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Topic 1: Intermolecular forces and gases

Explain the behaviour of gases using the kinetic theory of matter, and apply Boyle's law, Charles's law and the combined gas law to predict the effect of changing pressure, volume and temperature

A focused answer to the QCE Chemistry Unit 2 dot point on gas behaviour. States the assumptions of the kinetic theory of gases, derives Boyle's, Charles's and Gay-Lussac's laws qualitatively from particle behaviour, and works through combined gas law problems of the kind QCAA poses in EA Paper 1.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

QCAA wants you to describe gases using the kinetic theory of matter and predict the effect on a gas of changing pressure, volume or temperature using the named gas laws. The dot point sets up the ideal gas equation and gas stoichiometry that follow.

The answer

The kinetic theory of gases models gas particles as small, hard, freely moving objects whose macroscopic properties (pressure, volume, temperature) follow from the average behaviour of huge numbers of collisions. The gas laws (Boyle, Charles, Gay-Lussac, combined) are quantitative summaries of how these macroscopic properties trade off when one is held constant.

Kinetic theory of an ideal gas

Five assumptions:

  1. Gas particles are in continuous, random, straight-line motion.
  2. The particles have negligible volume compared with the volume of the container.
  3. The particles exert no intermolecular forces on each other except during instantaneous elastic collisions.
  4. Collisions between particles and with the walls are perfectly elastic (no kinetic energy lost).
  5. The average kinetic energy of the particles is directly proportional to the absolute (Kelvin) temperature.

Real gases approximate ideal behaviour best at low pressure and high temperature, where the particles are far apart and moving fast enough that intermolecular forces are negligible. They deviate from ideal behaviour at high pressure and low temperature, where particle volume and intermolecular attractions matter.

Pressure from the particle picture

Pressure is the total force exerted by particle-wall collisions divided by the wall area. Three things can change pressure:

  • More particles in the same volume: more collisions per second, higher pressure.
  • Smaller volume at the same particle count: collisions concentrated on a smaller area, higher pressure.
  • Higher temperature: particles move faster, collide harder and more often.

This intuition underwrites every gas law.

Boyle's law: P and V at constant T and n

At constant temperature, pressure is inversely proportional to volume for a fixed amount of gas.

P1V1=P2V2P_1 V_1 = P_2 V_2

Compressing a gas to half its volume doubles the pressure (twice as many wall collisions per second). A P-V graph is a hyperbola; P against 1/V is a straight line through the origin.

Charles's law: V and T at constant P and n

At constant pressure, volume is directly proportional to absolute temperature.

V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}

T must be in Kelvin (K = degrees C + 273.15). A V-T graph extrapolates to zero volume at 0 K (absolute zero), which is the experimental basis for the Kelvin scale.

Gay-Lussac's law: P and T at constant V and n

At constant volume, pressure is directly proportional to absolute temperature.

P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}

A sealed container heated from 300 K to 600 K doubles its internal pressure. This is why aerosol cans warn "do not heat".

Combined gas law

When more than one variable changes simultaneously (and n is fixed):

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Use this when a gas sample moves between two different sets of conditions. T must always be in Kelvin; P and V can use any units provided they are consistent on both sides.

Worked example: combined gas law

A 2.50 L sample of air at 20 degrees C and 100 kPa is heated to 80 degrees C and compressed to 1.00 L. Find the new pressure.

Convert: T_1 = 293 K, T_2 = 353 K.

P2=P1×V1V2×T2T1=100×2.501.00×353293=301 kPaP_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 100 \times \frac{2.50}{1.00} \times \frac{353}{293} = 301 \text{ kPa}

Sanity check: compression should increase pressure (it did, factor of 2.5), heating should also increase pressure (it did, factor of 1.20). Final factor is about 3.0.

Units and conventions used in QCE

  • Pressure: kPa preferred (or Pa, atm, mmHg). 1 atm = 101.325 kPa = 760 mmHg.
  • Volume: L or mL. Convert mL to L by dividing by 1000.
  • Temperature: must be in Kelvin for gas-law calculations. K = degrees C + 273 (273.15 for exact).
  • Amount: mol.

QCAA's data sheet lists R = 8.314 J/(mol K) for use with P in kPa and V in L (yielding J units). Get used to the unit set early.

