States of Matter

1. Kinetic Molecular Theory

Postulates

1. Matter consists of particles in continuous random motion.
2. The average kinetic energy of particles is proportional to the absolute temperature: $\bar{E}_k = \frac{3}{2}k_BT$.
3. Collisions between particles are perfectly elastic (total kinetic energy is conserved).
4. Particles exert no forces on each other except during collisions.
5. The volume of individual particles is negligible compared to the total volume of the gas.

Consequences

• Gases are compressible (particles are far apart).
• Gases fill their containers (particles move freely).
• Gases have low density compared to liquids and solids.
• Mixtures of gases are always homogeneous.

Maxwell-Boltzmann Distribution

At a given temperature, molecular speeds follow a distribution:

$$f(v) = 4\pi \left(\frac{m}{2\pi k_BT}\right)^{3/2} v^2 \exp\!\left(-\frac{mv^2}{2k_BT}\right)$$

Key features:

• The most probable speed ($v_p$) is the peak of the distribution.
• The mean speed is higher than $v_p$.
• As temperature increases, the distribution broadens and shifts to higher speeds.
• All molecules in the sample do not have the same speed.

Effect of Temperature

Increasing temperature:

• Increases the average kinetic energy.
• Broadens the Maxwell-Boltzmann distribution.
• Shifts the peak to higher velocities.
• Increases the proportion of molecules with energy exceeding the activation energy.

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2. Ideal Gas Law

Derivation from Empirical Laws

Law	Relationship	Condition held constant
Boyle's	$P \propto 1/V$	$n$, $T$
Charles's	$V \propto T$	$n$, $P$
Gay-Lussac's	$P \propto T$	$n$, $V$
Avogadro's	$V \propto n$	$P$, $T$

Combining: $PV \propto nT$, giving the ideal gas equation:

$$PV = nRT$$

Symbol	Meaning	SI Unit
$P$	Pressure	$\mathrm{Pa}$ ($\mathrm{N/m}^2$)
$V$	Volume	$\mathrm{m}^3$
$n$	Amount of substance	$\mathrm{mol}$
$R$	Universal gas constant	$8.314\mathrm{ J/(mol \cdot K)}$
$T$	Temperature	$\mathrm{K}$

Molar Volume

At STP ($0\degree\mathrm{C}$, $100\mathrm{ kPa}$): $V_m = 22.7\mathrm{ L/mol}$

At RTP ($25\degree\mathrm{C}$, $100\mathrm{ kPa}$): $V_m = 24.8\mathrm{ L/mol}$

Dalton's Law of Partial Pressures

$$P_{\mathrm{total}} = P_1 + P_2 + P_3 + \cdots$$

The partial pressure of component $i$:

$$P_i = x_i \cdot P_{\mathrm{total}}$$

where $x_i = \dfrac{n_i}{n_{\mathrm{total}}}$ is the mole fraction.

Graham's Law of Effusion

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

Lighter gases effuse faster. This is the basis for gas separation by diffusion and for detecting leaks.

Common Pitfalls

• Forgetting to convert temperature to Kelvin.
• Using the wrong value of $R$ (check units of $P$, $V$, $T$).
• Confusing STP definitions (the IB uses $100\mathrm{ kPa}$, not $1\mathrm{ atm}$).

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3. Real Gases

Deviations from Ideal Behaviour

Real gases deviate from ideal behaviour when:

• Pressure is high: molecules are close together, so their finite volume matters and intermolecular forces become significant.
• Temperature is low: molecules move slowly, so intermolecular forces have a greater effect.

Van der Waals Equation

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

Parameter	Meaning
$a$	Corrects for intermolecular attractions
$b$	Corrects for the finite volume of molecules

• The $a$ term increases the effective pressure (attractions reduce the pressure exerted on the walls).
• The $b$ term reduces the effective volume (molecules occupy space).

Comparing Gases

Gas	$a$ ($\mathrm{L}^2\cdot\mathrm{atm/mol}^2$)	$b$ ($\mathrm{L/mol}$)	Behaviour
He	$0.034$	$0.024$	Closest to ideal
$\mathrm{N}_2$	$1.39$	$0.039$	Moderate deviations
$\mathrm{CO}_2$	$3.59$	$0.043$	Largest deviations

Gases with larger $a$ values (stronger intermolecular forces) and larger $b$ values (larger molecules) deviate more from ideal behaviour.

