# 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**:

[latex]log_{10}(p*)=A-\frac{B}{T+C}[/latex]

where

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**:

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

\begin{align*}

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

\end{align*}

For T=200°C: We first convert the temperature to 473K, which is in the range for the first row of data.

\begin{align*}

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

\end{align*}

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

**References**

[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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