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Chapter 11: Problem 59

Referring to Figure 11.28 , describe all the phase changes that would occur in each of the following cases: (a) Water vapor originally at \(500 \mathrm{~Pa}\) and \(-0.5^{\circ} \mathrm{C}\) is slowly compressed at constant temperature until the final pressure is \(2000 \mathrm{kPa}\). (b) Water originally at \(100.0^{\circ} \mathrm{C}\) and \(50 \mathrm{kPa}\) is cooled at constant pressure until the temperature is \(-10^{\circ} \mathrm{C}\).

Short Answer

Expert verified
In case (a), water vapor initially at \(500 \: \mathrm{Pa}\) and \(-0.5^{\circ} \mathrm{C}\) undergoes a phase change from vapor to liquid as it is compressed at constant temperature until the final pressure is \(2000 \: \mathrm{kPa}\). In case (b), water initially at \(100.0^{\circ} \mathrm{C}\) and \(50 \: \mathrm{kPa}\) undergoes three phase changes as it is cooled at constant pressure to \(-10^{\circ} \mathrm{C}\): first, the liquid water turns into water vapor as it cools below the boiling point; then, the water vapor turns back into liquid water as it cools below the condensation point; finally, the liquid water solidifies into ice as the temperature goes below the freezing point.

Step by step solution

01

Identify the initial state of water vapor

In this case, water vapor starts at a pressure of \(500\) Pa and a temperature of \(-0.5^{\circ} \mathrm{C}\).
02

Determine the final state of the water vapor

The water vapor is compressed at a constant temperature, so the temperature remains the same, \(-0.5^{\circ} \mathrm{C}\), and the pressure increases to \(2000 \mathrm{kPa}\) or \(2 \times 10^6 \mathrm{Pa}\).
03

Describe the phase changes that occur in this case

As the water vapor is compressed, it undergoes a phase change from vapor to liquid. As the pressure increases to the final value of \(2 \times 10^6 \mathrm{Pa}\) at a constant temperature of \(-0.5^{\circ}\mathrm{C}\), the water vapor becomes liquid water. #Phase Changes in Case (b)#
04

Identify the initial state of water

In this case, water starts at a temperature of \(100^{\circ} \mathrm{C}\) and a pressure of \(50 \mathrm{kPa}\) or \(5 \times 10^4 \mathrm{Pa}\).
05

Determine the final state of the water

The water is cooled at a constant pressure, so the pressure remains the same, \(5 \times 10^4 \mathrm{Pa}\), and the temperature decreases to \(-10^{\circ}\mathrm{C}\).
06

Describe the phase changes that occur in this case

As the water cools at a constant pressure, the following phase changes occur: 1. The liquid water starts to boil and turn into water vapor as it cools below the boiling point at this pressure (\(100^{\circ} \mathrm{C}\)). 2. After all the liquid water becomes vapor, the temperature keeps decreasing, and the water vapor turns back into liquid water as it cools below the condensation point at this pressure. 3. Finally, the temperature reaches \(-10^{\circ}\mathrm{C}\), which is below the freezing point of water at this pressure, and the liquid water solidifies into ice.

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Key Concepts

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

Water Vapor
Water vapor is the gaseous phase of water, and it plays a critical role in the water cycle and various environmental processes. At -0.5掳C, the water vapor is a low-energy state compared to higher temperatures, meaning the molecules move relatively slowly.
  • The pressure given in the problem, 500 Pa at the start, is quite low, indicating a less dense vapor.
  • Compression at a constant temperature has a significant effect on vapor molecules, potentially leading to phase changes such as condensation.
Understanding water vapor behavior is crucial not only in physics but also in meteorology and engineering applications. It influences weather patterns and is essential for various industrial processes, such as in power plants and refrigeration systems.
Constant Temperature
A constant temperature implies that while other variables like pressure or volume might change, the temperature remains unchanged throughout the process. This is also known as isothermal conditions. Isothermal conditions are common assumptions when analyzing gas or vapor behavior.
  • In the problem, during compression, the temperature stays fixed at -0.5掳C, even though the pressure increases significantly.
  • This condition is vital for deriving relationships through the Ideal Gas Law, although slight deviations from ideal behavior should be considered for real gases like water vapor.
When dealing with constant temperature, changes in pressure or volume can often lead to phase changes, especially when dealing with substances near their boiling or melting points.
Compression
Compression involves decreasing the volume available to a gas or vapor, which results in increased pressure. This is relevant in various thermodynamic processes.
  • In this exercise, water vapor experiences compression from 500 Pa to 2,000 kPa. Such significant pressure increase forces the vapor to become denser.
  • When compressed at constant temperature, the increased proximity of water vapor molecules results in condensation, turning vapor into liquid water.
Understanding compression is essential in various fields, such as engineering and physics, where controlling the state of a substance through pressure is often necessary, such as in hydraulic systems and HVAC (heating, ventilation, and air conditioning) technologies.
Cooling Process
The cooling process refers to the removal of heat from a substance, resulting in a decrease in its temperature. This change can lead to phase changes if the temperature crosses the substance's melting or boiling point thresholds.
  • In the exercise, water is cooled from 100掳C to -10掳C at a constant pressure of 50 kPa.
  • This cooling leads to several phase changes: liquid to vapor, vapor back to liquid, and eventually liquid to solid (ice).
By understanding the cooling process, one can predict how substances will behave when subjected to lower temperatures. This is crucial in applications involving heat exchangers, refrigeration, and controlled environments for chemical reactions.

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Most popular questions from this chapter

(a) What is the significance of the critical point in a phase diagram? (b) Why does the line that separates the gas and liquid phases end at the critical point?

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) and pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) are both liquids at room temperature and room pressure, and have about the same molecular weight. (a) One of these liquids is much more viscous than the other. Which one do you predict is more viscous? (b) One of these liquids has a much lower normal boiling point \(\left(36.1^{\circ} \mathrm{C}\right)\) compared to the other one \(\left(198^{\circ} \mathrm{C}\right)\). Which liquid has the lower normal boiling point? (c) One of these liquids is the major component in antifreeze in automobile engines. Which liquid would you expect to be used as antifreeze? (d) One of these liquids is used as a "blowing agent" in the manufacture of polystyrene foam because it is so volatile. Which liquid would you expect to be used as a blowing agent?

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

(a) Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH} ?\) (b) Explain. [Section 11.3\(]\)

Suppose you have two colorless molecular liquids A and B whose boiling points are \(78^{\circ} \mathrm{C}\) and \(112^{\circ} \mathrm{C}\) respectively and both are at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) Both A and \(B\) are liquids with identical vapor pressure at room temperature of \(25^{\circ} \mathrm{C} .(\mathbf{b})\) Liquid A must consist of nonpo- (c) Both lar molecules with lower molecular weight than B. liquids A and \(B\) have higher total intermolecular forces than water. (d) Liquid \(\mathrm{A}\) is more volatile than liquid \(\mathrm{B}\) because it has a lower boiling point. (e) At \(112^{\circ} \mathrm{C}\) both liquids have a vapor pressure of \(1 \mathrm{~atm}\).
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