CHEM1011 · Chemistry 1a
Intermolecular Forces and States of Matter
Week 4 connects molecular polarity to the intermolecular forces — dispersion, dipole–dipole and hydrogen bonding — and uses their relative strength to explain boiling point, viscosity, surface tension and vapour pressure. The quantitative side pairs the ideal gas law pV = nRT and Dalton's law with Raoult's law for vapour pressure over mixtures, both of which are examinable calculation items, with van der Waals and deviations flagged Mastery.
What this chapter covers
- 01Bond polarity from electronegativity difference; molecular polarity as the vector sum over the VSEPR shape
- 02Symmetric molecules (CO₂, CH₄, BF₃) can be non-polar despite polar bonds; bent SO₂/H₂O are polar
- 03Intermolecular forces, weakest to strongest: dispersion (London) < dipole–dipole < hydrogen bonding (H on N/O/F)
- 04Property rationalisation: stronger IMFs → higher boiling point/viscosity/surface tension, lower vapour pressure
- 05Vapour pressure and boiling: boiling point is where vapour pressure equals external pressure
- 06Ideal gas law pV = nRT and Dalton's law of partial pressures p_total = Σpᵢ, pᵢ = xᵢ·p_total
- 07Raoult's law p_A = x_A·p°_A; total pressure and vapour-phase mole fraction of an ideal binary mixture (Mastery)
- 08Deviations from ideality and the van der Waals equation (a for attractions, b for molecular volume) (Mastery)
Total vapour pressure and vapour composition by Raoult's law
- +1Raoult's law gives each partial pressure from the liquid mole fraction: p_A = x_A·p°_A = 0.40 × 95 = 38 kPa.
- +1Likewise for B: p_B = x_B·p°_B = 0.60 × 29 = 17.4 kPa.
- +1Total vapour pressure is the sum (Dalton's law): p_total = p_A + p_B = 38 + 17.4 = 55.4 kPa.
- +1Vapour-phase mole fraction of A: y_A = p_A/p_total = 38/55.4 = 0.69. The vapour (0.69) is richer in A than the liquid (0.40), because A is the more volatile component.
Key terms
- Dispersion (London) forces
- The weakest intermolecular force, present in all species; arises from instantaneous and induced dipoles and grows with the number of electrons, molecular size and polarisability.
- Hydrogen bonding
- A strong dipole–dipole attraction when H bonded to N, O or F interacts with a lone pair on N/O/F of a neighbour; it raises boiling points and underlies water's unusual properties.
- Vapour pressure
- The pressure of vapour in equilibrium with its liquid at a given temperature; independent of the amount of liquid, higher for more volatile liquids (weaker IMFs), and increasing with temperature.
- Ideal gas law
- pV = nRT, with R = 8.314 J mol⁻¹ K⁻¹ = 0.08206 L atm mol⁻¹ K⁻¹ and T in kelvin; relates the pressure, volume, amount and temperature of a gas.
- Dalton's law of partial pressures
- The total pressure of a gas mixture is the sum of the partial pressures, p_total = Σpᵢ, and each partial pressure is pᵢ = xᵢ·p_total where xᵢ is the mole fraction.
- Raoult's law
- For an ideal mixture the partial vapour pressure of a component is p_A = x_A·p°_A (liquid mole fraction × pure-component vapour pressure); the total is the sum over components (Mastery).
Intermolecular Forces and States of Matter FAQ
How do I decide which intermolecular forces a molecule has?
Work from structure. First decide polarity: get the shape from VSEPR, then take the vector sum of the bond dipoles — a symmetric molecule (CO₂, CH₄) is non-polar even with polar bonds. Every molecule has dispersion forces; polar molecules add dipole–dipole; and if an H is bonded to N, O or F and a lone pair is available on a neighbouring N/O/F, add hydrogen bonding. Then rank strength to explain the physical property in question.
Why does stronger intermolecular attraction raise boiling point but lower vapour pressure?
Both follow from how tightly molecules are held in the liquid. Stronger IMFs mean more energy is needed to pull molecules into the gas, so fewer escape at a given temperature — that is a lower vapour pressure — and a higher temperature is needed before the vapour pressure reaches the external pressure, which is a higher boiling point. The same logic gives higher viscosity and surface tension for strongly interacting liquids.
What is the difference between the liquid and vapour mole fractions in a mixture?
The liquid mole fraction x is what you mix; the vapour mole fraction y = p_component/p_total is what sits above it. Because Raoult's law weights each component by its own pure-component vapour pressure, the vapour is always richer in the more volatile component than the liquid. That difference is exactly what a distillation exploits, and it is a common Mastery exam point.
Can Sia help me with gas-law and Raoult's-law problems?
Yes. Sia can set up a pV = nRT calculation with consistent units, apply Dalton's law for partial pressures, and run a Raoult's-law mixture to the total pressure and vapour composition, checking each step. It can also explain positive and negative deviations and the van der Waals corrections. It explains the method and checks your reasoning; it does not do graded assessment, and UNSW academic-integrity rules apply.
Exam move
Build the qualitative chain first — shape → polarity → intermolecular forces → physical property — because most short-answer marks here come from correctly identifying the dominant force and reasoning to a boiling point, viscosity or vapour-pressure comparison. Memorise the strength order (dispersion < dipole–dipole < hydrogen bonding) and the hydrogen-bond condition (H on N/O/F meeting a lone pair on N/O/F). On the quantitative side, keep the ideal gas law reflexive with strict unit discipline (T in kelvin, R matched to your pressure unit), and drill Raoult's-law mixtures as a fixed routine: partial pressures from x·p°, sum for the total, then y = p/p_total for the vapour, noting the enrichment in the volatile component. For the Mastery layer, be able to sketch positive and negative deviation curves and state the van der Waals meanings of a and b. When a gas-law unit conversion bites, ask Sia to walk it through.
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