Exposure to high concentration of gaseous ammonia can cause lung damage. The acceptable shortterm ammonia exposure level set by the Occupational Safety and Health Administration (OSHA) is 35 ppm for 15 minutes. Consider a vessel filled with gaseous ammonia at \(30 \mathrm{~mol} / \mathrm{L}\), and a 10 -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.

Step by step solution

01

Calculate the area of the plastic plug

We'll first find the area of the circular plastic plug: \(A = \pi r^2\) where A is the area, and r is the radius of the circle. Convert the diameter to meters and divide by 2 to get the radius: \( r = 10 \mathrm{\,cm} \cdot(1 \mathrm{~m}/ 100 \mathrm{\,cm}) \cdot (1/2) = 0.05 \mathrm{~m}\) Now, calculate the area: \(A = \pi (0.05 \mathrm{~m})^2 = 0.00785 \mathrm{~m}^2\)

02

Calculate the concentration gradient

The concentration gradient is the difference in concentration between the inside and the outside of the vessel divided by the thickness of the plug: \(\Delta C = \frac{C_{inside} - C_{outside}}{d}\) Since the outside concentration is negligible, we can assume it is 0 mol/L. Convert the thickness of the plug to meters: \(d = 2 \mathrm{\,mm} \cdot (1 \mathrm{~m} / 1000 \mathrm{\,mm}) = 0.002 \mathrm{~m}\) Now, calculate the concentration gradient: \(\Delta C = \frac{30 \mathrm{~mol} / \mathrm{L}}{0.002 \mathrm{~m}}\) Convert mol/L to mol/m³: \(30 \mathrm{~mol} / \mathrm{L} \cdot (1000 \mathrm{~L} / \mathrm{m}^3) = 30000 \mathrm{~mol} / \mathrm{m}^3\) Calculate the concentration gradient: \(\Delta C = \frac{30000 \mathrm{~mol} / \mathrm{m}^3}{0.002 \mathrm{~m}} = 1.5 \times 10^{7} \mathrm{~mol} / \mathrm{m}^4\)

03

Apply Fick's first law to calculate the diffusion rate

Fick's first law for diffusion is given by the equation: \(J = -D \cdot A \cdot \Delta C\) where J is the diffusion rate (in mol/s), D is the diffusion coefficient (1.3 x 10⁻¹⁰ m²/s), A is the area of the plug (0.00785 m²), and ΔC is the concentration gradient (1.5 x 10⁷ mol/m⁴). Now, we will calculate the diffusion rate: \(J = -(1.3 \times 10^{-10} \mathrm{~m}^2 / \mathrm{s})(0.00785 \mathrm{~m}^2)(1.5 \times 10^{7} \mathrm{~mol} / \mathrm{m}^4) = -1.535 \times 10^{-5} \mathrm{~ mol/s}\) Since we want only the magnitude of the diffusion rate: \( J = 1.535 \times 10^{-5} \mathrm{~ mol/s}\)

04

Convert the diffusion rate to mg/s

To determine whether the ventilation system will be sufficient, we need to convert the ammonia diffusion rate from mol/s to mg/s. The molar mass of ammonia (NH3) is approximately 17 g/mol. Convert the diffusion rate to mg/s: \(J = (1.535 \times 10^{-5} \mathrm{~ mol/s}) \cdot (17 \mathrm{~g} / \mathrm{mol}) \cdot (1000 \mathrm{~mg} / \mathrm{g}) = 0.26095 \mathrm{~ mg / s}\)

05

Compare the calculated rate with the acceptable exposure rate

Now, we compare our calculated diffusion rate (0.26095 mg/s) to the acceptable exposure rate given by OSHA and the ventilation system (0.2 mg/s). Since 0.26095 mg/s > 0.2 mg/s, the plastic plug will not safely contain the ammonia inside the vessel according to the given standard.

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