Calculators/Physics/Ideal gas law (pV = nRT)

Ideal gas law (pV = nRT)

Equation of state pV = nRT · gas process

Ideal gas law (pV = nRT) — illustration

Leave one field empty — I’ll compute it. R = 8.314 J/(mol·K).

V =
0.0224 m³

Temperature always in kelvin (K = °C + 273.15). Standard conditions: p = 101,325 Pa, T = 273.15 K → 1 mol occupies 22.4 dm³.

The ideal gas law (Clapeyron equation) relates pressure, volume, amount of substance and temperature:

pV=nRTpV=nRT

where R=8.314 Jmol⋅KR=8.314\ \tfrac{\text{J}}{\text{mol·K}}. Enter three of the four quantities (pp in Pa, VV in m³, nn in moles, TT in kelvin) and the calculator finds the missing one.

The second mode solves a gas process for a fixed amount of gas:

p1V1T1=p2V2T2\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}

From it come the special processes: isothermal (TT = const → p1V1=p2V2p_1V_1=p_2V_2), isobaric (pp = const) and isochoric (VV = const). Remember: temperature is always in kelvin.

Examples

1 mol of gas, T = 273.15 K, p = 101,325 Pa
→ V = nRT/p ≈ 0.0224 m³ = 22.4 dm³
Isothermal compression: p₁=100 kPa, V₁=1 m³, V₂=0.5 m³
→ p₂ = p₁V₁/V₂ = 200 kPa

Frequently asked questions

Why must temperature be in kelvin?+

Because the ideal gas law relies on absolute temperature. Using °C would give a wrong result — always convert: K = °C + 273.15.

How much volume does 1 mol of gas occupy at standard conditions?+

22.4 dm³ (at p = 101,325 Pa and T = 273.15 K). This follows directly from the ideal gas law.

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