Boiling Point Calculation: Formula and Examples

A boiling point calculation estimates the temperature at which a liquid’s vapor pressure equals the surrounding pressure. Enter the liquid’s normal boiling point, its enthalpy of vaporization, and the pressure you want to evaluate; the calculator estimates the boiling temperature in kelvins and degrees Celsius.

Formula

This estimate uses the integrated Clausius–Clapeyron equation, assuming the enthalpy of vaporization stays constant over the temperature range:

1/T₂ = 1/T₁ − (R/ΔHᵥₐₚ) ln(P₂/P₁)

• T₁ is the known boiling temperature in kelvins at reference pressure P₁.
• T₂ is the estimated boiling temperature in kelvins at the target pressure P₂.
• ΔHᵥₐₚ is the molar enthalpy of vaporization in kilojoules per mole.
• R is the gas constant, 0.008314462618 kJ/(mol·K).
• ln means the natural logarithm. The pressure ratio must use the same pressure units in numerator and denominator.

The calculator takes the normal boiling point as the reference temperature and 101.325 kPa (one standard atmosphere) as the reference pressure. Its first result is degrees Celsius; the second gives the same estimate in kelvins.

Worked Example

Using the calculator’s default values—normal boiling point 373.15 K, vaporization enthalpy 40.7 kJ/mol, and target pressure 101.325 kPa—gives:

1. Calculate the pressure ratio: P₂/P₁ = 101.325/101.325 = 1.
2. Take its natural logarithm: ln(1) = 0.
3. Substitute into the equation: 1/T₂ = 1/373.15 − (0.008314462618/40.7) × 0.
4. Therefore T₂ = 373.15 K, or 373.15 − 273.15 = 100.0 °C.

The result is 100.0 °C (373.2 K when the kelvin output is rounded to one decimal place). Because the target pressure equals the reference pressure, this example returns the reference boiling point.

What the Calculation Means

A liquid boils when bubbles of vapor can form and persist within it. That happens when the liquid’s vapor pressure reaches the pressure pushing on its surface. Vapor pressure increases with temperature, so reducing the surrounding pressure generally lowers the boiling temperature; increasing pressure generally raises it.

The normal boiling point is the temperature at which a substance boils at one standard atmosphere, 101.325 kPa, approximately 1.013 bar or 14.696 psi. It is a reference condition, not a claim that every sample boils at the same temperature in every location. For example, water’s normal boiling point is 100 °C, but water boils at a lower temperature at high elevation, where atmospheric pressure is lower.

When to Use This Estimate

This calculation is useful for a quick estimate when you know a substance’s normal boiling point and approximate enthalpy of vaporization, and want to see how a change in pressure affects its boiling temperature. It illustrates the direction and approximate size of the pressure effect without requiring a full vapor-pressure table.

For a more accurate value, use measured vapor-pressure data or a suitable thermodynamic model for the particular substance and temperature range. The equation is especially convenient for modest pressure changes near the reference condition; it is not a universal substitute for experimental data.

Assumptions and Common Errors

• Use kelvins in the equation. Celsius temperatures cannot be substituted directly into the reciprocal-temperature terms. Convert with K = °C + 273.15.
• Keep energy units consistent. The calculator expects vaporization enthalpy in kJ/mol because R is entered in kJ/(mol·K). If a source gives J/mol, divide by 1,000 first.
• Use absolute pressure. The target should be pressure measured relative to a vacuum, not gauge pressure. Gauge pressure must be converted by adding the local atmospheric pressure.
• Compare like pressure units. The ratio P₂/P₁ has no units, but both pressures must use the same units. The calculator uses kPa and a reference of 101.325 kPa.
• Do not overinterpret decimals. The output is rounded to one decimal place, but the model’s assumptions can cause greater uncertainty than that final digit suggests.

The approximation treats the enthalpy of vaporization as constant and assumes vapor behaves approximately as an ideal gas. In reality, vaporization enthalpy changes with temperature, and the equation becomes less reliable for large pressure or temperature changes, especially near a critical point. It also describes a pure substance in equilibrium; mixtures can boil over a range of temperatures, and their behavior depends on composition.

Frequently Asked Questions

How do you calculate boiling point from pressure?
Use a vapor-pressure relation such as the integrated Clausius–Clapeyron equation. You need a known boiling temperature and pressure, plus the substance’s enthalpy of vaporization. For higher accuracy, use reliable vapor-pressure data.

Does lower pressure lower the boiling point?
Yes, for a given liquid under ordinary conditions. Lower surrounding pressure means the liquid’s vapor pressure reaches the surrounding pressure at a lower temperature.

Why does water boil below 100 °C at high altitude?
Atmospheric pressure is lower at higher elevations. Water therefore reaches the pressure needed to boil at a temperature below its normal boiling point of 100 °C.

Can I use Celsius in the boiling-point formula?
No. The equation uses reciprocal absolute temperature, so temperatures must be in kelvins. Convert Celsius to kelvins by adding 273.15.

Is this calculation accurate for every liquid and pressure?
No. It is an estimate based on simplifying assumptions, including constant enthalpy of vaporization. Large pressure changes, mixtures, and conditions near a critical point require more appropriate data or models.

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in bp | Normal boiling point | K | 373.15 | 0
in dh | Enthalpy of vaporization | kJ/mol | 40.7 | 0
in p | Surrounding pressure | kPa | 101.325 | 0
out t | Estimated boiling point | °C | (1 / (1 / bp - 0.008314462618 * ln(p / 101.325) / dh) - 273.15) | 1
out tk | Estimated boiling point | K | 1 / (1 / bp - 0.008314462618 * ln(p / 101.325) / dh) | 1