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NotesPhysicsTopic 2.3Pressure, volume and temperature relationships
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2.3.16 min read

Pressure, volume and temperature relationships

IB Physics • Unit 2

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Contents

  • The three gas laws
  • The combined gas law
  • Exam-style question
The big idea: Pump up a bike tyre and the pump gets warm; squeeze a gas into less space and its pressure climbs. Those links between a gas's pressure P, volume V and temperature T are the gas laws.

Hold one of them fixed and the other two are linked in a simple way — there are three of these links, one for each thing you keep fixed.
LawWhat is held fixedThe linkIn words
Boyle'stemperature TP V = constantsqueeze it (V down) → pressure up
Charles'pressure PV ÷ T = constantheat it → it expands
Gay-Lussac'svolume VP ÷ T = constantheat a sealed can → pressure up
New word — absolute temperature: Absolute temperature means temperature measured in kelvin (K), counted from absolute zero (the coldest possible, −273 °C).

Convert: T (K) = θ (°C) + 273. So 27 °C = 300 K.

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Temperature is ALWAYS in kelvin: Charles' and Gay-Lussac's laws only work with temperature in kelvin, not °C.

Always do T (K) = °C + 273 before you put a temperature into a gas-law formula.

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The three laws are really one rule. For a fixed amount of gas the quantity P V ÷ T stays the same — so its value at one moment equals its value at another. This is the combined gas law, and it is given in the data booklet.

Given in the data booklet (combined gas law). T is the ABSOLUTE temperature, in kelvin.
pressure of the gas (Pa)
volume of the gas (m³)
absolute temperature (K) — always in kelvin
How to use it for two states: Compare a 'before' (state 1) and an 'after' (state 2) of the same gas:

P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂.

If one quantity is held fixed it cancels: fixed T → P₁V₁ = P₂V₂ (Boyle); fixed P → V₁ ÷ T₁ = V₂ ÷ T₂ (Charles).

Each law is a special case

  • Fixed T → P V = constant (Boyle)
  • Fixed P → V ÷ T = constant (Charles)
  • Fixed V → P ÷ T = constant (Gay-Lussac)

Always, before you start

  • Temperature in kelvin (°C + 273)
  • Same units on both sides
  • Pressure and volume units just have to match
IB-style questionCalculate[2 marks]

A gas has a volume of 6.0 m³ at a pressure of 100 kPa. At the same temperature it is compressed to 2.0 m³. Find the new pressure.

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How this is tested — gas-law questions almost always involve a graph or a before/after pair:

Paper 1B (data)

  • Plot P against 1/V at fixed temperature — a straight line through the origin.
  • Read its slope = the constant K and state K's SI unit.

Paper 2

  • Use P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to find a percentage change in pressure, volume or amount of gas.
The classic trap: Leaving the temperature in °C — every gas-law T must be in kelvin (add 273).
Why P against 1/V is a straight line: Boyle says P V = K. Divide by V: P = K × (1/V).

That is the form y = (slope) × x, so a graph of P (y-axis) against 1/V (x-axis) is a straight line through the origin whose slope is K.

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IB-style questionDetermine[4 marks]

At constant temperature a student plots pressure P (in Pa) against 1/V (in m⁻³) for a fixed mass of gas. The points lie on a straight line through the origin passing through (0.10, 3.0) and (0.40, 12). Find the constant K and state its SI unit.

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A fixed mass of gas has a volume of 8.0 L at a pressure of 250 kPa.

At the same temperature it is slowly compressed to a volume of 5.0 L.

the new pressure of the gas.
[2 marks]

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