Chemistry Notes: Vapor Pressure, Solubility, and Colligative Properties

Key ideas from the transcript

Instructor emphasizes proactive study habits and keeping pace with the syllabus
- Start studying early to avoid being surprised by material not covered in depth
- Homework extension: due date moved to Thursday
- Use of multiple practice resources: PowerPoints, handouts, Classtime, textbook problems, and a practice quiz created by Devin
- Strategy: test yourself without notes or external aids to gauge readiness, then study the gaps
Emphasis on consistent practice and building study habits
- Suggests 30 minutes per day of focused practice rather than long cramming sessions
- Practice problems: cover the solution part, try the problem, then check or reveal solutions later
- Use of iPad vs pen/paper: if exam is on paper, prefer physical notebooks; if comfortable with digital, that’s fine too
Class structure and materials
- References to IMF/Clausius–Clapeyron problem in PowerPoints (slide 33) and Chapter 12 material
- Three handouts to be posted before the exam; these contain extra problems with solutions
- Class time will be used to review material from the week (handout 3 as part of class time)
Specific topics introduced in the session
- Clausius–Clapeyron equation for vapor pressure and vaporization heat
- Problems involving labeling of P and T, and switching which term is unknown for easier solving
- Basic math concepts needed for log problems: natural log (ln) versus base-10 log, and the meaning of the logarithm base
- Brief intro to solubility, Henry’s law, and factors affecting solubility (temperature, pressure, solvent similarity)
- Real-world tie-ins: solubility trends with temperature, carbonation, and dissolution processes (solid solutes vs gaseous solutes)
Solubility and gas-liquid equilibria (conceptual recap)
- Solubility depends on the solute/solvent pair and the temperature/pressure conditions
- For solids, solubility generally increases with temperature; for gases, solubility can decrease with higher temperature
- Henry’s law: C = kH P, where C is concentration, P is partial pressure, and kH is Henry’s constant (depends on the solute-solvent pair)
- If solvents are chemically similar, dissolution is favored (light “like dissolves like” concept); poor miscibility (e.g., oil and water) reduces solubility
- Stirring and mixing enhance solubility by increasing the rate of dissolution (mass transfer considerations)
Colligative properties preview (to be covered next class)
- Freezing point depression: ΔTf = i Kf m
- Boiling point elevation: ΔTb = i Kb m
- Osmotic pressure: π = i M R T (or π = i c R T in molarity terms); i is the van’t Hoff factor, m is molality, M is molarity
- Noting that these formulas depend on mole fraction or molality depending on the context; familiarity with the symbols helps on quizzes
Practical exam tips mentioned
- Read the equation carefully (watch subscripts and signs)
- Label variables clearly (P1, P2, T1, T2) and decide which quantity to solve for first to simplify algebra
- When working with ln, remember the base is e (ln is natural log)
- For log problems, base-10 log is different from natural log; use the appropriate function on the calculator
- Practice with limited-idea problems and verify answers by estimating magnitude and sign
Key learning outcomes tied to the upcoming topics
- Accurate use of the Clausius–Clapeyron equation and proper unit handling
- Ability to convert temperatures to Kelvin and handle unit conversions in the context of ratios where units cancel (e.g., mmHg to atm may cancel in a ratio)
- Skill in solving for either vapor pressures or temperatures given ΔH_vap and a pair of p, T values
- Understanding of how temperature and pressure drive phase equilibrium in real systems (vaporization, carbonation, etc.)

