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

Convert pressure to bar

To convert the given pressure (in kPa) to bar, we can follow the conversion factor: 1 bar = 100 kPa So, the pressure in bar will be: \(\text{Pressure} (bar) = \frac{250 \mathrm{kPa}}{100 \mathrm{kPa/bar}} = 2.5 \mathrm{bar}\)

Calculate mole density of nitrogen

Using the given solubility of nitrogen in rubber, we can calculate the mole density: \(\text{Mole density}\ (\mathrm{kmol}/\mathrm{m}^3) = \text{Solubility} \times \text{Pressure}\) \(\text{Mole density}\ (\mathrm{kmol}/\mathrm{m}^3) = 0.00156\ \mathrm{kmol}/\mathrm{m}^3.\text{bar} \times 2.5\ \text{bar} = 0.0039\ \mathrm{kmol}/\mathrm{m}^3\)

Calculate mass density of nitrogen

The molar mass of nitrogen gas \((N_2)\) is 28.02 g/mol. To find the mass density, we can multiply the mole density by the molar mass, then convert it to kg by dividing by 1000: \(\text{Mass density}\ (\mathrm{kg}/\mathrm{m}^3) = \text{Mole density}\ (\mathrm{kmol}/\mathrm{m}^3) \times \text{Molar mass}\ (\mathrm{g}/\mathrm{mol}) \times \frac{1\ \mathrm{kg}}{1000\ \mathrm{g}}\) \(\text{Mass density}\ (\mathrm{kg}/\mathrm{m}^3) = 0.0039\ \mathrm{kmol}/\mathrm{m}^3 \times 28.02\ \mathrm{g}/\mathrm{mol} \times \frac{1\ \mathrm{kg}}{1000\ \mathrm{g}} = 0.109 \ \mathrm{kg}/\mathrm{m}^3\)

Select the correct answer

Comparing the mass density value (0.109 \(\mathrm{kg}/\mathrm{m}^3\)) with the provided options, we find that: The mass density of nitrogen at the interface is (e) 0.109 \(\mathrm{kg}/\mathrm{m}^3\)