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Chapter 14: Problem 156

Determine the mole fraction of the water vapor at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\) and compare it to the mole fraction of water in the lake. Take the atmospheric pressure at lake level to be \(92 \mathrm{kPa}\).

Short Answer

Expert verified
Answer: The mole fraction of water vapor in the air above the lake is approximately 0.0185.

Step by step solution

01

Calculate the saturation vapor pressure

At first, we need to determine the saturation vapor pressure of water at the given temperature, \(15^{\circ} \mathrm{C}\). We can use Antoine's equation for this purpose: \(P_s = 10^{\frac{A - B}{C + T}}\) Where \(P_s\) is the saturation vapor pressure in \(\mathrm{kPa}\), \(T\) is the temperature in \(^{\circ} \mathrm{C}\), and \(A\), \(B\), and \(C\) are constants. For water, we have the constants \(A = 7.96681\), \(B = 1668.21\), and \(C = 228.0\). Now, we can plug in the temperature and calculate the saturation pressure. \(P_s = 10^{\frac{7.96681 - 1668.21}{228.0 + 15}}\)
02

Calculate the saturation vapor pressure

After plugging in the values, we have: \(P_s = 10^{(7.96681 - 1668.21) / (228.0 + 15)} \approx 1.705\, \mathrm{kPa}\) So, the saturation vapor pressure of water at \(15^{\circ} \mathrm{C}\) is approximately \(1.705\, \mathrm{kPa}\).
03

Calculate the mole fraction of water vapor in the air

Now, we will use the calculated saturation vapor pressure to find the mole fraction of water vapor in the air (\(Y_{H_2O}\)). This can be done by using the following equation: \(Y_{H_2O} = \frac{P_s}{P}\) Where \(P\) is the atmospheric pressure at the lake level (\(92\, \mathrm{kPa}\)). Plugging in the values, we get: \(Y_{H_2O} = \frac{1.705}{92} \approx 0.0185\) Therefore, the mole fraction of water vapor in the air above the lake is approximately \(0.0185\).
04

Calculate the mole fraction of water in the lake

In the lake, assuming that it is purely water, the mole fraction of water is \(1\). As water in the lake represents the pure substance without any other moles of components, the mole fraction will be \(1\).
05

Compare the mole fractions

Now, we compare the mole fraction of water vapor in the air with the mole fraction of water in the lake: Mole fraction of water vapor in the air: \(0.0185\) Mole fraction of water in the lake: \(1\) The mole fraction of water vapor in the air is significantly lower than the mole fraction of water in the lake. This is expected as the atmosphere contains air, which is mainly nitrogen and oxygen, making the mole fraction of water vapor lower than the mole fraction of water in the lake.

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Most popular questions from this chapter

A glass of milk left on top of a counter in the kitchen at \(15^{\circ} \mathrm{C}, 88 \mathrm{kPa}\), and 50 percent relative humidity is tightly sealed by a sheet of \(0.009-\mathrm{mm}\)-thick aluminum foil whose permeance is \(2.9 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). The inner diameter of the glass is \(12 \mathrm{~cm}\). Assuming the air in the glass to be saturated at all times, determine how much the level of the milk in the glass will recede in \(12 \mathrm{~h}\). Answer: \(0.0011 \mathrm{~mm}\)

What is the relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) known as? Under what conditions is it valid? What is the practical importance of it? \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering?

An 8-cm-internal-diameter, 30-cm-high pitcher halffilled with water is left in a dry room at \(15^{\circ} \mathrm{C}\) and \(87 \mathrm{kPa}\) with its top open. If the water is maintained at \(15^{\circ} \mathrm{C}\) at all times also, determine how long it will take for the water to evaporate completely.

Consider a shallow body of water. Is it possible for this water to freeze during a cold and dry night even when the ambient air and surrounding surface temperatures never drop to \(0^{\circ} \mathrm{C}\) ? Explain.

The average heat transfer coefficient for air flow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton-Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at \(1 \mathrm{~atm}\) at a free stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) over a body covered with a layer of naphthalene. The surface area of the body is \(0.75 \mathrm{~m}^{2}\), and it is observed that \(100 \mathrm{~g}\) of naphthalene has sublimated in \(45 \mathrm{~min}\). During the experiment, both the body and the air were kept at \(25^{\circ} \mathrm{C}\), at which the vapor pressure and mass diffusivity of naphthalene are \(11 \mathrm{~Pa}\) and \(D_{A B}=0.61 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry.
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