Clausius–Clapeyron equation: vapor pressure and ΔH_vap problem walkthrough

Core formula (Clausius–Clapeyron in logarithmic form)
- oxed{
  rac{
  ext{d} igl( ext{ln} pigr)}{ ext{d}T} ext{ (not shown here)}
  }
- The working form used in the lecture for discrete data points is:
- oxed{ ext{ln}iggl(rac{p1}{p2}iggr) = rac{ riangle H{ ext{vap}}}{R} iggl(rac{1}{T2} - rac{1}{T_1}iggr) }
- Here p1 and p2 are vapor pressures at temperatures T1 and T2, respectively; ΔH_vap is the molar enthalpy of vaporization; R is the gas constant.
Practical steps used in the example
- Identify values from the problem: p1 = 271 mmHg at T1 = 30 °C; p2 = 565 mmHg at T2 = 50 °C
- Convert temperatures to Kelvin: T1 = 303.15\ ext{K},\ T2 = 323.15\ ext{K}
- Note that the ratio p1/p2 can be used without converting mmHg to atm because the ratio cancels units:
- \frac{p1}{p2} = \frac{271}{565}
- Compute the logarithm term: \ln\left(\frac{p1}{p2}\right) = \ln\left(\frac{271}{565}\right) \approx -0.736
- Compute the temperature term: \left( \frac{1}{T2} - \frac{1}{T1} \right) = \frac{1}{323.15} - \frac{1}{303.15} \approx -2.06 \times 10^{-4}\ \text{K}^{-1}
- Solve for ΔH_vap:
- \Delta H{ ext{vap}} = R \cdot \frac{\ln(p1/p2)}{(1/T2 - 1/T_1)}
- With R = 8.314 J mol^{-1} K^{-1}:
- \Delta H_{ ext{vap}} \approx 8.314 \times \frac{-0.736}{-2.06 \times 10^{-4}} \approx 2.97 \times 10^{4}\ \text{J mol}^{-1} = 29.7\ \text{kJ mol}^{-1}
- Note: The signs cancel to give a positive ΔH_vap, as expected since vaporization is endothermic
Important numerical checks and notes
- Ensure temperatures are in Kelvin for the 1/T terms
- If solving for p1 or p2, rearrange the equation accordingly, keeping track of which temperature pairs correspond to which pressure
- If you get a number like ~2.9×10^4 J/mol, convert to kJ/mol by dividing by 1000 when needed
- Realistic ΔH_vap values for diethyl ether are on the order of ~29–30 kJ/mol, which aligns with the calculation above

Second example: vapor pressure at a new temperature

Problem setup: Vapor pressure of diethyl ether is 401 mmHg at 18 °C; ΔH_vap = 26 kJ/mol. Find p at 32 °C.
Steps
- Label knowns: p2 = 401 mmHg at T2 = 18 °C (291.15 K); unknown p1 at T1 = 32 °C (305.15 K)
- Convert ΔHvap to J/mol: ΔHvap = 26,000 J/mol; R = 8.314 J mol^{-1} K^{-1}
- Use the same equation with p1 unknown:
- ext{ln}iggl(\frac{p1}{p2}\biggr) = \frac{\triangle H{ ext{vap}}}{R} \left( \frac{1}{T2} - \frac{1}{T_1} \right)
- Compute the temperature term:
- \left( \frac{1}{T2} - \frac{1}{T1} \right) = \frac{1}{291.15} - \frac{1}{305.15} \approx 0.000158\
- Compute the prefactor: \frac{\triangle H_{ ext{vap}}}{R} = \frac{26000}{8.314} \approx 3126.3
- Multiply: 3126.3 \times 0.000158 \approx 0.493
- Therefore: \ln\left(\frac{p_1}{401}\right) \approx 0.493
- Solve for p1: \frac{p1}{401} \approx e^{0.493} \approx 1.637 \Rightarrow p1 \approx 1.637 \times 401 \approx 657\ \text{mmHg}
- Result: p1 ≈ 650–660 mmHg (rounded). This is consistent with the expectation that higher temperature yields higher vapor pressure for a liquid with a given ΔH_vap.

