Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury $T=435.5 \times 10^{-3} \mathrm{Nm}^{-1}$.
Consider the diagram.
Radii of mercury droplets
$$\begin{aligned} & r_1=0.1 \mathrm{~cm}=1 \times 10^{-3} \mathrm{~m} \\ & r_2=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} \end{aligned}$$
Surface tension $(T)=435.5 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
Let the radius of the big drop formed by collapsing be $R$.
$\therefore$ Volume of big drop $=$ Volume of small droplets
$$\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^2$$
$$\begin{aligned} \text{or}\quad R^3 & =r_1^3+r_2^3 \\ & =(0.1)^3+(0.2)^3 \\ & =0.001+0.008 \\ & =0.009 \end{aligned}$$
or $$\quad R=0.21 \mathrm{~cm}=2.1 \times 10^{-3} \mathrm{~m}$$
$$\begin{aligned} &\therefore \text { Change in surface area }\\ &\begin{aligned} \Delta A & =4 \pi R^2-\left(4 \pi r_1^2+4 \pi r_2^2\right) \\ & =4 \pi\left[R^2-\left(r_1^2+r_2^2\right)\right] \end{aligned} \end{aligned}$$
$\therefore \quad$ Energy released $=T \cdot \Delta A \quad$ (where $T$ is surface tension of mercury)
$$\begin{aligned} = & T \times 4 \pi\left[R^2-\left(r_1^2+r_2^2\right)\right] \\ = & 435.5 \times 10^{-3} \times 4 \times 3.14\left[\left(2.1 \times 10^{-3}\right)^2\right. \\ & \left.\quad-\left(1 \times 10^{-6}+4 \times 10^{-6}\right)\right] \\ = & 435.5 \times 4 \times 3.14[4.41-5] \times 10^{-6} \times 10^{-3} \\ = & -32.23 \times 10^{-7} \quad \text { (Negative sign shows absorption) } \end{aligned}$$
Therefore, $3.22 \times 10^{-6} \mathrm{~J}$ energy will be absorbed.
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.
When a big drop of radius R, breaks into N droplets each of radius r, the volume remains constant.
$$\begin{aligned} \therefore \quad &\text { Volume of big drop }=N \times \text { Volume of each small drop }\\ &\begin{aligned} \frac{4}{3} \pi R^3 & =N \times \frac{4}{3} \pi r^3 \\ \text{or}\quad R^3 & =N r^3 \\ \text{or}\quad N & =\frac{R^3}{r^3} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { Now, } \quad \quad \text { change in surface area } & =4 \pi R^2-N 4 \pi r^2 \\ & =4 \pi\left(R^2-N r^2\right) \\ \text { Energy released }=T \times \Delta A & =S \times 4 \pi\left(R^2-N r^2\right) \quad \text{[T = Surface tension]} \end{aligned}$$
Due to releasing of this energy, the temperature is lowered. If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\Delta \theta$, then energy released $=m s \Delta \theta$ [ $s=$ specific heat $\Delta \theta=$ change in temperature]
$$T \times 4 \pi\left(R^2-N r^2\right)=\left(\frac{4}{3} \times R^3 \times \rho\right) s \Delta \theta \quad\left[\therefore m=v \rho=\frac{4}{3} \pi R^3 \rho\right]$$
$$\begin{aligned} \Rightarrow \quad \Delta \theta & =\frac{T \times 4 \pi\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho \times s} \\ & =\frac{3 T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{\left(R^3 / r^3\right) \times r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{1}{r}\right] \end{aligned}$$
The surface tension and vapour pressure of water at $20^{\circ} \mathrm{C}$ is $7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and $2.33 \times 10^3 \mathrm{~Pa}$, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{\circ} \mathrm{C}$ ?
Given, surface tension of water
$$(S)=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}$$
Vapour pressure $(p)=2.33 \times 10^3 \mathrm{~Pa}$
The drop will evaporate, if the water pressure is greater than the vapour pressure. Let a water droplet or radius $R$ can be formed without evaporating.
Vapour pressure $=$ Excess pressure in drop.
$\therefore p=\frac{2 S}{R}$
$$\begin{aligned} \text{or}\quad R & =\frac{2 S}{p}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^3} \\ & =6.25 \times 10^{-5} \mathrm{~m} \end{aligned}$$
(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure $d p$ over a differential height $d h$ ?
