CLEP Chemistry Study Guide: The States of Matter

Ideal Gas Laws and Assumptions

• Definition of Ideal Gases: Ideal gases are gases that behave according to an approximation that includes the following two assumptions:

  1. The volume of the gas molecule is negligible compared to the space between the molecules.

  2. There is negligible intermolecular attraction between gas molecules.

• Accuracy of the Ideal Gas Approximation: The ideal gas approximation is most accurate for gases at low pressure and high temperature.

Classical Gas Laws

• Boyle's Law: This law states that the volume of a gas is inversely proportional to pressure, when temperature is constant.

  • Mathematical Expression: $P_1 V_1 = P_2 V_2$

  • Example: A $4.0\,\text{liter}$ elastic weather balloon travels from sea level, at $1.0\,\text{atm}$ pressure, to a higher altitude, where the pressure is $0.20\,\text{atm}$. What is the new volume of the balloon?

  • Solution:         $V_2 = \frac{P_1 V_1}{P_2}$         $V_2 = \frac{1.0\,\text{atm} \times 4.0\,L}{0.20\,\text{atm}}$         $V_2 = 20.0\,L$

• Charles's Law: This law states that the volume of a given amount of gas is directly proportional to temperature, when pressure is constant.

  • Mathematical Expression: $V_1 T_2 = V_2 T_1$

  • Example: A gas occupies $2.0\,L$ at $300\,K$. What is the volume of the gas at $200\,K$, assuming that the pressure is constant?

  • Solution:         $V_2 = \frac{V_1 T_2}{T_1}$         $V_2 = \frac{2.0\,L \times 200\,K}{300\,K}$         $V_2 = 1.3\,L$

• Law of Gay-Lussac: This law states that at constant volume, the pressure exerted by a given mass of gas varies directly with the absolute temperature.

  • Mathematical Expression: $P_1 T_2 = P_2 T_1$

  • Example: A gas in a rigid container exerts $6.0\,\text{atm}$ at $300\,K$. What is the pressure that the gas exerts at $500\,K$?

  • Solution:         $P_2 = \frac{P_1 T_2}{T_1}$         $P_2 = \frac{6.0\,\text{atm} \times 500\,K}{300\,K}$         $P_2 = 10\,\text{atm}$

• Gay-Lussac's Law of Combining Volumes: This law states that when reactions take place in a gaseous state at constant temperature and pressure, the volume of reactants and products can be expressed as the ratios given by the stoichiometric coefficients in the balanced reaction.

• Avogadro's Law: This law states that under conditions of constant temperature and pressure, the volume of a gas is proportional to the number of moles of gas present.

  • Mathematical Expression: $V_1 n_2 = V_2 n_1$

  • Example: Suppose you were given $8.00\,\text{moles}$ of a gas occupying a volume of $4.00\,L$ at constant pressure and temperature. What volume of gas would $16.0\,\text{moles}$ occupy at the same temperature and pressure?

  • Solution:         $V_1 = 4.00\,L$         $n_1 = 8.00\,\text{mol}$         $n_2 = 16.0\,\text{mol}$         $V_2 = \frac{V_1 n_2}{n_1}$         $V_2 = \frac{(4.00\,L)(16.0\,\text{mol})}{8.00\,\text{mol}}$         $V_2 = 8.00\,L$

Dalton's Law of Partial Pressures

• Definition: Dalton's law states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.

  • Mathematical Expression: $P_{\text{total}} = P_1 + P_2 + P_3 + \dots$

• Gases Collected Over Water: Dalton's law comes into play when a gas is collected over water, where the total pressure measured is equal to the pressure exerted by the collected gas plus the water vapor pressure at the temperature of the system.

  • Example: A sample of methane gas is collected over water at an ambient pressure of $0.972\,\text{atm}$. The vapor pressure of water at this temperature is $0.025\,\text{atm}$. What is the pressure exerted by the methane?

