question_answer 53)

                  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 of vaporization for water \[{{L}_{v}}\,=\,540\,k\,cal\,\,k{{g}^{-1}},\] the mechanical equivalent of heat \[J=4.2\,J\,ca{{l}^{-1}},\]density of water \[{{\rho }_{w}}=\,{{10}^{3}}\,kg\,{{l}^{-1}},\] Avagadro?s No \[{{N}_{A}}=6.0\times {{10}^{26}}\,k\,mol{{e}^{-1}}\] and the molecular weight of water \[{{M}_{A}}=18\,kg\] for 1 k mole.                 (a) Estimate the energy required for one molecule of water to evaporate.                 (b) Show that the inter-molecular distance for water is \[d={{\left[ \frac{{{M}_{A}}}{{{N}_{A}}}\times \,\frac{1}{\rho \omega } \right]}^{1/3}}\] and find its value.                 (c) 1 g of water is the vapour state at 1 atm occupies \[1601\,c{{m}^{3}}\]. Estimate the intermolecular distance at boiling point, in the vapour state.                 (d) During vaporization a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d?. Estimate the value of F.                 (e) Calculate Fid, which is a measure of the surface tension.                

Answer:

                  (a) \[{{L}_{\upsilon }}=540\times {{10}^{3}}\,\text{cal}\,\text{k}{{\text{g}}^{-1}}\,=540\,\times \,{{10}^{3}}\]                 \[\times 4.2=2.27\times {{10}^{6}}\,J\,k{{g}^{-1}}\]                 Energy required to evaporate 1 molecule                 \[=\,\left( \frac{{{M}_{A}}}{{{N}_{A}}} \right)\,\times \,{{L}_{\upsilon }}\,=\,\frac{18}{6\times \,{{10}^{26}}}\,\times \,2.27\,\times \,{{10}^{6}}\]                 \[=6.81\times {{10}^{-20}}\,J\]                 (b) Volume around one molecule,                 \[{{x}^{3}}\,=\,\left( \frac{{{M}_{A}}}{{{N}_{A}}} \right)\,\times \,\frac{1}{{{\rho }_{w}}}\]                 \[\therefore \]\[x=\,{{\left[ \frac{{{M}_{A}}}{{{N}_{A}}{{\rho }_{w}}} \right]}^{1/3}}\]                 \[=\,{{\left( \frac{18}{6\times \,{{10}^{26}}\,\times \,{{10}^{3}}} \right)}^{1/3}}\]                 \[=3.1\times {{10}^{-10}}\,m\]                 (c) Volume of 1g vapor \[=1601\,\,c{{m}^{3}}\]                 \[=1601\times {{10}^{-6}}\,{{m}^{3}}\]                 or volume of 1 kg vapour \[=1601\times {{10}^{3}}{{m}^{3}}\]                         \[\therefore \] volume of 18 kg vapour \[\text{=}1601\times {{10}^{3}}\times 18\]                 \[=28.82\text{ }{{m}^{3}}\]                 Now volume of 1 molecule                 \[=\,\frac{28.82}{{{N}_{A}}}\,=\,\frac{28.82}{6\,\times \,{{10}^{26}}}\,=4.8\,\times \,{{10}^{-26}}\,{{m}^{3}}\]                 If \[x\] be the intermolecular distance                 \[\therefore \] \[x_{1}^{3}\,=\,4.8\,\,\times \,{{10}^{-26}}\,{{m}^{3}}\]                 or \[{{x}_{1}}=36.34\times {{10}^{-10}}\,m\]                 (d) Work done to change the distance from d to \[d'=F(d'-d)\]                 This work done is equal to the energy required to evaporate 1 molecule                 \[\therefore \] \[F(d'd)=6.81\times {{10}^{20}}\]                 Since \[d=\,x=3.1\,\,\times \,{{10}^{-10}}m\]                 and \[d'={{x}_{1}}=36.34\times {{10}^{-10}}\,m\]                 \[\therefore \]\[F=\,\frac{6.81\,\times \,{{10}^{-20}}}{(36.34\,\times \,{{10}^{-10}}-3.1\,\times \,{{10}^{-10}})}\]                 \[=\,\frac{6.81\,\,\times {{10}^{-20}}}{33.24\,\times \,{{10}^{-10}}}=2.05\,\,\times \,{{10}^{-11}}N\]                 (e) \[T=\,\frac{F}{d}\,=\frac{2.05\,\times \,{{10}^{-11}}}{3.1\times \,{{10}^{-10}}}\,=\,0.066\,N{{m}^{-1}}\]