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 vaporisation for water $L_v=540 \mathrm{k} \mathrm{cal} \mathrm{kg}^{-1}$, the mechanical equivalent of heat $J=4.2 \mathrm{~J} \mathrm{cal}^{-1}$, density of water $\rho_w=10^3 \mathrm{~kg} \mathrm{l}^{-1}$, Avagardro's number $N_A=6.0 \times 10^{26} \mathrm{k} \mathrm{mole}^{-1}$ and the molecular weight of water $M_A=10 \mathrm{~kg}$ 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 $d=\left[\frac{M_A}{N_A} \times \frac{1}{\rho_w}\right]^{1 / 3}$ and find its value.
(c) 1 g of water in the vapour state at 1 atm occupies $1601 \mathrm{~cm}^3$. Estimate the inter-molecular 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 $d^{\prime}$. Estimate the value of $F$.
(e) Calculate $F / d$, which is a measure of the surface tension.
(a) Given, $L_v=540 \mathrm{kcal} \mathrm{kg}^{-1}$
$$=540 \times 10^3 \mathrm{cal} \mathrm{kg}^{-1}=540 \times 10^3 \times 4.2 \mathrm{~J~kg}^{-1}$$
$\because \quad$ Energy required to evaporate 1 kg of water $=L_v \mathrm{~kcal}$
$\therefore \quad M_A \mathrm{~kg}$ of water requires $M_A L_V \mathrm{~kcal}\quad [\because Q=mL]$
Since, there are $N_{\mathrm{A}}$ molecules in $M_{\mathrm{A}} \mathrm{kg}$ of water the energy required for 1 molecule to evaporate is
$$\begin{aligned} U & =\frac{M_A L_v}{N_A} \mathrm{~J} \quad\left[\text { where } N_A=6 \times 10^{26}=\text { Avogadro number }\right] \\ & =\frac{18 \times 540 \times 4.2 \times 10^3}{6 \times 10^{26}} \mathrm{~J} \\ & =90 \times 18 \times 4.2 \times 10^{-23} \mathrm{~J} \\ & =6.8 \times 10^{-20} \mathrm{~J} \end{aligned}$$
(b) Let the water molecules to be points and are separated at distanced from each other.
Volume of $N_A$ molecule of water $=\frac{M_A}{\rho_w}$ $\left[\because V=\frac{M}{\rho}\right]$
Thus, the volume around one molecule is $=\frac{M_A}{N_A \rho_w}$
$$\begin{aligned} &\text { The volume around one molecule is }\\ &\begin{aligned} d^3 & =\left(M_A / N_A \rho_w\right) \\ \therefore \quad d & =\left(\frac{M_A}{N_A \rho_w}\right)^{1 / 3}=\left(\frac{18}{6 \times 10^{26} \times 10^3}\right)^{1 / 3} \\ \left(30 \times 10^{-30}\right)^{1 / 3} \mathrm{~m} & \approx 3.1 \times 10^{-10} \mathrm{~m} \end{aligned} \end{aligned}$$
(c) $\because 1 \mathrm{~kg}$ of vapour occupies volume $=1601 \times 10^{-3} \mathrm{~m}^3$
$\therefore 18 \mathrm{~kg}$ of vapour occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^3$
$6 \times 10^{26}$ molecules occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^3$
$\therefore 1$ molecule occupies $\frac{18 \times 1601 \times 10^{-3}}{6 \times 10^{26}} \mathrm{~m}^3$
$$ \begin{aligned} &\text { If } d \text { is the inter- molecular distance, then }\\ &\begin{aligned} d_1^3 & =\left(3 \times 1601 \times 10^{-29}\right) \mathrm{m}^3 \\ \therefore \quad d_1 & =(30 \times 1601)^{1 / 3} \times 10^{-10} \mathrm{~m} \\ & =36.3 \times 10^{-10} \mathrm{~m} \end{aligned} \end{aligned}$$
(d) Work done to change the distance from to $d_1$ is $=F\left(d_1-d\right)$ This work done is equal to energy required to evaporate 1 molecule. $$ \begin{aligned} \therefore \quad F\left(d_1-d\right) & =6.8 \times 10^{-20} \\ \text { or } \quad F & =\frac{6.8 \times 10^{-20}}{d_1-d} \\ & =\frac{6.8 \times 10^{-20}}{\left(36.3 \times 10^{-10}-3.1 \times 10^{-10}\right)} \\ & =2.05 \times 10^{-11} \mathrm{~N} \end{aligned}$$
$$\text { (e) Surface tension }=\frac{F}{d}=\frac{2.05 \times 10^{-11}}{3.1 \times 10^{-10}}=6.6 \times 10^{-2} \mathrm{~N} / \mathrm{m} \text {. }$$
A hot air balloon is a sphere of radius 8 m . The air inside is at a temperature of $60^{\circ} \mathrm{C}$. How large a mass can the balloon lift when the outside temperature is $20^{\circ} \mathrm{C}$ ? Assume air in an ideal gas, $R=8.314 \mathrm{~J}$ mole ${ }^{-1} \mathrm{~K}^{-1}, 1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa}$, the membrane tension is $5 \mathrm{~Nm}^{-1}$.
Let the pressure inside the balloon be $p_i$ and the outside pressure be $p_0$, then excess pressure is $p_i-p_0=\frac{2 S}{r}$.
where, $S=$ Surface tension
$r=$ radius of balloon
Considering the air to be an ideal gas $p_i V=n_i R T_i$ where, $V$ is the volume of the air inside the balloon, $n_i$ is the number of moles inside and $T_i$ is the temperature inside, and $p_0 V=n_0 R T_0$ where $V$ is the volume of the air displaced and $n_0$ is the number of moles displaced and $T_0$ is the temperature outside.
So, $n_i=\frac{p_i V}{R T_i}=\frac{M_i}{M_A}$
where, $M_i$ is the mass of air inside and $M_A$ is the molar mass of air and
$$n_0=\frac{p_0 V}{R T_0}=\frac{M_0}{M_A}$$
where, $M_0$ is the mass of air outside that has been displaced. If $w$ is the load it can raise, then $w+M_i g=M_o g$
$$\Rightarrow \quad w=M_o g-M_i g$$
As in atmosphere $21 \% \mathrm{O}_2$ and $79 \% \mathrm{~N}_2$-is present
$\therefore$ Molar mass of air
$$M_i=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}$$
$\therefore$ Weight raised by the balloon
$$w=\left(M_o-M_i\right) \mathrm{g}$$
$$\begin{aligned} \Rightarrow \quad \mathrm{w} & =\frac{M_A V}{R}\left(\frac{p_0}{T_0}-\frac{p_i}{T_i}\right) g \\ & =\frac{0.02884 \times \frac{4}{3} \pi \times 8^3 \times 9.8}{8.314} \quad\left(\frac{1.013 \times 10^5}{293}-\frac{1.013 \times 10^5}{333}-\frac{2 \times 5}{8 \times 313}\right) \\ & =\frac{0.02884 \frac{4}{3} \pi \times 8^3}{8.314} \times 1.013 \times 10^5\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8 \\ & =3044.2 \mathrm{~N} \end{aligned}$$
$$\begin{aligned} \therefore \quad \text { Mass lifted by the balloon } & =\frac{w}{g}=\frac{3044.2}{10} \approx 304.42 \mathrm{~kg} . \\ & \approx 305 \mathrm{~kg} . \end{aligned}$$