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Surafce tension is exhibited by liquids due to force of attraction between the molecules of the liquid .The surface tension decreases with increases in temperature and vanishes at boiling point Given that the latent heat of vaporization for water L_(v)=540(kcal)/(kg)the mechanical equivalent of heat J=4.2(J)/(cal) density of water rho_(w)=10^(3)kgl^(-1),Avogadro'sno.N_(A)=6.0xx10^(26)k " mole"^(-1).Molecular weight ofwater M_(A)=10kg, for 1k mole .(a)Estimate the energyrequired for one molecule of water to evaporated. (b)Show that the inter molecular distance for water is d=((M_(A))/(N_(A))xx(1)/(rho_(w)))^((1)/(3)) and find its value. (c ) 1 g of water in the vapour state at 1 atm occupies 1601cm^(3) , estimate the intermolecular distance at boiling point in the vapour state. (d) During vaporisation a molecule overcomes a force F, assumed constant to go from an inter molecular distance d to d .Estimate the value of F.(e ) Caculate (F)/(d) , which isa measure of the surface tension. |
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Answer» Solution :(a)`L_(V)=540kcal kg^(-1)` `=540xx10^(3)calkg^(-1)` `=540xx10^(3)xx(4.2J)kg^(-1)` 1kg of water requires `L_(v)Kcal`. `thereforeM_(A)kg` of water requires `M_(A)L_(v)Kcal.` `therefore` Mass of 1 mol water is `M_(A)`.There are `N_(A)` molecules in it . The energy required for 1 molecules in evaporate is `M_(A)L_(v)`. `U=(M_(A)L_(v))/(N_(A))J` `=((18)(540)(4.2xx16^(3)))/(6XX10^(26))J` `=6.8xx10^(-20)J` (b)Consider the water molecules to be points at a distanced from each other. Volume of `N_(A)` molecules of water `=(M_(A))/(rho_(w))` Volume of one molecule `d^(3)=(M_(A))/(N_(A)rho_(w))` `therefored=[(M_(A))/(N_(A)rho_(w))]^((1)/(3))` `therefored=[(18)/(6xx10^(26)xx10^(3))]^((1)/(3))` `therefored=(30xx10^(-30))^((1)/(3))` `d=3.1xx10^(-10)m` (c)The volume occupied by 1kg of vapour = `16.1cm^(3)` . The volume occupiedby `10^(-3)kg` of vapour=`1601xx10^(-6)m^(3)` `therefore` The volume occupied by 1kg of vapour `=1601xx10^(-3)m^(3)`....(1) Volumeoccupied by 1 kg of vapour `=1601xx10^(3)m^(3)` `therefore` Volume occupied by 18 kg of vapour `=18xx1601xx10^(-3)m^(3)` (But `M_(A)=18kg//kmol)`that is `18kg=6xx10^(26)` molecules`therefore` The volume of mass occupied by `6xx10^(26)=18xx1601xx10^(-3)m^(3)` `therefore` The volume occupied by 1 molecules occupies =? `=(18xx1601xx10^(-3))/(6xx10^(26))` `=(3xx1601xx10^(-29))m^(3)` If `d_(1)`is the inter molecular distance , then `d_(1)^(3)=(3xx1601xx10^(-29))m^(3)` `therefored_(1)^(3)=30xx1601xx10^(-30)` `therefored_(1)=(30xx1601)^((1)/(3))xx10^(-10)m` `d_(1)=36.3xx10^(-10)m` (d)Distance between molecules of water =d Distance between molecules of vapour `=d_(1)` (For vapour `d_(1)gtd`, more intermolecular distance then water) Work done from distance d to `d_(1)` `W=F(d_(1)-d)` This work done (energy)is required to 1 MOLEOF water into vapour. `F(d_(1)-d)=6.8xx10^(-20)` `F=(6.8xx16^(-20))/(d_(1)-d)` `F=(6.8xx10^(-20))/((36.3xx10^(-10)-3.1xx16^(10)))` `=2.05xx10^(-11)N` (e )Surface tension `=(F)/(d)` `=(2.05xx10^(-11))/(3.1xx10^(-10))` `=6.6xx10^(-2)N//m` |
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