States of Matter

(i) Molecules at the surface are attracted by neighbours only from the sides and below, so they are held less strongly. Those with kinetic energy above the average can overcome these attractions and escape into the vapour phase, so evaporation occurs only at the surface.

(ii) Surface tension makes a liquid try to have the smallest possible surface area. For a given volume a sphere has the least surface area, so free drops become spherical.

A hygroscopic substance absorbs moisture from the air but remains solid and does not dissolve in it. Example: quicklime (CaO) or conc. H$_2$SO$_4$ (as a liquid it simply gets diluted).

A deliquescent substance absorbs so much moisture from the air that it dissolves in it and turns into a solution. Example: anhydrous CaCl$_2$ or NaOH pellets.

Boyle's law: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure: $V \propto \frac{1}{P}$, i.e. $PV = \text{constant}$.

Charles' law: At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature: $V \propto T$, i.e. $\frac{V}{T} = \text{constant}$.

Derivation of $PV = nRT$: For $n$ moles of gas,

Boyle's law: $V \propto \frac{1}{P}$ (constant $T$, $n$)

Charles' law: $V \propto T$ (constant $P$, $n$)

Avogadro's law: $V \propto n$ (constant $P$, $T$)

Combining, $V \propto \frac{nT}{P}$, so

$V = R\,\frac{nT}{P} \quad\Rightarrow\quad PV = nRT$

where $R$ is the universal gas constant.

Numerical:

Mass of liquid $= 148 - 50 = 98$ g, so volume of vessel $= \frac{98}{0.98} = 100$ mL $= 0.100$ L.

Mass of gas $= 50.5 - 50 = 0.5$ g.

$P = 760$ mmHg $= 1$ atm, $T = 27 + 273 = 300$ K.

$n = \frac{PV}{RT} = \frac{1 \times 0.100}{0.0821 \times 300} = 4.06\times10^{-3}\text{ mol}$

$M = \frac{0.5}{4.06\times10^{-3}} \approx 123\text{ g/mol}$

Molecular mass of the gas $\approx 123$

Number of molecules in 12.3 g:

$\text{moles} = \frac{12.3}{123} = 0.1\text{ mol}$

$N = 0.1 \times 6.022\times10^{23} = 6.02\times10^{22}\text{ molecules}$

Surface tension is the force acting per unit length perpendicular to a line drawn on the surface of a liquid (equivalently, the energy required to increase the surface area by one unit):

$\gamma = \frac{F}{l}$

Its SI unit is $\mathrm{N\,m^{-1}}$ (equivalently $\mathrm{J\,m^{-2}}$).

Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its density (or molar mass). $r \propto \frac{1}{\sqrt{d}} \quad\Rightarrow\quad \frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{M_2}{M_1}}$ So a lighter gas diffuses faster than a heavier one. For example, H₂ ($M=2$) diffuses 4 times as fast as O₂ ($M=32$), since $\sqrt{32/2}=4$.

Molecular weight

$P = 760\ \mathrm{mm} = 1\ \mathrm{atm}$, $V = 126.4\ \mathrm{cc} = 0.1264\ \mathrm{L}$, $T = 27+273 = 300\ \mathrm{K}$, $w = 0.23\ \mathrm{g}$.

$PV=\frac{w}{M}RT \;\Rightarrow\; M=\frac{wRT}{PV}=\frac{0.23\times 0.0821\times 300}{1\times 0.1264}=44.8\ \mathrm{g\,mol^{-1}}$

Number of molecules

$n=\frac{0.23}{44.8}=5.13\times10^{-3}\ \mathrm{mol}$ $N = 5.13\times10^{-3}\times 6.022\times10^{23}\approx 3.09\times10^{21}\ \text{molecules}$

Evaporation needs latent heat, which the liquid takes from itself and its surroundings, so they cool.

1. Sweating: sweat evaporates from our skin by absorbing heat from the body, so we feel cool.
2. Earthen pot (matka): water seeps through the pores and evaporates from the outer surface, so the water inside stays cool. (Spilled ether or spirit on the palm also feels cold.)

Efflorescent substance: a hydrated salt that loses its water of crystallisation on exposure to air, becoming a powder. Example: washing soda, $\mathrm{Na_2CO_3\cdot10H_2O}$.

Deliquescent substance: a substance that absorbs so much moisture from the air that it dissolves in it and becomes a solution. Example: anhydrous $\mathrm{CaCl_2}$ (or NaOH).

Dalton's law of partial pressures. At constant temperature, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases:

$P_{\text{total}} = p_1 + p_2 + p_3 + \cdots$

The partial pressure of a gas is the pressure it would exert if it alone occupied the whole container at that temperature. If gas $A$ has mole fraction $x_A = \dfrac{n_A}{n_A+n_B+\cdots}$, then $p_A = x_A\,P_{\text{total}}$. For example, in a mixture of 2 mol $\mathrm{N_2}$ and 3 mol $\mathrm{O_2}$ at 5 atm, $p_{N_2} = \tfrac{2}{5}\times5 = 2$ atm and $p_{O_2} = 3$ atm.

