Question

It is found experimentally that the volume of a gas that dissolves in a given amount of water is independent of the pressure of the gas; that is, if \(5 \mathrm{~cm}^{3}\) of a gas dissolves in \(100 \mathrm{~g}\) of water at 1 atm pressure, \(5 \mathrm{~cm}^{3}\) will dissolve at a pressure of 2 atm, 5 atm, 10 atm, .... Show that this relationship follows logically from Henry's law and the ideal gas law.

Short answer

Question: Prove that the volume of a gas that dissolves in a given amount of water is independent of the pressure of the gas, using Henry's law and the ideal gas law. Answer: Following the step-by-step analysis and solution using Henry's law and the ideal gas law, we found that the volume of the gas that dissolves in a given amount of water is independent of the pressure of the gas because the volumes are equal regardless of the pressure (\(V_2 = V_1\)).

Step-by-step solution

Find the moles of the gas dissolved at 1 atm

Using the ideal gas law, we can find the moles of gas dissolved at 1 atm pressure: $$n_1 = \frac{P_1V}{RT}$$ For this exercise, the given values are: - \(P_1 = 1\) atm - \(V = 5 \mathrm{~cm}^3 = 5 \times 10^{-3} \mathrm{~L}\) - \(R = 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\) - \(T\): constant temperature We have: $$n_1 = \frac{(1 \mathrm{~atm})(5 \times 10^{-3} \mathrm{~L})}{(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}) T}$$

Find the moles of the gas dissolved at a higher pressure

Repeating the same process as before, we can find the moles of gas dissolved at a higher pressure, \(P_2\): $$n_2 = \frac{P_2V}{RT}$$ For this exercise, the given values are: - \(P_2 > 1\) atm - \(V\): same volume as before - \(R\): same ideal gas constant as before - \(T\): same temperature as before We have: $$n_2 = \frac{(P_2 \mathrm{~atm})(5 \times 10^{-3} \mathrm{~L})}{(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}) T}$$

Use Henry's law to find the relationship between pressures and moles

According to Henry's law: $$C = k_H P$$ Since the concentration of gas dissolved in the liquid is the ratio of the number of moles of the gas to the volume of the liquid, we have: $$\frac{n_1}{100 \mathrm{g}} = k_H P_1$$ $$\frac{n_2}{100 \mathrm{g}} = k_H P_2$$ Dividing the two equations, we get the relationship between the pressures and moles: $$\frac{n_2}{n_1} = \frac{P_2}{P_1}$$

Prove the volume of dissolved gas is independent of pressure

From Step 3, we have: $$\frac{\frac{P_2V}{RT}}{\frac{P_1V}{RT}} = \frac{P_2}{P_1}$$ After simplification, we get: $$V_2 = V_1$$ This result shows that the volume of the gas that dissolves in a given amount of water is independent of the pressure of the gas, as the volumes are equal regardless of the pressure. Thus, the relationship follows logically from Henry's law and the ideal gas law, and we have completed the exercise.