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

Here `L_(upsilon) = 540" kcal "kg^(-1) = 540xx10%3" cal "kg^(-1) = 540 xx 10^3 x 4.2" j "kg^(-1)` Energy required to eveporte 1 kg of water = `L_(upsilon) k cal.` Energy required to evaporate 1 k mole (il.e. `M_A kg` of water)` =M_A L_(upsilon)kcal.` There are `N_A` molecules in `M_A` kg of water. The energy reqiored to evaporate 1 molecule of water ` =((M_A)/(N_A)) L_(upsilon) J = (18xx540 xx 10^3 xx4.2)/(6xx10^(26)) = 6.8 xx10^(-20) J` (b) Water molecules can be considered as points. Let d be the distance amongst them in water. Then volume around `N_A` molecules of water `= (M_A)/(rho_w)` litres. Volume around one molecule of water is, `d^3 = (M_A)/(rho_w xx N_A)`
or `d = ((M_A)/(rho_w xx N_A))^(1//3) =[(18)/(10^3 xx 6 xx10^(26))]^(1//3) = 3.1xx 10^(-10)m`
(c ) Volume oc cupied by 1 g vapour ` = 1601 cm^3 = 1601 xx 10^(-6) m^3` volume oc cuppied by 1 kg vapour `=(1601 xx 10^(-6) xx10^3 = 1601 xx10^(-3) m^3`
volume oc cupied by 18 kg of vapur `= 1601 xx 10^(-3)) xx18 m^3`
Volume oc cupied by 1 molecule of vapour ` = (1601 xx 10^(-3) xx 18)/(6xx10^(26))m^3`
if `d_1` is the intermolecular distance, then `d_1^3 =(1601 xx 10^(-3) xx18)/(6xx10^(26)) m^3`
`d_1 = [(1601 xx10^(-3) xx18)/(6xx10^(26))]^(1//3) = 36.3 xx10^(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. `:. F(d_1 - d) = 6.8 xx10^(-20) or F = (6.8xx10^(20))/((36.3 xx10^(10)) - 3.1 xx10^(10)) = 2.05 xx 10 ^(11)N`
(e ) Surface tension ` =(F)/(d) = (2.05xx10^(-11))/(3.1 xx10^(-10)) = 6.6 xx 10^(-2) Nm^(-1)`