acalculator

What is the vapor pressure?

Enter a liquid’s vapor pressure at one temperature and its enthalpy of vaporization, and the calculator gives the vapor pressure at another temperature with the Clausius–Clapeyron equation. It also finds the temperature for a vapor pressure, ΔHvap from two points, or the vapor pressure from Antoine constants.

Your numbers

Units
Find
Pressure unit
Answer
P₂ = 30.3128 kPa

P₂ = 30.3128 kPa.

Vapor pressure
30.3128
Vapor pressure (bar)
0.303128
ΔHvap (kJ/mol)
31.3

Answer: P₂ = 30.3128 kPa. P₂ = 30.3128 kPa.

How to calculate

Finds a liquid’s vapor pressure at a new temperature with the Clausius–Clapeyron equation, the temperature for a vapor pressure, or the enthalpy of vaporization from two points, or the vapor pressure from Antoine constants.

Example with the default inputs (Find Vapor pressure, ΔHvap (kJ/mol) 31.3, Temperature T₁ 56.5 °C, Vapor pressure P₁ 101.3, Temperature T₂ 25 °C, Pressure unit kPa): P₂ = 30.3128 kPa.

Method: ln(P₂ ÷ P₁) = (ΔHvap ÷ R)(1 ÷ T₁ − 1 ÷ T₂), so P₂ = P₁ × e^((ΔHvap ÷ R)(1 ÷ T₁ − 1 ÷ T₂)), 1 ÷ T₂ = 1 ÷ T₁ − (R ÷ ΔHvap) ln(P₂ ÷ P₁) and ΔHvap = R ln(P₂ ÷ P₁) ÷ (1 ÷ T₁ − 1 ÷ T₂); Antoine: log₁₀(P ÷ bar) = A − B ÷ (T + C).

  • ΔHvap does not change with temperature between T₁ and T₂, and the vapor behaves as an ideal gas.
  • Antoine constants hold only in the temperature range their source gives; the page does not check it.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Find ΔHvap, Temperature T₁ 34 °C, Vapor pressure P₁ 10, Temperature T₂ 98.8 °C, Vapor pressure P₂ 100, Pressure unit kPa gives ΔHvap (kJ/mol) 33.7528, Answer ΔHvap = 33.7528 kJ/mol.Source: OpenStax, Chemistry 2e, §10.3 Phase Transitions (ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂)), https://openstax.org/books/chemistry-2e/pages/10-3-phase-transitions, Example 10.7 Estimating Enthalpy of Vaporization: 33.8 kJ/mol
  2. Find ΔHvap, Temperature T₁ 20 °C, Vapor pressure P₁ 5.95, Temperature T₂ 63.5 °C, Vapor pressure P₂ 53.3, Pressure unit kPa gives ΔHvap (kJ/mol) 41.3582.Source: OpenStax, Chemistry 2e, §10.3 Phase Transitions (ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂)), https://openstax.org/books/chemistry-2e/pages/10-3-phase-transitions, Example 10.7 Check Your Learning: 41,360 J/mol
  3. Find Temperature, ΔHvap (kJ/mol) 30.8, Temperature T₁ 80.1 °C, Vapor pressure P₁ 101.3, Vapor pressure P₂ 83.4, Pressure unit kPa gives Temperature T₂ (K) 346.819, Temperature T₂ (°C) 73.6694.Source: OpenStax, Chemistry 2e, §10.3 Phase Transitions (ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂)), https://openstax.org/books/chemistry-2e/pages/10-3-phase-transitions, Example 10.8 Estimating Temperature (or Vapor Pressure): 346.9 K (73.8 °C) from 353.3 K
  4. Find Vapor pressure, ΔHvap (kJ/mol) 31.3, Temperature T₁ 56.5 °C, Vapor pressure P₁ 101.3, Temperature T₂ 25 °C, Pressure unit kPa gives Vapor pressure 30.3128, Answer P₂ = 30.3128 kPa.Source: OpenStax, Chemistry 2e, §10.3 Phase Transitions (ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂)), https://openstax.org/books/chemistry-2e/pages/10-3-phase-transitions, Example 10.8 Check Your Learning: 30.1 kPa
  5. Find Antoine, Antoine A 4.6543, Antoine B 1,435.264, Antoine C -64.848, Temperature T 25 °C, Pressure unit kPa gives Vapor pressure (bar) 0.0317938, Vapor pressure 3.17938, Answer P = 3.17938 kPa (0.0317938 bar).Source: NIST Chemistry WebBook, SRD 69, Water: Antoine equation parameters (log₁₀(P) = A − B/(T + C), P in bar, T in K; Stull 1947, 255.9 to 373 K: A = 4.6543, B = 1435.264, C = −64.848), https://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Mask=4&Type=ANTOINE&Plot=on

