One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will find that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

One thing you are missing is that there air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).