The Ideal Gas Law Calculator: Pressure From PV = nRT

This calculator applies the ideal gas law, PV = nRT, to find the pressure of a gas. Enter the amount of gas in moles, the temperature in degrees Celsius and the volume in litres; you get the pressure in kilopascals, atmospheres and bar.

Formula

P = n × R × T ÷ V, with T in kelvin: T = t + 273.15.

• P: pressure; with V in litres the result is in kilopascals (1 J/L = 1 kPa)
• n: amount of gas in moles
• R = 8.31446261815324 J/(mol·K), the molar gas constant (exact in the SI)
• T: absolute temperature in kelvin; t: temperature in °C
• V: volume in litres (L)
• 1 atm = 101.325 kPa; 1 bar = 100 kPa

Worked Example

1 mol of gas at 25 °C in a 24.5 L container.

1. T = 25 + 273.15 = 298.15 K
2. n × R × T = 1 × 8.31446 × 298.15 = 2,478.96 J
3. P = 2,478.96 ÷ 24.5 = 101.18 kPa
4. In atmospheres: 101.18 ÷ 101.325 = 0.9986 atm; in bar: 101.18 ÷ 100 = 1.0118 bar

Assumptions

• The gas behaves ideally: no attraction between molecules and no molecular volume. This is close for most gases near room temperature and pressures up to a few bar.
• The temperature is converted to kelvin by adding 273.15; the law does not work with °C directly.
• The amount n is the total of all gases in a mixture (Dalton’s law); for one gas in a mixture you get its partial pressure.

How to read the result

At 25 °C and 1 atm one mole of an ideal gas fills about 24.47 L, so the example lands just under one atmosphere. The pressure rises in proportion to the amount and to the absolute temperature, and falls as the volume grows. For gases near condensation or at high pressure, compare with the van der Waals calculator.

Limits and common mistakes

• Leaving the temperature in °C instead of kelvin gives a nonsense result; the calculator converts for you.
• Mixing units of R: 8.314 J/(mol·K) goes with kPa and litres here; 0.08206 L·atm/(mol·K) gives atmospheres directly.
• Real gases deviate at high pressure and low temperature; the error can reach several percent for CO₂ at room temperature above about 10 bar.

Frequently Asked Questions

What is the pressure of 1 mol of gas at 25 °C in 24.5 L?

About 101.18 kPa, or 0.9986 atm.

Which value of R should I use?

Use 8.314 J/(mol·K) with pascals and cubic metres (or kPa and litres), or 0.08206 L·atm/(mol·K) with atmospheres and litres. They are the same constant in different units.

Why must temperature be in kelvin?

The law is a proportion to absolute temperature; 0 °C is not zero thermal energy, 0 K is.

Can I solve for volume or moles instead?

Yes: V = nRT ÷ P and n = PV ÷ RT. The same constants apply.

in n | Amount of gas | mol | 1 | 0
in t | Temperature | °C | 25 | -273.15
in v | Volume | L | 24.5 | 0.000001
out p | Pressure | kPa | n * 8.31446261815324 * (t + 273.15) / v | 2
out patm | Pressure in atmospheres | atm | p / 101.325 | 4
out pbar | Pressure in bar | bar | p / 100 | 4