Chapter Twelve
Thinking It Through
T12.1 It is necessary only to compare the densities of hot and cool water in order to answer this question. The cool water, being more dense, sinks to the bottom.
T12.2 The rate of evaporation on these two types of days must be compared. The relative humidity is the key to the question, because evaporation is more efficient on a dry day.
T12.3 The reason for the hiss must be the presence of vapors, which appear when gasoline vapors establish an equilibrium with liquid gasoline in the container.
T12.4 The molecular mass of the wax is much larger than that of water. If the comparison of boiling points is to be used in order to determine the relative strengths of intermolecular forces, it is important to compare substances having approximately the same molecular masses. Also, London forces have been shown to increase with increased molecular masses.
T12.5 First we look for the possibility for hydrogen bonding in either of the two substances, since this is the strongest of the intermolecular forces. The second compound, (CH3CH2)2NH, is an amine and can have hydrogen bonds. The first compound is an ether, and cannot have hydrogen bonds, since it does not possess an OH group. Other intermolecular forces may exist, but having found this difference in hydrogen–bonding, we can proceed to answer the question. The first compound (the ether) has the weaker intermolecular forces and, therefore, the higher vapor pressure.
T12.6 Since the boiling point depends on the pressure, we would need to know the pressure inside the flask in order to answer this question. Obviously the pressure inside the flask at the temperature of the ice water is lower than that of the vapor pressure of the trapped water.
T12.7 Air with 100% humidity is saturated with water vapor, meaning that the partial pressure of water in the air has become equal to the vapor pressure of water, at whatever temperature we may have. Since vapor pressure decreases with decreasing temperature, the total amount of water vapor in saturated air should decrease with decreasing temperature. When cold air is brought inside, the low water content constitutes only a small amount of the water content that is possible in the warm room.
T12.8 It is necessary to compare the water content of the two air masses. A cloud of tiny liquid water dropletswill form if there is enough water in the humid air mass to cause the relative humidity to exceed 100 % at the lower temperature.
T12.9 Here we have two quantities that differ in one important way. Vapor pressure does not depend on the amount of a sample, whereas the rate of evaporation does depend on the amount of surface area.
T12.10 Water will vaporize into such a container until the vapor pressure becomes as large as the equilibrium value, provided there is some amount of liquid water after equilibrium is reached. Evidently all of the liquid evaporated before this equilibrium was reached.
T12.11 We must compare two enthalpy values, DHvap and DHfus. The process of vaporization (and of condensation) generally involves more energy than the process of melting (and of crystallization). This is because a greater disruption of intermolecular forces is required when a substance is vaporized than when it is melted.
T12.12 A large molar heat of vaporization will correspond with the substance having the stronger intermolecular forces of attraction. We can make this comparison straightforwardly, since the chain lengths are the same for the two substances. In this case, we choose the O–containing analog, because of the high strength of its hydrogen bonds.
T12.13 Hydrogen bonding is not an issue here. The chain lengths are different, and we expect a difference in polarity between the two. The greater chain length of the first compound causes it to have the greater heat of vaporization.
T12.14 Since room temperature is already above the critical temperature, we conclude that condensation is not possible, regardless of the pressure.
T12.15 In order to answer this question, we must compare the strengths only of two types of forces, dipole–dipole and London forces, since these are the only two possibilities. We expect a difference in the permanent dipole moment of these two substances as well as a difference in the London forces, since chlorine is a polarizable atom. Based on this, CHCl3 has a higher vapor pressure.
T12.16 The fact that the sample undergoes a cooling tells us that some energy must have been "spent" in the process of expansion into a vacuum. This loss of energy is only necessary if intermolecular forces have to be overcome in order to move the gas molecules apart.
T12.17 The metal cation with the larger charge is more able to polarize the electron cloud of the anion towards itself. As the electron cloud of the anion becomes more polarized towards the cation, we have, by definition, a more covalent situation. The bond between cation and anion becomes more covalent because the cation is able to attract and share the anion's electrons more completely.
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