  • Solution:         $P_{\text{total}} = P_{\text{water}} + P_{\text{methane}}$         $P_{\text{methane}} = P_{\text{total}} - P_{\text{water}} = 0.972\,\text{atm} - 0.025\,\text{atm} = 0.947\,\text{atm}$

• Mole Fractions and Partial Pressure: Partial pressures of individual gases in a gas mixture are proportional to the mole fraction of the gas in the mixture.

  • Mathematical Expression: $P_{\text{gas a}} = P_{\text{total}} \times X_{\text{gas a}}$

  • Example: A rigid container with a combination of nitrogen and oxygen gas is at a pressure of $2.4\,\text{atm}$. If the mole fraction of nitrogen gas is $0.16$, what is the partial pressure exerted by the nitrogen gas?

  • Solution:         $P_{\text{nitrogen}} = P_{\text{total}} \times X_{\text{nitrogen}}$         $P_{\text{nitrogen}} = 2.4\,\text{atm} \times \frac{0.16\,\text{moles nitrogen}}{1.0\,\text{total moles}}$         $P_{\text{nitrogen}} = 0.38\,\text{atm}$

Ideal Gas Law and Density

• The Combined Ideal Gas Law: All the classical gas laws combine into the Ideal Gas Law.

  • Formula: $PV = nRT$

  • Variables:

    • $P$ = Pressure of the gas ($\text{atm}$)

    • $V$ = Volume of the gas ($L$)

    • $n$ = Number of gas moles ($\text{mol}$)

    • $R$ = Ideal gas constant, $0.082\,L\,\text{atm}/K\,\text{mol}$

    • $T$ = Absolute temperature ($K$)

  • Example: What is the pressure exerted by $3.0\,\text{moles}$ of gas at $200\,K$ in a $2.0\,\text{liter}$ container?

  • Solution:         $P = \frac{nRT}{V}$         $P = \frac{(3.0\,\text{mole} \times 0.082\,L\,\text{atm} \times 200\,K)}{2.0\,L \times K\,\text{mol}}$         $P = 24.6\,\text{atm}$

• Standard Temperature and Pressure (STP): STP is defined as $273\,K$ and $1.0\,\text{atm}$.

  • Note: This is distinct from thermodynamically standard conditions, which are $298\,K$ and $1.0\,\text{atm}$.

  • Molar Volume at STP: $1.0\,\text{mole}$ of gas at STP occupies a volume of $22.4\,L$, regardless of the identity of the gas.

• Ideal Gas Law in Terms of Molar Mass and Density:

  • Formula: $P(mm) = dRT$

  • Variables:

    • $P$ = Pressure of the gas ($\text{atm}$)

    • $mm$ = Molar mass of gas molecule ($g/mol$)

    • $d$ = Density of the gas ($g/L$)

    • $R$ = Ideal gas constant ($0.082\,L\,\text{atm}/K\,\text{mol}$)

    • $T$ = Absolute temperature ($K$)

  • Example (Molar Mass Calculation): A gas sample with a density of $1.67\,g/L$ exerts a pressure of $2.0\,\text{atm}$ at a temperature of $299\,K$. What is the molar mass of the gas?

  • Solution:         $\text{Molar mass} = \frac{dRT}{P}$         $\text{Molar mass} = \frac{(1.67\,g/L)(0.082\,L\,\text{atm}/K\,\text{mol})(299\,K)}{2.0\,\text{atm}}$         $\text{Molar mass} = 20.5\,g/mol$

  • Example (Density Comparison): Compare the density of hydrogen gas with that of water vapor, at constant temperature and pressure.

  • Solution:         $\frac{(mm)_{\text{water}}}{(mm)_{\text{hydrogen}}} = \frac{d_{\text{water}}}{d_{\text{hydrogen}}}$         $\frac{d_{\text{water}}}{d_{\text{hydrogen}}} = \frac{18}{2} = 9$         Therefore, water vapor is nine times more dense than hydrogen gas at any given temperature and pressure.

Kinetic Theory of Matter

• Kinetic Molecular Theory Postulates:

  1. Gases are composed of tiny particles that are separated from each other by otherwise empty space.

  2. Gas molecules are in constant, continuous, random, and straight-line motion.

  3. The molecules collide with one another in perfectly elastic collisions (there is no net loss in energy).

  4. The pressure of the gas is a result of the collisions between the gas molecules and the walls of the container.

  5. The average kinetic energy of all the molecules of gas is directly proportional to the absolute temperature of the gas.

