Cambridge International AS & A Level Chemistry - States of Matter and the Gaseous State

States of Matter and Particulate Properties

• Macroscopic and Microscopic Characteristics of the Three States of Matter:

  • Gases:

    • Possess no fixed shape and no fixed volume.

    • Particles are positioned far apart from one another, enabling gases to be easily compressed.

    • Particles are arranged completely randomly.

    • Particles move freely, rapidly, and independently from place to place in all directions.

  • Liquids:

    • Take the shape of the container they occupy while retaining a fixed volume.

    • Particles are close together with minimal free space, allowing only slight compressibility.

    • Particles are arranged mostly randomly.

    • Particles exhibit limited movement from place to place in all directions, sliding past one another.

  • Solids:

    • Possess both a fixed shape and a fixed volume.

    • Particles are touching one another, preventing solids from being compressed.

    • Particles are arranged in a regular, ordered lattice structure.

    • Particles cannot change positions with each other and are restricted to vibrating about fixed positions.

• Structural Types and Chemical Bonding:

  • The physical state of an element or compound at room temperature and pressure is governed by its structure and bonding.

  • Four primary structural classifications exist across elements and compounds:

    • Simple molecular or simple atomic: Characterized by discrete molecules or individual atoms held together by weak intermolecular attractions (e.g., carbon dioxide, $CO_2$, and argon, $Ar$).

    • Giant ionic: Characterized by electrostatic lattices of alternating cations and anions (e.g., sodium chloride, $NaCl$).

    • Giant metallic: Characterized by positive metal ions in a sea of delocalized electrons (e.g., iron, $Fe$, and copper, $Cu$).

    • Giant molecular (macromolecular): Characterized by extensive covalent networks (e.g., silicon(IV) oxide, $SiO_2$).

  • Noble Gas Atomicity:

    • Noble gases exist as isolated, single atoms (simple atomic structures).

    • They display physical properties analogous to simple molecular gases.

    • Noble gas atoms can conceptually be treated as having a molecular structure due to their shared physical behavior.

The Kinetic Theory of Gases and the Ideal Gas Model

• Definition and Scope of Kinetic Theory:

  • The kinetic theory states that particles in gases and liquids are in constant movement.

  • It provides the theoretical framework to explain how temperature and pressure influence gas volume, and explains the collision mechanics governing the rates of chemical reactions.

• Fundamental Assumptions of the Kinetic Theory of Gases:

  • Gas molecules move rapidly and randomly in straight lines in all directions.

  • The physical distance between gas molecules is significantly greater than the actual diameter of the molecules; hence, the actual volume occupied by the gas molecules themselves is negligible (approaching zero volume relative to the container).

  • No intermolecular forces of attraction or repulsion exist between the gas molecules.

  • All collisions between particles, as well as collisions between particles and container walls, are perfectly elastic, meaning zero kinetic energy is lost during collisions.

  • The absolute temperature of the gas is directly proportional to the average kinetic energy of its molecules.

• The Ideal Gas Concept:

  • An ideal gas is a theoretical construct that completely obeys all postulates of the kinetic theory under all conditions.

  • By definition, an ideal gas has particles of zero volume, zero intermolecular forces of attraction, and a volume that varies in direct proportion to absolute temperature and in inverse proportion to pressure.

• Real Gases versus Ideal Gas Behavior:

  • Real gases represent actual chemical gases that do not adhere strictly to the ideal gas law, showing pronounced non-ideal behavior at low temperatures and high pressures.

  • Noble gases with small atomic radii, such as helium ($He$) and neon ($Ne$), approach ideal gas behavior most closely because their electron clouds are tiny, causing their intermolecular dispersion forces to be extraordinarily small.

Fundamental Gas Relationships and Proportionality

• State Variables Governing Gas Volume:

  • The volume occupied by a gas is determined by its pressure ($p$), measured in pascals ($\text{Pa}$), and its absolute thermodynamic temperature ($T$), measured in kelvin ($\text{K}$).

  • Temperature Conversion:

    • The kelvin temperature equals the Celsius temperature plus $273$:

    • $T = \theta + 273$

    • Example conversion: $100\,^\circ\text{C} = 100 + 273 = 373\,\text{K}$.

• Microscopic Origin of Gas Pressure:

  • Gas pressure is generated by the continuous collisions of gas molecules against the interior walls of their container.

• Pressure-Volume Relationship (Boyle's Behavior):

  • At constant temperature, reducing the volume compresses the gas molecules into a smaller region of space.

  • The molecules strike the walls of the container with greater frequency (more collisions per unit area per second), resulting in an increase in pressure.

