Question

Exposure to high concentrations of gaseous short-term ammonia exposure level set by the Occupational Safety and Health Administration (OSHA) is $35 \mathrm{ppm}\( for \)15 \mathrm{~min}$. Consider a vessel filled with gaseous ammonia at \(30 \mathrm{~mol} / \mathrm{L}\), and a \(10-\mathrm{cm}\)-diameter circular plastic plug with a thickness of \(2 \mathrm{~mm}\) is used to contain the ammonia inside the vessel. The ventilation system is capable of keeping the room safe with fresh air, provided that the rate of ammonia being released is below \(0.2 \mathrm{mg} / \mathrm{s}\). If the diffusion coefficient of ammonia through the plug is $1.3 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}$, determine whether or not the plug can safely contain the ammonia inside the vessel.

Short answer

Answer: No, the plastic plug cannot safely contain ammonia gas inside the vessel, as the calculated leakage rate (0.52 mg/s) is higher than the safe rate as per OSHA guidelines (0.2 mg/s).

Step-by-step solution

Calculate the surface area of the plug

First, we need to calculate the surface area of the circular plug. We will use the formula for the area of a circle: A = πr^2, where r is the radius of the circle (half of the diameter). The radius of the plug is 5 cm, therefore, the surface area A in square meters is: A = π × (5 × 10^{-2})^2 A ≈ 0.00785 m^2

Calculate the ammonia concentration difference

We are given the ammonia concentration inside the vessel, which is C_i = 30mol/L. We can assume that the concentration outside the vessel is zero, C_o = 0mol/L.

Use the diffusion equation to calculate the rate of ammonia leakage

Using Fick's first law of diffusion, we have: J = -D × (C_i - C_o) / d where J is the molar flux (mol/m^2s), D is the diffusion coefficient (1.3 x 10^{-10} m^2/s), C_i and C_o are the inside and outside concentrations (30 mol/L and 0 mol/L, respectively), and d is the thickness of the plug (2 mm). J = - [(1.3 × 10^{-10}) × (30 × 10^3 - 0)] / (2 × 10^{-3}) J ≈ 1.95 × 10^{-6} mol/m^2s

Calculate the total leakage rate

Now, we need to multiply the molar flux by the surface area of the plug and the molar mass of ammonia to obtain the mass flux of ammonia (mg/s). Leakage rate = J × A × M Where M is the molar mass of ammonia (17 g/mol, but we will use 17,000 mg/mol for our calculation in milligrams). Leakage rate ≈ (1.95 × 10^{-6})(0.00785)(17000) Leakage rate ≈ 0.52 mg/s

Determine if the plug can safely contain the ammonia

The calculated leakage rate (0.52 mg/s) is higher than the safe rate as per OSHA guidelines (0.2 mg/s). Therefore, the answer is no - the plug cannot safely contain the ammonia inside the vessel.