Lessons · Engineering · the ideal gas law
The ideal gas law: pressure, volume, amount and temperature in one line
P V = n R T, with P in pascals, V in cubic metres, n in moles, T in kelvin and R = 8.314 J/(mol·K).
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.
What it is for
A sealed nitrogen bottle sat in the sun and the relief valve lifted. Nobody had asked what 20 °C to 60 °C does to the pressure inside. In kelvin it is a 14 % rise, and the valve was set at 10 %.
How to think about it
Kelvin first: add 273.15 to the Celsius. Pascals and cubic metres. Then rearrange for the unknown. For a sealed rigid tank, n and V are fixed, so P_2 / P_1 = T_2 / T_1 and the whole law collapses to a ratio.
Worked example
P V = n R T, with R = 8.314 J/(mol·K), P in Pa, V in m³, T in KThe law and the units it wants.
A 0.1 m³ cylinder at 500 kPa and 25 °C: T = 25 + 273.15 = 298.15 KKelvin first, always.
n = P V / (R T) = 500,000 × 0.1 / (8.314 × 298.15) = 50,000 / 2478.8 = 20.17 molRearranged for the amount of gas.
Warm it to 50 °C with the valve shut: P_2 = P_1 × T_2 / T_1 = 500 × 323.15 / 298.15 = 541.9 kPaSealed and rigid: pressure scales with kelvin temperature.
Your turn
1 m³ at 200 kPa and 300 K. Write n.
n = 200,000 × 1 / (8.314 × ) = 80.2 mol
Solve one, graded on the server
The trap
Leaving the temperature in Celsius. 25 °C in the equation instead of 298 K is a factor of twelve, and a doubling in Celsius is nowhere near a doubling in pressure.