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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.

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 K
The law and the units it wants.
A 0.1 m³ cylinder at 500 kPa and 25 °C: T = 25 + 273.15 = 298.15 K
Kelvin first, always.
n = P V / (R T) = 500,000 × 0.1 / (8.314 × 298.15) = 50,000 / 2478.8 = 20.17 mol
Rearranged 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 kPa
Sealed 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

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.

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