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Walk-through

How to Use This Calculator

3 steps

1

Pick what you want to solve for

On the Solve tab, choose P₁, T₁, P₂, T₂, or ΔHvap with the chip row. The calculator hides that field — fill in the other four and it solves automatically. If you just want a new vapor pressure or a ΔHvap value, use the dedicated Vapor Pressure or Enthalpy tabs instead.

2

Enter your two-point data

Vapor pressure and temperature must each come from the same physical state — a pressure paired with the temperature it was measured at. Temperatures must be in Kelvin (add 273.15 to a Celsius value); pressures can be in any unit as long as P₁ and P₂ use the same one.

3

Read the solved value and interpretation

The result card shows the solved variable with its equation, and the interpretation line explains what the number means — for example, that vapor pressure rises exponentially with temperature, or how sensitive that rise is to ΔHvap.

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Reference

Formula & Methodology

3 formulas

Clausius-Clapeyron equation

ln(P₂/P₁) = −ΔHvap/R · (1/T₂ − 1/T₁)

P₁ and P₂ are the vapor pressures (any consistent unit) at absolute temperatures T₁ and T₂ (Kelvin). ΔHvap is the molar enthalpy of vaporization (J/mol) and R = 8.314 J/(mol·K) is the universal gas constant. The equation assumes ΔHvap is constant over the temperature range, which holds well for a modest span but drifts for a very wide one.

Solving for a pressure

P₂ = P₁ · exp[ (ΔHvap/R)·(1/T₁ − 1/T₂) ]

Rearranging for P₂ (or P₁, by swapping the 1 and 2 subscripts) turns the log relationship into a direct exponential — this is the form used on the Vapor Pressure tab to project a known pressure to a new temperature.

Solving for ΔHvap

ΔHvap = R · ln(P₂/P₁) / (1/T₁ − 1/T₂)

Given two measured vapor-pressure/temperature points, this isolates the enthalpy of vaporization — the basis of the Enthalpy tab. It requires T₁ ≠ T₂, since the denominator vanishes otherwise.

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Glossary

Key Terms Explained

7 terms

Clausius-Clapeyron equation A relationship, derived from the Clapeyron equation plus the ideal-gas approximation for the vapor phase, that links a substance's vapor pressure to its temperature and its enthalpy of vaporization along the liquid-vapor coexistence line.

Vapor pressure The pressure exerted by a substance's vapor when it is in equilibrium with its liquid (or solid) at a given temperature. It rises with temperature — exponentially, not linearly — for every pure substance.

Enthalpy of vaporization The energy (usually in kJ/mol) needed to convert one mole of a liquid into vapor at constant pressure, without a temperature change. It reflects how strongly a substance's molecules attract each other in the liquid phase.

Phase change A transition between physical states of matter (e.g. liquid to gas) at which the substance absorbs or releases latent heat while its temperature stays constant. Vaporization is the liquid-to-gas phase change.

Gas constant The universal gas constant R = 8.314 J/(mol·K), a proportionality constant that appears in the ideal gas law and, through it, in the Clausius-Clapeyron equation.

Boiling point The temperature at which a liquid's vapor pressure equals the surrounding (usually atmospheric) pressure, so it can vaporize throughout the bulk liquid rather than only at the surface. It drops at lower ambient pressure, which is why water boils below 100°C at altitude.

Temperature In this calculator, always the absolute (Kelvin) temperature. The Clausius-Clapeyron equation's 1/T terms make it invalid to substitute Celsius or Fahrenheit directly — 0°C is not the same fraction of 0 K as 0°F is of a Fahrenheit-based absolute scale.

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Scenarios

Real-World Examples

3 worked examples

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Chemistry student

Vapor pressure at a new temperature

P₁, T₁ 1 atm at 373.15 K (water's normal boiling point) T₂, ΔHvap 350 K, 40.7 kJ/mol

P₂ = 1 × exp[(40700/8.314)·(1/373.15 − 1/350)] ≈ 0.42 atm. Cooling water by about 23 K below its boiling point cuts its vapor pressure by more than half — the exponential form means pressure is far more sensitive to temperature near the boiling point than a linear guess would suggest.

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Lab researcher

Solving for ΔHvap from two data points

P₁, T₁ 1 atm, 350 K P₂, T₂ 2 atm, 375 K

ΔHvap = 8.314 × ln(2/1) / (1/350 − 1/375) ≈ 30,300 J/mol ≈ 30.3 kJ/mol. Measuring vapor pressure at just two temperatures is enough to back out the enthalpy of vaporization without any calorimetry.

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Hiker planning a high-altitude trip

Boiling point at reduced atmospheric pressure

P₁, T₁ 1 atm at 373.15 K (sea-level boiling point) P₂, ΔHvap 0.6 atm (≈ 4,000 m altitude), 40.7 kJ/mol

Solving for T₂ on the Solve tab gives roughly 359 K (≈ 86°C) — water boils about 14°C cooler at that altitude because it needs less vapor pressure to match the lower ambient pressure, which is also why high-altitude cooking times run longer.