Compressibility Factor

Z = \frac`\{PV}``\{nRT}`

• $Z = 1$: ideal gas
• $Z \lt 1$: intermolecular attractions dominate (low $T$, moderate $P$)
• $Z \gt 1$: molecular volume dominates (high $P$)

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4. Intermolecular Forces

Types and Relative Strengths

Force	Strength	Present in
Covalent/ionic bonds	Strong	Within molecules/ionic lattices
Ion-dipole	Moderate	Ion + polar molecule
Hydrogen bonding	Moderate	H bonded to N, O, or F
Dipole-dipole	Weak	Polar molecules
London dispersion	Weakest	All molecules (temporary dipoles)

London Dispersion Forces

Definition. London dispersion forces arise from instantaneous dipoles created by the uneven distribution of electrons. These induce dipoles in neighbouring molecules.

Factors affecting strength:

1. Number of electrons: more electrons = larger electron cloud = greater polarizability = stronger dispersion forces.
2. Molecular shape: larger surface area of contact = stronger forces. Linear molecules have stronger dispersion forces than branched isomers.

Comparison	Stronger LDF	Reason
$\mathrm{F}_2$ vs $\mathrm{Cl}_2$	$\mathrm{Cl}_2$	More electrons
$\mathrm{C}_5\mathrm{H}_{12}$ isomers	n-pentane	Greater surface area than branched isomers

Dipole-Dipole Forces

Polar molecules have permanent dipoles. The positive end of one molecule attracts the negative end of another:

$$\mathrm{HCl}\cdots\mathrm{HCl}$$

Hydrogen Bonding

Definition. A hydrogen bond is a strong dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and a lone pair on another electronegative atom.

Requirements:

1. Hydrogen bonded directly to N, O, or F.
2. A lone pair available on the acceptor atom.

Effects of Hydrogen Bonding

Property	Effect	Example
Boiling point	Much higher than expected	$\mathrm{H}_2\mathrm{O}$ (bp $100\degree\mathrm{C}$) vs $\mathrm{H}_2\mathrm{S}$ (bp $-60\degree\mathrm{C}$)
Viscosity	Higher	Glycerol, $\mathrm{H}_2\mathrm{O}$
Surface tension	Higher	Water has high surface tension
Solubility	Enhances solubility of polar species	Glucose dissolves in water
Density anomaly	Ice less dense than liquid water	Open hydrogen-bonded structure of ice
Biological specificity	DNA base pairing, protein folding	Complementary H-bond donors/acceptors

Anomalous Properties of Water

Water's maximum density is at $4\degree\mathrm{C}$, not $0\degree\mathrm{C}$. In ice, each water molecule forms four hydrogen bonds in a tetrahedral arrangement, creating an open lattice with lower density than liquid water.

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5. Phase Diagrams

Features

A phase diagram shows the conditions of temperature and pressure at which each phase of a substance is stable.

Key features:

• Triple point: the unique temperature and pressure at which all three phases coexist in equilibrium.
• Critical point: the highest temperature and pressure at which a liquid can exist. Above $T_c$, the distinction between liquid and gas disappears (supercritical fluid).
• Normal melting point: the temperature at which solid and liquid are in equilibrium at $1\mathrm{ atm}$.
• Normal boiling point: the temperature at which liquid and gas are in equilibrium at $1\mathrm{ atm}$.

Phase Boundaries

Boundary	Process	Condition
Solid--liquid	Melting/freezing	Equilibrium curve
Liquid--gas	Vaporisation	Vapour pressure = external pressure
Solid--gas	Sublimation	Direct transition

Water's Phase Diagram

The solid--liquid boundary for water has a negative slope (unlike most substances). This is because ice is less dense than liquid water, so increasing pressure favours the denser liquid phase and lowers the melting point.

$\mathrm{CO}_2$ Phase Diagram

The triple point of $\mathrm{CO}_2$ is at $-57\degree\mathrm{C}$ and $5.1\mathrm{ atm}$. At $1\mathrm{ atm}$, $\mathrm{CO}_2$ sublimates directly from solid to gas — hence the name "dry ice."

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6. States of Matter: Condensed Phases

Liquids

• Particles are close together but can move past one another.
• Have a fixed volume but no fixed shape.
• Evaporation occurs when molecules at the surface have sufficient kinetic energy to escape.
• Boiling occurs when the vapour pressure equals the external pressure.

Solids

Property	Ionic Solids	Molecular Solids	Covalent Network	Metallic Solids
Particles	Ions	Molecules	Atoms	Metal cations + delocalised electrons
Bonding	Electrostatic	Intermolecular	Covalent	Metallic
Melting point	High	Low	Very high	Variable
Conductivity	Conduct when molten	Insulators	Insulators (except Si)	Good conductors
Solubility	Often in polar solvents	Varies	Insoluble	Insoluble

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Practice Problems

Problem 1

Explain why the boiling point of propan-1-ol ($97\degree\mathrm{C}$) is much higher than that of butane ($-1\degree\mathrm{C}$), even though butane has a larger molar mass.

