Boiling Point Calculator Under Vacuum: How It Works

A boiling point calculator under vacuum estimates the temperature at which a specified liquid boils at a specified absolute pressure. Enter the liquid’s normal boiling point and its enthalpy of vaporization, along with the vacuum pressure; the calculator returns an estimated boiling temperature and the pressure converted to familiar units.

Formula

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

ln(P₂/P₁) = −(ΔHvap/R)(1/T₂ − 1/T₁)

Rearranged for the unknown boiling temperature:

T₂ = 1 / (1/T₁ − (R/ΔHvap) ln(P₂/P₁))

• T₁ is the liquid’s normal boiling temperature in kelvins, found by adding 273.15 to its normal boiling point in °C.
• P₁ is the reference pressure, 101.325 kPa (one standard atmosphere).
• P₂ is the entered absolute pressure in kPa.
• ΔHvap is the molar enthalpy of vaporization in kJ/mol.
• R is the gas constant, 0.008314 kJ/(mol·K).
• T₂ is the estimated boiling temperature in kelvins; the calculator subtracts 273.15 to display °C.

The pressure output is p × 7.50062, converting kPa to mmHg. The displayed temperature is rounded to a whole degree, so it is an estimate rather than a precision laboratory value.

Worked Example

Using the calculator’s default values—normal boiling point 100°C, enthalpy of vaporization 40.7 kJ/mol, and absolute pressure 20 kPa:

1. Convert the reference boiling point: T₁ = 100 + 273.15 = 373.15 K.
2. Use the reference pressure P₁ = 101.325 kPa and P₂ = 20 kPa.
3. Substitute: T₂ = 1 / (1/373.15 − (0.008314/40.7) × ln(20/101.325)).
4. This gives approximately 333.7 K, or 333.7 − 273.15 = 60.5°C.
5. Convert pressure: 20 × 7.50062 = 150.0 mmHg.

The calculator rounds the temperature to the nearest whole degree, so its headline result is 61°C. The formula’s unrounded estimate is about 60.5°C; rounding conventions can differ at an exact half-degree.

Why Vacuum Lowers a Liquid’s Boiling Point

A liquid boils when its vapor pressure reaches the pressure exerted on its surface. At ordinary atmospheric pressure, water boils near 100°C at sea level. Reduce the surrounding pressure, and the liquid can reach that pressure match at a lower temperature. This is why vacuum distillation can separate or process some substances without heating them to their normal-pressure boiling temperatures.

“Vacuum” does not mean zero pressure in ordinary equipment. It describes pressure below the surrounding atmospheric pressure. The calculator needs absolute pressure: pressure measured from a perfect vacuum. A gauge vacuum reading instead describes how far pressure is below local atmospheric pressure, so it must be converted before use. For example, a vacuum gauge reading of 80 kPa below an atmosphere of 101.3 kPa corresponds to about 21.3 kPa absolute—not 80 kPa absolute.

How to Choose the Inputs

Use data for the specific pure liquid, not just a similar material. A normal boiling point is conventionally referenced to one atmosphere; the calculator uses 101.325 kPa as that reference. Enthalpy of vaporization must be entered in kJ/mol. It varies with temperature, so use a value appropriate to the substance and, where available, a reliable reference temperature near the range of interest.

• Enter absolute pressure in kPa. To convert other readings: 1 bar = 100 kPa; 1 atm = 101.325 kPa; 1 mmHg is approximately 0.133322 kPa.
• For a US customary pressure reading, 1 psi is approximately 6.89476 kPa. Convert before entering it.
• Use a positive pressure. A reading of zero is not a usable operating pressure for this equation.
• For mixtures, solutions, or changing compositions, a single normal boiling point and vaporization enthalpy may not represent the system.

At 20 kPa, the estimate for water using the defaults is about 61°C. That is a useful illustration, not a universal temperature for every liquid at that pressure: different substances have different vapor-pressure curves.

Accuracy, Assumptions, and Limits

The equation is a convenient approximation. It assumes vapor behaves ideally, the liquid’s enthalpy of vaporization is constant across the temperature interval, and the reference boiling data describe the same pure substance. In reality, enthalpy changes with temperature, and real substances can depart from these simplifying assumptions. The estimate may therefore differ from measured data, especially over a large pressure or temperature range.

Use a substance-specific vapor-pressure table, published correlation, or validated process model when accuracy matters. The calculator does not account for mixtures, dissolved substances, chemical decomposition, heat-transfer limits, pressure drops through equipment, or local changes in pressure. It also cannot tell whether a substance will remain chemically stable at the calculated temperature.

Common Mistakes and Safe Use

The most frequent error is entering gauge vacuum as though it were absolute pressure. Another is mixing units—for example, entering a value in Pa where the field expects kPa, or entering vaporization enthalpy in J/mol rather than kJ/mol. Such mistakes can shift the result dramatically. Check conversions and use unrounded inputs for any further calculations.

A calculated boiling point is not a complete operating instruction. Vacuum equipment can implode if it is not designed for external pressure, and heated liquids can cause burns or bumping (sudden, vigorous boiling). Use equipment rated for the intended pressure and temperature, follow appropriate laboratory procedures, and do not rely on this estimate alone to select operating conditions. Lower pressure can also make a liquid boil vigorously before it reaches a temperature expected at atmospheric pressure.

Frequently Asked Questions

How do I calculate boiling point at a given vacuum pressure?
Convert the vacuum reading to absolute pressure, then use a vapor-pressure relation such as the Clausius–Clapeyron equation with the liquid’s reference boiling point and enthalpy of vaporization. This calculator makes that approximate calculation.

Does a stronger vacuum always mean a lower boiling point?
For a given liquid in the usual range, reducing absolute pressure lowers its boiling temperature. “Stronger vacuum” means lower absolute pressure, not a larger absolute-pressure number.

Is vacuum pressure the same as absolute pressure?
No. Absolute pressure is measured from zero pressure. A vacuum gauge commonly reports the difference between atmospheric pressure and the system pressure; convert that gauge reading to absolute pressure before calculating.

Can I use this calculator for a mixture?
Not reliably with one normal boiling point and one enthalpy value. Mixtures can boil over a temperature range, and their composition and vapor pressure change during boiling. Use mixture-specific vapor–liquid equilibrium data or a suitable model.

Why might the actual boiling temperature differ?
The formula uses simplifying assumptions, and input data may not match the actual substance or conditions. Pressure measurement error, non-ideal behavior, mixtures, and temperature-dependent vaporization enthalpy can all affect the result.

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in tb | Normal boiling point | °C | 100 | -
in dh | Enthalpy of vaporization | kJ/mol | 40.7 | 0
in p | Absolute pressure | kPa | 20 | 0
out t | Estimated boiling point | °C | 60.5 | 0
out mm | Pressure | mmHg | p * 7.50062 | 1