Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat o vaporisation for water `L_(v)=540 "k cal kg"^(-1)`, the mechanical equivalent of heat `J. 4.2 "J cal"^(-1)` , density of water `rho_(w)=10^(3) kg l^(-1)` , Avagardro's number `N_(A)=6.0xx 10^(26)K "mole"^(-6)` and the molecular weight of water `M_(A)=10` kg for 1 k mole.
(a) Estimate the energy required for one molecules of water to evaporate.
(b) Show that the inter-molecular distance for water is `d=[(M_(A))/(N_(A))xx(1)/(rho_(w))]^(1//3)` and find its values.
(c) 1 g of water in the vapour state at 1 atm occupies `1601 cm^(3)` . Estimate the inter-molecules distance at boiling point, in the vapour state.
(d) During vaporisation a molecules overcomes a force F, assumed constant, to go from an inter-molecules distance d to d'. Estimate the value of F.
(e) Calculate F/d , which is a measure of the surface tension.

Explanation

Given, `L_(v)=540 "kcal kg"^(-1)`
`=540xx10^(3) "cal kg"^(-1) =540xx10^(3)xx4.2J kg ^(-1)`
`therefore` Energy required to evaporate 1 kg of water `=L_(v)` kcal
`therefore M_(A)` kg of water requires `M_(A)L_(v)` kcal `" "[because Q=mL]`
Since, there are `N_(A)` molecules in `M_(A)` kg of water the energy required for 1 molecule to evaporate is
`U=(M_(A)L_(V))/(N_(A))J" "`[Where `N_(A)=6xx10^(26)=` Avogadro number]
`=(18xx540xx4.2xx10^(3))/(6xx10^(26))J`
`=90xx18xx4.2xx10^(-23)J`
`=6.8xx10^(-20) J`
(b) Let the water molecules to be points and are separated at distance d from each other.
Volume of `N_(A)` molecules of water `(M_(A))/(rho_(w))" "[because V=(M)/(rho)]`
Thus, the volume around one molecules is `=(M_(A))/(N_(A)rho_(w))`
The volume around one molecules is
`d^(3)=(M_(A)//N_(A)rho_(w))`
`d=((M_(A))/(N_(A)rho_(w)))^(1//3=((18)/(6xx10^(26)xx10^(3)))^(1//3)`
`(30xx10^(-30))^(1//3)m~~3.1xx10^(-10)m`
(c) `because` 1 kg of vapour occupies volume `=1601 xx 10^(-3)m^(3)`
`therefore 18 `kg of vapour occupies `18xx 1601 xx 10^(-3) m^(3)`
`6xx10^(26)` molecules occupies `18xx1601xx10^(-3)m^(3)`
`therefore` 1 molecules occupies `(18xx1601xx10^(-3))/(6xx10^(26))m^(3)`
If d is the inter-molecular distance, then
`d_(1)^(3)=(3xx1601xx10^(-29))m^(3)`
`therefore d_(1)=(30xx1601)^(1//3)xx10^(-10)m`
`=36.3xx10^(-10)m`
(d) Work done to change the distance from d to `d_(1)` is `=F(d_(1)-d)`
This work done is equal to energy required to evaporate 1 molecule.
`therefore F(d_(1)-d)=6.8xx10^(-20)`
or `F=(6.8xx10^(-20))/(d_(1)-d)`
`=(6.8xx10^(-20))/((36.6xx10^(-10)-3.1xx10^(-10)))`
`=2.05xx10^(-11)`N
(e) Surface tension `=(F)/(d)=(2.05xx10^(11))/(3.1xx10^(-10))=6.6xx10^(-2)` N/m.