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"The molar enthalpy of vaporisation of a liquid is the enthalpy change when one mole of the liquid changes into one mole of its gas at the boiling temperature." The part which i am not sure of is the one highlighted above. Does it mean that the temperature of the flame that is supplying the energy needed to vaporise a substance is at the boiling temperature of the liquid ? Or is the liquid itself at its boiling point ? Because i cant imagine how the temperature of a liquid could be maintained at its boiling temperature so i guess its the temperature of the flame right ?

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    $\begingroup$ The temperature of the flame has absolutely nothing to do with this except that the flame must be hotter than the boiling point of the liquid. // The enthalpy of vaporization is the amount of heat necessary to convert one mole of liquid to one mole of gas given that the liquid is already at the boiling point. In other words the heat to rasie the liquid from room temperature to the boiling point doesn't count in the heat necessary for the liquid to vapor transition. $\endgroup$ – MaxW Feb 25 '17 at 17:26
  • $\begingroup$ but if it is already at its boiling point then the liquid is already boiling and doesnt need any energy to turn into vapor so the definition is "the amount of energy needed to keep a mole of a liquid at its boiling temperature until it is converted to a mole of gaseous molecules" ? $\endgroup$ – user41793 Feb 25 '17 at 18:37
  • $\begingroup$ That is essentially correct. // You have to realize that all of the thermodynamics terminology hangs together. There are all sorts of problems hiding in simple explanations. $\endgroup$ – MaxW Feb 25 '17 at 18:45
  • $\begingroup$ It doesn't have to be the boiling point. The enthalpy of vaporization applies to converting a saturated liquid to a saturated vapor at any temperature and corresponding vapor pressure. $\endgroup$ – Chet Miller Feb 26 '17 at 1:03
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I'll give it a try :-) There is always a transition of liquid to gas till the system gets to equilibrium by rault's law which is true at every temperature.But at a specific temperature all the liquid will go to the gas phase, that is the boiling point. So, the molar enthlapy of vaporisation of a liquid refers to the enthalpy change of one mole of liquid that changes into gas at the boiling temperature of that specific solution.

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  • $\begingroup$ I don't think the OP was referring to a chemical mixture. $\endgroup$ – Chet Miller Feb 26 '17 at 1:00
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You need to understand that the thermodynamic quantity being measured (the heat of vaporisation) is a separate idea from the heat capacity of the liquid.

If you took a liquid at well below its boiling point and applied heat until it was turned to vapour, some of the energy would be used to heat the liquid. The energy required to vapourise the liquid would be overestimated if you just measured the total energy input.

The point of the sentence is to separate the two thermodynamic concepts. It take energy to warm a liquid and it takes energy (a different amount) to vaporise the liquid. These quantities are different and should not be confused.

The thermodynamic concepts are ideal concepts and not strictly an explanation of what happens when you boil a liquid (unless you do it very carefully and slowly). In the real world a boiling liquid will not be evenly heated and some of it won't be at the boiling point. the thermodynamic concept is about the ideal experiment where you do everything slowly and ensure perfect mixing. In this case the liquid will reach the boiling point and any extra energy added will cause vaporisation rather than a mixture of warming and vaporisation.

The temperature that matters here is the temperature of the liquid.

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