Examples in context

Example 1. Hot-air ballooning over the Atherton Tablelands. Cairns operators inflate balloon envelopes by burning propane at the burner, heating internal air from 20C20^{\circ}\text{C} to 120C120^{\circ}\text{C}. Applying Charles's law at constant pressure, V1/T1=V2/T2V_1/T_1 = V_2/T_2. Ratio V2/V1=393.15/293.15=1.34V_2/V_1 = 393.15 / 293.15 = 1.34. The 34%34\% volume increase pushes excess air out the open throat, reducing internal density (now 0.92kg m30.92 \, \text{kg m}^{-3} vs cold 1.21kg m31.21 \, \text{kg m}^{-3}); the buoyant force from the displaced cooler outside air lifts the balloon. Kinetic theory frames the result: hotter molecules collide more energetically with the envelope walls, requiring greater volume to maintain ambient pressure.

Example 2. Compressed-air storage at Mount Isa mines. Mount Isa underground compressed-air systems run at 700kPa700 \, \text{kPa} for pneumatic rock drills. When operators bleed air into a 50L50 \, \text{L} surface tank at 25C25^{\circ}\text{C} and then the tank rolls into 45C45^{\circ}\text{C} summer sun, Gay-Lussac's law at constant volume gives p2/T2=p1/T1p_2/T_2 = p_1/T_1, so p2=700×318/298=747kPap_2 = 700 \times 318/298 = 747 \, \text{kPa}. Engineering tolerance must cover this. Boyle's law also bites: if the same air leaks into a wider 100L100 \, \text{L} chamber, pressure halves to 350kPa350 \, \text{kPa}. Kinetic theory accounts for all three laws via molecular collisions with walls.

Try this

Q1. State Boyle's law and explain it using kinetic molecular theory. [3 marks]

  • Cue. At constant TT, pV=pV = constant. Smaller volume means more frequent wall collisions per unit area \rightarrow higher pressure.

Q2. A 4.00L4.00 \, \text{L} vessel at 300K300 \, \text{K} contains gas at 150kPa150 \, \text{kPa}. The gas is compressed to 1.00L1.00 \, \text{L} and warmed to 400K400 \, \text{K}. Calculate the new pressure. [4 marks]

  • Cue. Combined gas law: p2=150×(4.00/1.00)×(400/300)=800kPap_2 = 150 \times (4.00/1.00) \times (400/300) = 800 \, \text{kPa}.

Q3. Kinetic theory assumes (i) negligible molecular volume and (ii) no intermolecular forces. (a) State a condition where each assumption breaks down. (b) Predict whether real NH3\text{NH}_3 at 25C25^{\circ}\text{C} deviates above or below ideal. (c) Justify. [2+1+3 marks]

  • Cue. (a) High pp, low TT. (b) Below: H-bonding pulls molecules together so observed V<VidealV < V_{ideal}. ISMG analysis + judgement.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA-style4 marksA sample of helium gas occupies 250 mL at 27 degrees C and 101.3 kPa. (a) Calculate the new volume if the pressure is increased to 152.0 kPa at the same temperature. (b) Calculate the new volume if, instead, the temperature is raised to 127 degrees C at constant pressure.
Show worked answer →

A 4-mark answer needs both calculations with correct laws stated and Kelvin conversions.

(a) Boyle's law (constant T and n). P_1 V_1 = P_2 V_2.

V_2 = (P_1 V_1) / P_2 = (101.3 x 250) / 152.0 = 166.6 mL.

Pressure rose by a factor of 1.5, so volume falls to about two-thirds of the original.

(b) Charles's law (constant P and n). V_1 / T_1 = V_2 / T_2.

Convert: T_1 = 300 K, T_2 = 400 K. V_2 = V_1 x (T_2 / T_1) = 250 x (400 / 300) = 333 mL.

Temperature ratio is 4/3, so volume increases by the same ratio.

Markers reward the correct law named, Kelvin conversions, and units. Forgetting to convert degrees C to K is the most common single-mark loss.

2023 QCAA-style3 marksExplain, in terms of the kinetic theory of gases, why the pressure of a fixed mass of gas in a rigid container increases when its temperature is raised.
Show worked answer →

A 3-mark answer needs three linked claims.

Particle speed
Temperature is a measure of the average kinetic energy of the particles. Heating the gas increases the average kinetic energy, so particles move faster.
Collision frequency and force
Faster particles strike the container walls more often per second and with greater momentum per collision.
Pressure
Pressure is force per unit area delivered by these collisions. More frequent and harder collisions deliver more force per unit area, so pressure rises.

In a rigid container the volume is fixed, so the increase in particle activity has nowhere to go except into pressure. This is Gay-Lussac's law: P / T = constant at fixed V and n.

Markers reward the kinetic-energy link, the collision-rate-and-force link, and the pressure conclusion.

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