• Implication: All gases at the same temperature will have the same kinetic energy.

Graham's Law of Effusion

• Foundation: Graham's law results from kinetic theory; two different gases at the same temperature and pressure ($\text{gas a}$ and $\text{gas b}$) share the same kinetic energy:     $\frac{1}{2}(m_a v_a^2) = \frac{1}{2}(m_b v_b^2)$     Where $m$ = mass and $v$ = velocity.

• Effusion: The process by which a gas escapes from one chamber to another by moving through a small hole. The rate of effusion is proportional to the velocity of the gas.

• Graham's Law Formula: Gas molecules of smaller molar mass move faster than those of larger molar mass.     $\frac{r_a}{r_b} = \sqrt{\frac{M_b}{M_a}}$

  • $r_a$ = rate of effusion of gas a

  • $r_b$ = rate of effusion of gas b

  • $M_a$ = molar mass of gas a ($g/mol$)

  • $M_b$ = molar mass of gas b ($g/mol$)

• Example: A mixture of helium and carbon dioxide form a mixture in a rigid container. A small leak is created in the container; how much faster will the helium exit the container than the carbon dioxide?

  • Solution:         $\frac{r_{\text{helium}}}{r_{\text{carbon dioxide}}} = \sqrt{\frac{M_{\text{carbon dioxide}}}{M_{\text{helium}}}}$         $\frac{r_{\text{helium}}}{r_{\text{carbon dioxide}}} = \sqrt{\frac{44}{4}} = \sqrt{11} \approx 3.3$         Therefore, helium leaves the container $3.3$ times faster than carbon dioxide.

Deviations from Ideal Gas Laws

• Non-ideal (Real) Behavior: Occurs when the volume of the molecule is significant relative to the space between molecules, or when there is significant intermolecular attraction.

  • Volume Deviation: If the space occupied by molecules is significant, the measured volume will be larger than the ideal calculation.

  • Pressure Deviation: If intermolecular attraction is significant, the measured pressure will be less than the ideal calculation.

• Conditions for Real Behavior: High pressures or low temperatures (when molecules are closer together).

• van der Waals Equation:     $\left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT$

  • Variables:

    • $P$ = Pressure of the real gas ($\text{atm}$)

    • $V$ = Volume of the gas molecules + space between them ($L$)

    • $n$ = number of gas moles

    • $R$ = Ideal gas constant

    • $T$ = Absolute temperature

    • $a$ = Intermolecular attraction constant for the gas

    • $b$ = Space occupied by one mole of gas molecules

Phase Diagrams and Phase Changes

• Phase Diagram Components:

  • Lines: Indicate where a substance exists simultaneously in two phases.

  • Critical Point: The temperature and pressure above which a substance must exist as a gas.

  • Triple Point: The temperature and pressure at which the substance may exist in all three phases: solid, liquid, and gas.

  • Vapor Pressure Curve: Boundary between liquid and gas phases; determines partial pressure of gas in the vapor phase at a given temperature.

• Phase Change Terminology and Enthalpy ($\Delta H$):

  • Melting (Fusion): Solid to liquid; \Delta H > 0

  • Freezing: Liquid to solid; \Delta H < 0

  • Vaporization: Liquid to gas; \Delta H > 0

  • Condensation: Gas to liquid; \Delta H < 0

  • Sublimation: Solid to gas; \Delta H > 0

  • Deposition: Gas to solid; \Delta H < 0

  • Specific Enthalpy for Water: $\Delta H_{\text{fusion}} = 6.0\,kJ/mol$, $\Delta H_{\text{vaporization}} = 40.7\,kJ/mol$.

• Heating and Cooling Curves: Graph of Temperature vs. Time (or heat added).

  • Procedure 1 (Change of State): Multiply enthalpy change by the amount of material.

