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Enthalpy of vaporization
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== Thermodynamic background == [[Image:Heat Content of Zn(c,l,g).PNG|thumb|right|350px|'''Molar enthalpy of zinc''' above 298.15{{nbsp}}K and at 1{{nbsp}}atm pressure, showing discontinuities at the melting and boiling points. The enthalpy of melting (Δ''H''°m) of zinc is 7323{{nbsp}}J/mol, and the enthalpy of vaporization (Δ''H''°v) is {{val|115330|u=J/mol}}.]] The enthalpy of vaporization can be written as :<math>\Delta H_\text{vap} = \Delta U_\text{vap} + p\,\Delta V</math> It is equal to the increased [[internal energy]] of the vapor phase compared with the liquid phase, plus the work done against ambient pressure. The increase in the internal energy can be viewed as the energy required to overcome the [[Chemical bond#Intermolecular interactions|intermolecular interactions]] in the liquid (or solid, in the case of [[Sublimation (chemistry)|sublimation]]). Hence [[helium]] has a particularly low enthalpy of vaporization, 0.0845 kJ/mol, as the [[van der Waals force]]s between helium [[atom]]s are particularly weak. On the other hand, the [[molecule]]s in liquid [[Water (molecule)|water]] are held together by relatively strong [[hydrogen bond]]s, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C ([[Heat capacity|''c''<sub>p</sub>]] = 75.3 J/K·mol). Care must be taken, however, when using enthalpies of vaporization to ''measure'' the strength of intermolecular forces, as these forces may persist to an extent in the gas phase (as is the case with [[hydrogen fluoride]]), and so the calculated value of the [[bond strength]] will be too low. This is particularly true of metals, which often form [[Covalent bond|covalently bonded]] molecules in the gas phase: in these cases, the [[enthalpy of atomization]] must be used to obtain a true value of the [[bond energy]]. An alternative description is to view the enthalpy of condensation as the heat which must be released to the surroundings to compensate for the drop in [[entropy]] when a gas condenses to a liquid. As the liquid and gas are in [[Chemical equilibrium|equilibrium]] at the boiling point (''T''<sub>b</sub>), [[Gibbs free energy|Δ<sub>v</sub>''G'']] = 0, which leads to: :<math>\Delta_\text{v} S = S_\text{gas} - S_\text{liquid} = \frac{\Delta_\text{v} H}{T_\text{b}}</math> As neither entropy nor [[enthalpy]] vary greatly with temperature, it is normal to use the tabulated standard values without any correction for the difference in temperature from 298 K. A correction must be made if the [[pressure]] is different from 100 [[pascal (unit)|kPa]], as the entropy of an [[ideal gas]] is proportional to the logarithm of its pressure. The entropies of liquids vary little with pressure, as the [[Thermal expansion|coefficient of thermal expansion]] of a liquid is small.<ref>Note that the rate of change of entropy with pressure and the rate of thermal expansion are related by the [[Maxwell Relations|Maxwell Relation]]: :<math>\left(\frac{\partial S}{\partial P}\right)_T = \left(\frac{\partial V}{\partial T}\right)_P.</math></ref> These two definitions are equivalent: the boiling point is the temperature at which the increased entropy of the gas phase overcomes the intermolecular forces. As a given quantity of matter always has a higher entropy in the gas phase than in a condensed phase (<math>\Delta_\text{v} S</math> is always positive), and from :<math>\Delta G = \Delta H - T\Delta S</math>, the [[Gibbs free energy]] change falls with increasing temperature: gases are favored at higher temperatures, as is observed in practice.
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