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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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Key Concepts

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

Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. It's a simple yet powerful concept in thermodynamics. The mole fraction is the ratio of the number of moles of a particular component to the total number of moles in the mixture. This can be expressed with the formula:
  • For a component "A" in a mixture, the mole fraction (\(X_A\) ) is given by: \(X_A = \frac{n_A}{n_{total}}\)
where \(n_A\) is the moles of component "A" and \(n_{total}\) is the total moles in the mixture.

In the case of the exercise, we find two mole fractions:
  • The mole fraction of water vapor in the air above the lake.
  • The mole fraction of water in the lake itself.
The mole fraction of water vapor is considerably lower because the air around the lake contains not only water vapor but also other gases like nitrogen and oxygen, while the water in the lake is pure and hence has a mole fraction of 1.
Saturation Vapor Pressure
Saturation vapor pressure is an important concept in understanding how vapor behaves in equilibrium with its liquid or solid phase. It represents the pressure exerted by a vapor in balance with its liquid or solid form at a specific temperature.

For water, as the temperature rises, the saturation vapor pressure also increases. This is because more water molecules have enough energy to escape from the liquid phase into the vapor phase. When a liquid is in a closed system and equilibrated at a certain temperature, the saturation vapor pressure is reached.

In the exercise, the saturation vapor pressure of water at 15°C is calculated to be approximately 1.705 kPa. This informs us about the tendency of water to evaporate at this temperature.
Antoine's Equation
Antoine's Equation is a useful formula for calculating the saturation vapor pressure of a substance as a function of temperature. It is an empirical relation expressed in the form:
  • \(P_s = 10^{\frac{A - B}{C + T}}\)
in which \(P_s\) represents the saturation vapor pressure in kPa, \(T\) is the temperature in °C, and \(A\), \(B\), and \(C\) are empirical constants specific to each substance.

In this exercise, the constants for water are \(A = 7.96681\), \(B = 1668.21\), and \(C = 228.0\). By substituting these into the equation along with the temperature, we can find out the saturation vapor pressure, which is essential for determining the mole fraction of water vapor in the air.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the weight of the atmosphere above a point on the Earth's surface. It plays a significant role in various thermodynamic calculations, including the determination of the mole fraction of gases in a mixture.

At the lake level in the given exercise, the atmospheric pressure is taken to be 92 kPa. This value is used to calculate the mole fraction of water vapor in the air using the formula:
  • \(Y_{H_2O} = \frac{P_s}{P}\)
where \(P_s\) is the saturation vapor pressure and \(P\) is the atmospheric pressure. This pressure is vital as it determines how much water vapor the air can potentially hold, which is why it's lower at high altitudes and varying weather conditions.

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Most popular questions from this chapter

In an experiment, a sphere of crystalline sodium chloride \((\mathrm{NaCl})\) was suspended in a stirred tank filled with water at \(20^{\circ} \mathrm{C}\). Its initial mass was \(100 \mathrm{~g}\). In 10 minutes, the mass of sphere was found to have decreased by 10 percent. The density of \(\mathrm{NaCl}\) is \(2160 \mathrm{~kg} / \mathrm{m}^{3}\). Its solubility in water at \(20^{\circ} \mathrm{C}\) is \(320 \mathrm{~kg} / \mathrm{m}^{3}\). Use these results to obatin an average value for the mass transfer coefficient.

Consider a brick house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(85 \mathrm{kPa}\). The walls of the house are made of 20 -cm-thick brick whose permeance is \(23 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 5-\mathrm{m}\) section of a wall during a 24-h period.

Show that for an ideal gas mixture maintained at a constant temperature and pressure, the molar concentration \(C\) of the mixture remains constant but this is not necessarily the case for the density \(\rho\) of the mixture.

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: \(0.0011 \mathrm{~mm}\)

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 \(\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \mathrm{Pr}^{0.4}\) where \(\mathrm{Re}=10,550\) and \(\operatorname{Pr}=0.731\). 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) \(0.0022 \mathrm{~m} / \mathrm{s}\)

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