Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 14: Problem 74

Consider a tank that contains moist air at 3 atm and whose walls are permeable to water vapor. The surrounding air at \(1 \mathrm{~atm}\) pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain.

Short Answer

Expert verified
Answer: No, water vapor cannot flow into the tank from the surroundings because the partial pressure of water vapor in the tank is greater than the partial pressure of water vapor in the surroundings. Instead, water vapor will flow from the tank towards the surrounding air.

Step by step solution

01

Understand the concept of partial pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases, like the moist air in the tank. It depends on the mole fraction of that gas and the total pressure. For our problem, we need to see which side has a higher water vapor partial pressure.
02

Calculate the partial pressure of water vapor for both the tank and surroundings

Let \(P_{H_2O-tank}\) be the partial pressure of water vapor in the tank and \(P_{H_2O-surroundings}\) be the partial pressure of water vapor in the surroundings. We can't calculate these values directly with the given information, but we can compare both partial pressures. If \(P_{H_2O-tank} > P_{H_2O-surroundings}\), the water vapor will flow from the tank to the surroundings. If \(P_{H_2O-tank} < P_{H_2O-surroundings}\), the water vapor will flow from the surroundings into the tank. If \(P_{H_2O-tank} = P_{H_2O-surroundings}\), there is no net flow of water vapor.
03

Determine the relation between the partial pressures

We know that the total pressure inside the tank is 3 atm, and the total surrounding pressure is 1 atm. We can assume that both the tank and surroundings are at the same temperature. The partial pressure of water vapor in the tank (being a part of the total pressure) will also be higher than the partial pressure of water vapor in the surroundings. Therefore, the relationship becomes: \(P_{H_2O-tank} > P_{H_2O-surroundings}\)
04

Conclusion

Based on the relationship \(P_{H_2O-tank} > P_{H_2O-surroundings}\), we can conclude that it is not possible for the water vapor to flow into the tank from the surroundings. Instead, water vapor will flow from the tank towards the surrounding air.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was $1800 \mathrm{~m}^{2}\(. The skin of this balloon is \)2 \mathrm{~mm}$ thick and is made of a material whose helium diffusion coefficient is $1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}$. The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

The top section of an 8-ft-deep, 100-ft \(\times 100-\mathrm{ft}\) heated solar pond is maintained at a constant temperature of \(80^{\circ} \mathrm{F}\) at a location where the atmospheric pressure is \(1 \mathrm{~atm}\). If the ambient air is at \(70^{\circ} \mathrm{F}\) and 100 percent relative humidity and wind is blowing at an average velocity of \(40 \mathrm{mph}\), determine the rate of heat loss from the top surface of the pond by \((a)\) forced convection, \((b)\) radiation, and (c) evaporation. Take the average temperature of the surrounding surfaces to be \(60^{\circ} \mathrm{F}\).

Reconsider Prob. 14-82. In order to reduce the migration of water vapor through the wall, it is proposed to use a \(0.051-\mathrm{mm}\)-thick polyethylene film with a permeance of $9.1 \times 10^{-12} \mathrm{~kg} / \mathrm{s}^{2} \mathrm{~m}^{2}$.Pa. Determine the amount of water vapor that will diffuse through the wall in this case during a \(24-\mathrm{h}\) period. Answer: \(26.4 \mathrm{~g}\)

Air flows in a 4-cm-diameter wet pipe at \(20^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}\( with an average velocity of \)4 \mathrm{~m} / \mathrm{s}$ in order to dry the surface. The Nusselt number in this case can be determined from \(\mathrm{Nu}=0.023 \operatorname{Re}^{0.8} \mathrm{Pr}^{0.4}\) where \(\operatorname{Re}=10,550\) and \(\mathrm{Pr}=\) \(0.731\). Also, the diffusion coefficient of water vapor in air is $2.42 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$. Using the analogy between heat and mass transfer, the mass transfer coefficient inside the pipe for fully developed flow becomes (a) \(0.0918 \mathrm{~m} / \mathrm{s}\) (b) \(0.0408 \mathrm{~m} / \mathrm{s}\) (c) \(0.0366 \mathrm{~m} / \mathrm{s}\) (d) \(0.0203 \mathrm{~m} / \mathrm{s}\) (e) \(0.0022 \mathrm{~m} / \mathrm{s}\)

Heat convection is expressed by Newton's law of cooling as $\dot{Q}=h A_{s}\left(T_{s}-T_{\infty}\right)$. Express mass convection in an analogous manner on a mass basis, and identify all the quantities in the expression and state their units.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Engineering Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Thermodynamics

Read Explanation

Dynamics

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free