Topic 5 · ~11% of Regents exam · Lots of reference tables

Physical Behavior of Matter

This is the topic where you actually use the reference tables hardest. Heating curves (Table B), vapor pressure (Table H), solubility curves (Table G), gas laws, and solution math all live here. The unifying idea is kinetic molecular theory: matter is made of particles in motion, and almost every "physical" change is about how that motion changes.

1. Kinetic Molecular Theory (KMT)

KMT is a model that describes the behavior of an ideal gas. It is built on five assumptions:

  1. Gas particles are in constant, random, straight-line motion.
  2. Particles are point-sized: their actual volume is negligible compared to the space they occupy.
  3. Collisions between particles (and with the container walls) are perfectly elastic: no kinetic energy is lost.
  4. There are no attractive or repulsive forces between particles.
  5. The average kinetic energy of gas particles is directly proportional to the Kelvin temperature of the gas.

Real gases are not ideal

Real gases behave most like an ideal gas at high temperature and low pressure. That is when particles are moving fast (overcoming attractions) and spread out (their own volume is negligible). Gases deviate from ideal behavior at low temperature and high pressure. H₂ and He are the gases that behave most ideally because they are tiny and have weak IMFs.

2. Phases of matter

PhaseSymbolShapeVolumeCompressible?Particle motion
Solid(s)DefiniteDefiniteNoVibrate in place
Liquid(ℓ)Takes container's shapeDefiniteBarelySlide past each other
Gas(g)Fills containerFills containerYesFast, random, far apart
Aqueous(aq)Dissolved in water (a solution)

3. Phase changes

ChangeFrom → ToEnergy
Melting (fusion)Solid → LiquidAbsorbed (endothermic)
Freezing (solidification)Liquid → SolidReleased (exothermic)
Vaporization (boiling/evaporation)Liquid → GasAbsorbed (endothermic)
CondensationGas → LiquidReleased (exothermic)
SublimationSolid → GasAbsorbed (endothermic)
DepositionGas → SolidReleased (exothermic)

The pattern

Going "up" the phase ladder (solid → liquid → gas) requires energy input. Going "down" the ladder (gas → liquid → solid) releases energy. Same energy in, same energy out. This is why ice cubes cool a drink: melting ice absorbs heat from the surrounding liquid.

4. Heating and cooling curves

Temperature (°C) Heat added → 0°C 100°C solid melting (Hf) liquid boiling (Hv) gas Flat segments: phase change, KE constant — energy breaks IMFs.
Heating curve of water — slopes warm; plateaus change phase
SegmentWhat's happeningFormulaVariables & units
Slope (solid / liquid / gas)Temperature rises; KE increasesq = m c ΔTq (J); m (g); c (J/g·°C, Table B); ΔT (°C)
Plateau at melting ptPhase change s ↔ l; PE rises, T constantq = m HfHf = 334 J/g for water (Table B)
Plateau at boiling ptPhase change l ↔ g; PE rises, T constantq = m HvHv = 2260 J/g for water (Table B)
Cooling curveMirror image — same equations, q is releasedsame as aboveq negative (system loses heat)

A heating curve plots temperature vs heat added as a substance is warmed from solid to gas. It has five segments and two plateaus.

SegmentWhat's happeningTemperature?KE or PE changing?
1. Warming solidSolid heats upRisingKE increasing
2. Melting plateauSolid → liquidConstantPE increasing
3. Warming liquidLiquid heats upRisingKE increasing
4. Boiling plateauLiquid → gasConstantPE increasing
5. Warming gasGas heats upRisingKE increasing

The two big ideas

  • Sloped segments: temperature changes, kinetic energy (KE) changes. Particles speed up.
  • Flat plateaus: temperature is constant, but heat is still being added. That energy is going into breaking attractive forces between particles, which raises potential energy (PE). The boiling plateau is longer than the melting plateau because more energy is needed to fully separate particles into a gas.

Regents classic

During a phase change, temperature does not change even though heat is being added. Average kinetic energy stays constant; potential energy is what increases. The exam asks this in many forms.

5. Heat energy calculations

Use Reference Table T for these three formulas. The right one depends on whether temperature is changing or a phase change is happening.

Warming or cooling (sloped segment):

q = mCΔT

where q = heat (J), m = mass (g), C = specific heat capacity (J/g·°C), ΔT = change in temperature (°C).

Melting or freezing (melting plateau):

q = mHf

where Hf = heat of fusion (334 J/g for water, from Table B).

Boiling or condensing (boiling plateau):

q = mHv

where Hv = heat of vaporization (2260 J/g for water, from Table B).

Worked example

How much heat is needed to melt 50.0 g of ice at 0 °C?

q = mHf = (50.0 g)(334 J/g) = 16,700 J

Notice we use Hf because we are at the melting plateau (no temperature change). If the ice were below 0 °C, you would first use q = mCΔT to warm it up to 0 °C, then q = mHf to melt it.

