How to Calculate Boiling Point in Chemistry

This calculator determines the boiling point of a liquid at a given pressure using the Clausius–Clapeyron equation. You enter the normal boiling point (at 1 atm), the enthalpy of vaporization, and the desired pressure; you get back the new boiling point in degrees Celsius. It’s a practical tool for chemistry labs, process design, and understanding how pressure affects phase changes.

Formula

The Clausius–Clapeyron equation relates the vapor pressure of a liquid to temperature:

ln(P₂/P�) = -ΔHvap/R × (1/T₂ – 1/T�)

Where:

• Pâ‚� = initial pressure (usually 1 atm, the standard boiling point pressure)
• Pâ‚‚ = new pressure (atm)
• Tâ‚� = normal boiling point (in Kelvin)
• Tâ‚‚ = new boiling point (in Kelvin)
• ΔHvap = molar enthalpy of vaporization (J/mol, but often given in kJ/mol – convert to J/mol by multiplying by 1000)
• R = universal gas constant = 8.314 J/(mol·K)

To solve for Tâ‚‚, rearrange:

T₂ = 1 / (1/T� – (R × ln(P₂/P�)) / ΔHvap)

All temperatures must be in Kelvin (K = °C + 273.15). Pressures must be in the same unit – here we use atm.

Worked Example

Using the default values: water with a normal boiling point of 100°C, ΔHvap = 40.65 kJ/mol, and a new pressure of 0.5 atm.

1. Convert T� to Kelvin: T� = 100 + 273.15 = 373.15 K
2. Convert ΔHvap to J/mol: 40.65 kJ/mol × 1000 = 40650 J/mol
3. Calculate the ratio P₂/P� = 0.5 / 1 = 0.5
4. Take natural log: ln(0.5) = -0.6931
5. Compute the exponent term: (R × ln(P₂/P�)) / ΔHvap = (8.314 × -0.6931) / 40650 = -5.761 / 40650 = -0.0001417 K�¹
6. Compute 1/T� = 1/373.15 = 0.002679 K�¹
7. Add the exponent term: 0.002679 + (-0.0001417) = 0.002537 K�¹
8. Take reciprocal: Tâ‚‚ = 1 / 0.002537 = 394.2 K
9. Convert to °C: 394.2 – 273.15 = 121.0°C

So water boils at about 121°C when the pressure is 0.5 atm (about 380 mmHg). The calculator gives 121.0°C, matching.

When to Use This Calculation

Use this when you need to predict the boiling point at a pressure other than 1 atm. Common situations include:

• Adjusting a reaction temperature for high-altitude labs (where atmospheric pressure is lower).
• Designing distillation processes under vacuum or elevated pressure.
• Estimating the boiling point of a solvent in a closed system.

The result tells you the temperature at which the liquid’s vapor pressure equals the external pressure. Below that, the liquid won’t boil; above it, it will.

Common Mistakes

• Using Celsius instead of Kelvin: Always convert to Kelvin before plugging into the equation. Forgetting to add 273.15 leads to large errors.
• Mixing pressure units: The equation requires consistent pressure units. If you use mmHg, you must use the same for both Pâ‚� and Pâ‚‚. Here we use atm, but you can convert: 1 atm = 760 mmHg = 101.325 kPa.
• Forgetting to convert kJ to J: ΔHvap is often given in kJ/mol. The gas constant R is in J/(mol·K), so you must multiply kJ by 1000.
• Using the wrong ΔHvap: The enthalpy of vaporization depends on temperature, but the Clausius–Clapeyron equation assumes it’s constant over the range. Use the value at the normal boiling point for best accuracy.
• Assuming the equation works for all conditions: It’s valid for ideal gases and over moderate pressure ranges. Near the critical point, it fails.

Tips

• Always double-check your units: Kelvin for temperature, J/mol for ΔHvap, and matching pressure units.
• For a quick estimate, you can use the rule of thumb that boiling point decreases by about 0.5°C for every 1 mmHg decrease in pressure near 1 atm, but this is only approximate.
• If you don’t know ΔHvap, look up a table of values for common solvents (water, ethanol, acetone). For water, it’s about 40.65 kJ/mol at 100°C.
• When the pressure is very low (vacuum), the boiling point can be far below the normal value; the equation handles that well as long as the liquid doesn’t freeze.

Troubleshooting

• Result is negative or absurdly low: Check that you converted ΔHvap to J/mol and that temperatures are in Kelvin. Also, ensure the pressure ratio Pâ‚‚/Pâ‚� is positive (pressures are always positive).
• Result is higher than the critical temperature: The equation becomes unreliable near the critical point. Look up the critical temperature for your substance; if your result is above it, the substance is a supercritical fluid, not a boiling liquid.
• Getting a divide-by-zero error: This happens if 1/Tâ‚� equals (R*ln(Pâ‚‚/Pâ‚�))/ΔHvap, which would imply Tâ‚‚ is infinite. That occurs when the pressure is extremely low; physically, the boiling point approaches 0 K, but the equation breaks down.
• Calculator gives different result than a table: Tables often use more precise ΔHvap values that vary with temperature. The equation is an approximation; differences of a few degrees are normal.

Frequently Asked Questions

What is the normal boiling point?

The normal boiling point is the temperature at which a liquid boils at 1 atm (101.325 kPa) pressure. For water, it’s 100°C.

Why does boiling point decrease with altitude?

At higher altitudes, atmospheric pressure is lower, so the liquid’s vapor pressure reaches the external pressure at a lower temperature. For example, on Mount Everest (about 0.33 atm), water boils around 71°C.

Can I use this for any liquid?

Yes, as long as you have the normal boiling point and enthalpy of vaporization. The equation works for pure substances; for mixtures, it’s more complex.

What if I only know the boiling point at one pressure?

You need two data points (P and T) to use the Clausius–Clapeyron equation. If you only have one, you can’t determine ΔHvap independently. However, if you know ΔHvap from tables, you can use the normal boiling point as the reference.

How accurate is the Clausius–Clapeyron equation?

It’s accurate to within a few percent for many liquids over a moderate pressure range (0.1 to 10 atm). The main assumption is that ΔHvap is constant and the vapor behaves as an ideal gas. For polar liquids or high pressures, errors increase.

What is the difference between boiling point and evaporation?

Evaporation occurs at any temperature when molecules at the surface escape into the gas phase. Boiling occurs when the vapor pressure equals the external pressure, forming bubbles throughout the liquid. The boiling point is the temperature at which boiling occurs.

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in t1 | Normal boiling point | °C | 100 | 0
in Hvap | Enthalpy of vaporization | kJ/mol | 40.65 | 0
in P | New pressure | atm | 0.5 | 0
out T2 | Boiling point at new pressure | °C | (1/(1/(t1+273.15) - (8.314 * ln(P) / (Hvap*1000)))) - 273.15 | 2