Solution:

Propan-1-ol can form hydrogen bonds between the O--H group and lone pairs on neighbouring molecules. Hydrogen bonds are much stronger than the London dispersion forces that are the only intermolecular forces in butane. Despite butane having more electrons (stronger dispersion forces), the hydrogen bonding in propan-1-ol dominates and requires more energy to overcome.

Problem 2

A gas occupies $5.00\mathrm{ L}$ at $2.00\mathrm{ atm}$ and $300\mathrm{ K}$. Calculate the volume at $5.00\mathrm{ atm}$ and $400\mathrm{ K}$.

Solution:

Using the combined gas law:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$$$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{2.00 \times 5.00 \times 400}{5.00 \times 300} = \frac{4000}{1500} = 2.67\mathrm{ L}$$

Problem 3

Explain why $\mathrm{NH}_3$ has a higher boiling point ($-33\degree\mathrm{C}$) than $\mathrm{PH}_3$ ($-88\degree\mathrm{C}$), but the boiling point of $\mathrm{H}_2\mathrm{O}$ ($100\degree\mathrm{C}$) is higher than that of $\mathrm{NH}_3$ despite both having hydrogen bonding.

Solution:

$\mathrm{NH}_3$ vs $\mathrm{PH}_3$: $\mathrm{NH}_3$ forms hydrogen bonds (H bonded to N), which are much stronger than the dipole-dipole and dispersion forces in $\mathrm{PH}_3$ (P is not electronegative enough for H-bonding).

$\mathrm{H}_2\mathrm{O}$ vs $\mathrm{NH}_3$: Each water molecule can form up to two hydrogen bonds (two O--H donors and two lone pair acceptors), whereas each $\mathrm{NH}_3$ molecule can form only one hydrogen bond (one N--H donor, but the lone pair on N is partly delocalised). Additionally, water forms a more extensive three-dimensional hydrogen-bond network.

Problem 4

Using the van der Waals equation, calculate the pressure exerted by $1.00\mathrm{ mol}$ of $\mathrm{CO}_2$ in a $5.00\mathrm{ L}$ container at $300\mathrm{ K}$. Compare with the ideal gas prediction. For $\mathrm{CO}_2$: $a = 3.59\mathrm{ L}^2\cdot\mathrm{atm/mol}^2$, $b = 0.043\mathrm{ L/mol}$.

Solution:

Ideal gas law:

P_{\mathrm{ideal}} = \frac`\{nRT}`{V} = \frac{1.00 \times 0.0821 \times 300}{5.00} = \frac{24.63}{5.00} = 4.93\mathrm{ atm}

Van der Waals:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$$$\left(P + \frac{3.59 \times 1}{25.0}\right)(5.00 - 0.043) = 24.63$$$$(P + 0.1436)(4.957) = 24.63$$$$P + 0.1436 = \frac{24.63}{4.957} = 4.970$$$$P = 4.970 - 0.1436 = 4.83\mathrm{ atm}$$

The van der Waals pressure ($4.83\mathrm{ atm}$) is lower than the ideal gas pressure ($4.93\mathrm{ atm}$) because intermolecular attractions reduce the effective pressure.

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Worked Examples

Worked Example: Ideal Gas Law with Unit Consistency

A sample of nitrogen gas occupies $250.0\mathrm{ mL}$ at $25\degree\mathrm{C}$ and $98.5\mathrm{ kPa}$. Calculate the number of moles of nitrogen.

Solution

Convert all quantities to SI units:

• $V = 250.0\mathrm{ mL} = 0.2500\mathrm{ L} = 2.500 \times 10^{-4}\mathrm{ m}^3$
• $T = 25\degree\mathrm{C} = 298\mathrm{ K}$
• $P = 98.5\mathrm{ kPa} = 98\,500\mathrm{ Pa}$

Using $R = 8.314\mathrm{ J/(mol \cdot K)}$ (SI units):

n = \frac`\{PV}``\{RT}` = \frac{98\,500 \times 2.500 \times 10^{-4}}{8.314 \times 298} = \frac{24.63}{2477.6} = 0.00994\mathrm{ mol}

Alternatively, using $R = 8.314 \times 10^{-2}\mathrm{ L \cdot kPa/(mol \cdot K)}$ and keeping $V$ in litres and $P$ in kPa:

$$n = \frac{98.5 \times 0.2500}{8.314 \times 10^{-2} \times 298} = \frac{24.63}{247.8} = 0.00994\mathrm{ mol}$$

Both approaches give $n = 0.00994\mathrm{ mol}$ (approximately $0.01\mathrm{ mol}$).