6. Gas laws

Gas laws describe how pressure (P), volume (V), and temperature (T) of a fixed amount of gas are related. Temperature must always be in Kelvin in gas law problems.

The Kelvin rule

Convert Celsius to Kelvin by adding 273: K = °C + 273. The formula is on Table T. Never plug Celsius into a gas law. If you forget this, you will get an absurd answer.

Boyle's Law (constant T, n)

Pressure and volume are inversely proportional. Squeeze the volume, the pressure goes up.

P1V1 = P2V2

Charles's Law (constant P, n)

Volume and temperature are directly proportional. Heat a gas at constant pressure and it expands.

V1/T1 = V2/T2

Combined Gas Law (constant n)

Combines Boyle's, Charles's, and Gay-Lussac's into one formula. This is the only one on Table T. If a problem changes only two variables, just set the third equal on both sides and cancel.

(P1V1)/T1 = (P2V2)/T2

Worked example

A gas occupies 2.0 L at 1.0 atm and 273 K. What volume will it occupy at 2.0 atm and 546 K?

(P1V1)/T1 = (P2V2)/T2
(1.0)(2.0)/(273) = (2.0)(V2)/(546)
V2 = (1.0)(2.0)(546) / [(273)(2.0)] = 2.0 L

Pressure doubled (would halve volume) but temperature also doubled (would double volume). The two effects cancel and volume stays the same.

STP: Standard Temperature and Pressure

From Reference Table A:

  • Standard temperature: 273 K (0 °C)
  • Standard pressure: 101.3 kPa (1 atm)
  • Molar volume at STP: 22.4 L/mol (any ideal gas)

7. Vapor pressure (Reference Table H)

Vapor pressure is the pressure exerted by a gas above its own liquid at equilibrium in a closed container. It increases with temperature.

A liquid boils when its vapor pressure equals the surrounding atmospheric pressure. At standard atmospheric pressure (101.3 kPa), the boiling point of water is 100 °C. At higher altitudes (lower atmospheric pressure), water boils at a lower temperature, which is why it takes longer to cook pasta in Denver.

How to read Table H

Table H plots vapor pressure (y-axis) vs temperature (x-axis) for four liquids: propanone (acetone), ethanol, water, and ethanoic acid.

  • To find the boiling point at standard pressure: find 101.3 kPa on the y-axis, draw a horizontal line until it hits the curve, drop down to the x-axis.
  • To find the vapor pressure at a given temperature: go up from that temperature to the curve, then left to the y-axis.
  • The curve to the far left (propanone) boils at the lowest temperature, so it has the weakest intermolecular forces. The curve to the far right (ethanoic acid) has the strongest IMFs.

8. Solutions and solubility (Reference Table G)

A solution is a homogeneous mixture: solute(s) dissolved in solvent. In an aqueous solution, water is the solvent. Reference Table G plots how much of each substance dissolves in 100 g of water at various temperatures.

Three types of solutions

Unsaturated

Less solute dissolved than the maximum. Below the curve on Table G. Will dissolve more solute if added.

Saturated

Exactly at the maximum. Right on the curve on Table G. Adding more solute results in it settling out.

Supersaturated

Above the curve. Unstable; achieved by carefully cooling a saturated hot solution. Disturbance causes solute to crash out.

Factors affecting solubility

Solute typeEffect of increasing temperatureEffect of increasing pressure
Most solids (sugar, KNO₃)Solubility increasesNegligible
Some salts (NaCl)Solubility barely changesNegligible
Gases (CO₂, O₂)Solubility decreasesSolubility increases

Why gases are weird

Warming a soda makes it go flat because dissolved CO₂ escapes (solubility drops with temperature). Capping a soda bottle keeps pressure high, which keeps more CO₂ dissolved. Open it, pressure drops, gas comes out, fizz.

9. Molarity (concentration)

Molarity (M) is the most important way to measure concentration. It is moles of solute per liter of solution.

molarity = moles of solute / liters of solution

From Reference Table T.

Worked example

What is the molarity of a solution containing 4.0 mol of NaCl in 2.0 L of solution?

M = 4.0 mol / 2.0 L = 2.0 M

"2.0 M" reads "two molar" or "2.0 moles per liter."

Other concentration measures

  • Parts per million (ppm) = (grams of solute / grams of solution) × 1,000,000. Used for very dilute solutions. On Table T.
  • Percent by mass = (mass of solute / mass of solution) × 100.

10. Colligative properties

Colligative properties depend on the number of solute particles dissolved in a solvent, not on what those particles are. Adding solute to a solvent does two things:

  • Boiling point elevation: the solution boils at a higher temperature than the pure solvent.
  • Freezing point depression: the solution freezes at a lower temperature than the pure solvent.

Why we salt icy roads

Salt (NaCl) dissolved in water lowers water's freezing point below 0 °C, so ice melts even when the air temperature is below 0 °C. The same idea is why antifreeze is added to car radiators: it prevents the coolant from freezing in winter and from boiling in summer.

