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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 gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(5.55 \times 10^{-4} \mathrm{M}\), and the solubility of nitrogen gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(9.0 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Identify given data

We are given the Henry's law constants for helium gas and nitrogen gas in water at 30°C. The constant for helium is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and for nitrogen gas is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\). The partial pressure of both gases is given as 1.5 atm.
02

Apply Henry's law

According to Henry's law, the solubility of a gas (C) in a liquid is directly proportional to the partial pressure of the gas (P) above the liquid. Mathematically, it can be represented as: \[ C = k_H \times P \] where, \(C\) is the solubility of the gas, \(k_H\) is the Henry's law constant and \(P\) is the partial pressure of the gas.
03

Calculate solubility of helium gas

Using the Henry's law constant and the partial pressure given for helium gas, let's calculate the solubility. We will use the following formula: \[ C_\mathrm{He} = k_\mathrm{He} \times P_\mathrm{He} \] Substitute the given values into the equation: \[ C_\mathrm{He} = (3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm}) \] After performing the calculations, we get: \[ C_\mathrm{He} = 5.55 \times 10^{-4} \mathrm{M} \]
04

Calculate solubility of nitrogen gas

Similarly, we can calculate the solubility of nitrogen gas using the Henry's law constant and the partial pressure given for nitrogen gas. We will use the following formula: \[ C_\mathrm{N_2} = k_\mathrm{N_2} \times P_\mathrm{N_2} \] Substitute the given values into the equation: \[ C_\mathrm{N_2} = (6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm}) \] After performing the calculations, we get: \[ C_\mathrm{N_2} = 9.0 \times 10^{-4} \mathrm{M} \]
05

Write the final answer

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

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Most popular questions from this chapter

A solution is made containing 14.6 \(\mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in 184 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate (a) the mole fraction of \(\mathrm{CH}_{3} \mathrm{OH},\) (b) the mass percent of \(\mathrm{CH}_{3} \mathrm{OH},(\mathbf{c})\) the molality of \(\mathrm{CH}_{3} \mathrm{OH}\) .

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