H2 Physics Ideal Gas Notes: Kinetic Theory, pV = NkT, Kelvin | 9478

Study guideUpdated 21 Aug 2026
Q: What should I revise for H2 Physics ideal gases and kinetic theory?
A: For SEAB 9478 Topic 12, revise Kelvin temperature, pV=nRT pV=nRT , pV=NkT pV=NkT , the Boltzmann constant, Avogadro constant, kinetic-theory assumptions, pressure derivation, rms speed, and the link between temperature and mean translational kinetic energy.
TL;DR
These H2 Physics notes tie Kelvin temperature, gas laws, pV=nRTpV = nRT, pV=NkTpV=NkT, and kinetic theory into one revision workflow. Master the macro-to-micro links and the usual kelvin/unit traps so ideal-gas questions become much more systematic.

Concrete example: how to use this page

If pressure, volume, and temperature change, convert Celsius to kelvin first, then decide whether the amount of gas is constant. That check tells you whether to use a combined gas law or pV=nRTpV=nRT.

Temperature and gas-law decision map

Question clueFirst checkEquation familyTrap to avoid
Same sealed sample before and after a changenn is constantp1V1T1=p2V2T2\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}

Misconception check: Kelvin is not just a nicer unit for Celsius. It is an absolute temperature scale, so doubling TT in kelvin doubles the average translational kinetic energy of an ideal-gas particle. Doubling a Celsius reading does not mean the same physical thing.

If you searched for a formula sheet

Search intentUse this part of the pageThen route to
ideal gases a level physics notesStart with the decision map, then the ideal-gas equation section.Topic 13 Thermodynamic Systems for First Law and heat-transfer questions.
kinetic theory of gases formula sheetUse the pressure derivation checkpoint and rms-speed mass checkpoint.H2 Physics Data and Formulae 2026 for what is printed in the official data pages.
boltzmann constant formulaCompare pV=NkT pV=NkT , pV=nRT pV=nRT

Need the rest of the Thermal Physics sequence? Browse the H2 Physics notes hub for Topic 13 Thermodynamic Systems and the rest of the 9478 notes. For the official document route and paper weightings, see our H2 Physics syllabus guide.


1 Thermodynamic temperature: why Kelvin rules

A thermodynamic scale fixes its zero at absolute zero (0 K)(0 \space \text{K}), a universal anchor that does not depend on mercury, platinum or any other material indicator.

Hence the SI defines the kelvin (K)(\text{K}) by setting the Boltzmann constant to an exact value k=1.380649×1023 JK1k = 1.380 649 \times 10^{-23} \space \pu{J.K-1}

1.1 °C ↔ K switch

Convert the lab thermometer reading with

TK=TC+273.15. T_{\text{K}} = T_{^{\circ}\text{C}} + 273.15.

Mini-drill: Liquid nitrogen boils at 196 C-196 \space ^{\circ} \text{C}. What is this in kelvin? Answer: 77.15 K77.15 \space \text{K}.


2 Empirical gas laws (Boyle, Charles, Gay-Lussac)

LawWhat stays constantProportionalityEquation form
BoyleT,nT,np1/Vp \propto 1/Vp1V1=p2V2p_1V_1 = p_2V_2

Exam cue: Quote temperatures in kelvin or the proportionalities break - a common IP trap in Paper 1 MCQ.


3 Ideal-gas equation: macro ↔ micro

The empirical laws blend into a single statement

pV=nRT=NkT, pV = nR T = N k T,

where nn is moles, NN is particle count, R=8.314 Jmol1K1R = 8.314 \space \pu{J.mol-1.K^-1}

3.1 Avogadro constant

One mole contains exactly 6.02214076×10236.022 140 76 \times 10^{23} entities, giving the bridge Nk=nRNk = nR. Parents: this fixed particle count is why chem-physics cross-topic conversions always cancel neatly.

Macro-micro gas equation checkpoint

Before substituting into an ideal-gas equation, identify whether the question describes the gas as a sample in moles or as individual particles.

Given quantityFirst conversionEquation to useCommon trap
Moles of gas, nn, are givenKeep amount in mol.Use pV=nRTpV=nRT.Multiplying by Avogadro constant when the question already gives mol.
Particle count, NN, is given

Misconception check: RR is the gas constant per mole, while kk is the gas constant per particle. The equation changes because the amount-counting unit changes, not because the gas behaves differently.


4 Kinetic-theory model of an ideal gas

4.1 Core assumptions

  1. Particles are point masses with negligible volume.
  2. Motion is random and obeys Newtonian mechanics.
  3. Collisions are perfectly elastic.
  4. No intermolecular forces except during collisions.