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Deep Dive

Understanding the Clausius-Clapeyron Equation

3 chapters · 3 min read

The Clausius-Clapeyron equation is the standard tool for connecting a substance's vapor pressure to its temperature. Because that relationship is exponential rather than linear, knowing just two pressure-temperature points — plus the enthalpy of vaporization — lets you predict the pressure at any other temperature, or work backward to find ΔHvap itself.

How the equation works

The equation comes from combining the exact Clapeyron equation (which relates the slope of the liquid-vapor coexistence curve to the entropy and volume change of vaporization) with the ideal-gas approximation for the vapor and the assumption that the liquid's volume is negligible next to the vapor's. The result is a clean log-linear relationship: plotting ln(P) against 1/T gives a straight line whose slope is −ΔHvap/R. That's exactly what the two-point form used here captures — subtract the equation at one point from the equation at another, and the integration constant cancels out.

Because ΔHvap is treated as constant, the two-point form is most accurate over a modest temperature range. Over a very wide span, ΔHvap itself drifts with temperature (it goes to zero at the critical point), so a single average ΔHvap will introduce some error the further T₂ is from T₁.

Inputs and what they mean

Pressure (P₁, P₂) can be entered in any unit — atm, mmHg, kPa, torr — as long as both use the same one, since only their ratio P₂/P₁ enters the equation. Temperature (T₁, T₂) must always be in Kelvin; add 273.15 to a Celsius reading first. Enthalpy of vaporization (ΔHvap) is usually quoted in kJ/mol in reference tables (water is about 40.7 kJ/mol near its boiling point) — the calculator converts to J/mol internally to match R = 8.314 J/(mol·K).

Limits and edge cases

The two-point form breaks down if T₁ equals T₂ (there's no temperature change to solve ΔHvap from) or if ΔHvap is exactly zero while solving for a temperature (the equation can't isolate T). It also assumes the vapor behaves ideally and that ΔHvap doesn't change much between T₁ and T₂ — both approximations get worse near a substance's critical point, where the distinction between liquid and vapor disappears entirely. For everyday ranges (well below the critical point), the equation is accurate enough for lab and classroom use.

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Questions

Frequently Asked Questions

6 questions

What is the formula for the Clausius-Clapeyron equation?+

ln(P₂/P₁) = −ΔHvap/R · (1/T₂ − 1/T₁), where P₁ and P₂ are vapor pressures at absolute temperatures T₁ and T₂, ΔHvap is the molar enthalpy of vaporization, and R = 8.314 J/(mol·K).

How do I solve for ΔHvap?+

Given two vapor-pressure/temperature points, ΔHvap = R · ln(P₂/P₁) / (1/T₁ − 1/T₂). The Enthalpy tab does this directly from four inputs: P₁, T₁, P₂, T₂.

Why does the boiling point drop at high altitude?+

Boiling happens when a liquid's vapor pressure equals the surrounding atmospheric pressure. At altitude, atmospheric pressure is lower, so the liquid needs a lower vapor pressure — and therefore a lower temperature — to start boiling.

Does temperature have to be in Kelvin?+

Yes. The equation's 1/T terms only work on an absolute temperature scale, so Celsius or Fahrenheit values must be converted to Kelvin first (K = °C + 273.15) before entering them.

What is R in this equation?+

R is the universal gas constant, 8.314 J/(mol·K). It's the same constant used in the ideal gas law, since the Clausius-Clapeyron equation assumes the vapor phase behaves as an ideal gas.

What units does the calculator use?+

Temperatures are always Kelvin. Pressure can be any unit you like as long as P₁ and P₂ share it, since only their ratio matters. Enthalpy of vaporization can be entered in kJ/mol (the common convention) or J/mol.

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Clausius-Clapeyron Calculator

Inputs

Result

Find the vapor pressure at a new temperature

Result

Find ΔHvap from two measured points

Result

How to Use This Calculator

Pick what you want to solve for

Enter your two-point data

Read the solved value and interpretation

Formula & Methodology

Key Terms Explained

Real-World Examples

Chemistry student

Lab researcher

Hiker planning a high-altitude trip

Understanding the Clausius-Clapeyron Equation

How the equation works

Inputs and what they mean

Limits and edge cases

Frequently Asked Questions

Get a branded PDF of your results

Clausius-Clapeyron Calculator

Inputs

Result

Find the vapor pressure at a new temperature

Result

Find ΔHvap from two measured points

Result

How to Use This Calculator

Pick what you want to solve for

Enter your two-point data

Read the solved value and interpretation

Formula & Methodology

Key Terms Explained

Real-World Examples

Understanding the Clausius-Clapeyron Equation

How the equation works

Inputs and what they mean

Limits and edge cases

Frequently Asked Questions

Get a branded PDF of your results

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