Answer to Question #129722 in Physics for Bontle

Question

Calculate the quantity of oxygen in a bucket of fresh water (volume10litre) at two different altitudes .sea level and 5000 m.a.s.l use Henry's law and assume a temperature of 15°c

Answer 1

Henry's law:

$$x = \dfrac{p}{H}$$

where $x$ is the mole fraction of the oxigen in the water, $p$ is the partial oxygen preassure in air (in atm) and $H = 3.64\times 10^{4}$ is Henry's constant for oxygen at 15°c.

1. The partial oxygen preassure in air at sea level is:

$$p_0 = 1\space atm\times 0.21 = 0.21\space atm$$

where $0.21$ represents the fraction of oxygen in air.

Thus:

$$x_0 = \dfrac{0.21}{3.64\times 10^{4}} \approx 5.8\times 10^{-6}$$

One liter of water contains 55.56 moles of water. Thus, 10 liters will contain 555.6 moles. Thus, the quantity of oxygen will be:

$$n_0 = x_0\times 555.6 \space moles = 5.8\times 10^{-6}\times 555.6\space moles \approx 3.22\times 10^{-3}\space moles$$

2. The partial oxygen preassure in air at 5000 m.a.s is:

$$p_0 = 0.53\space atm\times 0.21 = 0.11\space atm$$

where $0.21$ represents the fraction of oxygen in air.

Thus:

$$x_0 = \dfrac{0.11}{3.64\times 10^{4}} \approx 3.02\times 10^{-6}$$

One liter of water contains 55.56 moles of water. Thus, 10 liters will contain 555.6 moles. Thus, the quantity of oxygen will be:

$$n_0 = x_0\times 555.6 \space moles = 3.02\times 10^{-6}\times 555.6\space moles \approx 1.68\times 10^{-3}\space moles$$

Answer. Sea level: 3.22*10^-3 moles, 5000 m.a.s: 1.68*10^-3 moles.