4.2 Microscopic origin of pressure

A molecule hitting a wall reverses its xx-momentum (2mux)(2mu_x). Summing impulses over collision rate gives the pressure relation below.

Pressure derivation checkpoint

Use this map to keep the one-particle collision story connected to the final gas equation.

one molecule hits wall
  -> momentum change in x-direction
  -> force from impulse per time
  -> pressure from force per area
  -> sum over all molecules
  -> replace x-direction average by one-third of total mean-square speed
Derivation stepQuantity to writeWhy it belongs thereCommon trap
One wall collisionMomentum change is 2mux2mu_x.The molecule reverses its x-component of velocity after an elastic collision.Using total speed before choosing the wall direction.
Time between hits on the same wallTime is 2Lux\dfrac{2L}{u_x}

Worked check: the factor 13\dfrac{1}{3} does not come from the shape of the container. It comes from random motion in three perpendicular directions, so only one third of the mean-square speed contributes to pressure on a chosen pair of walls.

Misconception check: pressure is not caused by molecules "pushing" continuously on the wall. It comes from many tiny momentum changes during collisions, averaged over time and area.

pV=13Nmv2, pV = \dfrac13 N m v^2,

where v2v^2 is the mean-square speed.

4.3 Temperature as kinetic energy

Equating the kinetic result with pV=NkTpV = NkT gives

12mv2=32kT, \dfrac12 m v^2 = \dfrac32 kT,

so temperature measures average translational kinetic energy.

rms-speed mass checkpoint

For speed questions, decide whether the mass in the question describes one particle or one mole of particles. That choice decides whether the energy equation should use kk or RR.

Given mass informationUse this formMass unit to checkCommon trap
Mass of one molecule or atom, mmvrms=3kTmv_\text{rms} = \sqrt{\dfrac{3kT}{m}}

Worked check: helium has M=4.00 gmol1=4.00103 kgmol1M=\pu{4.00 g.mol-1}=\pu{4.00e-3 kg.mol-1}

vrms=3RTM=3(8.31)(300)4.00×1031.37103 ms1. v_\text{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.31)(300)}{4.00\times10^{-3}}}\approx \pu{1.37e3 m.s-1}.

Misconception check: kk and RR are not interchangeable constants. They match different ways of counting gas: one particle versus one mole.

Worked example: At 300 K300 \space \text{K} the rms speed of helium is 1.36×103 ms1\approx 1.36 \times 10^3 \space \pu{m.s-1} - faster than any IP badminton smash!


5 IP-style marks maximiser

  1. Unit discipline - always state “K” or “Pa” before substituting numbers.
  2. Boyle-Charles combo Qs - re-write into pV/T=constantpV/T = \text{constant} early to avoid 2-line algebra traps.
  3. Kinetic-theory derivation - memorise the one-dimension proof and the final three-dimensional averaging step because Topic 12 explicitly names that extension.
  4. Graph WA - plot pp v. TT to extrapolate to 273.15  C\pu{-273.15 \space ^\circ \text{C}}

6 Bridging ahead: link to latent heat & real gases

Knowing that KavgTK_{\text{avg}} \propto T explains why water vapour deviates from the ideal-gas equation near condensation - intermolecular attractions become non-negligible as energy drops. Flag this for Section V (Phase Equilibria) revision.


Need structured practice on Temperature and Ideal Gases? Our H2 Physics tuition programme covers this topic with weekly problem sets and practical data-handling drills.


Comprehensive revision pack

9478 Section IV, Topic 12 Syllabus outcomes

Candidates should be able to:

  • (a) show an understanding that a thermodynamic scale of temperature has an absolute zero and is independent of the property of any particular substance.
  • (b) convert temperatures measured in degrees Celsius to kelvin: T/K=T/C+273.15 T/\text{K} = T/^\circ\text{C} + 273.15 .
  • (c) recall and use the equation of state for an ideal gas expressed as pV=NkT pV = NkT

Concept map (in words)

Base everything on the Kelvin scale. Combine empirical gas laws into pV=nRTpV = nRT and translate to particle language with NkTNkT. Use kinetic theory to show that temperature measures average kinetic energy. Experimentally, plot graphs to verify straight-line behaviour and extrapolate to absolute zero.

