laws. Useful when pressure, volume, and temperature all change.
Avogadro's Law
V₁/n₁ = V₂/n₂ (constant P and T)
At constant temperature and pressure, volume is directly proportional to moles. Equal volumes contain equal numbers of molecules.
Example Problems
Example 1: Ideal Gas Law
Problem: What volume does 2.0 moles of gas occupy at 1.0 atm and 273.15 K?
Solution:
PV = nRT
V = nRT/P
V = (2.0 mol)(0.08206 L·atm/mol·K)(273.15 K) / (1.0 atm)
V = 44.8 L
Example 2: Boyle's Law
Problem: A gas occupies 5.0 L at 2.0 atm. What volume will it occupy at 1.0 atm?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁/P₂
V₂ = (2.0 atm)(5.0 L) / (1.0 atm)
V₂ = 10.0 L
Example 3: Charles's Law
Problem: A balloon has a volume of 2.0 L at 300 K. What will its volume be at 400 K?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = (2.0 L)(400 K) / (300 K)
V₂ = 2.67 L
Common Unit Conversions
Pressure
- • 1 atm = 101,325 Pa = 101.325 kPa
- • 1 atm = 760 mmHg = 760 torr
- • 1 atm = 1.01325 bar
- • 1 atm = 14.696 psi
Volume
- • 1 L = 1000 mL = 1 dm³
- • 1 L = 0.001 m³
- • 1 mL = 1 cm³
- • 1 m³ = 1000 L
Temperature
- • K = °C + 273.15
- • °C = K - 273.15
- • °F = (9/5)°C + 32
- • °C = (5/9)(°F - 32)
Gas Constant (R)
- • 0.08206 L·atm/(mol·K)
- • 8.314 J/(mol·K)
- • 8.314 kPa·L/(mol·K)
- • 62.36 L·mmHg/(mol·K)
Real-World Applications
1. Scuba Diving
Boyle's Law explains why scuba divers must ascend slowly. As pressure decreases, gas expands. Rapid ascent can cause decompression sickness ("the bends").
2. Hot Air Balloons
Charles's Law explains hot air balloon operation. Heating air causes expansion, decreasing density and creating lift.
3. Tire Pressure
Gay-Lussac's Law explains why tire pressure increases in summer and decreases in winter. Pressure is directly proportional to temperature.
4. Respiratory System
Boyle's Law governs breathing. Diaphragm expansion increases lung volume, decreasing pressure and drawing air in.
5. Industrial Processes
Gas laws are essential in chemical engineering for reactor design, gas storage, and process optimization.
References
The gas law calculations are based on fundamental physical chemistry principles from reputable sources:
Note: These calculations assume ideal gas behavior. Real gases deviate from ideal behavior at high pressures and low temperatures. For precise calculations with real gases, use the van der Waals equation or other equations of state. Always use absolute temperature (Kelvin) in gas law calculations.
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