- #1
FireBones
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I'm trying to understanding "Heat of Vaporization" in fundamental terms.
People say that processes like evaporation and boiling are endothermic because chemical bonds are being broken, but for actual cooling to occur there must a loss in translational kinetic energy of the molecules involved. So I'm asking myself "how can the breaking of bonds lead to a decrease in molecular (translational) kinetic energy?"
The obvious answer is that neighboring molecules slow down a molecule as it is escaping a liquid. This means energy is lost by the system as a whole. This does not explain directly how evaporation cools water (since the energy being lost is lost by the escaping molecule, not the molecules left behind), but it at least explains how a closed system can be cooled by evaporation/boiling going on inside it.
One could also argue that the molecules that escape are, on average, the faster moving molecules, so the average kinetic energy of the sample goes down because in the process of phase change it is losing its "hotter" particles.
With these notions in mind, I decided to look at the question of "how much energy is actually lost when a molecule escapes?" In other words, how much does the tugging by nearby molecules slow down an escapee? That is when I got very confused.
I looked at typical water simulation models, and it appears that it is far easier to escape nearby molecules than I had thought. Taking the Lennard-Jones as a rough approximation, the potential well formed by a pair of molecules has a depth of about 650 J/mole. If we mentally put 3 neighbors around the molecule, it would seem that the amount of energy necessary for water molecules to escape their neighbors is only about 2kJ/mole, which is well below the enthalpy of vaporization [40 kJ/mole].
In fact, the average water molecule at any temperature [0C to 100C] has more than enough kinetic energy to escape its neighbors according to the L-J model. Even taking into consideration issues like direction of motion, etc., one would expect an unsettling percentage of molecules (about 25 percent) on the surface of a puddle at room temperature to be moving in the right direction and with sufficient momentum to escape at every nanosecond.
Something is not adding up here.
Could someone explain to me:
A. Why is the depth of the L-J potential such a tiny fraction of the energy of vaporization?
B. Why don't puddles evaporate into dry air practically instantaneously since about a quarter of the molecules on their surface at any given instant are moving fast enough and in the right direction to escape from their L-J potential wells?
Thanks
People say that processes like evaporation and boiling are endothermic because chemical bonds are being broken, but for actual cooling to occur there must a loss in translational kinetic energy of the molecules involved. So I'm asking myself "how can the breaking of bonds lead to a decrease in molecular (translational) kinetic energy?"
The obvious answer is that neighboring molecules slow down a molecule as it is escaping a liquid. This means energy is lost by the system as a whole. This does not explain directly how evaporation cools water (since the energy being lost is lost by the escaping molecule, not the molecules left behind), but it at least explains how a closed system can be cooled by evaporation/boiling going on inside it.
One could also argue that the molecules that escape are, on average, the faster moving molecules, so the average kinetic energy of the sample goes down because in the process of phase change it is losing its "hotter" particles.
With these notions in mind, I decided to look at the question of "how much energy is actually lost when a molecule escapes?" In other words, how much does the tugging by nearby molecules slow down an escapee? That is when I got very confused.
I looked at typical water simulation models, and it appears that it is far easier to escape nearby molecules than I had thought. Taking the Lennard-Jones as a rough approximation, the potential well formed by a pair of molecules has a depth of about 650 J/mole. If we mentally put 3 neighbors around the molecule, it would seem that the amount of energy necessary for water molecules to escape their neighbors is only about 2kJ/mole, which is well below the enthalpy of vaporization [40 kJ/mole].
In fact, the average water molecule at any temperature [0C to 100C] has more than enough kinetic energy to escape its neighbors according to the L-J model. Even taking into consideration issues like direction of motion, etc., one would expect an unsettling percentage of molecules (about 25 percent) on the surface of a puddle at room temperature to be moving in the right direction and with sufficient momentum to escape at every nanosecond.
Something is not adding up here.
Could someone explain to me:
A. Why is the depth of the L-J potential such a tiny fraction of the energy of vaporization?
B. Why don't puddles evaporate into dry air practically instantaneously since about a quarter of the molecules on their surface at any given instant are moving fast enough and in the right direction to escape from their L-J potential wells?
Thanks