Views: 98 Author: Site Editor Publish Time: 2024-07-03 Origin: Site
The two gases with the highest content in the air are N2 and O2, both of which are non-polar gases.
Such non-polar molecules with very low solubility must exist in the gaps between water molecules, so gap filling is a way for air to dissolve in water.
In addition, we know that as the temperature rises, the amount of gas dissolved in water will decrease, so it is also believed that there is a specific chemical reaction between water molecules and gas molecules called hydration. Therefore, gas dissolution in water can be summarized into two dissolution modes: gap filling and hydration.
Gap filling is the entry of gas molecules into the gap space of liquid molecules, which is a physical change; hydration is the chemical reaction between liquid molecules and one or more gas molecules to generate hydrated molecules while releasing heat, which is a chemical reaction. The two dissolution modes are shown in Figure 1-1.
(a) Gap filling
(b) Hydration
(c) Joint action
Figure 1-1 Gap filling and hydration dissolution
Here, we need to introduce the concept of "effective interstitial space." Adequate interstitial space means that when the temperature is constant, the space in a certain mass of water that allows a specific gas molecule to enter is called the adequate interstitial volume of the gas. This volume's ratio to the corresponding water's total volume is called effective interstitial space. Formula 1-2 is the expression of adequate interstitial space:
（1-2）
In the formula, Vi represents the adequate interstitial volume of gas I at a specific temperature; Vw represents the corresponding total volume of water.
A large number of studies and experiments have shown that when helium is dissolved in water, the relationship between the maximum effective interstitial volume of water and temperature (℃) is as follows (1-3):
λhe = 0.009696829 + 3.1639178 × 10-5t - 1.257929 × 10-6t2 + 2.129631 × 10-8t3 (1-3)
If you want to get the maximum effective gap of other gases, you only need to correct it based on λhe, which can be expressed as formula (1-4):
（1-4）
Where λi is the maximum effective interstitial density of gas i and αi is the correction factor, which is the ratio of the van der Waals volume of helium to that of gas i.
The substances contained in the air and their proportions are shown in Table 2-1.
| Air composition | N2 | O2 | Ar、O3 | CO2etc. |
| Volume Percent | 78% | 21% | 0.97% | 0.03% |
Table 2-1 Air composition ratio
It can be seen from Table 2-1 that nitrogen and oxygen in the air account for about 99% of the total air.
By consulting the data, we can find that the van der Waals volumes of the three gases are: νmhe = 2.370 × 10-5, νmN2 = 3.913 × 10-5 , νmO2 = 3.183 ×10-5 ( m3/mol)
It can be calculated by formula 1-3 and 1-4:
When nitrogen is dissolved in water, the relationship between the maximum effective interstitial density of water and temperature is as follows (1-5):
λi = 0.606(0.009696829 + 3.1639178 × 10-5t - 1.257929 ×10-6t2 + 2.129631 × 10-6t3 (1-5)
When oxygen is dissolved in water, the relationship between the maximum effective interstitial density of water and temperature is as follows (1-6):
λi = 0.745(0.009696829 + 3.1639178 × 10-5t - 1.257929 ×10-6t2 + 2.129631 × 10-6t3 (1-6)
According to van der Waals's idea of dealing with natural gases, the ideal gas state equation is shown in equations 1-7:
Pv = nRT (1-7)
Assuming that the van der Waals volume of the gas molecule is vm (m3/mol), according to formula (1-8), the corrected gas state equation can be obtained as formula (1-8).
p(Vj - njvm) = njRT (1-8)
In the formula, nj represents the amount of gas entering the interstitial space of water in mol. Solving formula (1-3) and formula (1-8) together can yield formula (1-9).
（1-9）
Therefore, formula (1-9) is the amount of gas i dissolved in the gaps in water.
Studies have shown that the solubility of gas in water is affected by temperature. The higher the temperature, the lower the solubility. Therefore, there is a chemical reaction between water molecules and gas molecules. This type of chemical reaction is called hydration.
Hydration is another form of dissolved gas in water. Hydrate is a special enveloping compound. Hydrogen bonds connect water molecules as the main body, and the main molecules and guest molecules interact through van der Waals forces.