Application. Used to find the pressure of a dry gas collected over water: $P_{\text{dry gas}} = P_{\text{atm}} - \text{aqueous tension}$.

Numerical.

Volume of one balloon: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\times3.1416\times(10.5)^3 = 4849\ \text{cm}^3 = 4.849\ \text{L}$

Hydrogen in the cylinder: $P_1 = 20$ atm, $V_1 = 2.82$ L, $T_1 = 300$ K. At NTP ($P_2 = 1$ atm, $T_2 = 273$ K):

$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{20\times2.82\times273}{300\times1} = 51.32\ \text{L}$

When the pressure in the cylinder falls to 1 atm, the cylinder still holds 2.82 L of hydrogen (at NTP) which cannot be used. Volume available for balloons:

$51.32 - 2.82 = 48.50\ \text{L}$

$\text{Number of balloons} = \frac{48.50}{4.849} \approx 10$

So 10 balloons can be filled.

(i) The molecules of a gas are in constant, rapid, random motion with high kinetic energy, and the attractive forces between them are negligible. This kinetic energy is enough to overcome gravity, so the molecules keep moving through the whole vessel and do not settle at the bottom.

(ii) A deliquescent substance (e.g. $\mathrm{CaCl_2}$, $\mathrm{NaOH}$) has a very low vapour pressure of its saturated solution compared with the partial pressure of water vapour in air. So it absorbs moisture from the air and dissolves in it, turning into a solution.

Boyle's law: At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.

$V \propto \frac{1}{P} \quad\text{or}\quad PV = \text{constant}\quad (T,\ n\ \text{constant})$

For two states: $P_1V_1 = P_2V_2$.

Graphs (all at constant temperature):

• (i) $P$ vs $V$: a rectangular hyperbola, with $P$ falling as $V$ rises. Each curve is an isotherm, and higher temperature gives a curve farther from the axes.
• (ii) $V$ vs $\frac{1}{P}$: a straight line passing through the origin.
• (iii) $PV$ vs $P$: a straight line parallel to the $P$-axis, since $PV$ is constant.

Practical significance: Gases are easily compressed, so large amounts can be stored in small cylinders (LPG, oxygen). It explains breathing (the lungs' volume change alters pressure), the working of syringes and pumps, and why bubbles expand as they rise in water. It is also used in the design of pneumatic devices.

Balloon problem

At $10^\circ\mathrm{C}$ ($T_1 = 283$ K), $V_1 = 960$ cc, which is $\frac{7}{8}$ of the maximum capacity.

$V_{max} = 960\times\frac{8}{7} = 1097.1\ \text{cc}$

The balloon is a flexible container, so the pressure stays constant (atmospheric) and Charles's law applies. At $25^\circ\mathrm{C}$ ($T_2 = 298$ K):

$V_2 = V_1\frac{T_2}{T_1} = 960\times\frac{298}{283} = 1010.9\ \text{cc}$

Since $1010.9 < 1097.1$ cc, the balloon will not burst at $25^\circ\mathrm{C}$.

Bursting temperature: the balloon bursts when $V = 1097.1$ cc.

$T = T_1\times\frac{V_{max}}{V_1} = 283\times\frac{8}{7} = 323.4\ \text{K}$

$t = 323.4 - 273 \approx \mathbf{50.4^\circ C}$

(d) Isomorphism

Isomorphism is the property in which two substances have the same crystalline form and a similar chemical composition (e.g. ZnSO₄·7H₂O and MgSO₄·7H₂O). Allotropism and isotopism deal with different forms of the same element, not with similar chemical composition.

Derivation of the ideal gas equation, $PV = nRT$

The ideal gas equation is obtained by combining the three gas laws:

1. Boyle's law (at constant $T$ and $n$): $V \propto \frac{1}{P} \quad \text{... (i)}$

2. Charles's law (at constant $P$ and $n$): $V \propto T \quad \text{... (ii)}$

3. Avogadro's law (at constant $T$ and $P$): $V \propto n \quad \text{... (iii)}$

Combining (i), (ii) and (iii): $V \propto \frac{nT}{P}$ $V = R\,\frac{nT}{P}$ where $R$ is the universal gas constant.

Rearranging: $\boxed{PV = nRT}$

Here $P$ = pressure, $V$ = volume, $n$ = number of moles, $T$ = absolute temperature, and $R$ = universal gas constant.

Conditions under which a real gas approaches ideal behaviour

1. At low pressure — the volume is large, so the volume occupied by the gas molecules themselves becomes negligible compared with the total volume, and intermolecular forces become very weak.
2. At high temperature — the molecules have high kinetic energy, so the attractive forces between them become negligible.

(Under these conditions the compressibility factor $Z = PV/nRT$ approaches 1.)