Boiling Point vs Pressure Calculator

This calculator estimates the boiling point of a liquid at a given pressure using the Clausius–Clapeyron equation. You enter the pressure (in atmospheres, bar, or mmHg) and the normal boiling point of your liquid, and it returns the boiling point at that pressure. It works for water and other liquids when you know their heat of vaporization.

Formula

The calculator uses the integrated Clausius–Clapeyron equation, which relates the change in boiling point to pressure:

ln(Pâ‚‚ / Pâ‚�) = -(ΔH_vap / R) × (1/Tâ‚‚ – 1/Tâ‚�)

Where:

• Pâ‚� = reference pressure (usually 1 atm, the normal boiling point condition)
• Pâ‚‚ = the pressure you enter (same unit as Pâ‚�)
• Tâ‚� = normal boiling point (in Kelvin), the boiling point at 1 atm
• Tâ‚‚ = boiling point at the new pressure (in Kelvin, output converted to °C)
• ΔH_vap = molar heat of vaporization (J/mol; the calculator converts kJ/mol to J/mol)
• R = universal gas constant = 8.314 J/(mol·K)

For water, ΔH_vap ≈ 40.65 kJ/mol at its normal boiling point. This value changes slightly with temperature, but the equation assumes it stays constant, which is a good approximation over modest pressure ranges.

Worked Example

Using the defaults: pressure = 1 atm, normal boiling point = 100 °C, heat of vaporization = 40.65 kJ/mol.

1. Convert T� to Kelvin: 100 + 273.15 = 373.15 K
2. Convert ΔH_vap to J/mol: 40.65 × 1000 = 40,650 J/mol
3. Calculate the exponent: (8.314 × ln(1)) / 40,650 = 0 (since ln(1) = 0)
4. Compute 1/Tâ‚‚ = 1/373.15 – 0 = 0.002680
5. Tâ‚‚ = 1 / 0.002680 = 373.15 K
6. Convert to °C: 373.15 – 273.15 = 100.0 °C

At 1 atm, the boiling point is exactly the normal boiling point, as expected. If you lower the pressure to 0.5 atm, the calculator would give about 81 °C for water.

When to Use This Calculator

Use this tool whenever you need to know the boiling point of a liquid under non-standard pressure. Common situations include:

• High-altitude cooking (e.g., at 2,000 m, atmospheric pressure is about 0.8 atm, water boils near 93 °C)
• Vacuum distillation in a lab or industrial process
• Pressure cooking (at 2 atm, water boils near 121 °C)
• Designing equipment that operates at different altitudes

The calculator works for any liquid as long as you know its normal boiling point and heat of vaporization. For many common solvents, these values are readily available in reference tables.

How to Read the Result

The output is the temperature at which the liquid’s vapor pressure equals the external pressure you entered. At that temperature, bubbles of vapor form throughout the liquid—that’s the visual definition of boiling.

Remember that the result is an estimate. The Clausius–Clapeyron equation assumes ideal gas behavior and constant heat of vaporization. For most liquids over a pressure range of about 0.1 to 10 atm, the error is typically less than 1–2 °C. For extreme pressures or near the critical point, the error grows.

Common Mistakes and Misconceptions

• Mixing units: Pressure must be in the same unit for both Pâ‚� and Pâ‚‚. If you input pressure in mmHg, the reference must also be in mmHg (760 mmHg for 1 atm). The calculator uses a consistent unit, but if you manually compute, keep units uniform.
• Forgetting Kelvin: Always use Kelvin in the equation. Adding 273.15 to Celsius is essential; forgetting it gives wildly wrong results.
• Using the wrong ΔH_vap: Heat of vaporization varies with temperature. Use the value at the normal boiling point for best accuracy.
• Thinking boiling point is fixed: Many people believe water always boils at 100 °C. In reality, it changes with pressure—this is why recipes need adjustment at high altitudes.

Limitations of the Formula

The Clausius–Clapeyron equation is a simplification. It assumes:

• The vapor behaves as an ideal gas
• The molar volume of the liquid is negligible compared to the vapor
• ΔH_vap is constant over the temperature range

These assumptions hold well for pressures between about 0.01 and 10 atm for most liquids. Outside that range, more complex equations (like Antoine’s equation) give better results. For water, the Antoine equation is accurate from 1 to 100 °C, but the Clausius–Clapeyron approach is simpler and still quite useful.

Frequently Asked Questions

Why does water boil at lower temperature at high altitude?

At higher altitudes, atmospheric pressure is lower. Boiling occurs when vapor pressure equals external pressure. Lower external pressure means the liquid reaches that point at a lower temperature. For example, at 3,000 m (about 0.7 atm), water boils at roughly 90 °C instead of 100 °C.

Can I use this for any liquid?

Yes, as long as you know the normal boiling point and the heat of vaporization. For many common liquids like ethanol (78.4 °C, 38.6 kJ/mol) or acetone (56.1 °C, 29.1 kJ/mol), the calculator will give good estimates. Just replace the default values.

What is the heat of vaporization and how do I find it?

It’s the energy required to turn one mole of liquid into vapor at constant temperature. You can find it in chemistry reference books, safety data sheets, or online databases. For water, it’s about 40.65 kJ/mol at 100 °C. The value changes slightly with temperature, but using the normal boiling point value is usually sufficient.

Does the calculator work for vacuum pressures?

Yes, for pressures down to about 0.01 atm (7.6 mmHg). Below that, the ideal gas assumption breaks down and the estimate becomes less accurate. For very low pressures, you’d need more sophisticated models.

How accurate is this for pressure cooking?

Pressure cookers typically operate at about 1.5–2 atm absolute pressure. The calculator gives water boiling points of about 111–121 °C in that range, which matches real cooking temperatures. The accuracy is within a few degrees, which is fine for cooking purposes.

What if I don’t know the heat of vaporization?

For water, you can use the default 40.65 kJ/mol. For other liquids, you can estimate it using Trouton’s rule: ΔH_vap ≈ 88 × T_normal (in Kelvin) J/mol. This gives a rough value, but for better accuracy, look up the actual value.

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in p | Pressure | atm | 1 | 0.01
in t0 | Normal boiling point | °C | 100 | -273
in hvap | Heat of vaporization | kJ/mol | 40.65 | 1
out tb | Boiling point at pressure | °C | (1 / (1 / (t0 + 273.15) - (8.314 * ln(p)) / (hvap * 1000))) - 273.15 | 1