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IB DP Physics · Theme B The particulate nature of matter · B.3 · ~7 min read

B.3 Gas laws

Pressure, the mole and Avogadro's number, the three gas laws and the ideal gas equation in moles and molecules, the kinetic model of an ideal gas, P = ⅓ρv², the internal energy of a monatomic gas, and when real gases are not ideal.

PressureMoles & AvogadroBoyle, Charles & pressure lawsPV = nRTKinetic modelP = ⅓ρv²Internal energyReal gases
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Pressure is force per unit area, P = F/A. In a gas it comes from molecules hitting the walls.
  • For a fixed amount of ideal gas, PV/T = constant. The full ideal gas law is PV = nRT (moles) or PV = NkBT (molecules), with T in kelvin.
  • The kinetic model links pressure to molecular speeds, P = ⅓ρ⟨v2⟩, and gives the internal energy of an ideal monatomic gas, U = 32nRT.
  • Real gases behave like ideal gases at low pressure, high temperature and low density.
  • B.3 is the same at SL and HL.

By the end you can

  • Use P = F/A and n = N/NA.
  • Apply the gas laws and PV = nRT or PV = NkBT, with temperatures in kelvin.
  • State the assumptions of the kinetic model and explain pressure with it.
  • Use P = ⅓ρ⟨v2⟩ and U = 32NkBT.
  • Explain when and why real gases differ from ideal gases.

1Pressure and the mole

P = FA DATA BOOKLET
pressure in pascals (Pa = N m−2)
n = NNA DATA BOOKLET
amount of substance in moles; NA = 6.02 × 1023 mol−1
  • Molar mass M is the mass of one mole: about 0.028 kg mol−1 for N2, 0.032 for O2, 0.004 for He. Mass of gas m = nM.

2The gas laws

VPT₁T₂ > T₁isotherms: PV = constanthotter gas: curvefurther from the axes
At constant temperature, halving the volume doubles the pressure.
LawKept constantRelationship
Boyle's lawT and nPV = constant
Charles' lawP and nV/T = constant
Pressure (Gay-Lussac) lawV and nP/T = constant
PVT = constant DATA BOOKLET
for a fixed mass of ideal gas, so P1V1/T1 = P2V2/T2
θ / °CP−273 °C = 0 KmeasuredextrapolatedEvery gas extrapolates to zero pressure at the same temperature: absolute zero.
Straight lines that all point back to −273 °C are how absolute zero was found.
Worked example

A tyre is at 2.0 × 105 Pa at 20 °C. After a drive its temperature is 40 °C. Find the new pressure (volume constant).

  1. T in kelvin: 293 K and 313 K M1
  2. P2 = 2.0 × 105 × 313/293 M1
  3. = 2.14 × 105 Pa A1
Worked example

A weather balloon holds 2.0 L of gas at 1.0 × 105 Pa and 290 K. It rises to 0.50 × 105 Pa and 250 K. Find its volume.

  1. V2 = V1 × (P1/P2) × (T2/T1) M1
  2. = 2.0 × 2.0 × (250/290) M1
  3. = 3.4 L A1
Common mistakePutting °C into P/T or V/T. 40 °C is not "double" 20 °C: in kelvin the rise is only 313/293 = 1.07.

3The ideal gas law

PV = nRT DATA BOOKLET
R = 8.31 J mol−1 K−1
PV = NkBT DATA BOOKLET
kB = R/NA = 1.38 × 10−23 J K−1
Worked example

How many moles and molecules of gas are in 0.50 m3 at 1.0 × 105 Pa and 300 K?

  1. n = PV/RT = (1.0 × 105 × 0.50)/(8.31 × 300) M1
  2. n = 20 mol A1
  3. N = nNA = 20 × 6.02 × 1023 = 1.2 × 1025 A1

4The kinetic model of an ideal gas

  • Assumptions: a very large number of identical molecules in random motion; the molecules' own volume is negligible compared with the container; collisions (with each other and the walls) are elastic and last a negligible time; no forces between molecules except during collisions; between collisions they move in straight lines at constant speed.
Each collision with a wallreverses the momentumcomponent: Δp = 2mv.Many collisions per secondgive a steady force, soa pressure P = F/A.P = ⅓ρ⟨v²⟩
Pressure is the average force per unit area from huge numbers of collisions with the walls.
  • Explaining the laws: heat the gas at constant volume → faster molecules → harder and more frequent collisions → higher pressure. Halve the volume at constant temperature → twice as many collisions per second on each m2 → double the pressure.
P = 13ρ⟨v2⟩ DATA BOOKLET
⟨v2⟩ is the mean square speed of the molecules
Worked example

Find the root mean square speed of nitrogen molecules at 300 K.

  1. combine P = ⅓ρ⟨v2⟩ with PV = nRT, ρ = nM/V: ⟨v2⟩ = 3RT/M M1
  2. = 3 × 8.31 × 300/0.028 = 2.67 × 105 m2 s−2 M1
  3. vrms = 517 m s−1 A1
U = 32NkBT = 32nRT DATA BOOKLET
internal energy of an IDEAL MONATOMIC gas: all of it is kinetic
Worked example

Find the internal energy of 2.0 mol of helium at 300 K.

  1. U = 1.5 × 2.0 × 8.31 × 300 M1
  2. = 7.5 kJ A1
  • Why no potential energy? In an ideal gas there are no intermolecular forces, so the intermolecular potential energy is zero. The internal energy depends only on temperature.

5Real gases

  • Real gases deviate at high pressure and low temperature: the molecules are close together, so their own volume and the attractive forces between them are no longer negligible. Near condensation the ideal gas law fails completely.
  • Good approximation: low pressure, high temperature, low density (for example air at room conditions).
Exam tip"Explain in terms of molecules" questions want: molecular speed or kinetic energy, the rate of collisions with the walls, and the change of momentum per collision. Name all three.
Linking questionThe ideal gas is the working substance for heat engines in B.4, and U = 32nRT is exactly the ΔU in the first law there.
Idea for your IA / explorationTest Boyle's law with a syringe and pressure sensor, or the pressure law with a flask in a water bath. Plot P against 1/V (or P against θ) and extrapolate to estimate absolute zero.

Test yourself — cover the answers first

Quick check

  1. What is 1 Pa in base units?
  2. State Charles' law.
  3. Why must T be in kelvin in PV = nRT?
  4. Name two assumptions of the kinetic model.
  5. If the absolute temperature of a gas doubles, what happens to vrms?
  6. When does a real gas behave least like an ideal gas?

Exam-style

  1. A 0.020 m3 cylinder holds oxygen (M = 0.032 kg mol−1) at 1.5 × 107 Pa and 290 K. Find the mass of oxygen.
  2. Air has density 1.2 kg m−3 at 1.0 × 105 Pa. Estimate the rms speed of its molecules.
Show answers
  1. kg m−1 s−2
  2. at constant pressure, V ∝ T (in kelvin) for a fixed mass of gas
  3. the equation needs T ∝ average kinetic energy, which is zero at 0 K, not at 0 °C
  4. elastic collisions; negligible molecular volume (also: no intermolecular forces, random motion)
  5. it increases by a factor of √2
  6. at high pressure and low temperature
  7. n = PV/RT = 1.5 × 107 × 0.020/(8.31 × 290) = 124.5 mol
    m = nM = 124.5 × 0.032
    = 4.0 kg
  8. ⟨v2⟩ = 3P/ρ = 3 × 1.0 × 105/1.2 = 2.5 × 105
    vrms = 500 m s−1
Practise this chapter
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Next chapterB.4 Thermodynamics →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.