How it works

Clausius–Clapeyron equation (OpenStax §10.3). For one liquid at two temperatures,

ln(P₂ ÷ P₁) = (ΔHvap ÷ R)(1 ÷ T₁ − 1 ÷ T₂)

with T in kelvins (°C + 273.15; °F converted exactly), ΔHvap typed in kJ/mol and used in J/mol (× 1,000), and R = 8.31446261815324 J/(mol·K) exactly (NIST). OpenStax rounds R to 8.3145.

  • Vapor pressure: P₂ = P₁ × e^((ΔHvap ÷ R)(1 ÷ T₁ − 1 ÷ T₂)).
  • Temperature: 1 ÷ T₂ = 1 ÷ T₁ − (R ÷ ΔHvap) × ln(P₂ ÷ P₁); the page shows T₂ in K and °C (K − 273.15).
  • ΔHvap: ΔHvap = R × ln(P₂ ÷ P₁) ÷ (1 ÷ T₁ − 1 ÷ T₂), in kJ/mol.

P₁ and P₂ use one unit, picked below them. 1 ÷ T₁ − 1 ÷ T₂ = (T₂ − T₁) ÷ (T₁ × T₂) is worked exactly from the typed temperatures; logarithms and powers of e are double-precision numbers. In the temperature mode, 1 ÷ T₂ and T₂ − 273.15 are worked exactly from the logarithm, so a T₂ near 0 °C keeps its digits (ΔHvap 1,000 kJ/mol, T₁ 273.15 K, P₂ ÷ P₁ = 1.0000000001: T₂ = 273.15 K, 6.2035 × 10⁻¹¹ °C).

Antoine equation (NIST Chemistry WebBook form).

log₁₀(P ÷ bar) = A − B ÷ (T + C)

with T in kelvins and A, B and C as NIST lists them (the default is water, 255.9 to 373 K). The page converts P from bar into the unit picked, and shows it in bar too. The page does not check the constants’ temperature range.

Pressure units (NIST SP 811). 1 kPa = 1,000 Pa; 1 atm = 101,325 Pa; 1 torr = 101,325 ÷ 760 Pa; 1 mmHg = 133.322387415 Pa; 1 bar = 100,000 Pa; 1 psi = 4.4482216152605 N ÷ (0.0254 m)² = 6,894.76 Pa.

Rules the page checks.

  • Temperatures from 1 K to 100,000 K; pressures from 10⁻¹² to 10¹² in their unit; ΔHvap from 0.001 to 1,000 kJ/mol; A from −100 to 100, B from −100,000 to 100,000, C from −1,000 to 1,000.
  • ΔHvap mode: the two temperatures must differ, and the result must be above 0 (the higher temperature needs the higher vapor pressure); otherwise no answer.
  • Temperature mode: 1 ÷ T₂ must be above 0, and T₂ at most 100,000 K; otherwise no answer.
  • Vapor pressure mode: a P₂ above 10³⁰⁰ or below 10⁻³⁰⁰ (in the unit picked) has no answer.
  • Antoine mode: T + C must be above 0, and A − B ÷ (T + C) between −300 and 300; otherwise no answer.

