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The melting point of a substance is the temperature at which the solid form's vapor pressure equals the liquid form's vapor pressure. I read that the vapor pressure of something depends ONLY on the temperature. So as the external pressure changes, the substance's vapor pressure should stay the same if temperature is constant. In that case, the melting point should not be affected by external pressure.

I understand that intuitively, melting point increasing with pressure makes sense, as more energy is needed to overcome larger external pressure in order to liquefy. But how can this be explained with the vapor pressure definition?

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    $\begingroup$ This is a duplicate. All 1st order phase changes have a volume change, hence there will be a PV term in the free energy. Thus, the pressure matters. $\endgroup$ – Jon Custer Nov 20 '15 at 0:38
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    $\begingroup$ Please go back and check your source for the opening statement. Melting point is a transition between the liquid and solid phases of a substance and not its vapor phase. $\endgroup$ – iad22agp Nov 20 '15 at 14:49
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    $\begingroup$ @iad22agp I know that melting point is the transition between liquid and solid phases. What I was saying was that it is defined as the point where the liquid and solid forms have identical vapor pressures. For reference, this is from pg. 827 of Chemical Principles 7th ed. by Zumdahl/Decoste. $\endgroup$ – carbenoid Nov 20 '15 at 23:52
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    $\begingroup$ @swenger Thanks for the reference. I can see how at the melting point, liquid and solid are in equilibrium, so maybe it follows that they would have to have identical vapor pressures at that temperature. But at a lower temperature, you only have the solid phase present, so how can you measure the vp of the liquid form? $\endgroup$ – iad22agp Nov 21 '15 at 5:03
  • $\begingroup$ @iad22agp I think you can measure the vp of the liquid below its normal melting point by changing the external pressure. Here's a diagram showing the vp of ice and water as the temperature approaches 273K. link $\endgroup$ – carbenoid Nov 22 '15 at 0:33

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