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Chapter 13: Problem 45

Natural gas consists of about \(90 \%\) methane, \(\mathrm{CH}_{4}\) Assume that the solubility of natural gas at \(20^{\circ} \mathrm{C}\) and 1 atm gas pressure is about the same as that of \(\mathrm{CH}_{4}\) \(0.02 \mathrm{g} / \mathrm{kg}\) water. If a sample of natural gas under a pressure of 20 atm is kept in contact with \(1.00 \times 10^{3} \mathrm{kg}\) of water, what mass of natural gas will dissolve?

Short Answer

Expert verified
The mass of natural gas that will dissolve in the water is 400 g.

Step by step solution

01

Understand Henry's Law

Henry's law states that 'At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.' In this case, we are given the solubility at 1 atm (0.02g in 1kg of water) and we are asked to find the solubility at 20 atm.
02

Apply the Law

Since the solubility of the gas is directly proportional to its partial pressure, the solubility of gas at 20 atm would be 20 times the solubility at 1 atm. This is because the pressure has increased by a factor of 20. So, the solubility of gas at 20 atm is \(20 \times 0.02 \, g/kg = 0.4 \, g/kg\). This means 0.4 g of gas will dissolve in every kg of water at 20 atm.
03

Calculate the total mass of dissolved gas

To find the total mass of the gas that will dissolve, multiply the solubility of the gas at 20 atm by the total mass of the water. In this case, the water mass is \(1.00 \times 10^{3} \, kg\), so \(0.4 \, g/kg \times 1.00 \times 10^{3} \, kg = 400 \, g\). Therefore, 400 g of natural gas will dissolve in the water.

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Most popular questions from this chapter

What are the partial and total vapor pressures of a solution obtained by mixing 35.8 g benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) and \(56.7 \mathrm{g}\) toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\) at \(25^{\circ} \mathrm{C} ? \mathrm{At} 25^{\circ} \mathrm{C}\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{6}=95.1 \mathrm{mm} \mathrm{Hg} ;\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}=28.4 \mathrm{mmHg}\).

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