Although it is difficult for oxygen and nitrogen to hydrate at room temperature, this reaction must exist. The reaction equation is (1-10):
（1-10）
In the formula, G(g) represents a gas molecule; Q represents heat.
Assuming that the amount of water is large and the concentration is approximately constant, when gas i reaches dissolution equilibrium at a certain pressure and temperature, its equilibrium constant Kp can be expressed by formula 1-11:
（1-11）
In the formula, Nh is the number of hydrated molecules; Nm is the Avogadro constant.
The amount of substance of the hydrated molecule can be obtained from formula 1-12:
（1-12）
Combining equations (1-11) and (1-12) gives the gas hydration constant mass calculation equation (1-13):
（1-13）
In the formula, nhH is the amount of substance of the hydrated molecule; Vw is the volume of water; Kp is the dissolved gas pressure; Kp is the equilibrium constant under a specific temperature and pressure.
Generally, it is believed that gap filling and hydration jointly affect the solubility of gas in water.
The gaps between water molecules are very small, so the volume of the gaps between water molecules occupied by hydration cannot be ignored. Assuming that the gap volume occupied by hydration is Vh, the volume of the remaining space is Equation 1-14:
（1-14）
Gap filling only occurs in the remaining space, so combining equations 1-2 and 1-9, replacing Vi with Vs, we get equations 1-15
（1-15）
The total dissolved amount of gas is the sum of nj and nh, which can be expressed by formula 2-16:
（1-16）
In the above formula, Nh can be obtained from formula 1-12 and 1-13, and then 1-16 is substituted into formula 2-17:
（1-17）
Formula 1-17 divided by the volume of water is the molar concentration of dissolved gas, see formula 1-18
（1-18）
Based on the total solubility equation of gas (Formula 1-18), the solubility equation expressed by the ratio of gas volume to water volume under standard conditions can be calculated (Formula 1-19):
（1-19）
According to equations 1-18 and 1-19, the expression 1-20 of the hydration equilibrium constant Kp can be derived:
（1-20）
In the formula: Kp is the hydration equilibrium constant, vm is the van der Waals volume of the gas (m3/mol), p is the pressure (pa), T is the temperature (K), R is the gas constant, and λi is the effective interstitial density of the gas.
The value of Rs can be obtained from the literature and can be substituted into formula 1-20 to obtain the Kp value for oxygen and nitrogen.
In imperative forms 1-20, 
Assume the pressure is 101325pa. Formula 1-20 can be written as Formula 1-21:
（1-21）
(1) Calculation of nitrogen hydration equilibrium constant is shown in Table 2-2
| Temperature/℃ | A | B | C | K P |
| 0 | 1.050893 | 0.261728 | 176.593324 | 7.80205E-06 |
| 5 | 0.931250 | 0.260461 | 173.424331 | 6.63152E-06 |
| 10 | 0.830804 | 0.258018 | 170.367069 | 5.66248E-06 |
| 15 | 0.752232 | 0.254867 | 167.415732 | 4.91674E-06 |
| 20 | 0.689732 | 0.251446 | 164.564908 | 4.33258E-06 |
| 25 | 0.640179 | 0.248165 | 161.809548 | 3.87507E-06 |
| 30 | 0.599107 | 0.245404 | 159.144936 | 3.49627E-06 |
| 35 | 0.560714 | 0.243521 | 156.566662 | 3.1353E-06 |
| 40 | 0.528571 | 0.242849 | 154.070596 | 2.82415E-06 |
| 45 | 0.504464 | 0.243703 | 151.652869 | 2.57737E-06 |
| 50 | 0.485714 | 0.246376 | 149.309849 | 2.36558E-06 |
Table 2-2 Parameters of nitrogen hydration equilibrium constant
(2) Calculation of oxygen hydration equilibrium constant is shown in Table 2-3
| Temperature/℃ | A | B | C | K P |
| 0 | 2.182589 | 0.321866 | 143.695213 | 1.839E-05 |
| 5 | 1.913839 | 0.320307 | 141.115758 | 1.57489E-05 |
| 10 | 1.697321 | 0.317300 | 138.627277 | 1.36384E-05 |
| 15 | 1.524554 | 0.313423 | 136.225040 | 1.1969E-05 |
| 20 | 1.384821 | 0.309215 | 133.904641 | 1.06295E-05 |
| 25 | 1.263839 | 0.305178 | 131.661967 | 9.47356E-06 |
| 30 | 1.164286 | 0.301781 | 129.493178 | 8.52315E-06 |