Worked Example: Graham's Law of Effusion

A sample of an unknown gas effuses through a small opening at a rate that is $0.387$ times the rate of effusion of $\mathrm{O}_2$ under identical conditions. Determine the molar mass of the unknown gas and identify it.

Solution

Graham's law:

$$\frac{r_{\mathrm{unknown}}}{r_{\mathrm{O}_2}} = \sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{unknown}}}}$$

Given $r_{\mathrm{unknown}}/r_{\mathrm{O}_2} = 0.387$ and $M_{\mathrm{O}_2} = 32.0\mathrm{ g/mol}$:

$$0.387 = \sqrt{\frac{32.0}{M_{\mathrm{unknown}}}}$$$$0.387^2 = \frac{32.0}{M_{\mathrm{unknown}}}$$$$0.1498 = \frac{32.0}{M_{\mathrm{unknown}}}$$$$M_{\mathrm{unknown}} = \frac{32.0}{0.1498} = 213.5\mathrm{ g/mol}$$

The molar mass is approximately $214\mathrm{ g/mol}$. This is close to that of mercury vapour ($\mathrm{Hg}$, $A_r = 200.6$) or radon ($\mathrm{Rn}$, $A_r = 222$). Given typical IB problem contexts, this is most consistent with radon gas.

Worked Example: Dalton's Law of Partial Pressures

A mixture of $2.00\mathrm{ g}$ of $\mathrm{H}_2$ and $14.0\mathrm{ g}$ of $\mathrm{N}_2$ is confined in a $10.0\mathrm{ L}$ vessel at $27\degree\mathrm{C}$. Calculate the partial pressure of each gas and the total pressure.

Solution

$$n(\mathrm{H}_2) = \frac{2.00}{2.02} = 0.990\mathrm{ mol}$$$$n(\mathrm{N}_2) = \frac{14.0}{28.0} = 0.500\mathrm{ mol}$$$$n_{\mathrm{total}} = 0.990 + 0.500 = 1.490\mathrm{ mol}$$$$T = 27\degree\mathrm{C} = 300\mathrm{ K}$$

Using the ideal gas law for the total mixture ($R = 8.314 \times 10^{-2}\mathrm{ L \cdot kPa/(mol \cdot K)}$):

$$P_{\mathrm{total}} = \frac{n_{\mathrm{total}}RT}{V} = \frac{1.490 \times 8.314 \times 10^{-2} \times 300}{10.0} = \frac{371.6}{10.0} = 371.6\mathrm{ kPa}$$

Mole fractions:

$$x_{\mathrm{H}_2} = \frac{0.990}{1.490} = 0.664$$$$x_{\mathrm{N}_2} = \frac{0.500}{1.490} = 0.336$$

Partial pressures:

$$P_{\mathrm{H}_2} = 0.664 \times 371.6 = 247\mathrm{ kPa}$$$$P_{\mathrm{N}_2} = 0.336 \times 371.6 = 125\mathrm{ kPa}$$

Check: $247 + 125 = 372\mathrm{ kPa}$ (rounding difference from $371.6\mathrm{ kPa}$).

Worked Example: Interpreting a Phase Diagram

On the phase diagram of water, describe what happens when solid ice at $-10\degree\mathrm{C}$ is subjected to increasing pressure at constant temperature. Explain why this behaviour differs from that of most substances.

Solution

On the phase diagram of water, starting at $-10\degree\mathrm{C}$ (solid phase) and increasing pressure at constant temperature, you move horizontally to the right. You cross the solid--liquid boundary, meaning the ice melts into liquid water.

This is unusual because the solid--liquid boundary on water's phase diagram has a negative slope. For nearly all other substances, increasing pressure at constant temperature in the solid region does not cross the melting curve — instead, the solid remains solid or the melting point increases.

The negative slope for water arises because ice is less dense than liquid water. Increasing pressure favours the phase with the smaller volume (higher density), which is liquid water. Le Chatelier's principle predicts that the system shifts toward the denser phase when pressure increases.

Worked Example: Boiling Point Comparison Using IMF Analysis

Explain the boiling point trend: $\mathrm{CH}_4$ ($-162\degree\mathrm{C}$) $\lt$ $\mathrm{SiH}_4$ ($-112\degree\mathrm{C}$) $\lt$ $\mathrm{GeH}_4$ ($-88\degree\mathrm{C}$).

