Question

A rubber object is in contact with nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(298 \mathrm{~K}\) and \(250 \mathrm{kPa}\). The solubility of nitrogen gas in rubber is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3}\).bar. The mass density of nitrogen at the interface is (a) \(0.049 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(0.064 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(0.077 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(0.092 \mathrm{~kg} / \mathrm{m}^{3}\) (e) \(0.109 \mathrm{~kg} / \mathrm{m}^{3}\)

Short answer

Answer: (e) 0.109 kg/m³

Step-by-step solution

Identify the given information

We are given the following information: - Temperature, T = 298 K - Pressure, P = 250 kPa - Solubility of nitrogen in rubber = 0.00156 kmol/m³.bar - Nitrogen gas molar mass, M = 28.02 g/mol (approximately)

Convert pressure to bars

The pressure is given in kPa, but solubility is given in kmol/m³.bar. We need to convert the pressure to bars. 1 bar = 100 kPa So, 250 kPa = 250/100 = 2.5 bars

Calculate the solubility at the given pressure

We need to find the solubility of nitrogen in the rubber at the given pressure. Use the given solubility formula and the pressure in bars: Solubility = (0.00156 kmol/m³.bar) * (2.5 bars) = 0.0039 kmol/m³

Convert solubility to mass density

Now, convert the solubility from kmol/m³ to kg/m³ using the molar mass of nitrogen gas: Mass density = (0.0039 kmol/m³) * (28.02 g/mol) * (1 kg/1000 g) = 0.109 kg/m³.

Match the result to the options

Compare the calculated mass density to the given options: (a) 0.049 kg/m³ (b) 0.064 kg/m³ (c) 0.077 kg/m³ (d) 0.092 kg/m³ (e) 0.109 kg/m³ The correct option is (e) 0.109 kg/m³.

Key concepts

Solubility

The concept of solubility refers to the ability of a substance, often a gas in this context, to dissolve in a solvent. Here we consider nitrogen gas dissolving in rubber. Solubility is expressed in units that relate concentration to pressure, specifically \( ext{kmol/m}^3 ext{.bar}\).
The given solubility value indicates the amount of nitrogen that can dissolve in a cubic meter of rubber per bar of pressure.
This value, \0.00156 \text{kmol/m}^3 ext{.bar}\, tells us how much nitrogen can be held in the rubber if subjected to conditions that exert a pressure of 1 bar on the nitrogen gas.

• Increased pressure generally increases solubility, as more nitrogen molecules are pushed into the rubber.


• The temperature, which in this problem is kept constant at 298 K, can also influence solubility to change in some contexts.

Understanding solubility is crucial for predicting how gases like nitrogen will behave in materials, impacting processes such as gas storage or manufacturing materials with specific gas-holding capacities.

Gas-Liquid Interface

The gas-liquid interface is the boundary between the gas and the liquid, or similar boundaries in solid solutions. In this exercise, the focus is on the interaction between nitrogen gas and the rubber material.
This interface is crucial because it's where the solubility dynamic occurs; gases dissolve into the solids like rubber from this interface.

At this boundary, nitrogen molecules collide and interact with the rubber, governed by principles like diffusion and solubility.
Factors affecting the gas-liquid interface include:

• Pressure: Higher pressures increase the number of molecules colliding at the interface, enhancing dissolvability.


• Temperature: Impacts the energy and mobility of gas molecules at the interface.
• Nature of material: Some materials are inherently more absorptive than others for specific gases.

The gas-liquid interface is vital in calculating how much nitrogen is absorbed since it represents the point where nitrogen molecules transition from the gas phase into the rubber, leveraging the solubility parameter.

Molar Mass Conversion

Molar mass conversion is crucial when translating between chemical quantities like moles and physical quantities like mass. For nitrogen, given its molar mass \(M = 28.02 \text{g/mol}\), this means we need to convert the amount of substance represented in moles to kilograms to find properties related to mass density.
Molar mass helps convert solubility (in terms of kmol) into a mass concentration (kg/m³), which is what's needed to compare to the options provided.
Here's the conversion process:

• Calculate moles from solubility: \(0.0039 \text{kmol/m}^3\).


• Convert these moles to grams using nitrogen's molar mass: \(0.0039 \text{kmol/m}^3 \times 28.02 \text{g/mol}\).


• Transform grams into kilograms by dividing by 1000.

The conversion allows us to express the gas concentration in terms we can measure (mass over volume) and match with options like 0.109 kg/m³. Understanding molar mass conversions helps tackle various chemistry problems where substance amounts need translating between different units for practicality or analysis.