Vapor Pressure: What Affects Gasoline Evaporation?

AI Thread Summary
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature, influencing how liquids like gasoline evaporate. Gasoline, with a vapor pressure ranging from 20 kPa to 80 kPa at room temperature, can still form vapor clouds when released into the atmosphere due to the energy of individual molecules allowing some to escape. The phenomenon is similar to water, which evaporates at room temperature despite having a lower vapor pressure than its boiling point. The vapor pressure of gasoline varies because it is not a pure substance, containing various volatile components that affect its evaporation rate. Understanding vapor pressure is essential for grasping how different substances behave under varying temperatures and pressures.
Ask1122
Messages
51
Reaction score
0
Hi all,

I have a question on vapor pressure. As far as I understood, a liquid will flashes into gas when the absolute pressure it is experiencing (in a pipe / in the atmosphere) is lower than the liquid's vapor pressure (assuming temperature is constant).
Everyone knows that gasoline forms a vapor cloud when it is released into the atmosphere. But a lots of sources I came across on the internet states that the vapor pressure of gasoline ranges between 20kPa - 80 kPa at room temperature on average, so if that is true, how does gasoline form a vapor in the atmosphere?

Thank you~
 
Physics news on Phys.org
The way I have understood it, is that temperature is just an average. Some of the molecules will have enough energy by chance to escape as vapor at any temperature. The numbers you state are average, and if the average is high enough for vaporization, you get a large scale effect, ie boiling.
 
Ask1122 said:
Hi all,

I have a question on vapor pressure. As far as I understood, a liquid will flashes into gas when the absolute pressure it is experiencing (in a pipe / in the atmosphere) is lower than the liquid's vapor pressure (assuming temperature is constant).
Everyone knows that gasoline forms a vapor cloud when it is released into the atmosphere. But a lots of sources I came across on the internet states that the vapor pressure of gasoline ranges between 20kPa - 80 kPa at room temperature on average, so if that is true, how does gasoline form a vapor in the atmosphere?

Thank you~

Well then, how does water from a vapour in air at room temperature?
The vapor pressure of water at 20 deg C or room temperature is 17.5 mm Hg, yet liquid water still evaporates at atmospheric pressure.

But boiling water at 100 deg C has a vapour vapor pressure is 760.0 mm Hg - ie atmospheric pressure.

The vapour pressure is a function of temperature. If you put a substance in a closed container, an equilibrium will be reached after some time where the pressure of the gas of the substance in the container reaches a value depending upon the temperature. That is the vapour pressure at that temperature. Since gasoline is not a pure substance its vapour pressure varies depending upon its volatile constituents.

Here is an excerpt fom Wiki regarding dalton's Law
Dalton's law of partial pressuresMain article: Dalton's law
The partial pressure of an ideal gas in a mixture is equal to the pressure it would exert if it occupied the same volume alone at the same temperature. This is because ideal gas molecules are so far apart that they don't interfere with each other at all. Actual real-world gases come very close to this ideal.

A consequence of this is that the total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the individual gases in the mixture as stated by Dalton's law.[2] For example, given an ideal gas mixture of nitrogen (N2), hydrogen (H2) and ammonia (NH3):

where:
= total pressure of the gas mixture
= partial pressure of nitrogen (N2)
= partial pressure of hydrogen (H2)
= partial pressure of ammonia (NH3)

It takes some time to get your head around all of this vapour pressure stuff I know.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top