Solution

All three molecules are tetrahedral and non-polar, so the only intermolecular forces present are London dispersion forces (LDFs). LDF strength depends on the number of electrons and the polarizability of the electron cloud.

Going down Group 14: $\mathrm{C}$ (6 electrons) $\to$ $\mathrm{Si}$ (14 electrons) $\to$ $\mathrm{Ge}$ (32 electrons). The increasing number of electrons produces larger, more polarizable electron clouds, which generate stronger instantaneous dipoles and thus stronger LDFs.

Since boiling requires overcoming intermolecular forces, stronger LDFs mean a higher boiling point. The trend is entirely consistent with increasing LDF strength down the group.

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Common Pitfalls

• Using Celsius instead of Kelvin in gas calculations: The ideal gas law requires absolute temperature. A gas at $27\degree\mathrm{C}$ has $T = 300\mathrm{ K}$, not $27$. Forgetting this conversion typically produces answers that are off by a factor of $\approx 10$.

• Choosing the wrong value of $R$: $R = 8.314\mathrm{ J/(mol \cdot K)}$ requires SI units ($\mathrm{Pa}$, $\mathrm{m}^3$). $R = 0.0821\mathrm{ L \cdot atm/(mol \cdot K)}$ requires pressure in atm and volume in litres. Mixing unit systems (e.g., kPa with $0.0821$) yields incorrect results.

• Confusing STP definitions: The IB uses STP as $0\degree\mathrm{C}$ and $100\mathrm{ kPa}$, giving $V_m = 22.7\mathrm{ L/mol}$. Many textbooks use the older IUPAC definition ($1\mathrm{ atm} = 101.3\mathrm{ kPa}$, $V_m = 22.4\mathrm{ L/mol}$). Using the wrong molar volume introduces a $\approx 1.3\%$ error.

• Claiming hydrogen bonding without checking all criteria: A molecule must have H bonded directly to N, O, or F. $\mathrm{CH}_4$, $\mathrm{HCl}$, and $\mathrm{PH}_3$ do not exhibit hydrogen bonding. $\mathrm{H}$ bonded to C is never sufficient, regardless of molecular polarity.

• Misapplying Graham's law to mixtures: Graham's law compares rates of effusion for two pure gases. It does not directly give the composition of a gas mixture after partial effusion (the lighter component enriches preferentially, changing the composition over time).

• Ignoring molecular shape when comparing LDFs: $n$-pentane and neopentane (2,2-dimethylpropane) have the same molar mass and number of electrons, but $n$-pentane (linear) has a higher boiling point ($36\degree\mathrm{C}$) than neopentane (spherical, $9.5\degree\mathrm{C}$) because the linear shape allows greater surface-area contact and stronger LDFs.

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Exam-Style Problems

1. [Medium] A sealed container holds $0.200\mathrm{ mol}$ of $\mathrm{CO}_2$ at $50.0\degree\mathrm{C}$ and a pressure of $250\mathrm{ kPa}$. Calculate the volume of the container in litres.

2. [Medium] Explain why $\mathrm{HF}$ has a lower boiling point ($19.5\degree\mathrm{C}$) than $\mathrm{H}_2\mathrm{O}$ ($100\degree\mathrm{C}$), even though fluorine is more electronegative than oxygen.

3. [Hard] Using the van der Waals equation, calculate the pressure of $2.00\mathrm{ mol}$ of $\mathrm{NH}_3$ in a $5.00\mathrm{ L}$ container at $400\mathrm{ K}$. Given $a = 4.17\mathrm{ L}^2 \cdot \mathrm{atm/mol}^2$ and $b = 0.037\mathrm{ L/mol}$. Compare with the ideal gas prediction and explain the direction of the deviation.

4. [Hard] A gas mixture contains $\mathrm{He}$, $\mathrm{Ne}$, and $\mathrm{Ar}$ with mole fractions $0.40$, $0.35$, and $0.25$ respectively. The total pressure is $120\mathrm{ kPa}$. Calculate the partial pressure of each component and the total mass of gas in a $10.0\mathrm{ L}$ container at $300\mathrm{ K}$.

5. [Medium] On a phase diagram, label the triple point, critical point, and all phase boundaries. Explain why the critical point represents the temperature above which a gas cannot be liquefied by pressure alone.

