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Clausius–Clapeyron calculator

Find the temperature where a liquid reaches a target vapor pressure: from water at 101325 Pa and 373.15 K, with 40650 J/mol, you get the temperature for 2339 Pa. Every substitution step is listed so you can check it. Free in your browser.

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If you enter
P1
101,325
T1
373.15
P2
2,339
Delta h
40,650
You get
Resistance (Ω)
8.314462618
Gas constant
8.314462618
Ln p2 over p1
-3.7686096849
Inv t1
0.0026798874

A Clausius–Clapeyron equation calculator turns two vapor pressures, one known absolute temperature, and the enthalpy of vaporization into the second boiling or condensation temperature on the vapor-pressure curve. Enter P1 at T1, the target vapor pressure P2, and ΔHvap (with an optional gas constant R that defaults to the SI value); the tool rearranges the two-point form ln of P2 over P1 equals ΔHvap over R times one over T1 minus one over T2, then returns T2 in kelvin with inverse temperatures, the log pressure ratio, a short message, and ordered substitution steps. Homework checks, lab estimates of boiling point under reduced pressure, and coding kata expected values all need that same deterministic T2 without a spreadsheet.

What the two-point Clausius–Clapeyron equation solves

The Clausius–Clapeyron relation links how equilibrium vapor pressure changes with temperature for a pure liquid or solid when the enthalpy of vaporization or sublimation can be treated as roughly constant over a modest temperature interval. In the integrated two-point form, the natural logarithm of the pressure ratio P2 over P1 equals the enthalpy of vaporization divided by the gas constant, times the difference of reciprocal absolute temperatures one over T1 minus one over T2.

Solving for the unknown temperature gives T2 equals one divided by the quantity one over T1 minus R over ΔHvap times ln of P2 over P1. Absolute temperatures must be in kelvin; pressures may be in any unit as long as both P1 and P2 use the same unit because only their ratio appears.

Enthalpy and R must share energy units: with the default R of about eight point three one four joules per mole per kelvin, enter ΔHvap in joules per mole; if your table lists kilojoules per mole, either multiply ΔH by one thousand or pass a matching R near zero point zero zero eight three. Zero vapor pressure is not allowed: ln of a zero or infinite ratio is undefined, and that failure mode is the documented error path for this capability.

The check path recomputes the same intermediate log and reciprocal temperatures with the same fixed rounding used for published fields, so a successful response always ships with steps you can assert in tests or worksheets.

Practical tips, units, limits, and related searches

Convert Celsius to kelvin before calling: add two hundred seventy-three point one five to °C; never pass bare Celsius into t1. Keep P1 and P2 in the same pressure unit so the ratio is dimensionless and unit-free. Prefer ΔHvap and R in matching energy units rather than mixing kJ and J without adjusting R.

Prefer this capability when you already have two pressures, one temperature, and ΔHvap and only need T2; if your source is a word problem in °C, convert first. Round-trip checks are easy: recompute ln of P2 over P1 by hand, form one over T2 from the published formula, invert it, and confirm it equals the published t2 within the same twelve-decimal rounding.

Homework checks for boiling-point shifts

Students enter textbook vapor pressures, T1, and ΔHvap and compare hand-written T2 and substitution steps against a deterministic result before submitting.

Lab estimates under reduced pressure

Process and organic labs estimate the temperature at which a pure liquid reaches a target vacuum pressure when ΔHvap is known from a handbook.

What is the Clausius–Clapeyron equation?

In two-point form, ln(P2/P1) = (ΔHvap/R)·(1/T1 − 1/T2). It relates two vapor pressures on the equilibrium curve when enthalpy of vaporization is treated as constant.

What units should I use?

T1 and T2 are absolute kelvin. P1 and P2 may use any shared pressure unit. Default R is 8.314462618 J/(mol·K), so enter ΔHvap in J/mol unless you override R.

Does this convert °C for me?

No. Convert Celsius to kelvin yourself (T(K) = t(°C) + 273.15) before sending t1.

