# Why does temperature affect the dissolved oxygen concentration?

I know that an increase in temperature decreases the solubility of oxygen in water, but I don't know why. Could anybody explain?

• Have you tried searching a bit on your own? You didn't say what you know so far or where you looked prior to posting here. Searching google for "oxygen dissolution temperature" yields this as the third result. Give it a try. Dec 21 '16 at 13:16
• @Don_S oh cool, it says it's because dissolving oxygen is exothermic, and according to Le Chatelier's principle, adding heat would favor the endothermic side, so adding heat shifts the equilibrium to the left. Thanks for the link, perfect! Dec 21 '16 at 22:34

## 3 Answers

The solubility of a many gases in a variety of liquids is observed to decrease with increase in temperature, for example oxygen, nitrogen and methane in water. This follows the qualitative predictions of Le Chatelier’s principle, which in thermodynamics is supplanted by the Clapeyron equation. As one of the two phases involved is a gas the Clausius–Clapeyron equation can be used.

This assumes that the molar volume of any gas/vapour is far greater than that of the solution. Assuming that all values correspond to one mole the Clausius–Clapeyron equation is

$$\frac{\mathrm d\ln(p)}{\mathrm dT}=\frac{\Delta H}{RT^2}$$ which in its integrated form between pressure and temperature $p_1, T_1$ to $p_2, T_2$, and by assuming that $\Delta H$ is independent of temperature gives

$$\ln\left( \frac{p_2}{p_1}\right)=- \frac{\Delta H}{R}\left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$

Despite the approximations involved in its derivation this equation does describe well the measurements of the vapour pressure over many solids and liquids, for example depression of freezing point, elevation of boiling point and solubility of solids and gases in liquids, although the latter is not as frequently described as the other two.

In the elevation of the boiling point the solution at a temperature $T_1$ has a lower vapour pressure $p_1$ than the pure solvent. The solution only boils when the temperature is increased to $T_2$ where its vapour pressure has increased to $p_2$ (normally 1 atm ), which is the pressure at which the pure solvent boils. By Raoult’s law $p_1=p_2x_{\mathrm s}$, where $x_{\mathrm s}$ is the mole fraction of solvent $\ln(p_2/p_1)=\ln(1/x_s)$ As $x_\mathrm s=1-x_\mathrm t$ where $x_\mathrm t$ is the mole fraction of solute, $$\ln\left( \frac{1}{1-x_t}\right)= -\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$ and as $x_\mathrm t$ is far smaller than unity $\ln(1-x_t) \approx -x_t -x_t^2/2 \ldots$ from which the second term can be ignored as it is insignificant compared to the first then, $$x_\mathrm t= \frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$

Thermodynamically the elevation of boiling point and solubility are very similar but we must swap solute and solvent over so what we formerly called solvent has now to be solute and vice versa.

Considering now solubility, the variation of the mole fraction dissolved $x_\mathrm t$ can be written as $$x_\mathrm t=\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2} - \frac{1}{T_\mathrm B} \right)$$

where what was $T_2$ is now the normal boiling temperature of the liquefied gas. Also $\Delta H_\text{vap}$ (a positive quantity) is now the (molar) heat of vaporisation of the liquefied gas and we approximate this to be the heat of vaporisation from solution.

This last equation indicates that the solubility decreases with increase in temperature and should be the same in all solvents in which it forms ideal solutions.

In the solution a gas molecule has an attractive interaction energy between itself and the solvent. (Experiment shows that $\Delta H_\text{vap}$ is positive thus the heat to dissolve is negative). The solvent molecules also have an attractive interaction between themselves otherwise they would not form a liquid. Thus for a gas-solvent interaction a solvent-solvent one has to be destroyed. Overall the difference in energy is probably small. However, there is an entropic effect to consider. Clearly this will have increased when a gas dissolves as now there is another species in the solvent and thus the number of ways the molecules can be arranged has increased and this will have increased the entropy. Together these effect make the free energy negative as expected for a spontaneous process.

As the temperature is increased the molecules gain in average energy and so are more able to overcome the attractive potential that keeps them in solution and thus the vapour pressure of the solvent increases and the solubility of the gas decreases. It is also worth noting that the average kinetic energy $(3/2)kT$ is the same in the solution and in the gas phase, in solution it is merely restricted to a smaller range on motion. Thus the notion that it is the kinetic energy that is acquired that causes a molecule to enter the gas phase in incorrect.

For ideal gases (probably in solution the physics are a little different) but anyway as oxygen is an ideal gas, the equations from physics that you need to keep in mind are ...

kinetic energy = (1 / 2)(mass)(velocity)^2 = (3 / 2)(Boltzmann constant)(Temperature),

so therefore temperature is a measure of the velocity of your molecules, where as temperature increases, so does the velocity of the molecules increase. Thus, all molecules in solution, including gaseous oxygen move more rapidly gaining kinetic energy to escape solvation energy, and so they are more prone to bubble up the surface and escape acqueous solution because oxygen gas is less dense than liquid water.

• Unfortunately this answer is not correct either. Dec 21 '16 at 20:08

Cold water holds more oxygen molecule than hot water because water molecules are closer together in cold water which makes it harder oxygen molecule to escape. At higher temperature, the oxygen molecule have greater amount of kinetic energy. So the gas molecule in hot water can more easily overce the weak binding forces within the water and escape through the surface.

• Unfortunately this answer is not correct Dec 21 '16 at 20:07