6. [Hard] Explain why $\mathrm{CCl}_4$ (molar mass $154\mathrm{ g/mol}$) is a liquid at room temperature (bp $76.7\degree\mathrm{C}$) while $\mathrm{Ar}$ (molar mass $40\mathrm{ g/mol}$) is a gas (bp $-186\degree\mathrm{C}$), even though both are non-polar and only experience London dispersion forces.

7. [Medium] A sample of gas effuses through an apparatus in $42\mathrm{ s}$. An equal number of moles of an unknown gas effuses through the same apparatus in $78\mathrm{ s}$. If the first gas is $\mathrm{CO}_2$, calculate the molar mass of the unknown gas.

8. [Hard] At high pressures, real gases can have $Z \gt 1$. Explain the molecular-level origin of this deviation. Under what conditions of temperature and molecular identity would you expect this deviation to be most significant?

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Worked Examples (Expanded)

Worked Example: RMS Speed and the Maxwell-Boltzmann Distribution

Calculate the root-mean-square speed of $\mathrm{O}_2$ molecules at $298\;\mathrm{K}$ and at $500\;\mathrm{K}$. By what factor does the rms speed increase? ($M_r(\mathrm{O}_2) = 32.0\;\mathrm{g/mol}$, $R = 8.314\;\mathrm{J/(mol \cdot K)}$)

Solution

$v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$

At $298\;\mathrm{K}$: $v_{\mathrm{rms}} = \sqrt{\frac{3 \times 8.314 \times 298}{0.0320}} = \sqrt{\frac{7429}{0.0320}} = \sqrt{232\,154} = 482\;\mathrm{m/s}$

At $500\;\mathrm{K}$: $v_{\mathrm{rms}} = \sqrt{\frac{3 \times 8.314 \times 500}{0.0320}} = \sqrt{\frac{12471}{0.0320}} = \sqrt{389\,719} = 624\;\mathrm{m/s}$

Factor of increase: $\frac{624}{482} = 1.29$

Since $v_{\mathrm{rms}} \propto \sqrt{T}$, the factor is $\sqrt{500/298} = \sqrt{1.678} = 1.29$, consistent with the calculation.

Worked Example: Critical Point and Supercritical Fluid

The critical point of $\mathrm{CO}_2$ is at $T_c = 304.3\;\mathrm{K}$ ($31.1\degree\mathrm{C}$) and $P_c = 7.38\;\mathrm{MPa}$ ($72.8\;\mathrm{atm}$). (a) Explain what happens at the critical point. (b) $\mathrm{CO}_2$ is commonly used as a supercritical fluid in decaffeination. If $\mathrm{CO}_2$ is held at $40\degree\mathrm{C}$ and $80\;\mathrm{atm}$, is it supercritical? (c) State two properties of supercritical $\mathrm{CO}_2$ that make it useful as a solvent.

Solution

(a) At the critical point, the densities of the liquid and gas phases become identical. Above $T_c$ and $P_c$, no phase boundary exists between liquid and gas — the substance exists as a single, dense phase called a supercritical fluid. The meniscus between liquid and gas disappears.

(b) The conditions are $T = 40\degree\mathrm{C} = 313\;\mathrm{K}$, which is above $T_c = 304\;\mathrm{K}$, and $P = 80\;\mathrm{atm}$, which is above $P_c = 72.8\;\mathrm{atm}$. Since both $T > T_c$ and $P > P_c$, the $\mathrm{CO}_2$ is in the supercritical state.

(c) Supercritical $\mathrm{CO}_2$ has a density similar to a liquid (good solvating power) but a viscosity similar to a gas (high diffusivity). This allows it to penetrate porous solids (like coffee beans) rapidly and dissolve non-polar organic compounds (like caffeine) efficiently. It is also non-toxic, non-flammable, and easily removed by depressurisation, leaving no solvent residue.

Worked Example: Clausius-Clapeyron Equation and Vapour Pressure

The vapour pressure of water is $2.34\;\mathrm{kPa}$ at $20\degree\mathrm{C}$ and $7.38\;\mathrm{kPa}$ at $40\degree\mathrm{C}$. Use the Clausius-Clapeyron equation to estimate the enthalpy of vaporisation of water.

