28 Estimating Vapour Pressure

Learning Objectives

By the end of this section, you should be able to:

Estimate the vapour pressure of pure compounds at a given temperature using Antoine’s equation


Vapour pressure([latex]p*[/latex]) is defined as the pressure at which a gas coexists with its solid or liquid phase. Vapour pressure is created by faster molecules that break away from the liquid or solid and enter the gas phase. The vapour pressure of a substance depends on both the substance and its temperature—an increase in temperature increases the vapour pressure. [latex]^{[1]}[/latex]

One empirical method to estimate [latex]p∗[/latex] is the Antoine equation:



A, B and C are constants depending on the type of substance

NOTE: Sometimes different sources of data can use different units for temperature, or used [latex]ln[/latex] (base e) instead of [latex]log_{10}[/latex] (base 10).


Finding Antoine Equation Parameters on NIST:

  1. Search for “NIST Chemistry Webbook” or go to https://webbook.nist.gov/chemistry/
  2. Scroll down to search options and choose, name, formula or other desired search criteria
  3. Enter search criteria (e.g. methanol, CH3OH…)
  4. Leave other options as default and search for substance
  5. Select desired substance from the list provided if not brought to substance homepage immediately
  6. On the substance homepage scroll down to the “Other data available” section and select “phase change data”.
  7. Scroll through this section to find desired data


Example: Estimating Vapour Pressure for Water

Temperature(K) A B C
379-573 3.55959 643.748 -198.043
273-303 5.40221 1838.675 -31.737
304-333 5.20389 1733.926 -39.485

data obtained from NIST

Based on the data above, what is the vapour pressure of water at 100°C and 200°C?

For T=100°C: Since the temperature is given in the Kelvin scale on the chart, we convert 100°C to 373K, which is closest to the range in the first row of data, so we will use that range.
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
&= 10^{3.55959-\frac{643.748}{373-198.043}}\\
& = 0.76 bar

For T=200°C: We first convert the temperature to 473K, which is in the range for the first row of data.
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
&= 10^{3.55959-\frac{643.748}{473-198.043}}\\
& = 16.53 bar


Consider the calculated result for vapour pressure at 100 °C, compare it with your knowledge of vapour pressure of water. Does it seem accurate?


We know the vapour pressure at 100 °C should be closer to 1 bar as water would boil at 100 °C at 1 atm, this shows the error that might be present in these correlations, especially when taking them outside of their T ranges. As shown below, the value calculated at T=200 °C is more accurate, likely because it is well within the range of temperatures provided for the A, B and C parameters (although there is still some error). Remember these correlations are generalizations, meaning it is rare they are 100% accurate.
Temperature(K) [latex]p*[/latex] calculated (bar) [latex]p*[/latex] actual(bar) % difference
373 0.758798 1.013 25.09403
473 16.53187 15.55 6.314304


[1] OpenStax University Physics Volume 2. 2016. 13.5 Phase Changes. [online] <https://openstax.org/books/college-physics/pages/13-5-phase-changes> [Accessed 15 May 2020].



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