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Library / Chemical / CHE-02

Chemical

Ideal gas law

Pressure of a gas from amount, temperature and volume.

Ref
CHE-02
Rev
v1
Field
Chemical
Result
p [kPa]

p absolute pressure, V volume, n amount in moles, Ru = 8.314 J/(mol·K) the universal gas constant, T absolute temperature in kelvin. Shown here for p in kPa and V in litres (1 J/L = 1 kPa).

For a gas, pressure rises when you add more gas, heat it, or squeeze it into a smaller space. This law ties those together.

Calculator

Solve for
kPa
mol
0.1100
K
200600
L
11,000
  • Pressure in bar0.9977 bar
  • Gas density for M = 29 g/mol1.16 kg/m³

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Absolute pressure (kPa) current value 99.77 kPa
nAmount of gasKey driver
4.994,990
VVolume
4.994,990
TTemperature
66.5200
  • At 0 °C and 101.325 kPa one mole of gas takes 22.4 L, a useful anchor value.
  • Always use absolute temperature. 0 °C is 273.15 K, not zero.
  • At high pressure or low temperature, real gases deviate; use a compressibility factor Z.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kPa)Local stepChange in result
n Amount of gas0.1 to 100 mol4.9884,9884,983+10%+10 %
V Volume1 to 1,000 L4.9884,9884,983+10%−9.09 %
T Temperature200 to 600 K66.51199.5133+10%+10 %

Sweep of Temperature

Temperature (K)Absolute pressure (kPa)
20066.51
25083.14
30099.77
350116.4
400133
450149.7
500166.3
550182.9
600199.5

Sweep one variable

Absolute pressure (kPa)Hover, touch, or use the arrow keys on the chart
100150200 200300400500600 Temperature (K) now 99.77

Variables and how each one contributes

p

Absolute pressure

Result · kPa

What the equation gives you. Everything else feeds it.

n

Amount of gas

mol · typical 0.1 to 100

Number of moles. Directly proportional.

T

Temperature

K · typical 200 to 600

Absolute temperature (°C + 273.15). Directly proportional: 10% hotter means 10% higher pressure.

V

Volume

L · typical 1 to 1,000

Container volume. Inversely proportional.

Worked example with the values above

  1. n R T = 2 × 8.314 × 300 = 4,988 J.
  2. Divide by the volume in litres: 4,988 / 50 = pressure in kPa.
  3. p = 99.77 kPa.

Assumptions and limits

  • Ideal gas behaviour (low pressure, moderate temperature).
  • Absolute pressure and absolute temperature.
  • Closed container.

Common mistakes

  • Using gauge pressure or °C.
  • Mixing litres and m³.
  • Using the law near condensation.

Statistics and uncertainty

Real gases deviate by 1% or less at atmospheric pressure but several per cent above about 10 bar. Temperature in kelvin carries a small relative error, so measurement uncertainty mostly comes from the volume and the moles.

Where it is used

  • Gas cylinder contents and pressure with temperature.
  • Pneumatic system sizing.
  • Combustion and propellant gas calculations.

Maths you need

  • Absolute temperature: kelvin
  • Moles: amount of gas

Related formulas

Practise this formula