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Chapter 11: Problem 49

Explain how each of the following affects the vapor pressure of a liquid: (a) volume of the liquid, (b) surface area, (c) intermolecular attractive forces, (d) temperature, (e) density of the liquid.

Short Answer

Expert verified
In summary, volume and surface area do not directly affect the vapor pressure of a liquid. Intermolecular attractive forces, temperature, and density significantly affect vapor pressure. Strong intermolecular forces and higher density result in lower vapor pressures, while higher temperature increases vapor pressure. The effects of these factors on vapor pressure can be understood in terms of the equilibrium between the liquid and its vapor phase, and the energy required for liquid molecules to overcome intermolecular forces and escape into the vapor phase.

Step by step solution

01

Vapor Pressure Definition

Vapor pressure is the pressure exerted by the vapor molecules in equilibrium with its liquid phase at a given temperature. The vapor pressure of a liquid depends on the properties of the liquid and the temperature at which it is measured.
02

(a) Volume of the liquid

The volume of the liquid has no direct impact on the vapor pressure of a liquid. This is because the vapor pressure depends on the equilibrium between the liquid and its vapor phase, which is mainly determined by the molecular characteristics of the liquid and the temperature. When the volume of the liquid increases or decreases, the number of liquid molecules that can evaporate into the vapor will still be the same at equilibrium.
03

(b) Surface area

An increase in the surface area of the liquid increases the rate at which liquid molecules can evaporate into the vapor phase, as there are more liquid molecules exposed to the vapor/air interface. However, at equilibrium, the vapor pressure remains constant since it originates from the inherent properties of the liquid and its temperature. Therefore, surface area does not affect the vapor pressure of a liquid, but it does affect how quickly equilibrium is reached.
04

(c) Intermolecular attractive forces

The vapor pressure of a liquid is greatly affected by the strength of the intermolecular attractive forces. Liquids with stronger intermolecular attractive forces need more energy for their molecules to overcome these forces and escape into the vapor phase. This means that a liquid with stronger intermolecular forces will have a lower vapor pressure at a given temperature compared to another liquid with weaker intermolecular forces. Examples of intermolecular forces include hydrogen bonding, dipole-dipole interactions, and van der Waals forces (also known as London dispersion forces).
05

(d) Temperature

Temperature has a significant effect on the vapor pressure of a liquid. When the temperature increases, the kinetic energy of the liquid molecules also increases, enabling them to overcome the intermolecular attractive forces more easily. This results in a higher number of liquid molecules escaping into the vapor phase, thereby increasing the vapor pressure. Typically, vapor pressure increases exponentially with temperature, which can be described by the Clausius-Clapeyron equation.
06

(e) Density of the liquid

Density of a liquid is defined as the mass per unit volume. Although density doesn't have a direct impact on the vapor pressure, it is related to the strength of the intermolecular attractive forces, which do affect vapor pressure. Denser liquids, which have more compact molecular structures, typically have stronger intermolecular attractive forces. As a result, denser liquids tend to have lower vapor pressures at a given temperature compared to less dense liquids. However, it is important to note that this is a general correlation, and there are more specific factors (such as the type of intermolecular forces between the liquid molecules) that determine the vapor pressure of a liquid.

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Most popular questions from this chapter

At standard temperature and pressure the molar volume of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are \(22.06 \mathrm{~L}\) and \(22.40 \mathrm{~L},\) respectively (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) \(\mathrm{On}\) cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are \(2.02 \mathrm{~g} / \mathrm{cm}^{3}\) and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

If \(42.0 \mathrm{~kJ}\) of heat is added to a 32.0 -g sample of liquid methane under 1 atm of pressure at a temperature of \(-170{ }^{\circ} \mathrm{C},\) what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is \(-161.5^{\circ} \mathrm{C}\). The specific heats of liquid and gaseous methane are 3.48 and \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. [Section 11.4\(]\)

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The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\).
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