Ideal gas law (PV = nRT)

Introduction

The ideal gas law links the four things that describe a gas in a container: its pressure, its volume, its temperature and how much of it there is. Know three and the fourth follows. With the gas's molar mass, the same equation gives its mass and its density, which is how engineers find the air in a room, the gas in a cylinder, the pressure rise in a closed vessel that heats up, or the density of a gas in a duct.

This calculator solves for whichever variable you need and converts the amount into kilograms for air, nitrogen, oxygen, carbon dioxide, methane, hydrogen, helium, argon, ammonia or a gas of your own. It is for mechanical, process and marine engineers, and for students of thermodynamics and chemistry.

Formulas

The ideal gas law:

P V = n R T

Rearranged for each unknown:

n = P V / (R T),   P = n R T / V,   V = n R T / P,   T = P V / (n R)

Mass, density and the specific gas constant:

m = n M
ρ = P M / (R T)
R_specific = R / M

• P — absolute pressure, Pa
• V — volume, m³
• n — amount of gas, mol
• T — absolute temperature, K (T = t + 273.15 for t in °C)
• R — molar gas constant, 8.314 462 618 J/(mol·K), exact
• M — molar mass, kg/mol (air 0.028 965, N₂ 0.028 013, CO₂ 0.044 010)

Assumptions and limitations

• The gas is ideal. Real gases follow the law to within about 1 % at pressures of a few bar and temperatures well above their boiling point. At high pressure, multiply n R T by the compressibility factor Z from a chart or an equation of state.
• Pressure must be absolute. A gauge reading of 7 bar is 8.01 bar absolute at sea level, and entering the gauge value gives an answer about 13 % low.
• Temperature must be absolute. The calculator converts °C and °F for you, but a ratio of temperatures in °C is meaningless.
• Molar masses are for the pure gas. Humid air is slightly lighter than dry air, and natural gas is heavier than pure methane.
• Near condensation, such as carbon dioxide below 31 °C under pressure or any gas close to its boiling point, the ideal gas law does not apply.

Worked example

How much air is in a room 5 m × 4 m × 2.5 m at normal atmospheric pressure and 20 °C?

1. Volume: V = 5 × 4 × 2.5 = 50 m³.
2. Absolute pressure: P = 101.325 kPa = 101 325 Pa. Absolute temperature: T = 20 + 273.15 = 293.15 K.
3. Amount of air: n = P V / (R T) = 101 325 × 50 / (8.314 × 293.15) = 2 079 mol.
4. Mass: m = n M = 2 079 × 0.028 965 = 60.2 kg.
5. Density: ρ = P M / (R T) = 1.204 kg/m³, the textbook density of air at 20 °C.
6. The specific gas constant of air is R / M = 8.314 / 0.028 965 = 287.1 J/(kg·K).

So an ordinary room holds about 60 kg of air. Heated to 30 °C in a sealed space, its pressure would rise in proportion to the absolute temperature, by 10 / 293 = 3.4 %, which is why a sealed, sunlit enclosure builds up pressure.

FAQ

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