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  Is Vaporization against Archimedes' principle and Newton's laws ?

+ 0 like - 1 dislike

Vapor is a is a substance in the gas phase at a temperature lower than its critical temperature. Which means in case of water: small particles of water ( lets say a collection of molecules held together having a mass ) leaves the water surface in to the air.

Archimedes' principle states that the upward buoyant force =  weight of the fluid that the body displaces

So the upward buoyant force on the water particle ( Which is vapor ) = weight of the air it displaces

We know that density of water > density of air. 

So weight of the water particle > weight of the air it displaces

Which means weight of the water particle > upward buoyant force on it

So when  you consider the equilibrium of the water particle; it going upwards is against Newton's laws. 

Can someone please enlighten me on this ?

Thanks a lot.

Closed as per community consensus as the post is Not graduate-level
asked Aug 1, 2019 in Closed Questions by dahamv (-5 points) [ no revision ] 1 flag
recategorized Dec 22, 2019 by Dilaton

not graduate+ level. Users with 500+ reputation may vote here to close.

1 Answer

+ 0 like - 0 dislike

A hot gas is less dense than a cold gas; for example in an ideal gas : $\frac{P}{k_b T}= \rho$ with $\rho$ the density, $P$ the pressure and $T$ the temperature, $k_b$ the Boltzmann constant. That's why even if an undercooled gas of $H_2 0$ particle were to have a lower density, there is still no contradiction between evaporation and gravity.

They speak about this question here : https://www.researchgate.net/post/Can_someone_advise_on_the_density_of_water_vapor.

And this is from wikipedia (https://en.wikipedia.org/wiki/Water_vapor) : "Water vapor is lighter or less dense than dry air. At equivalent temperatures it is buoyant with respect to dry air, whereby the density of dry air at standard temperature and pressure (273.15 K, 101.325 kPa) is 1.27 g/L and water vapor at standard temperature has a vapor pressure of 0.6 kPa and the much lower density of 4.85 mg/L. " And they explain how to get it.

answered Aug 1, 2019 by JA (20 points) [ revision history ]
edited Aug 2, 2019 by JA

@JA thanks for your answer.

I have this doubt. Now density of water is 997 kg/m³ and density of air is 1.225 kg/m3. So at vaporization water has to become 813 time less dense. That means if the Mass stays a constant, the volume of the water particle has to increase 813 times. How is that possible ? 

Besides vaporization happens at normal temperature so can we think that its heat that makes the water particle expand to increase volume and do the trick ? 

If Pressure is the culprit, it has to drop significantly which would I think make all the water disappear. How can we explain ( By considering Archimedes' principle and Newton's laws ) the phenomenon that only a fraction of the molecules in a liquid escape from the liquid into the air ?  

Thanks again.

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