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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