Why is the latent heat of vaporization greater at lower vaporization temperatures?
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Why is the latent heat of vaporization greater at lower vaporization temperatures? My foundation isn’t very strong; please help me out, expert
This post was last edited by qugd on 2012-5-28 at 16:23. Due to low temperatures, the thermal motion of molecules is reduced, which means that more kinetic energy is required for molecules to move from the surface of a liquid or solid state into the gas phase. Molecular motion is a bit like the *sexuality* of some animals in nature: when it’s cold, they tend to move less, and may even stop moving altogether.
Water at 99°C needs to absorb some heat in order to become water at 100°C. However, when water at 100°C turns into steam at 100°C, its temperature remains the same, but it still requires heat – and a lot of it. Similarly, the lower the vaporization temperature, the easier it is for the substance to vaporize; the more heat that needs to be absorbed, and the greater the latent heat will be. I’m not sure if this explanation is satisfactory? !
It seems to make sense in this way, but the logic isn’t very strong. Putting the mechanism aside, from a results perspective, this is because the specific heat capacity of the liquid phase is greater than that of the gas phase. The following relationship holds approximately (the effect of pressure on enthalpy is temporarily ignored – the pressures at states 1 and 2 are different): h2,gas = h1,liquid + hfg1 + Cp,gas×(T2–T1) = h1,liquid + Cp,liquid×(T2–T1) + hfg2. Since Cp,liquid > Cp,gas, it follows that hfg1 > hfg2