Key relations

RelationComment
Celsius-Kelvin conversionTK=TC+273.15T_{\pu{K}} = T_{\pu{^\circ C}} + 273.15

Derivations & reasoning to master

  1. Kinetic theory pressure derivation: start with one molecule colliding elastically with a wall and sum over particles.
  2. Connection between Celsius and Kelvin: use linear extrapolation of pp vs TT graph to show intercept at 273.15 C\pu{-273.15 ^\circ C}.
  3. Root-mean-square speed: derive from pV=13Nmv2=NkT pV = \dfrac{1}{3} N m \langle v^2 \rangle = N k T

Worked example 1 - combined gas law

A 2.5 L\pu{2.5 L} sample of nitrogen at 150 kPa\pu{150 kPa} and 22 C\pu{22 ^\circ C} is heated to 80 C\pu{80 ^\circ C}

Solution sketch: Convert to kelvin, apply V2=V1T2/T1V_2 = V_1 T_2/T_1

Convert temperatures: T1=295.15 KT_1 = \pu{295.15 K}, T2=353.15 KT_2 = \pu{353.15 K}.

V2=2.5(353.15295.15)=2.99 L  (3.0 L). V_2 = 2.5\left(\dfrac{353.15}{295.15}\right) = \pu{2.99 L} \; (\approx \pu{3.0 L}).

Using n=pV/RTn = pV/RT with p=150 kPap=\pu{150 kPa}, V=2.5 LV=\pu{2.5 L}, T=295.15 KT=\pu{295.15 K}

n=(1.50×105)(2.5×103)8.314(295.15)=0.153mol  N=nNA9.2×1022molecules. n = \dfrac{(1.50\times 10^{5})(2.5\times 10^{-3})}{8.314(295.15)} = 0.153\,\pu{mol} \Rightarrow\; N = nN_A \approx 9.2\times 10^{22}\,\text{molecules}.

Worked example 2 - rms speed

Calculate the rms speed of oxygen molecules at 320 K. Compare with nitrogen at the same temperature.

Key steps: Use crms=3RTM c_{\text{rms}} = \sqrt{\frac{3RT}{M}}

crms(O2)=3(8.314)(320)0.0325.00102 ms1. c_{\text{rms}}(\text{O}_2) = \sqrt{\dfrac{3(8.314)(320)}{0.032}} \approx \pu{5.00e2 m.s-1}.

crms(N2)=3(8.314)(320)0.0285.34102 ms1. c_{\text{rms}}(\text{N}_2) = \sqrt{\dfrac{3(8.314)(320)}{0.028}} \approx \pu{5.34e2 m.s-1}.

So N2\text{N}_2 molecules move faster at the same TT because MM is smaller.

Practical & data tasks

  • Perform a pressure vs temperature experiment with a sealed syringe and digital sensor; extrapolate to absolute zero.
  • Use data logger to record pressure vs volume under constant temperature; confirm inverse proportionality.
  • Model kinetic theory using computer simulations (e.g., PhET Gas Properties) and note deviations when assumptions break.

Common misconceptions & exam traps

  • Plugging Celsius values directly into proportional equations.
  • Forgetting that Boltzmann constant k links to R through Avogadro constant.
  • Assuming real gases obey ideal behaviour near liquefaction; mention limitations when conditions deviate.
  • Mixing mass and molar mass when computing rms speeds.

Quick self-check quiz

  1. Why must temperature be in kelvin for gas laws?
    Because proportional relationships rely on an absolute scale with zero at 0 K0 \space \text{K} (absolute zero).
  2. What happens to pressure if volume halves at constant temperature?
    It doubles (Boyle's law).
  3. State two ideal gas assumptions.
    Point particles with negligible volume;
    collisions perfectly elastic; no intermolecular forces between collisions.
  4. How is average kinetic energy related to temperature?
    E=3/2kTE = 3/2 kT per molecule.
  5. Write the ideal gas equation in molecular form.
    pV=NkTpV = NkT

Revision workflow

  1. Re-derive the kinetic theory expression for pressure weekly.
  2. Create flashcards for each gas law with typical exam question types.
  3. Solve one mixed gas problem (changing p, V, T) and one rms speed calculation per study session.
  4. Summarise real-gas limitations and note where A-level examiners expect you to mention them.

Practice Quiz

Test yourself on the key concepts from this guide.


7 Further reading


8 Call-to-action

Parents: book a 60-min Thermodynamics booster to pre-empt Term 4 WA slippage. Students: try recasting every gas MCQ into pV=NkT pV = NkT form - even Section B “real-world” stories shrink into a one-line calculation once constants are parked.

Last updated 14 Jul 2025. Next review when SEAB publishes a newer H2 Physics syllabus document.

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Chee Wei Jie
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Chee Wei Jie·Academic Advisor (Physics)