Which of the following is true regarding the latent heat of vaporization during condensation?
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If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. The Heat of Vaporization (also called the Enthalpy of Vaporization) is the heat required to induce this phase change. Since vaporization requires heat to be added to the system and hence is an endothermic process, therefore \( \Delta H_{vap} > 0\) as defined: \[ \Delta H_{vap} = H_{vapor} - H_{liquid}\] where Heat is absorbed when a liquid boils because molecules which are held together by intermolecular attractive interactions and are jostled free of each other as the gas is formed. Such a separation requires energy (in the form of heat). In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the Table T5 bears this out.
Example \(\PageIndex{1}\) If a liquid uses 50 Joules of heat to vaporize one mole of liquid, then what would be the enthalpy of vaporization? Solution The heat in the process is equal to the change of enthalpy, which involves vaporization in this case \[q_{tot} = \Delta_{vap}\] so \[q_{tot} = 50 \; J= \Delta_{vap}\] So the enthalpy of vaporization for one mole of substance is 50 J.
Kinetic energy does not change The kinetic energy of the molecules in the gas and the silquid are the same since the vaporization process occues at constant temperature. However, the add thermal energy is used to break the potential energies of the intermolecular forces in the liquid, to generate molecules in the gas that are free of potential energy (for an ideal gass). Thus, while \(H_{vapor} > H_{liquid}\), the kinetic energies of the molecules are equal.
Condensation is the opposite of vaporization, and therefore \( \Delta H_{condensation}\) is also the opposite of \( \Delta H_{vap}\). Because \( \Delta H_{vap}\) is an endothermic process, where heat is lost in a reaction and must be added into the system from the surroundings, \( \Delta H_{condensation}\) is an exothermic process, where heat is absorbed in a reaction and must be given off from the system into the surroundings. \[\begin{align} ΔH_{condensation} &= H_{liquid} - H_{vapor} \\[4pt] &= -ΔH_{vap} \end{align}\] Because \(ΔH_{condensation}\), also written as \(ΔH_{cond}\), is an exothermic process, its value is always negative. Moreover, \(ΔH_{cond}\) is equal in magnitude to \(ΔH_{vap}\), so the only difference between the two values for one given compound or element is the positive or negative sign.
Example \(\PageIndex{2}\) 2.055 liters of steam at 100°C was collected and stored in a cooler container. What was the amount of heat involved in this reaction? The \(ΔH_{vap}\) of water = 44.0 kJ/mol. Solution 1. First, convert 100°C to Kelvin. °C + 273.15 = K 2. Find the amount involved (in moles). \[\begin{align*} n_{water} &= \dfrac{PV}{RT} \\[4pt] &= \dfrac{(1.0\; atm)(2.055\; L)}{(0.08206\; L\; atm\; mol^{-1} K^{-1})(373.15\; K)} \\[4pt] &= 0.0671\; mol \end{align*}\] 3. Find \(ΔH_{cond}\) \[ΔH_{cond} = -ΔH_{vap} \nonumber\] so \[ΔH_{cond} = -44.0\; kJ/ mol \nonumber\] 4. Using the \(ΔH_{cond}\) of water and the amount in moles, calculate the amount of heat involved in the reaction. To find kJ, multiply the \(ΔH_{cond}\) by the amount in moles involved. \[\begin{align*} (ΔH_{cond})(n_{water}) &= (-44.0\; kJ/mol)(0.0671\; mol) \\[4pt] &= -2.95\; kJ \end{align*} \]
Vaporization (or Evaporation) the transition of molecules from a liquid to a gaseous state; the molecules on a surface are usually the first to undergo a phase change.
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Contributors and Attributions
When a substance changes phase, the arrangement of its molecules changes, but its temperature does not change. If the new arrangement has a higher amount of thermal energy, then the substance absorbs thermal energy from its environment in order to make the phase change. If the new arrangement has a lower amount of thermal energy, the substance releases thermal energy to its environment.
Latent heat of vaporization is a physical property of a substance. It is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. It is expressed as kg/mol or kJ/kg. When a material in liquid state is given energy, it changes its phase from liquid to vapor; the energy absorbed in this process is called heat of vaporization. The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol.
The vaporization is the opposite process of condensation. The heat of condensation is defined as the heat released when one mole of the...
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