Question

The solubility of nitrogen gas in rubber at \(25^{\circ} \mathrm{C}\) is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3} \cdot\) bar. When phase equilibrium is established the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at \(300 \mathrm{kPa}\) is \((a) 0.005 \mathrm{kg} / \mathrm{m}^{3}\) \((b) 0.018 \mathrm{kg} / \mathrm{m}^{3}\) \((c) 0.047 \mathrm{kg} / \mathrm{m}^{3}\) \((d) 0.13 \mathrm{kg} / \mathrm{m}^{3}\) \((e) 0.28 \mathrm{kg} / \mathrm{m}^{3}\)

Short answer

Answer: (d) 0.13 kg/m³

Step-by-step solution

Convert the given pressure to bar

Given pressure is \(300 \mathrm{kPa}\). To convert this pressure to bar, divide by 100: $$\frac{300 \mathrm{kPa}}{100} = 3 \mathrm{bar}$$

Find the amount of nitrogen dissolved in rubber at the given pressure

Use the given solubility value to find the amount of nitrogen dissolved in rubber at \(3 \mathrm{bar}\) pressure: $$0.00156 \mathrm{kmol} / \mathrm{m}^{3}\cdot \mathrm{bar} \times 3 \mathrm{bar} = 0.00468 \mathrm{kmol}/\mathrm{m}^3$$

Convert the amount from kmol to mol

To convert the amount of nitrogen in rubber from kmol to mol, multiply by 1000: $$0.00468 \mathrm{kmol}/\mathrm{m}^3 \times 1000 = 4.68 \mathrm{mol}/\mathrm{m}^3$$

Convert molar amount to mass

Since the molar mass of nitrogen is approximately \(28 \mathrm{g/mol}\), we can convert the molar amount to mass (in kg) as follows: $$4.68 \mathrm{mol}/\mathrm{m}^3 \times 28 \mathrm{g/mol} \times \frac{1 \mathrm{kg}}{1000 \mathrm{g}} = 0.131 \mathrm{kg}/\mathrm{m}^3$$

Compare the result to the given options

The calculated density is \(0.131 \mathrm{kg}/\mathrm{m}^3\). Comparing this result to the given options, we can see that the closest option is \((d) 0.13 \mathrm{kg}/\mathrm{m}^3\). So, the answer is option \((d)\).

Key concepts

Phase Equilibrium

Understanding phase equilibrium is vital when learning about the behavior of substances in different states, such as gases, liquids, and solids. Phase equilibrium refers to the balance between different phases of matter in a system, which occurs when the processes of phase change are happening at equal rates.

For example, consider a closed container with water and its vapor. At equilibrium, the rate of water evaporating (liquid to vapor) is exactly equal to the rate of vapor condensing (vapor to liquid). This concept is crucial in the exercise, as it helps to determine the solubility of a gas within a solid. When rubber is placed in a nitrogen gas chamber, it's assumed to reach phase equilibrium, meaning the amount of nitrogen dissolving into the rubber equals the amount of nitrogen leaving the rubber.

Environmental factors like temperature and pressure greatly influence phase equilibrium. The exercise specifies a temperature of 25°C, and the pressure is given as 300 kPa. This information is critical because the solubility of gases often increases with higher pressure and decreases with higher temperature.

Gas Solubility in Solids

Gas solubility in solids is an intriguing phenomenon with practical applications in industries ranging from food packaging to tire manufacturing. For gases to dissolve in solids, molecular interactions must occur between the gas and the solid material. These interactions are governed by factors like pressure and temperature, which is why the solubility of nitrogen in rubber at a specific temperature and pressure was emphasized in the exercise.

In the given scenario, the system reached phase equilibrium at 300 kPa, and we needed to calculate the density of nitrogen in rubber. The solubility of the nitrogen gas in rubber is given in terms of kilometers per cubic meter per bar. This represents how much nitrogen dissolves in rubber for each unit of pressure. Because the pressure in the system was increased, the solubility of nitrogen in the rubber is expected to be higher compared to a lower pressure. This is an example of Henry's Law, which states that the solubility of a gas in a liquid or solid is directly proportional to its partial pressure above the material.

Molar Conversions

Molar conversions are essential for chemists and engineers as they provide a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. In the exercise, we started with a solubility given in kilometers per cubic meter per bar, and through a series of conversions, we determined the density in kilograms per cubic meter.

Here's a summarized process of the molar conversions used:

• Convert pressure from kPa to bar to use with the given solubility.
• Use the solubility coefficient with the converted pressure to find the amount of nitrogen in kilometers per cubic meter.
• Convert from kilometers to moles, since there are 1000 moles in a kilomole.
• Finally, convert moles to mass using the molar mass of nitrogen (approximately 28 g/mol). Since the question requires the density in kg/m³, we also convert grams to kilograms.

This sequence of conversions is crucial in solving problems involving the solubility of gases in solids since it allows us to compute the actual mass of a gas that dissolves in a certain volume of solid at a specific pressure and temperature.