Question

Saturated water vapor at $25^{\circ} \mathrm{C}\left(P_{\text {stt }}=3.17 \mathrm{kPa}\right)\( flows in a pipe that passes through air at \)25^{\circ} \mathrm{C}$ with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a \(9-\mathrm{mm}\) internal-diameter tube that extends $10 \mathrm{~m}$ into the air. The diffusion coefficient of vapor through air is \(2.5 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The amount of water vapor lost to the atmosphere through this individual tube by diffusion is (a) \(1.7 \times 10^{-6} \mathrm{~kg}\) (b) \(2.3 \times 10^{-6} \mathrm{~kg}\) (c) \(3.8 \times 10^{-6} \mathrm{~kg}\) (d) \(5.0 \times 10^{-6} \mathrm{~kg}\) (e) \(7.1 \times 10^{-6} \mathrm{~kg}\)

Short answer

Answer: \(3.8 \times 10^{-6} \mathrm{~kg}\)

Step-by-step solution

Calculate the partial pressure of the water in the air

To calculate the partial pressure of water in the air, use the relative humidity value given. Since relative humidity is the ratio of the actual vapor pressure to the saturation vapor pressure, we can calculate the actual vapor pressure \(P_{v}\) as follows: \(P_{v}=Relative\ Humidity\% \times P_{stt} \) \(P_{v}=0.4 \times 3.17\ \mathrm{kPa} \) \(P_{v}=1.268\ \mathrm{kPa} \)

Calculate the molar concentration of water vapor at saturation and in the air

Using the Ideal Gas Law, we can calculate the molar concentration of water vapor \(n_{i}\): \(n_{i}=\frac{P_{i}}{R \times T}\) For saturated water vapor \(n_{stt}\): \(n_{stt}=\frac{3.17\ \mathrm{kPa}}{8.314\ \mathrm{\frac{J}{mol\ K}} \times (25+273.15)\ \mathrm{K}}\) \(n_{stt}=1.474 \times 10^{-2}\ \mathrm{\frac{mol}{m^3}}\) For water vapor in the air \(n_{air}\): \(n_{air}=\frac{1.268\ \mathrm{kPa}}{8.314\ \mathrm{\frac{J}{mol\ K}} \times (25+273.15)\ \mathrm{K}}\) \(n_{air}=5.888 \times 10^{-3}\ \mathrm{\frac{mol}{m^3}}\)

Calculate the molar flux of water vapor

Now, we use Fick's First Law of Diffusion to calculate the molar flux \((J)\). The molar flux is the rate of flow of water vapor per unit area, which depends on the diffusion coefficient \((D)\), the concentration difference, and the length of the tube: \(J = -D \times \frac{n_{stt}-n_{air}}{L}\) \(J = -(2.5 \times 10^{-5}\ \mathrm{\frac{m^2}{s}}) \times \frac{1.474 \times 10^{-2}\ \mathrm{\frac{mol}{m^3}} - 5.888 \times 10^{-3}\ \mathrm{\frac{mol}{m^3}}}{10\ \mathrm{m}}\) \(J = -6.42 \times 10^{-7}\ \mathrm{\frac{mol}{m^2\ s}}\)

Calculate the mass of water vapor lost

To find the mass of water vapor lost, we need to consider the molar flux, molecular weight (\(M_{w}\)) of water, and the tube's cross-sectional area: \(Mass = J × M_{w} × Area × Time\) \(Mass = (-6.42 \times 10^{-7}\ \mathrm{\frac{mol}{m^2\ s}}) × (18.015\ \mathrm{\frac{g}{mol}}) × \frac{\pi (9 \times 10^{-3}\ \mathrm{m})^2}{4} × (1\ \mathrm{s})\) \(Mass = 3.8 \times 10^{-6}\ \mathrm{kg}\) So, the correct answer is (c) \(3.8 \times 10^{-6} \mathrm{~kg}\).