Surface tension is exhibited by liquids due to the force of attraction between the molecules of the liquid. The surface tension decreases with an increase in temperature and vanishes at the boiling point. Given that the latent heat of vaporization for water $L_v = 540 \text{ kcal/kg}$,the mechanical equivalent of heat $J = 4.2 \text{ J/cal}$,density of water $\rho_w = 10^3 \text{ kg/m}^3$,Avogadro's number $N_A = 6.0 \times 10^{26} \text{ molecules/kmol}$,and the molecular weight of water $M_A = 18 \text{ kg/kmol}$.
$(a)$ Estimate the energy required for one molecule of water to evaporate.
$(b)$ Show that the intermolecular distance for water is $d = \left( \frac{M_A}{N_A \rho_w} \right)^{1/3}$ and find its value.
$(c)$ $1 \text{ g}$ of water in the vapour state at $1 \text{ atm}$ occupies $1601 \text{ cm}^3$. Estimate the intermolecular distance at the boiling point in the vapour state.
$(d)$ During vaporisation,a molecule overcomes a force $F$,assumed constant,to go from an intermolecular distance $d$ to $d'$. Estimate the value of $F$.
$(e)$ Calculate $\frac{F}{d}$,which is a measure of the surface tension.
$(a)$ Estimate the energy required for one molecule of water to evaporate.
$(b)$ Show that the intermolecular distance for water is $d = \left( \frac{M_A}{N_A \rho_w} \right)^{1/3}$ and find its value.
$(c)$ $1 \text{ g}$ of water in the vapour state at $1 \text{ atm}$ occupies $1601 \text{ cm}^3$. Estimate the intermolecular distance at the boiling point in the vapour state.
$(d)$ During vaporisation,a molecule overcomes a force $F$,assumed constant,to go from an intermolecular distance $d$ to $d'$. Estimate the value of $F$.
$(e)$ Calculate $\frac{F}{d}$,which is a measure of the surface tension.
(N/A) Energy per molecule $U = \frac{M_A L_v}{N_A} = \frac{18 \text{ kg/kmol} \times 540 \times 10^3 \text{ cal/kg} \times 4.2 \text{ J/cal}}{6.0 \times 10^{26} \text{ molecules/kmol}} = 6.8 \times 10^{-20} \text{ J}$.
$(b)$ Volume of $N_A$ molecules is $\frac{M_A}{\rho_w}$. Volume per molecule is $d^3 = \frac{M_A}{N_A \rho_w}$. Thus $d = (\frac{18}{6 \times 10^{26} \times 10^3})^{1/3} \approx 3.1 \times 10^{-10} \text{ m}$.
$(c)$ Volume per molecule in vapour $d'^3 = \frac{V}{N} = \frac{1601 \times 10^{-6} \text{ m}^3}{N_A / 18000} \approx 3.0 \times 10^{-9} \text{ m}$.
$(d)$ Work done $F(d' - d) = U$. Since $d' \gg d$,$F \approx \frac{U}{d'} = \frac{6.8 \times 10^{-20}}{3.0 \times 10^{-9}} \approx 2.3 \times 10^{-11} \text{ N}$.
$(e)$ $\frac{F}{d} = \frac{2.3 \times 10^{-11}}{3.1 \times 10^{-10}} \approx 0.074 \text{ N/m}$.
$(b)$ Volume of $N_A$ molecules is $\frac{M_A}{\rho_w}$. Volume per molecule is $d^3 = \frac{M_A}{N_A \rho_w}$. Thus $d = (\frac{18}{6 \times 10^{26} \times 10^3})^{1/3} \approx 3.1 \times 10^{-10} \text{ m}$.
$(c)$ Volume per molecule in vapour $d'^3 = \frac{V}{N} = \frac{1601 \times 10^{-6} \text{ m}^3}{N_A / 18000} \approx 3.0 \times 10^{-9} \text{ m}$.
$(d)$ Work done $F(d' - d) = U$. Since $d' \gg d$,$F \approx \frac{U}{d'} = \frac{6.8 \times 10^{-20}}{3.0 \times 10^{-9}} \approx 2.3 \times 10^{-11} \text{ N}$.
$(e)$ $\frac{F}{d} = \frac{2.3 \times 10^{-11}}{3.1 \times 10^{-10}} \approx 0.074 \text{ N/m}$.
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