How results are written. 6 significant figures, trailing zeros dropped, thousands commas, the minus sign −, and m × 10ⁿ below 0.000001 or from 10²¹. The sentence reads “P₂ = 30.3128 kPa”, “T₂ = 346.819 K (73.6694 °C)”, “ΔHvap = 33.7528 kJ/mol”, or “P = 3.17938 kPa (0.0317938 bar)” (just “P = 0.0317938 bar” when bar is picked).

Assumptions. ΔHvap is constant between the two temperatures, and the vapor is an ideal gas.

Worked examples by hand

Isooctane (OpenStax Example 10.7). T₁ = 34.0 + 273.15 = 307.15 K, T₂ = 98.8 + 273.15 = 371.95 K. ΔHvap = 8.31446261815324 × ln(100.0 ÷ 10.0) ÷ (1/307.15 − 1/371.95) = 8.31446 × 2.302585 ÷ 5.67205 × 10⁻⁴ = 33.7528 kJ/mol (OpenStax: 33.8 kJ/mol).

Ethanol (Check Your Learning). 5.95 kPa at 293.15 K and 53.3 kPa at 336.65 K: ΔHvap = 41.3582 kJ/mol (OpenStax: 41,360 J/mol).

Benzene at 83.4 kPa (Example 10.8). 1/T₂ = 1/353.25 − (8.31446261815324 ÷ 30,800) × ln(83.4 ÷ 101.3) = 0.00283086 + 0.0000524886 = 0.00288334, so T₂ = 346.819 K (73.6694 °C). OpenStax uses 353.3 K and gets 346.9 K (73.8 °C).

Acetone at 25.0 °C (Check Your Learning). (31,300 ÷ 8.31446261815324) × (1/329.65 − 1/298.15) = −1.20652, so P₂ = 101.3 × e^−1.20652 = 30.3128 kPa. OpenStax prints 30.1 kPa.

Water at 25 °C (NIST, Stull 1947). log₁₀ P = 4.6543 − 1435.264 ÷ (298.15 − 64.848) = −1.49767, so P = 0.0317938 bar = 3.17938 kPa.

Other questions people ask

How do I calculate vapor pressure at a different temperature?

Use the two-point Clausius–Clapeyron equation, ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂), with temperatures in kelvins. Acetone boils at 56.5 °C (329.65 K) at 101.3 kPa with ΔHvap = 31.3 kJ/mol; at 25.0 °C, P₂ = 101.3 × e^(−1.20652) = 30.31 kPa.

How do I find the enthalpy of vaporization from two vapor pressures?

Rearrange: ΔHvap = R ln(P₂/P₁) ÷ (1/T₁ − 1/T₂). Isooctane has 10.0 kPa at 34.0 °C and 100.0 kPa at 98.8 °C, so ΔHvap = 8.314 × ln 10 ÷ (1/307.15 − 1/371.95) = 33.75 kJ/mol (OpenStax: 33.8 kJ/mol).

How do I find the boiling point at a lower pressure?

A liquid boils when its vapor pressure equals the pressure above it. Solve for T₂ with P₂ = the new pressure: benzene boils at 80.1 °C at 101.3 kPa, and at 83.4 kPa (Denver) at about 346.8 K, or 73.7 °C.

What is the Antoine equation?

An empirical fit of vapor pressure: log₁₀ P = A − B ÷ (T + C). The NIST Chemistry WebBook lists A, B and C for many substances with P in bar and T in kelvins, each for a temperature range. For water from 255.9 to 373 K (Stull, 1947): A = 4.6543, B = 1435.264, C = −64.848, which gives 0.0318 bar (3.18 kPa) at 25 °C.

Why does my answer differ from the textbook’s?

Textbooks round temperatures and R along the way. OpenStax converts 80.1 °C to 353.3 K and gets 346.9 K for benzene; with 353.25 K the page gives 346.8 K. For acetone it gives 30.1 kPa where the unrounded equation gives 30.3 kPa.

Does the pressure unit matter?

No, as long as P₁ and P₂ share it: the equation uses their ratio. The page offers kPa, atm, mmHg, torr, bar, Pa and psi, and shows the Antoine result in bar as well.