Solution

The integrated Clausius-Clapeyron equation:

$\ln\!\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Substituting: $P_1 = 2.34\;\mathrm{kPa}$, $T_1 = 293\;\mathrm{K}$ $P_2 = 7.38\;\mathrm{kPa}$, $T_2 = 313\;\mathrm{K}$

$\ln\!\left(\frac{7.38}{2.34}\right) = -\frac{\Delta H_{\mathrm{vap}}}{8.314}\left(\frac{1}{313} - \frac{1}{293}\right)$

$\ln(3.154) = -\frac{\Delta H_{\mathrm{vap}}}{8.314}\left(0.003195 - 0.003413\right)$

$1.148 = -\frac{\Delta H_{\mathrm{vap}}}{8.314}(-0.000218)$

$1.148 = \frac{\Delta H_{\mathrm{vap}} \times 0.000218}{8.314}$

$\Delta H_{\mathrm{vap}} = \frac{1.148 \times 8.314}{0.000218} = 43800\;\mathrm{J/mol} = 43.8\;\mathrm{kJ/mol}$

The accepted literature value is $44.0\;\mathrm{kJ/mol}$. The close agreement validates the use of the Clausius-Clapeyron equation over this temperature range.

Worked Example: Heating Curve and Enthalpy Calculations

A $50.0\;\mathrm{g}$ sample of ice at $-20\degree\mathrm{C}$ is heated to $120\degree\mathrm{C}$ at standard pressure. Calculate the total energy required. Given: specific heat of ice $= 2.09\;\mathrm{J/(g \cdot \degree C)}$, specific heat of water $= 4.18\;\mathrm{J/(g \cdot \degree C)}$, specific heat of steam $= 2.01\;\mathrm{J/(g \cdot \degree C)}$, $\Delta H_{\mathrm{fusion}} = 334\;\mathrm{J/g}$, $\Delta H_{\mathrm{vap}} = 2260\;\mathrm{J/g}$.

Solution

Step 1: Heating ice from $-20\degree\mathrm{C}$ to $0\degree\mathrm{C}$ $q_1 = m \cdot c_{\mathrm{ice}} \cdot \Delta T = 50.0 \times 2.09 \times 20 = 2090\;\mathrm{J}$

Step 2: Melting ice at $0\degree\mathrm{C}$ $q_2 = m \cdot \Delta H_{\mathrm{fusion}} = 50.0 \times 334 = 16700\;\mathrm{J}$

Step 3: Heating water from $0\degree\mathrm{C}$ to $100\degree\mathrm{C}$ $q_3 = m \cdot c_{\mathrm{water}} \cdot \Delta T = 50.0 \times 4.18 \times 100 = 20900\;\mathrm{J}$

Step 4: Vaporising water at $100\degree\mathrm{C}$ $q_4 = m \cdot \Delta H_{\mathrm{vap}} = 50.0 \times 2260 = 113000\;\mathrm{J}$

Step 5: Heating steam from $100\degree\mathrm{C}$ to $120\degree\mathrm{C}$ $q_5 = m \cdot c_{\mathrm{steam}} \cdot \Delta T = 50.0 \times 2.01 \times 20 = 2010\;\mathrm{J}$

Total: $q_{\mathrm{total}} = 2090 + 16700 + 20900 + 113000 + 2010 = 154\,700\;\mathrm{J} = 155\;\mathrm{kJ}$

Note that vaporisation alone accounts for $\frac{113000}{154700} = 73\%$ of the total energy, demonstrating the dominance of the enthalpy of vaporisation.

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Exam-Style Problems (Expanded)

Problem 9: Quantitative -- Boltzmann Distribution and Activation Energy

For a reaction with activation energy $E_a = 75.0\;\mathrm{kJ/mol}$, calculate the ratio of molecules with sufficient energy to react at $350\;\mathrm{K}$ versus $300\;\mathrm{K}$. ($R = 8.314\;\mathrm{J/(mol \cdot K)}$). Use the Boltzmann factor $\frac{n(E > E_a)}{n_{\mathrm{total}}} \propto e^{-E_a/RT}$. Explain the significance of this ratio for the effect of temperature on reaction rate.

Problem 10: Extended Response -- Boiling Point Trends in Hydrogen Halides

The boiling points of the hydrogen halides are: $\mathrm{HF}$ ($19.5\degree\mathrm{C}$), $\mathrm{HCl}$ ($-85\degree\mathrm{C}$), $\mathrm{HBr}$ ($-67\degree\mathrm{C}$), $\mathrm{HI}$ ($-35\degree\mathrm{C}$). (a) Explain the anomalous boiling point of $\mathrm{HF}$. (b) Explain the trend from $\mathrm{HCl}$ to $\mathrm{HI}$. (c) Predict the boiling point of $\mathrm{H}_2\mathrm{O}$ relative to $\mathrm{H}_2\mathrm{S}$ and explain, given that the boiling point of $\mathrm{H}_2\mathrm{S}$ is $-60\degree\mathrm{C}$.

