Vapour Pressure of Pure Water
At 0°C (32°F), the vapour pressure of pure water is approximately 611.2 pascals (Pa). As the temperature increases, so does the vapour pressure. For instance, at 25°C (77°F), the vapour pressure rises to about 3,169 Pa, and at 100°C (212°F), it reaches 101,325 Pa, which is equivalent to the atmospheric pressure at sea level. This increase in vapour pressure with temperature is due to the greater kinetic energy of water molecules, which allows more of them to escape from the liquid into the vapour phase.
The relationship between temperature and vapour pressure is often described by the Clausius-Clapeyron equation. This equation provides a way to calculate the vapour pressure of a liquid at different temperatures by using the enthalpy of vaporization, which is the amount of heat required to convert the liquid into vapour.
To illustrate this concept further, the following table shows the vapour pressure of water at various temperatures:
Temperature (°C) | Vapour Pressure (Pa) |
---|---|
0 | 611.2 |
10 | 1,228 |
20 | 2,339 |
30 | 4,246 |
40 | 7,384 |
50 | 12,352 |
60 | 20,063 |
70 | 31,824 |
80 | 48,372 |
90 | 71,458 |
100 | 101,325 |
Understanding vapour pressure is essential in various applications. For instance, in meteorology, it helps in predicting weather patterns and humidity levels. In engineering, it is crucial for designing systems involving steam and other vapours. Moreover, in environmental science, vapour pressure data is used to study the effects of temperature changes on water bodies and atmospheric conditions.
In summary, the vapour pressure of pure water provides significant insights into the behaviour of water in different conditions and is essential for numerous scientific and practical applications.
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