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    $\begingroup$ I suspect the molar mass difference really is the biggest factor at work here. Assuming we're talking about the vapour phases as ideal gasses (which we are), the natural quantity to talk about is number of molecules. Two puddles of water and octane with the same volume (what "feels intuitively right" to compare) actually refers to amounts of molecules which differ by a factor of $\mathrm{\frac{114.23}{18.02}\frac{0.997}{0.703}=8.99}$ times - there is simply just much less octane to evaporate. $\endgroup$
    – Nicolau Saker Neto
    Commented Oct 6, 2023 at 10:40
  • $\begingroup$ @NicolauSakerNeto I do agree, assuming the effect is on the order molar mass > vapor saturation > evaporation enthalpy. The mentioning order was not meant as order of importance. I sil reorder that. $\endgroup$
    – Poutnik
    Commented Oct 6, 2023 at 10:51
  • $\begingroup$ Since molar enthalpies are similar it is the Number o molecules that is pertinent $\endgroup$
    – jimchmst
    Commented Oct 6, 2023 at 15:58
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    $\begingroup$ @BuckThorn I mean, you see 2 comparable drops of water and n-octane and you do not say "Look, the n-octane drop contains 8 times less molecules.". You say "Look, the n-octane drop has similar volume and mass as the water one.". // I also assume that in pseudostatic scenario is not the determining factor the rate of bidirectional mass transfer between phases, but mass concentration of saturated vapor. $\endgroup$
    – Poutnik
    Commented Oct 10, 2023 at 9:32
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    $\begingroup$ That too, it is quite high for water. Like 2 effects of 1 cause ( pM/RT and ΔH_evap,spec) of the molecular mass. $\endgroup$
    – Poutnik
    Commented Oct 10, 2023 at 9:36