Problem 11: Quantitative -- Vapour Pressure and Boiling Point at Altitude

At the summit of Mount Everest (altitude $8848\;\mathrm{m}$), atmospheric pressure is approximately $33\;\mathrm{kPa}$. Using the Clausius-Clapeyron equation with $\Delta H_{\mathrm{vap}} = 40.7\;\mathrm{kJ/mol}$ for water, estimate the boiling point of water at this altitude. ($R = 8.314\;\mathrm{J/(mol \cdot K)}$; normal boiling point $= 100\degree\mathrm{C}$ at $101.3\;\mathrm{kPa}$)

Problem 12: Extended Response -- Liquid Crystal Phases

Liquid crystals are materials that exhibit phases intermediate between isotropic liquids and crystalline solids. (a) Describe the nematic and smectic phases of liquid crystals, explaining the degree of molecular order in each. (b) Explain how the application of an electric field can change the optical properties of a nematic liquid crystal. (c) State one application of liquid crystals and explain how the phase transition temperature limits their use.

Problem 13: Data Analysis -- Compressibility Factor

A gas has a compressibility factor $Z = 0.87$ at $200\;\mathrm{K}$ and $10\;\mathrm{atm}$. (a) Calculate the actual molar volume of the gas and compare it with the ideal gas prediction. (b) Explain whether intermolecular attractions or molecular volume effects dominate at these conditions. (c) Predict qualitatively whether $Z$ would be greater than, less than, or equal to $1$ at $500\;\mathrm{K}$ and $100\;\mathrm{atm}$ for the same gas.

Problem 14: Extended Response -- Phase Diagram Interpretation

The phase diagram of a substance shows a triple point at $T = 150\;\mathrm{K}$, $P = 0.5\;\mathrm{atm}$, a normal melting point of $180\;\mathrm{K}$, and a critical point at $T = 350\;\mathrm{K}$, $P = 45\;\mathrm{atm}$. The solid-liquid boundary has a positive slope. (a) Sketch the phase diagram. (b) Describe what happens when the substance at $200\;\mathrm{K}$ and $1.0\;\mathrm{atm}$ is compressed to $50\;\mathrm{atm}$ at constant temperature. (c) Explain why the solid-liquid boundary has a positive slope for this substance but a negative slope for water. (d) At what minimum temperature can the substance be liquefied by pressure alone?

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Common Pitfalls (Expanded)

• Confusing rms speed, average speed, and most probable speed: For the Maxwell-Boltzmann distribution, $v_{\mathrm{mp}} < \bar{v} < v_{\mathrm{rms}}$. The relationships are $v_{\mathrm{rms}} = \sqrt{3RT/M}$, $\bar{v} = \sqrt{8RT/(\pi M)}$, and $v_{\mathrm{mp}} = \sqrt{2RT/M}$. Each differs by a constant factor.

• Using the Clausius-Clapeyron equation outside its validity range: The equation assumes constant $\Delta H_{\mathrm{vap}}$ over the temperature range, which is a reasonable approximation for narrow ranges but breaks down over large temperature intervals or near the critical point.

• Assuming the heating curve is linear: During a phase transition (melting or boiling), temperature remains constant while energy is absorbed or released. The heating curve is flat during these transitions — this is the latent heat, not a failure of heating.

• Confusing vapour pressure with gas pressure: Vapour pressure is the pressure exerted by the vapour in equilibrium with its liquid at a given temperature. It is a property of the substance and temperature only, not of the container volume or the amount of liquid.

• Applying Graham's law to diffusion through porous media: Graham's law gives the ratio of effusion rates through a small orifice. Diffusion through air is more complex because gas molecules collide with air molecules, not just the orifice wall. The calculated Graham's law ratio is an upper bound.

• Ignoring the role of unit conversions in gas density calculations: Gas density $\rho = PM/(RT)$ requires $P$ in $\mathrm{Pa}$, $M$ in $\mathrm{kg/mol}$, $R = 8.314\;\mathrm{J/(mol \cdot K)}$, and $T$ in $\mathrm{K}$ to get density in $\mathrm{kg/m}^3$. Using $M$ in $\mathrm{g/mol}$ gives a result that is off by a factor of 1000.

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If You Get These Wrong, Revise:

• Atomic structure and electron configuration → Review ./atomic-theory
• Bonding and intermolecular forces in depth → Review ./chemical-bonding-advanced
• Thermodynamics (energy changes in state transitions) → Review ./thermodynamics/thermochemistry
• Stoichiometry and the mole concept → Review ./stoichiometry/stoichiometry
• Periodic trends affecting physical properties → Review ./periodicity