Question

A wall made of natural rubber separates \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) gases at \(25^{\circ} \mathrm{C}\) and \(750 \mathrm{kPa}\). Determine the molar concentrations of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) in the wall.

Short answer

Question: Determine the molar concentrations of O2 and N2 in the wall made of natural rubber that separates them, given the pressure of 750 kPa and temperature of 25°C. Answer: The molar concentrations of both O2 and N2 in the wall made of natural rubber are 30.19 mol/m³.

Step-by-step solution

Convert temperature to Kelvin

We are given the temperature as 25°C. We can convert this to Kelvin by adding 273.15: T = 25°C + 273.15 = 298.15 K

Write down the given values

We are given the pressure, P = 750 kPa, which we can convert to Pa by multiplying it by 1000: P = 750 kPa * 1000 = 750000 Pa The Ideal Gas Constant, R = 8.314 J/(mol·K) Now we have all the necessary values to solve for the molar concentrations.

Apply the Ideal Gas Law

As both gases are separated and we assume that the total volume of the wall remains constant, we can apply the Ideal Gas Law separately for O2 and N2. The concentration (C) of a gas can be defined as the ratio of the number of moles (n) to the volume (V). C = n/V Applying the Ideal Gas Law for O2, and rearranging for n/V, we get: C_O2 = P/(R*T) Similarly, applying the Ideal Gas Law for N2: C_N2 = P/(R*T)

Solve for the molar concentrations

Now, we can plug in the values for the molar concentrations of O2 and N2: C_O2 = 750000 Pa / (8.314 J/(mol·K) * 298.15 K) = 30.19 mol/m³ C_N2 = 750000 Pa / (8.314 J/(mol·K) * 298.15 K) = 30.19 mol/m³ Both gases have the same molar concentration of 30.19 mol/m³ in the wall made of natural rubber.