  • Procedure 2 (Change of Temperature): Multiply mass by specific heat and by change in temperature ($q = mc\Delta T$).

  • Normal Points: "Normal" refers to the temperature of equilibrium at $1.0\,\text{atm}$ pressure.

Example Heating Curve Calculation

• Problem: Calculate total heat to raise $20.0\,g$ of frozen water at $-10\,^{\circ}C$ to steam at $115\,^{\circ}C$.

• Constants: $C_{\text{ice}} = 2.1\,J/g\,^{\circ}C$; $C_{\text{water}} = 4.2\,J/g\,^{\circ}C$; $C_{\text{steam}} = 1.8\,J/g\,^{\circ}C$; $\Delta H_{\text{fus}} = 6.0\,kJ/mol$; $\Delta H_{\text{vap}} = 40.7\,kJ/mol$.

• Steps:

  1. Raise ice to melting point: $20\,g \times 2.1\,J/g\,^{\circ}C \times 10\,^{\circ}C = 420\,J = 0.420\,kJ$

  2. Melt ice: $20.0\,g \times \frac{1.0\,\text{mol}}{18.0\,g} \times 6.0\,kJ/mol = 6.66\,kJ$

  3. Raise water to boiling point: $20\,g \times 4.2\,J/g\,^{\circ}C \times 100\,^{\circ}C = 8400\,J = 8.40\,kJ$

  4. Boil water: $20\,g \times \frac{1.0\,\text{mol}}{18.0\,g} \times 40.7\,kJ/mol = 45.2\,kJ$

  5. Raise steam to final temperature: $20\,g \times 1.8\,J/g\,^{\circ}C \times 15\,^{\circ}C = 540\,J = 0.54\,kJ$

  6. Total: $0.420 + 6.66 + 8.40 + 45.2 + 0.54 = 61.22\,kJ$

Properties of Liquids and Solids

• Liquids:

  • Definite volume, shape depends on container.

  • Molecules move constantly and randomly.

  • Incompressible; temperature changes cause only small volume changes.

  • Diffusion occurs more slowly than in gases; rate increases with temperature.

  • Surface Tension: Inward force of a liquid toward itself; decreases as temperature increases.

• Solids:

  • Retain shape and volume; virtually incompressible.

  • Particles held in fixed positions by strong attractive forces.

  • Crystalline Solids: Defined by specific geometric pattern; sharp melting points. Smallest repeating unit is the Unit Cell.

    • Simple cubic: One atom at each corner; total of one atom per unit cell (8 corners $\times$ 1/8 atom).

    • Face-centered: Simple cubic plus one atom shared on each face; total of four atoms per unit cell.

    • Body-centered: Simple cubic plus one atom in the center; total of two atoms per unit cell.

  • Amorphous Solids: No specific geometry or sharp melting point (e.g., glass).

Solutions and Solubility

• The Solution Process:

  • Solvation: Interaction of solvent with solute molecules/ions to form loosely bonded aggregates.

  • Condition for Dissolving: Attraction between solute and solvent must exceed the forces holding the solute together.

  • Hydration: Solvation when water is the solvent.

  • Miscible: One substance is soluble in all proportions with another.

• Solubility Factors:

  • Saturated Solution: Solid solute is in equilibrium with dissolved solute.

  • Supersaturated: Contains more solute than required for saturation.

  • Temperature Effects (Solids): Solubility increases with temperature if $\Delta H_{\text{solvation}}$ is endothermic; decreases if exothermic.

  • Temperature Effects (Gases): Solubility usually decreases with increasing temperature.

  • Pressure (Henry's Law): For gases in liquids, solubility ($C$) is directly proportional to partial pressure ($P$).

    • Henry's Law Formula: $C = kP$

    • Example: $P_{\text{CO}_2} = 0.0004\,\text{atm}$, $k = 32.0\,mol/(L\,\text{atm})$. Find concentration ($C$).

    • Solution: $C = (32.0\,mol/L\,\text{atm})(0.0004\,\text{atm}) = 0.0128\,M$

Solution Concentrations

• Molarity (M): $\frac{\text{Moles Solute}}{\text{Liters Solution}}$

  • Example: $10.0\,g\,HCl$ in $500\,mL$ water.