Is the Clausius–Clapeyron calculator free?

The browser tool is free and runs locally.

What if my enthalpy of vaporization is in kJ/mol?

Multiply it by 1000 to enter it in J/mol, or set the gas constant to 0.008314462618 so both use kilojoules. The default constant is 8.314462618 J/(mol·K), and mixing it with kJ gives a wrong temperature.

Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.

POSThttps://api.kit.forhosting.com/chem/clausius-clapeyron

Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.

curl -X POST https://api.kit.forhosting.com/chem/clausius-clapeyron \
  -H "Authorization: Bearer $KIT_KEY" \
  -H "Content-Type: application/json" \
  -d '{"p1":101325,"t1":373.15,"p2":2339,"delta_h":40650}'
{
  "p1": 101325,
  "t1": 373.15,
  "p2": 2339,
  "delta_h": 40650
}
{
  "task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
  "type": "chem.clausius_clapeyron",
  "status": "queued",
  "_links": {
    "result": "/tasks/tsk_…/result"
  }
}

The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.

Per request$0.002

Published price — no tokens, no invented credits. A failed task is never charged.

max_abs1000000000000
HTTPCodeMeaning
401unauthorizedMissing or invalid API key.
402insufficient_balanceYour balance doesn't cover the task price.
404unknown_typeThat task type doesn't exist.
429rate_limitedToo many requests. Use the webhook instead of polling.

When either vapor pressure is zero the logarithm is undefined, so the API returns a clear invalid_input error instead of NaN.

How to call the API and what comes back

Both pressures must be strictly positive; values at or below one times ten to the minus fifteen absolute are treated as zero and rejected with a message that mentions non-zero pressure. Send a JSON body with four required finite numbers: p1 (aliases P1, pressure1, vapor_pressure1, p_initial), t1 (aliases T1, temperature1, temp1, t_initial), p2 (aliases P2, pressure2, vapor_pressure2, p_final), and delta_h (aliases delta_h_vap, deltaH, dH, enthalpy_vaporization, hvap, Delta_H, DH). Optionally include r (aliases R, gas_constant, gasConstant); when omitted the solver uses eight point three one four four six two six one eight joules per mole per kelvin. Absolute values must stay within the published limit of one trillion so intermediate products remain finite on the Worker and in the browser. Temperature t1 must be strictly positive kelvin, and both delta_h and r must be non-zero. On success the response echoes rounded p1, t1, p2, delta_h, and r (also as gas_constant), sets ln_p2_over_p1, inv_t1, inv_t2, and t2, formula to the two-point rearrangement string, law to clausius_clapeyron, conditions noting constant ΔHvap, unit hints for pressure and temperature, message to a short English summary, and steps to an ordered array of substitution lines. Failed validation—missing fields, non-finite numbers, zero pressure, non-positive t1, zero ΔHvap or R, or an unphysical pair that would make T2 non-positive or non-finite—returns invalid_input and is never charged. Pricing is a flat $0.002 per successful request with unit item and zero marginal unit price.

Practical tips, units, limits, and related searches

When teaching, pair examples: water near one atmosphere with ΔHvap about forty thousand six hundred fifty joules per mole, a lower target pressure for reduced-pressure boiling, and a deliberate p1 or p2 equal to zero to demonstrate the hard invalid_input path. This landing targets people searching for a Clausius–Clapeyron equation calculator, vapor pressure temperature calculator, or enthalpy of vaporization temperature estimate; it is not a full Antoine equation fitter, not a multi-component VLE flash, and not a heat-of-vaporization estimator from two measured points (though you can rearrange the same algebra offline for that). Failed tasks that return invalid_input do not consume prepaid balance, which keeps exploratory scripts cheap when learners deliberately set a pressure to zero.

Unit tests for chemistry calculators

Generate known-good T2 cases plus deliberate zero-pressure rejections for coding kata and automated grading expected values.

What happens when a pressure is zero?

ln(P2/P1) is undefined. This tool returns invalid_input and never charges the request.

Read the full KIT documentation →