Quick recap: what to remember about the Clausius–Clapeyron problem

When using ln(p1/p2) = (ΔH_vap/R) (1/T2 - 1/T1):
- The sign of the temperature term depends on which temperature is T1 vs T2; keep T in Kelvin
- The natural log ln is base e; the base-10 log is log10; the calculator keys differ
- Units: ΔH_vap is in J/mol (convert to kJ/mol if needed); R = 8.314 J mol^{-1} K^{-1}
- Pressure units can be mmHg or atm; the ratio p1/p2 cancels the unit (since you’re taking a ratio)
- Solve for the unknown by algebraic rearrangement; you can solve for ΔH_vap, p1, or p2 depending on which quantities are given

Solubility, gas–liquid equilibria, and related formulas (overview)

Henry’s law (gas solubility in liquids):
- C = kH \, P where C is concentration, P is partial pressure, and kH is Henry’s constant (depends on solute and solvent)
- Real-world example: carbonation is due to increased pressure pushing more CO₂ into the beverage; when pressure is released, CO₂ escapes, producing fizz
Temperature effects on solubility
- Solids: increasing temperature often increases solubility (e.g., sugar dissolving faster in hot tea)
- Gases: increasing temperature generally decreases solubility in liquids (gas escapes more readily)
Effect of pressure on solubility of gases in liquids
- Higher pressure generally increases the amount of gas that dissolves (Le Châtelier’s principle context)
Structural similarity of solvent and solute matters
- “Like dissolves like”: similar polarity/structure improves solubility; oil and water are poorly miscible due to polarity differences
Practical takeaway
- Stirring and agitation improve dissolution rates by enhancing mass transfer
- Understanding solubility principles helps explain real-world phenomena (e.g., beverage carbonation, sugar in hot vs. cold drinks)

Colligative properties (preview and key formulas)

Freezing point depression
- \Delta Tf = i \; Kf \; m
- i = van’t Hoff factor; K_f = freezing-point depression constant; m = molality
Boiling point elevation
- \Delta Tb = i \; Kb \; m
- i = van’t Hoff factor; K_b = boiling-point elevation constant
Osmotic pressure
- \pi = i M R T (ideal solution form, M is molarity, T in Kelvin)
- For non-ideal cases, adjustments may be needed, but the basic idea is that solute concentration and temperature drive osmotic pressure
Practical note
- These relate to solute concentration and how solutes perturb colligative properties of solvents; they will appear in upcoming quizzes and exams

StudyStrategies and class activities (practical implications)

Active study habits
- Use a timer to simulate exam conditions; attempt problems without notes; then review solutions
- If you score poorly on a problem, identify the exact concept gap and target that with focused practice
Problem-solving approach
- Label variables clearly (P1, P2, T1, T2)
- Decide what you are solving for first to simplify algebra
- When using logs, know the base: ln is natural log (base e); log is base-10
Resource utilization
- Three new handouts before the exam with solutions for self-check
- PowerPoints and class time review (handout 3 will be used during class)
Real-world relevance
- Understanding vapor pressure, solubility, and colligative properties helps explain everyday phenomena (beverages, iced tea vs hot tea, carbonated drinks, etc.)

Connections to foundational concepts

Links to thermodynamics and phase equilibria
- Clausius–Clapeyron ties phase equilibrium to enthalpy of vaporization and temperature
- Temperature dependence of solubility reflects enthalpic and entropic contributions to dissolution
Relation to the ideal gas law and unit analysis
- Use of R and consistency of units in the log-based expressions
- Unit cancellations in p-ratio calculations (e.g., mmHg vs atm)
Methodological takeaways for exams
- Proper labeling and consistent use of Kelvin temperatures
- Clear algebraic steps, sign checks, and unit verification

Ethical, philosophical, and practical implications

Academic integrity
- Practice problems without notes helps gauge true understanding and reduces reliance on external aids during exams
Practical mathematics literacy
- Understanding when and how to apply log-based equations improves problem-solving confidence, reduces sign errors, and supports careful reading of problem statements
Real-world relevance and critical thinking
- Connecting abstract formulas to tangible phenomena (sugar dissolving in hot vs cold drinks, carbonation) strengthens intuition and retention