  • Solution: $[HCl] = \frac{10.0\,g \times (1\,mol/36.45\,g)}{0.5\,L} = 0.55\,M$

• pH: Defined as $-\log[H^+]$.

  • Example: For $0.55\,M\,HCl$, $pH = -\log(0.55) = 0.26$.

• Molality (m): $\frac{\text{Moles Solute}}{\text{Kilograms Solvent}}$

  • Example: $10.0\,g\,NaCl$ in $800\,g$ water.

  • Solution: $m = \frac{10.0\,g \times (1\,mol/58.45\,g)}{0.8\,kg} = 0.214\,molal$

• Mole Fraction ($X$): $\frac{\text{moles solute}}{\text{total solution moles}}$

  • Example: $1.00\,g$ ethanol ($46.07\,g/mol$) mixed with $100.0\,g$ water ($18.00\,g/mol$).

  • Solution:         $\text{Moles ethanol} = 0.0217\,mol$         $\text{Moles water} = 5.555\,mol$         $X_{\text{ethanol}} = \frac{0.0217}{0.0217 + 5.555} = 0.00389$

Colligative Properties

• Raoult's Law: Describes the depression of solvent vapor pressure by a solute.

  • Formula: $P = X P^o$

  • Example: $400.0\,g$ glucose in $500.0\,g$ water ($P^o = 0.031\,\text{atm}$ at $25\,^{\circ}C$).

  • Solution:         $\text{Moles glucose} = 2.22\,mol$         $\text{Moles water} = 27.77\,mol$         $X_{\text{water}} = \frac{27.77}{29.99} = 0.93$         $P = (0.93)(0.031\,\text{atm}) = 0.029\,\text{atm}$

• Boiling Point Elevation: $\Delta T = k_b m i$

  • Van't Hoff factor ($i$): Moles of particles per mole of solute (e.g., $1.0$ for non-electrolytes like glucose; $2.0$ for $NaCl$; $3.0$ for $MgCl_2$).

  • Example: Compare $10.0\,g$ sucrose ($158.0\,g/mol$) vs $10.0\,g\,MgCl_2$ ($95.2\,g/mol$) in $100.0\,g$ water ($k_b = 0.51\,^{\circ}C\,kg/mol$).

  • Solution:         $\Delta T_{\text{sucrose}} = 0.51 \times \frac{10.0/158.0}{0.10} \times 1.0 = 0.3\,^{\circ}C$         $\Delta T_{MgCl_2} = 0.51 \times \frac{10.0/95.2}{0.10} \times 3.0 = 1.6\,^{\circ}C$

• Freezing Point Depression: $\Delta T = k_f m i$

  • Example: $2.5\,molal\,MgCl_2$ solution on a road ($k_f = 1.86\,^{\circ}C\,kg/mol$, $i = 3.0$).

  • Solution:         $\Delta T = 1.86 \times 2.5 \times 3.0 = 14\,^{\circ}C$         New freezing point is $0.0\,^{\circ}C - 14\,^{\circ}C = -14\,^{\circ}C$.

• Osmotic Pressure: $\Pi V = nRTi$

  • Example (Molar Mass of Protein): $1.000\,g$ unknown protein in $500\,mL$ water gives $\Pi = 0.002\,\text{atm}$ at $298\,K$ ($i=1.0$).

  • Solution:         $\text{Molar mass} = \frac{\text{mass solute} \times RT}{\Pi V}$         $\text{Molar mass} = \frac{1.000\,g \times 0.082 \times 298}{0.002 \times 0.5} = 24,436\,g/mol$

Nonideal Solutions

• Negative Deviations from Raoult's Law: Occur when solute-solvent attractions (like hydrogen bonding) are stronger than pure component attractions, hindering evaporation. Also linked to large exothermic $\Delta H_{\text{solution}}$.

• Positive Deviations from Raoult's Law: Occur when solute and solvent are both very volatile, or when $\Delta H_{\text{solution}}$ is large and endothermic.