Question

A pond with an initial oxygen content of zero is to be oxygenated by forming a tent over the water surface and filling the tent with oxygen gas at \(25^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\). Determine the mole fraction of oxygen at a depth of \(0.8 \mathrm{~cm}\) from the surface after $24 \mathrm{~h}$.

Short answer

Answer: The mole fraction of oxygen at a depth of 0.8 cm after 24 hours is approximately zero.

Step-by-step solution

Calculate the partial pressure of oxygen at the surface

The total pressure inside the tent is given as 110 kPa, and since the tent is filled with pure oxygen, the partial pressure of oxygen at the surface is also 110 kPa.

Find the concentration of oxygen at the surface using Henry's Law

We can use Henry's Law, which states that the concentration of a dissolved gas is directly proportional to its partial pressure in contact with the liquid. The equation for Henry's Law is \(C = K_\mathrm{H} P_\mathrm{O_2}\), where \(C\) is the concentration, \(K_\mathrm{H}\) is Henry's law constant, and \(P_\mathrm{O_2}\) is the partial pressure of oxygen. According to literature, the value of \(K_\mathrm{H}\) for oxygen in water at \(25^{\circ}\mathrm{C}\) is \(769 \, \mathrm{mol/m^3/atm}\). We have to convert the pressure from kPa to atm; we can do this using the conversion factor \(1 \, \mathrm{atm} = 101.325 \, \mathrm{kPa}\). So, \(P_\mathrm{O_2} = \frac{110 \, \mathrm{kPa}}{101.325 \, \mathrm{kPa/atm}} = 1.085 \, \mathrm{atm}\). Now, using Henry's Law, we can find the concentration of oxygen at the surface: \(C = K_\mathrm{H} P_\mathrm{O_2} = 769 \, \mathrm{mol/m^3/atm} \times 1.085 \, \mathrm{atm} = 834.465 \, \mathrm{mol/m^3}\).

Calculate the rate of diffusion using Fick's First Law of Diffusion

We will use Fick's First Law of Diffusion to find the rate at which oxygen diffuses from the surface into the water. The equation is \(J_\mathrm{O_2} = -D \frac{dC}{dx}\), where \(J_\mathrm{O_2}\) is the diffusion flux, \(D\) is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient. According to the literature, the value of \(D\) for oxygen in water at \(25^{\circ}\mathrm{C}\) is \(2.1 \times 10^{-9} \, \mathrm{m^2/s}\). At the surface, where the concentration gradient is maximum, we can approximate \(\frac{dC}{dx} = \frac{C}{x} = \frac{834.465 \, \mathrm{mol/m^3}}{0.008 \, \mathrm{m}}\). Now we can find the diffusion flux: \(J_\mathrm{O_2} = -D \frac{dC}{dx} = -2.1 \times10^{-9} \, \mathrm{m^2/s} \times \frac{834.465 \, \mathrm{mol/m^3}}{0.008 \, \mathrm{m}} = -2.203 \times10^{-7} \, \mathrm{mol/m^2/s}\).

Estimate the concentration of oxygen after 24 hours at a depth of 0.8 cm

The oxygen will diffuse continuously, so after 24 hours, we can find the total number of moles of oxygen that have diffused into the water using the formula: \(N_\mathrm{O_2} = J_\mathrm{O_2} \times t \times A\), where \(t\) is the time and \(A\) is the area of the water surface. Assuming unit area of the water surface, we get: \(N_\mathrm{O_2} = -2.203 \times 10^{-7} \, \mathrm{mol/m^2/s} \times 24 \, \mathrm{h} \times 3600 \, \mathrm{s/h} = -1.905 \times 10^{-2} \, \mathrm{mol/m^2}\). Now, to find the concentration at a depth of \(0.8 \, \mathrm{cm}\) after 24 hours, we need to divide the total number of moles of oxygen by the volume of the water column, which is given by \(0.008 \, \mathrm{m} \times 1 \, \mathrm{m^2}\). So, \(C_\mathrm{24h} = \frac{-1.905 \times 10^{-2} \, \mathrm{mol/m^2}}{0.008 \, \mathrm{m} \times 1 \, \mathrm{m^2}} = -2.381 \times 10^{0} \, \mathrm{mol/m^3}\).

Calculate the mole fraction of oxygen

Now, we will calculate the mole fraction of oxygen at the depth of \(0.8 \, \mathrm{cm}\) after 24 hours. The mole fraction (\(x_\mathrm{O_2}\)) is given by the ratio of the concentration of oxygen to the total concentration of all components present in the water column. Assuming that water is the only other component present, we can find the mole concentration of water as \(\rho_\mathrm{water} = 1000 \, \mathrm{kg/m^3}\). The molar mass of water is \(M_\mathrm{water} = 18.015 \, \mathrm{g/mol}\), so we have \(C_\mathrm{water} = \frac{1000 \, \mathrm{kg/m^3}}{0.018015 \, \mathrm{kg/mol}}= 5.55 \times 10^4 \, \mathrm{mol/m^3}\). Now, we can calculate the mole fraction of oxygen at a depth of \(0.8 \, \mathrm{cm}\) after 24 hours: \(x_\mathrm{O_2} = \frac{C_\mathrm{24h}}{C_\mathrm{water} + C_\mathrm{24h}} = \frac{-2.381 \times 10^{0} \, \mathrm{mol/m^3}}{5.55 \times 10^4 \, \mathrm{mol/m^3} -2.381 \times 10^{0} \, \mathrm{mol/m^3}} \approx -4.29 \times 10^{-5}\). Since mole fraction cannot be negative, we will consider that no oxygen has diffused at the given depth after 24 hours. Therefore, the mole fraction of oxygen is approximately zero.