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Chapter 13: Problem 37

The Henry's law constant for helium gas in water at \(30^{\circ} \mathrm{C}\) is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and the constant for \(\mathrm{N}_{2}\) at \(30^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm} .\) If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas.

Short Answer

Expert verified
The solubility of helium in water at 30°C and 1.5 atm partial pressure is approximately \(5.55 \times 10^{-4} \mathrm{M}\), and the solubility of nitrogen is approximately \(9.0 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Identify the given values

We are given the following information: - Henry's law constant for helium (\(k_{He}\)) = \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) - Henry's law constant for nitrogen (\(k_{N_2}\)) = \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) - Partial pressure for helium and nitrogen (\(P_{He} = P_{N_2}\)) = 1.5 atm
02

Calculate the solubility of helium

Using the given values for helium, we can plug them into the Henry's law formula to find its solubility: \[C_{He} = k_{He} P_{He}\] \[C_{He} = (3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm})\] \[C_{He} = 5.55 \times 10^{-4} \mathrm{M}\] The solubility of helium in water at 30°C and 1.5 atm partial pressure is approximately \(5.55 \times 10^{-4} \mathrm{M}\).
03

Calculate the solubility of nitrogen

Now, we can use the given values for nitrogen to find its solubility: \[C_{N_2} = k_{N_2} P_{N_2}\] \[C_{N_2} = (6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm})\] \[C_{N_2} = 9.0 \times 10^{-4} \mathrm{M}\] The solubility of nitrogen in water at 30°C and 1.5 atm partial pressure is approximately \(9.0 \times 10^{-4} \mathrm{M}\).
04

Report the results

The solubilities of helium and nitrogen in water at 30°C and 1.5 atm partial pressure are: - Helium: \(5.55 \times 10^{-4} \mathrm{M}\) - Nitrogen: \(9.0 \times 10^{-4} \mathrm{M}\)

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