| 35 | 1.089286 | 0.299464 | 127.394681 | 7.80475E-06 |
| 40 | 1.029464 | 0.298637 | 125.363113 | 7.22164E-06 |
| 45 | 0.976339 | 0.299686 | 123.395324 | 6.68619E-06 |
| 50 | 0.933036 | 0.302971 | 121.488355 | 6.22572E-06 |
From chemical kinetics, we know that when a reaction reaches equilibrium, the rates of the forward and reverse reactions are equal. Its equilibrium constant can be expressed as formula 1-22:
（1-22）
Assume that the rates of both the forward and reverse reactions can be expressed using the Arrhenius formula, namely:
（1-23）
（1-24）
In the formula: K is the rate constant, A is the frequency factor, and E is the activation energy.
Assuming that the activation energy and frequency factor are not affected by temperature within a certain range, then:
（1-25）
Taking the logarithm of both sides of formula1-25, we can obtain:
（1-26）
In the formula, ΔH is the heat of reaction of formula 1-10, and C is a constant. According to the assumption, ΔH can be considered as a constant, so ln Kp is proportional to 1/T. With 1/T as the horizontal axis and ln Kp as the vertical axis, Figure 1-2 shows the relationship between nitrogen and oxygen hydration constants and temperature according to the data in Tables 2-2 and 2-3.
Figure 1-2 Relationship between nitrogen and oxygen hydration constants and temperature
As shown in Figures 1-2, the logarithms of the hydration constants of nitrogen and oxygen show an excellent linear relationship with 1/T. By fitting the above data using the polyfit function in Matlab, we can obtain the relationship between the hydration constants of the two gases and temperature.
Nitrogen: 
(1-27)
oxygen: 
(1-28)
In summary, both nitrogen and oxygen in the air will undergo gap-filling and hydration in water. Among them, the amount of nitrogen gap filling is formula 1-29:
(1-29)
The amount of nitrogen hydration is given by formula 1-30:
(1-30)
The amount of oxygen gap filling is given by Equation 1-31:
(1-31)
The amount of oxygen hydration is given by formula 1-32:
(1-32)
When the volume and pressure of water are the same, the temperature change will affect the values of a, b, c, and d in the formula. When the volume and pressure of water are the same, the temperature will affect the amount of gap filling and hydration. Assuming that the pressure is 0.5MPa and the water is 1L, the curves of a, b, c, and d affected by temperature are shown in Figure 2-3.
Figure 2-3 Relationship between gap filling and hydration affected by temperature
Figures a, b, c, and d represent the amount of nitrogen gap filling, the amount of nitrogen hydration, the amount of oxygen gap filling, and the amount of oxygen hydration, respectively.
As can be seen from the figure, under the conditions of 0.5MPa pressure and 1L water, the amount of gap filling increases with the increase in temperature, and the amount of hydration decreases with the temperature rise.
At around 50℃, the amount of gap filling by nitrogen is roughly equal to the amount of hydration; when the temperature is below 50℃, the amount of hydration by nitrogen gradually increases with the decrease of temperature; when the temperature is above 50℃, the amount of gap filling by nitrogen increases with the increase of temperature. For oxygen, at around 80℃, the amount of the two dissolution modes is equal.
This is because as the temperature rises, the distance between water molecules increases, making the effective gap larger so that more gas molecules can be filled, resulting in the gap-filling amount increasing with the increase in temperature.
The increase in temperature leads to a decrease in the amount of hydration because hydration releases heat. The increase in external temperature will prevent the reaction from proceeding in the forward direction and promote the reaction in the reverse direction, thereby reducing the dissolution amount of hydration.
