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Chapter 14: Problem 11

At a given temperature and pressure, do you think the mass diffusivity of air in water vapor will be equal to the mass diffusivity of water vapor in air? Explain.

Short Answer

Expert verified
Answer: No, the mass diffusivity of air in water vapor will not be equal to the mass diffusivity of water vapor in air, even when the temperature and pressure are constant, due to the difference in molecular weights of the two substances.

Step by step solution

01

Understanding Mass Diffusivity

Mass diffusivity (D) is a measure of how quickly one substance can diffuse through another. It depends on factors such as temperature, pressure, and the properties of the substances involved. In this exercise, we are dealing with air and water vapor, so we will focus on those two substances.
02

Factors Affecting Mass Diffusivity

Some factors that affect mass diffusivity are temperature, pressure, and molecular weight of the substances involved. In this case, we have two scenarios: 1. The mass diffusivity of air in water vapor. 2. The mass diffusivity of water vapor in air. Even though temperature and pressure are given to be constant, the molecular weights of air and water vapor are different.
03

Molecular Weight and Mass Diffusivity

The molecular weight of a substance, denoted as M, plays a role in determining the mass diffusivity. The larger the difference in molecular weights between the two substances, the lower the mass diffusivity. In this case, the molecular weights of air and water vapor are not equal, with air having a molecular weight of approximately 29 g/mol and water vapor having a molecular weight of approximately 18 g/mol.
04

Conclusion

Since the molecular weights of air and water vapor are not equal, the mass diffusivity of air in water vapor will not be equal to the mass diffusivity of water vapor in air, even when the temperature and pressure are constant. This is because the molecular weights of the substances have a direct impact on the mass diffusivity, regardless of the temperature and pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Weight
Molecular weight is an important factor when considering how substances interact with each other, particularly in diffusion processes. It is essentially the mass of one molecule of a substance, usually measured in grams per mole (g/mol). The molecular weight can influence how quickly a molecule will move, or diffuse, when mixed with another substance.
In our context, the difference in molecular weights between air and water vapor significantly affects their mass diffusivity - the rate at which molecules spread from areas of high concentration to areas of lower concentration.
  • The molecular weight of air is approximately 29 g/mol.
  • The molecular weight of water vapor is roughly 18 g/mol.
These differing values mean that the heavier molecules of air will diffuse through water vapor at a different rate compared to the lighter water vapor diffusing through air.
Diffusion in Gases
Diffusion in gases is a process where gas molecules spread out and mix with another gas. This mixing continues until a uniform distribution is achieved. The speed and efficiency of this process depend largely on factors like the molecular weight of the gases involved. In general, lighter gas molecules will diffuse more quickly than heavier ones.
In our case of air and water vapor, heterogeneity in molecular weight means each will diffuse at different rates. This is why, under identical external conditions, the mass diffusivity of air in water vapor cannot match that of water vapor in air.
The principle is similar to pouring milk into coffee; both liquids will eventually mix, but the process happens faster depending on certain properties, like temperature and density. For gases, it's molecular weight that plays a big role in how quickly they mix.
Temperature and Pressure Effects
While this exercise specifies that temperature and pressure are constant, it's crucial to understand how these factors normally affect diffusion.
  • Temperature: Higher temperatures generally increase diffusivity because molecules move faster, enhancing their ability to spread out quickly.
  • Pressure: At higher pressure, more molecules are present in a given space, which can influence how frequently molecules collide and thereby affect diffusion.
However, even at constant temperature and pressure, the intrinsic properties of the molecules, such as their molecular weight, remain a dominant factor in determining mass diffusivity. Hence, the mass diffusivity difference between air and water vapor persists, irrespective of these constant external conditions.

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Most popular questions from this chapter

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.

Benzene \((M=78.11 \mathrm{~kg} / \mathrm{kmol})\) is a carcinogen, and exposure to benzene increases the risk of cancer and other illnesses in humans. A truck transporting liquid benzene was involved in an accident that spilled the liquid on a flat highway. The liquid benzene forms a pool of approximately \(10 \mathrm{~m}\) in diameter on the highway. In this particular windy day at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with an average wind velocity of \(10 \mathrm{~m} / \mathrm{s}\), the liquid benzene surface is experiencing mass transfer to air by convection. Nearby at the downstream of the wind is a residential area that could be affected by the benzene vapor. Local health officials have assessed that if the benzene level in the air reaches \(500 \mathrm{~kg}\) within the hour of the spillage, residents should be evacuated from the area. If the benzene vapor pressure is \(10 \mathrm{kPa}\), estimate the mass transfer rate of benzene being convected to the air, and determine whether the residents should be evacuated or not.

Consider a carbonated drink in a bottle at \(37^{\circ} \mathrm{C}\) and \(130 \mathrm{kPa}\). Assuming the gas space above the liquid consists of a saturated mixture of \(\mathrm{CO}_{2}\) and water vapor and treating the drink as water, determine \((a)\) the mole fraction of the water vapor in the \(\mathrm{CO}_{2}\) gas and \((b)\) the mass of dissolved \(\mathrm{CO}_{2}\) in a 200-ml drink.

The composition of moist air is given on a molar basis to be 78 percent \(\mathrm{N}_{2}, 20\) percent \(\mathrm{O}_{2}\), and 2 percent water vapor. Determine the mass fractions of the constituents of air.

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) \(5.2 \mathrm{~kg} / \mathrm{h}\)
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