  • Volume is inversely proportional to pressure:

    • $V \propto \frac{1}{p}$

  • A graph plotting volume ($V$) on the vertical axis against the reciprocal of pressure ($\frac{1}{p}$) on the horizontal axis yields a straight line passing through the origin, confirming direct proportionality between $V$ and $\frac{1}{p}$.

• Temperature-Volume Relationship (Charles's Behavior):

  • When a gas is heated at constant pressure, its molecules absorb energy, move faster, and collide with the container walls with greater force and frequency.

  • To maintain constant pressure, the molecules must spread farther apart, forcing the container volume to expand.

  • At constant pressure, gas volume is directly proportional to temperature measured on the kelvin scale:

    • $V \propto T$

  • A plot of gas volume ($V$) against kelvin temperature ($T$) produces a straight line passing through absolute zero ($0\,\text{K}$).

• Mathematical Concept of Proportionality:

  • Two variables are strictly proportional if the ratio between them remains constant ($\frac{V}{T} = \text{constant}$).

  • Experimental data verification:

    • At $T = 136.5\,\text{K}$, $\text{Volume} = 11.2\,\text{dm}^3$ (ratio: $\frac{11.2}{136.5} \approx 0.08205\,\text{dm}^3\,\text{K}^{-1}$).

    • At $T = 273\,\text{K}$, $\text{Volume} = 22.4\,\text{dm}^3$ (ratio: $\frac{22.4}{273} \approx 0.08205\,\text{dm}^3\,\text{K}^{-1}$).

    • At $T = 546\,\text{K}$, $\text{Volume} = 44.8\,\text{dm}^3$ (ratio: $\frac{44.8}{546} \approx 0.08205\,\text{dm}^3\,\text{K}^{-1}$).

Limitations of the Ideal Gas Laws in Real Gases

• Kinetic Theory Violations in Real Gases:

  • Real gases fail to behave ideally because two central assumptions of the kinetic theory are violated in reality:

    • Molecules exhibit non-zero attractive intermolecular forces.

    • Gas molecules occupy a finite physical volume that cannot be ignored.

• Conditions Leading to Significant Deviations:

  • Deviations from ideality become most pronounced under conditions of very high pressures and very low temperatures.

  • High Pressure Effects:

    • Molecules are forced into close physical proximity.

    • The actual volume occupied by the gas molecules represents a substantial, non-negligible fraction of the total volume of the container.

  • Low Temperature Effects:

    • Molecules move much more slowly, having lower average kinetic energy.

    • Attractive forces (instantaneous dipole-induced dipole forces or permanent dipole-permanent dipole forces) become significant relative to kinetic energy.

    • These attractive forces pull molecules toward one another and away from the container walls.

  • Net Physical Consequences in Real Gases:

    • The measured pressure is lower than that predicted for an ideal gas due to attractive forces pulling particles inward, softening wall impacts.

    • The effective free volume available to the gas is smaller than expected for an ideal gas because molecular sizes cannot be compressed to zero.

The General Gas Equation

• Mathematical Formulation:

  • Combining the empirical gas relationships ($V \propto \frac{1}{p}$, $V \propto T$, and $V \propto n$) yields the general ideal gas equation:

    • $p V = n R T$

  • Algebraic rearrangements:

    • $n = \frac{p V}{R T}$

    • $V = \frac{n R T}{p}$

    • $p = \frac{n R T}{V}$

    • $T = \frac{p V}{n R}$

• Variables, Parameters, and Standard Units:

  • $p$: Pressure in pascals ($\text{Pa}$). Standard atmospheric pressure is defined as $1.0\,\text{atmosphere} = 101\,325\,\text{Pa}$.

  • $V$: Volume in cubic metres ($\text{m}^3$), where $1\,\text{m}^3 = 1000\,\text{dm}^3$.

  • $n$: Amount of gas in moles ($\text{mol}$), calculated via $n = \frac{m}{M_r}$.

  • $R$: Universal molar gas constant, having the value $8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$.

  • $T$: Thermodynamic temperature in kelvin ($\text{K}$).

• Unit Cancellation Principle in Computations:

  • If pressure is expressed in kilopascals ($\text{kPa}$) and volume is expressed in cubic decimetres ($\text{dm}^3$), these values can be substituted directly into $p V = n R T$ without manual unit conversion.

  • Mathematical proof: Converting $\text{kPa}$ to $\text{Pa}$ requires multiplying by $1000$, while converting $\text{dm}^3$ to $\text{m}^3$ requires dividing by $1000$. Because $p$ and $V$ are multiplied together on the same side of the equation, the two scaling factors cancel out exactly:

    • $(p \times 1000) \times \left(\frac{V}{1000}\right) = p \times V$

  • Substituting pressure in $\text{kPa}$ will always yield a calculated volume in $\text{dm}^3$.

  • Substituting pressure in $\text{Pa}$ will always yield a calculated volume in $\text{m}^3$.

• Worked Examples of General Gas Equation Calculations:

  • Worked Example 1: Volume Calculation:

    • Problem: Calculate the volume occupied by $0.500\,\text{mol}$ of carbon dioxide at a pressure of $150\,\text{kPa}$ and a temperature of $19\,^\circ\text{C}$ ($R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$).

    • Step 1: Unit Conversions:

      • $p = 150\,\text{kPa} = 150\,000\,\text{Pa}$

      • $T = 19 + 273 = 292\,\text{K}$

    • Step 2: Rearrangement:

      • $V = \frac{n R T}{p}$

    • Step 3: Computation:

      • $V = \frac{0.500 \times 8.31 \times 292}{150\,000} = 8.09 \times 10^{-3}\,\text{m}^3 = 8.09\,\text{dm}^3$

  • Worked Example 2: Pressure Calculation:

    • Problem: A flask of volume $5.00\,\text{dm}^3$ contains $4.00\,\text{g}$ of oxygen. Calculate the pressure exerted by the gas at a temperature of $127\,^\circ\text{C}$ ($R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$, $M_r(\text{oxygen}) = 32.0$).

    • Step 1: Unit Conversions and Mole Determination:

      • $T = 127 + 273 = 400\,\text{K}$

      • $V = 5.00\,\text{dm}^3 = \frac{5.00}{1000}\,\text{m}^3 = 5.00 \times 10^{-3}\,\text{m}^3$

      • $n = \frac{m}{M_r} = \frac{4.00}{32.0} = 0.125\,\text{mol}$

    • Step 2: Rearrangement:

      • $p = \frac{n R T}{V}$

    • Step 3: Computation:

      • $p = \frac{0.125 \times 8.31 \times 400}{5.00 \times 10^{-3}} = 8.31 \times 10^4\,\text{Pa}$

Determination of Relative Molecular Masses ($M_r$)

• Experimental Methodologies:

  • Mass Spectrometry: Highly precise analytical method used to measure exact relative molecular masses.

  • Laboratory Gas Density Method:

    • A less accurate but accessible school laboratory method that uses the general gas equation by determining the mass of a gas filling a large flask of known volume.

    • Weighing gases presents experimental difficulty because gases have extremely low masses and require buoyancy corrections for air, yet the method yields workable results.

  • Volatile Liquid Method:

    • Used to determine the relative molecular mass of a volatile liquid.

    • A liquid sample is injected into a gas syringe enclosed within a temperature-controlled syringe oven.

    • The liquid fully vaporizes, and the resulting volume of vapour is read directly from the syringe scale at known temperature and atmospheric pressure.

• Mathematical Derivation for Molecular Mass:

  • Substituting the mole expression $n = \frac{m}{M_r}$ into the general gas equation gives:

    • $p V = \frac{m}{M_r} R T$

  • Rearranging explicitly for relative molecular mass ($M_r$):

    • $M_r = \frac{m R T}{p V}$

• Worked Example 3: Determination of $M_r$ from Experimental Data:

  • Problem: A flask of volume $2.00\,\text{dm}^3$ was found to contain $5.28\,\text{g}$ of a gas. The pressure in the flask was $200\,\text{kPa}$ and the temperature was $20\,^\circ\text{C}$. Calculate the relative molecular mass of the gas ($R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$).

  • Step 1: Unit Conversions:

    • $p = 200\,\text{kPa} = 2.00 \times 10^5\,\text{Pa}$

    • $T = 20 + 273 = 293\,\text{K}$

    • $V = 2.00\,\text{dm}^3 = \frac{2.00}{1000}\,\text{m}^3 = 2.00 \times 10^{-3}\,\text{m}^3$

  • Step 2: Formula Expression:

    • $M_r = \frac{m R T}{p V}$

  • Step 3: Computation:

    • $M_r = \frac{5.28 \times 8.31 \times 293}{(2.00 \times 10^5) \times (2.00 \times 10^{-3})} = 32.14 \approx 32\,\text{g}\,\text{mol}^{-1}$

Questions & Discussion

• Phase Changes in Terms of Closeness and Particle Motion:

  • Question: Describe the changes that occur in the closeness and motion of the particles when: (a) a solid changes to a liquid, and (b) a liquid changes to a gas.

  • Response:

    • (a) Solid to liquid: In terms of closeness, particles remain close together with very little change in separation distance (slight expansion). In terms of motion, particles break free from their fixed positions in the ordered lattice; rather than simply vibrating about fixed points, they gain translational energy and can slide past one another in limited random movement.

    • (b) Liquid to gas: In terms of closeness, particles move very far apart so that the distance between molecules becomes many times greater than their diameters. In terms of motion, particles cease having limited localized movement and move freely, rapidly, and randomly in all directions.

• Intermolecular Forces in Helium and Neon:

  • Question: Explain why the intermolecular forces in a sample of helium and neon are very small.

  • Response: Helium and neon exist as very small, individual monoatomic atoms possessing very few electrons. Because their electron counts are minimal, their electron clouds are small and hold electrons tightly, resulting in minimal temporary dipole formation. Consequently, instantaneous dipole-induced dipole forces (London dispersion forces) are exceptionally weak.

• Temperature and Pressure Conversions:

  • Question 3(a): Convert the following temperatures into the kelvin temperature: (i) $245\,^\circ\text{C}$, (ii) $-45\,^\circ\text{C}$.

    • Calculation (i): $T = 245 + 273 = 518\,\text{K}$

    • Calculation (ii): $T = -45 + 273 = 228\,\text{K}$

  • Question 3(b): How many pascals are there in $15\,\text{kPa}$?

    • Calculation: $15\,\text{kPa} = 15 \times 1000\,\text{Pa} = 15\,000\,\text{Pa}$ (or $1.5 \times 10^4\,\text{Pa}$).

• Kinetic Theory and Heating of Sealed Glass Tubes:

  • Question: Some chemical reactions involving gases are performed in sealed glass tubes that do not melt at high temperatures. If the tubes have walls that are too thin, the tubes can easily break. Use the kinetic theory of gases to explain why these tubes should not be heated to high temperatures.

  • Response: According to the kinetic theory, heating a gas increases the average kinetic energy and velocity of its molecules. As molecular speed increases inside a rigid, fixed-volume sealed glass tube, molecules hit the interior glass walls more frequently and with substantially greater impact force. This causes a steep increase in internal gas pressure. If the glass tube walls are thin, they lack the tensile strength to counter the exerted outward force, leading to rupture and explosive breakage.

• Ideal Gas Characterization and Non-Ideal Conditions:

  • Question 5(a): What is meant by the term ideal gas?

    • Response: An ideal gas is a theoretical gas whose volume varies in direct proportion to absolute temperature and in inverse proportion to pressure ($p V = n R T$). Its particles have zero volume and exert zero intermolecular forces of attraction or repulsion.

  • Question 5(b): Under what conditions do real gases differ from ideal gases? Explain your answer.

    • Response: Real gases deviate most significantly under conditions of very high pressure and very low temperature. Under high pressure, particles are compressed close together, meaning the volume occupied by the molecules themselves is no longer negligible compared to container volume. Under low temperature, particles move slowly with low kinetic energy, enabling intermolecular attractive forces (dispersion or dipole forces) to pull molecules together and away from container walls, decreasing pressure and effective volume below ideal predictions.

• Calculations Involving the General Gas Equation:

  • Question 6(a): Calculate the volume occupied by $272\,\text{g}$ of methane at a pressure of $250\,\text{kPa}$ and a temperature of $54\,^\circ\text{C}$ ($R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$; $M_r\text{ of methane} = 16.0$).

    • Step 1: Calculate moles of methane:

      • $n = \frac{m}{M_r} = \frac{272}{16.0} = 17.0\,\text{mol}$

    • Step 2: Convert units:

      • $p = 250\,\text{kPa} = 250\,000\,\text{Pa}$

      • $T = 54 + 273 = 327\,\text{K}$

    • Step 3: Solve for volume:

      • $V = \frac{n R T}{p} = \frac{17.0 \times 8.31 \times 327}{250\,000} = \frac{46\,201.77}{250\,000} = 0.1848\,\text{m}^3 = 185\,\text{dm}^3$

  • Question 6(b): The pressure exerted by $0.25\,\text{mol}$ of carbon monoxide in a $10\,\text{dm}^3$ flask is $120\,\text{kPa}$. Calculate the temperature in the flask in kelvin.

    • Step 1: Identify given variables:

      • $n = 0.25\,\text{mol}$

      • $p = 120\,\text{kPa} = 120\,000\,\text{Pa}$

      • $V = 10\,\text{dm}^3 = 10 \times 10^{-3}\,\text{m}^3 = 0.010\,\text{m}^3$

      • $R = 8.31\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$

    • Step 2: Rearrange general gas equation for temperature:

      • $T = \frac{p V}{n R}$

    • Step 3: Calculate temperature:

      • $T = \frac{120\,000 \times 0.010}{0.25 \times 8.31} = \frac{1200}{2.0775} = 577.6\,\text{K} \approx 578\,\text{K}$