The "more particles" rule

Ionic compounds break into multiple ions when dissolved, so they have a bigger effect than a molecular compound at the same molarity. NaCl → Na⁺ + Cl⁻ (2 particles). CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles). At 1.0 M, CaCl₂ lowers the freezing point more than NaCl, which lowers it more than sugar (which doesn't dissociate at all).

Key terms

Kinetic Molecular TheoryModel describing ideal gas behavior. Particles in constant random motion.
Ideal gasHypothetical gas obeying all five KMT assumptions. Real gases approach this at high T, low P.
STPStandard Temperature and Pressure: 273 K and 101.3 kPa.
Heat of fusion (Hf)Energy to melt 1 g of a substance. For water: 334 J/g.
Heat of vaporization (Hv)Energy to vaporize 1 g of a substance. For water: 2260 J/g.
Specific heat capacity (C)Energy to raise 1 g by 1 °C. For liquid water: 4.18 J/g·°C.
Boiling pointTemperature at which vapor pressure of a liquid equals atmospheric pressure.
Vapor pressurePressure of gas above its own liquid in a closed container at equilibrium.
Saturated solutionSolution holding the maximum amount of dissolved solute at a given temperature.
Molarity (M)Moles of solute per liter of solution.
Colligative propertyProperty depending on number of solute particles (boiling point elevation, freezing point depression).
SublimationDirect change from solid to gas without becoming liquid (dry ice).

Practice questions

Q1. Which sample of gas would behave most like an ideal gas?
  1. O₂ at 273 K and 100 kPa
  2. H₂ at 500 K and 50 kPa
  3. Cl₂ at 200 K and 200 kPa
  4. CO₂ at 100 K and 500 kPa

Answer: (2) H₂ at 500 K and 50 kPa. Real gases behave most ideally at high temperature and low pressure, and small light gases (H₂, He) are closest to ideal. Option 2 has the highest temperature, lowest pressure, and the smallest gas molecule. Option 4 is the worst (cold, compressed, and CO₂ has stronger IMFs).

Q2. During a phase change at constant pressure, the temperature of a substance remains constant. What does this indicate about the energy added?
  1. It is increasing the average kinetic energy of the particles.
  2. It is being used to break the attractive forces between particles.
  3. It is being lost to the surroundings.
  4. It is converting matter to energy.

Answer: (2) It is being used to break the attractive forces between particles. Temperature is a measure of average kinetic energy, and that is constant during a phase change. The energy being added goes into increasing potential energy by overcoming intermolecular attractions.

Q3. A sample of gas occupies 6.0 L at 300 K and 100 kPa. What volume will it occupy at 600 K and 50 kPa?
  1. 3.0 L
  2. 6.0 L
  3. 12.0 L
  4. 24.0 L

Answer: (4) 24.0 L. Use the combined gas law: (P₁V₁)/T₁ = (P₂V₂)/T₂. (100)(6.0)/(300) = (50)(V₂)/(600). Solving: V₂ = (100)(6.0)(600) / [(300)(50)] = 360,000/15,000 = 24.0 L. Temperature doubled (volume × 2) and pressure halved (volume × 2), so volume × 4.

Q4. (Part B-2) Explain, in terms of particle motion and energy, what happens at the boiling point of a liquid. Reference both kinetic and potential energy in your answer.

Sample full-credit response: At the boiling point, the vapor pressure of the liquid equals the surrounding atmospheric pressure, and the liquid converts to gas throughout (not just at the surface). The average kinetic energy of the particles stays constant during this phase change, which is why the temperature does not rise even though heat is continuously being added. Instead, the added heat energy goes into increasing the potential energy of the particles by overcoming the intermolecular forces (such as hydrogen bonds in water) that hold them together as a liquid. Once these attractive forces are broken, the particles are free to move as gas molecules.

Q5. (Part C) A student dissolves 11.7 g of NaCl in enough water to make 250. mL of solution. (a) Calculate the molarity of the solution. Show all work and units. (b) Explain how dissolving the NaCl affects the freezing point of the water and why.

Sample full-credit response:
(a) Calculation:
Step 1: Find moles of NaCl. Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
Moles = 11.7 g ÷ 58.5 g/mol = 0.200 mol
Step 2: Convert volume to liters. 250. mL × (1 L / 1000 mL) = 0.250 L
Step 3: Apply molarity formula (Table T).
M = moles / liters = 0.200 mol / 0.250 L = 0.800 M

(b) Freezing point effect: Dissolving NaCl lowers the freezing point of the water. This is called freezing point depression and is a colligative property, meaning it depends on the number of dissolved particles in solution. NaCl dissociates into Na⁺ and Cl⁻ ions, so each formula unit produces two particles. These dissolved particles disrupt the formation of the ordered crystalline ice structure, so a lower temperature is required before the water can freeze.