Equilibrium
Equilibrium
Liquid and gas phases are in equilibrium at the boiling temperature. (See the figure below.) If a substance is in a closed container at the boiling point, then the liquid is boiling and the gas is condensing at the same rate without net change in their relative amount. Molecules in the liquid escape as a gas at the same rate at which gas molecules stick to the liquid, or form droplets and become part of the liquid phase. The combination of temperature and pressure has to be “just right”; if the temperature and pressure are increased, equilibrium is maintained by the same increase of boiling and condensation rates.
Equilibrium between liquid and gas at two different boiling points inside a closed container. (a) The rates of boiling and condensation are equal at this combination of temperature and pressure, so the liquid and gas phases are in equilibrium. (b) At a higher temperature, the boiling rate is faster and the rates at which molecules leave the liquid and enter the gas are also faster. Because there are more molecules in the gas, the gas pressure is higher and the rate at which gas molecules condense and enter the liquid is faster. As a result the gas and liquid are in equilibrium at this higher temperature.
Triple Point Temperatures and Pressures
| Substance | Temperature | Pressure | ||
|---|---|---|---|---|
| \(\text{K}\) | \(\text{º}\text{C}\) | \(\text{Pa}\) | \(\text{atm}\) | |
| Water | 273.16 | 0.01 | \(6\text{.}\text{10}×{\text{10}}^{2}\) | 0.00600 |
| Carbon dioxide | 216.55 | −56.60 | \(5\text{.}\text{16}×{\text{10}}^{5}\) | 5.11 |
| Sulfur dioxide | 197.68 | −75.47 | \(1\text{.}\text{67}×{\text{10}}^{3}\) | 0.0167 |
| Ammonia | 195.40 | −77.75 | \(6\text{.}\text{06}×{\text{10}}^{3}\) | 0.0600 |
| Nitrogen | 63.18 | −210.0 | \(1\text{.}\text{25}×{\text{10}}^{4}\) | 0.124 |
| Oxygen | 54.36 | −218.8 | \(1\text{.}\text{52}×{\text{10}}^{2}\) | 0.00151 |
| Hydrogen | 13.84 | −259.3 | \(7\text{.}\text{04}×{\text{10}}^{3}\) | 0.0697 |
One example of equilibrium between liquid and gas is that of water and steam at \(\text{100}\text{º}\text{C}\) and 1.00 atm. This temperature is the boiling point at that pressure, so they should exist in equilibrium. Why does an open pot of water at \(\text{100}\text{º}\text{C}\) boil completely away? The gas surrounding an open pot is not pure water: it is mixed with air. If pure water and steam are in a closed container at \(\text{100}\text{º}\text{C}\) and 1.00 atm, they would coexist—but with air over the pot, there are fewer water molecules to condense, and water boils.
What about water at \(\text{20}\text{.}0\text{º}\text{C}\) and 1.00 atm? This temperature and pressure correspond to the liquid region, yet an open glass of water at this temperature will completely evaporate. Again, the gas around it is air and not pure water vapor, so that the reduced evaporation rate is greater than the condensation rate of water from dry air. If the glass is sealed, then the liquid phase remains. We call the gas phase a vapor when it exists, as it does for water at \(\text{20}\text{.}0\text{º}\text{C}\), at a temperature below the boiling temperature.
Check Your Understanding
Explain why a cup of water (or soda) with ice cubes stays at \(0\text{º}\text{C}\), even on a hot summer day.
Solution
The ice and liquid water are in thermal equilibrium, so that the temperature stays at the freezing temperature as long as ice remains in the liquid. (Once all of the ice melts, the water temperature will start to rise.)
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws