4.3 States of Matter: Gas Laws, Liquid Properties & Solid Structures

4.3 States of Matter: Gas Laws, Liquid Properties & Solid Structures

Physical chemistry requires a clear understanding of the three classical states of matter. This section details gas laws, non-ideal gas corrections, kinetic theory, liquid transport properties, and solid-state crystallography.

---

Gas Laws & The Ideal Gas Equation

1. Boyle's Law: At constant temperature ($T$) and mole count ($n$), gas volume is inversely proportional to pressure: $P \propto \frac{1}{V} \implies P_1 V_1 = P_2 V_2$
2. Charles's Law: At constant pressure ($P$) and mole count ($n$), gas volume is directly proportional to absolute temperature ($T$ in Kelvin): $V \propto T \implies \frac{V_1}{T_1} = \frac{V_2}{T_2} \quad (T_{\text{Kelvin}} = ^\circ\text{C} + 273.15)$
   • Absolute Zero ($0 \text{ K} = -273.15^\circ\text{C}$) is the theoretical temperature at which gas volume shrinks to zero.
3. Avogadro's Law: Equal volumes of ideal gases at identical temperature and pressure contain equal numbers of moles ($V \propto n$).
   • Standard Molar Volume at STP ($0^\circ\text{C}$ / $273.15 \text{ K}$ and $1 \text{ atm}$): $V_m = 22.414 \text{ dm}^3/\text{mol}$ (or $\text{L/mol}$).

The Ideal Gas Equation & Gas Density

Combining these laws yields: $P V = n R T = \frac{m}{M} R T$

• Universal Gas Constant ($R$) Values:
  • $R = 0.0821 \text{ atm}\cdot\text{dm}^3\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
  • $R = 8.314 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} = 8.314 \text{ N}\cdot\text{m}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
  • $R = 62.4 \text{ mmHg}\cdot\text{dm}^3\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
• Gas Density ($d$) & Molar Mass ($M$) Formulas: $d = \frac{P M}{R T} \quad \text{and} \quad M = \frac{d R T}{P}$

Dalton's Law of Partial Pressures

Total pressure of a non-reacting gas mixture equals the sum of partial pressures: $P_{\text{total}} = P_A + P_B + P_C + \dots \quad \text{where} \quad P_A = x_A P_{\text{total}}$ ($x_A$ is the mole fraction of gas $A$).

---

Kinetic Molecular Theory & Graham's Law

Postulates of Kinetic Molecular Theory (KMT)

1. Gases consist of tiny, widely separated particles whose individual volumes are negligible compared to total volume.
2. Gas molecules are in continuous, random straight-line motion colliding elastically with each other and container walls.
3. There are no attractive or repulsive forces between ideal gas molecules.
4. Average Kinetic Energy ($\bar{E}_k$) is directly proportional to absolute temperature: $\bar{E}_k = \frac{3}{2} R T$.
5. Root Mean Square Velocity ($v_{rms}$): $v_{rms} = \sqrt{\frac{3 R T}{M}}$

Graham's Law of Diffusion / Effusion

At constant $T$ and $P$, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass ($M$) or density ($d$): $\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}} = \frac{t_2}{t_1}$ (where $t$ is effusion time for equal volumes).

---

Non-Ideal Gas Behavior & van der Waals Equation

Real gases deviate from ideal behavior at high pressure and low temperature because:

1. Intermolecular attractive forces become significant.
2. Actual molecular volume is no longer negligible compared to container volume.

Compressibility Factor ($Z$)

$Z = \frac{P V}{n R T}$

• For ideal gases, $Z = 1$.
• For real gases, $Z < 1$ at moderate pressures (attractive forces dominate) and $Z > 1$ at high pressures (repulsive volume effects dominate).

van der Waals Equation

$\left( P + \frac{n^2 a}{V^2} \right) (V - n b) = n R T$

• Constant $a$: Corrects for intermolecular attractive forces. Higher $a$ values indicate stronger IMFs and easier gas liquefaction (units: $\text{atm}\cdot\text{dm}^6\cdot\text{mol}^{-2}$).
• Constant $b$: Corrects for finite co-volume of gas molecules ($b = 4 \times \text{actual molar volume } V_m$, units: $\text{dm}^3\cdot\text{mol}^{-1}$).

Joule-Thomson Effect & Critical Phenomena

• Joule-Thomson Effect: The cooling produced when a compressed real gas expands adiabatically through a porous plug into a region of low pressure.
• Critical Temperature ($T_c$): Maximum temperature above which a gas cannot be liquefied, no matter how much pressure is applied (e.g., $T_c$ of $\text{CO}_2 = 31.1^\circ\text{C}$, $T_c$ of $\text{H}_2\text{O} = 374^\circ\text{C}$).

---

Liquid State Properties

1. Vapor Pressure: Pressure exerted by vapors in dynamic equilibrium with its liquid at a given temperature. Vapor pressure increases exponentially with temperature according to the Clausius-Clapeyron equation.
2. Boiling Point: Temperature at which liquid vapor pressure equals external atmospheric pressure.
   • Normal Boiling Point: Measured at $1 \text{ atm}$ ($760 \text{ mmHg} = 101.325 \text{ kPa}$).
   • Pressure cookers increase external pressure, raising boiling point above $100^\circ\text{C}$ to cook food faster. High altitudes lower atmospheric pressure, decreasing boiling point.
3. Surface Tension ($\gamma$): Energy required to increase liquid surface area by one unit (units: $\text{N/m}$ or $\text{J/m}^2$). Decreases with increasing temperature.
4. Viscosity ($\eta$): Internal resistance of a liquid to flow (units: $\text{N}\cdot\text{s/m}^2$ or Poise). Viscosity decreases as temperature rises due to weakened IMFs.

---

Solid State & Crystal Structures

Solids are categorized as Crystalline (sharp melting points, anisotropic, regular geometry) or Amorphous (range of melting temperatures, isotropic, pseudo-solids like glass).

Types of Crystalline Solids

Cubic Unit Cell Calculations

---

Worked Numerical Examples

Example 1: Graham's Law Effusion Ratio

Problem: Compare the rates of effusion of Methane ($\text{CH}_4$, $M = 16 \text{ g/mol}$) and Sulfur Dioxide ($\text{SO}_2$, $M = 64 \text{ g/mol}$) under identical conditions.

Solution: Using Graham's Law formula: $\frac{\text{Rate}_{\text{CH}_4}}{\text{Rate}_{\text{SO}_2}} = \sqrt{\frac{M_{\text{SO}_2}}{M_{\text{CH}_4}}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2.0$ Conclusion: Methane effuses 2.0 times faster than Sulfur Dioxide.

Example 2: Ideal Gas Density Calculation

Problem: Calculate the density of Carbon Dioxide gas ($\text{CO}_2$, $M = 44 \text{ g/mol}$) at STP ($P = 1.0 \text{ atm}, T = 273.15 \text{ K}$).

Solution: $d = \frac{P M}{R T} = \frac{1.0 \text{ atm} \times 44 \text{ g/mol}}{0.0821 \text{ atm}\cdot\text{dm}^3\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \times 273.15 \text{ K}} = \frac{44}{22.42} = 1.96 \text{ g/dm}^3$