Question

Consider a rubber plate that is in contact with nitrogen gas at \(298 \mathrm{K}\) and \(250 \mathrm{kPa}\). Determine the molar and mass density of nitrogen in the rubber at the interface.

Short answer

Answer: The molar density of nitrogen in the rubber at the interface is 100.891 mol/m³, and the mass density is 2.825 kg/m³.

Step-by-step solution

Ideal Gas Law Equation

Write down the ideal gas law equation and identify the known values: \(PV = nRT\) where P = 250 kPa, T = 298 K, R = 8.314 J/(mol K)

Convert Pressure to Pascals

Convert the given pressure from kPa to Pascals (Pa) as follow: 1 kPa = 1000 Pa 250 kPa = 250 * 1000 Pa = 250,000 Pa Now, P = 250,000 Pa

Calculate the Number of Moles

Solve for the number of moles, n: \(PV = nRT \Rightarrow n = \frac{PV}{RT}\) \(n = \frac{(250,000 \ \text{Pa}) (V)}{(8.314 \ \text{J/(mol K)}) (298 \ \text{K})}\) \(n = \frac{250,000V}{2477.972} = 100.891V\) moles

Calculate Molar Density

Divide the number of moles, n, by the volume, V, to find the molar density: Molar Density = \(\frac{n}{V} = \frac{100.891V}{V} = 100.891 \ \text{mol/m}^3\)

Calculate Mass Density

Now, we'll calculate the mass density. Use the molar mass of nitrogen, M_N2, which is 28.02 g/mol: Mass = n * M_N2 Converting the molar mass to kg/mol: 28.02 g/mol = 0.02802 kg/mol Mass = 100.891V * 0.02802 kg/mol = 2.825 V kg Now, divide the mass by the volume to get the mass density: Mass Density = \(\frac{\text{Mass}}{V} = \frac{2.825V}{V} = 2.825 \ \text{kg/m}^3\) The molar density of nitrogen in the rubber at the interface is 100.891 mol/m³, and the mass density is 2.825 kg/m³.