(b) Considering the pressure $p$ to be proportional to the density, find the pressure $p$ at a height $h$ if the pressure on the surface of the earth is $p_0$.
(c) If $p_0=1.03 \times 10^5 \mathrm{Nm}^{-2}, \rho_0=1.29 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=9.8 \mathrm{~ms}^{-2}$, at what height will be pressure drop to $(1 / 10)$ the value at the surface of the earth?
(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.
(a) Consider a horizontal parcel of air with cross-section A and height dh.
Let the pressure on the top surface and bottom surface be $p$ and $p+d p$. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.
$$\begin{aligned} & \text { i.e., } \\ & (p+d p) A-p A=-p g A d h \quad(\because \text { Weight }=\text { Density } \times \text { Volume } \times g) \\ & =-\rho \times A d h \times g \\ & \Rightarrow \quad d p=-\rho g d h . \quad(\rho=\text { density of air) } \end{aligned}$$
Negative sign shows that pressure decreases with height.
(b) Let $p_0$ be the density of air on the surface of the earth.
As per question, pressure $\propto$ density
$$\begin{aligned} \Rightarrow \quad \frac{p}{\rho_0} & =\frac{\rho}{\rho_0} \\ \Rightarrow \quad \rho & =\frac{\rho_0}{\rho_0} p \\ \therefore \quad d p & =-\frac{\rho_0 g}{P_0} p d h \quad[\because d p=-\rho g d h] \end{aligned}$$
$$\Rightarrow \quad \frac{d p}{p}=-\frac{\rho_0 g}{\rho_0} d h$$
$$\Rightarrow \quad \int_\limits{\rho_0}^p \frac{d p}{p}=-\frac{\rho_o g}{\rho_0} \int_\limits0^h d h \quad\left[\begin{array}{l} \because \text { at } h=0, r=p_0 \\ \text { and } \quad \text { at } h=h, p=p \end{array}\right]$$
$$\Rightarrow \quad \ln \frac{\rho}{p_0}=-\frac{\rho_0 g}{p_0} h$$
By removing log, $$p=p_0 e\left(-\frac{\rho_0 g h}{p_0}\right)$$
$$\begin{aligned} &\text { (c) As } p=p_0 e^{-\frac{p_0 g h}{p_o}} \text {, }\\ &\Rightarrow \quad \ln \frac{P}{P_0}=-\frac{\rho_0 g h}{P_0} \end{aligned}$$
By question,
$$p=\frac{1}{10} p_0$$
$$ \begin{array}{ll} \Rightarrow & \ln \left(\frac{\frac{1}{10} P_0}{P_0}\right)=-\frac{\rho_0 g}{P_0} h \\ \Rightarrow & \ln \frac{1}{10}=-\frac{\rho_o g}{P_0} h \rho_{\circ} \end{array}$$
$$\begin{aligned} \therefore \quad h & =-\frac{p_0}{\rho_o g} \ln \frac{1}{10}=-\frac{p_0}{p_0 g} \ln (10)^{-1}=\frac{p_0}{p_0 g} \ln 10 \\ & =\frac{p_0}{\rho_o g} \times 2.303 \quad \left[\because \ln (x)=2.303 \log _{10}(\mathrm{x})\right]\\ & =\frac{1.013 \times 10^5}{1.22 \times 9.8} \times 2.303=0.16 \times 10^5 \mathrm{~m} \\ & =16 \times 10^3 \mathrm{~m} \end{aligned}$$
(d) We know that $\quad \rho \propto \rho \quad$ (when $T=$ constant i.e., isothermal pressure) Temperature $(T)$ remains constant only near the surface of the earth, not at greater heights.
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. $$ \begin{aligned} \therefore \quad F\left(d_1-d\right) & =6.8 \times 10^{-20} \\ \text { or } \quad F & =\frac{6.8 \times 10^{-20}}{d_1-d} \\ & =\frac{6.8 \times 10^{-20}}{\left(36.3 \times 10^{-10}-3.1 \times 10^{-10}\right)} \\ & =2.05 \times 10^{-11} \mathrm{~N} \end{aligned}$$
$$\text { (e) Surface tension }=\frac{F}{d}=\frac{2.05 \times 10^{-11}}{3.1 \times 10^{-10}}=6.6 \times 10^{-2} \mathrm{~N} / \mathrm{m} \text {. }$$