A level-readyAll subjects · Physics · WPH15 (Unit 5)
The ideal gas equation
Pressure, volume and temperature of a gas are linked by one equation. It contains the gas laws you may already know.
pV = NkT
For an ideal gas:
pV = NkT
- p: pressure (Pa)
- V: volume (m3)
- N: number of molecules
- k: Boltzmann constant, 1.38 x 10-23 J/K
- T: absolute temperature (K)
Always use kelvin, and convert cm3 to m3 (1 cm3 = 10-6 m3).
The gas laws
For a fixed mass of gas (N constant), pV / T is constant:
- Boyle's law (constant T): pV = constant, so p is inversely proportional to V.
- Constant pressure: V / T = constant.
- Constant volume: p / T = constant.
Use p_1V_1 / T_1 = p_2V_2 / T_2 for before-and-after problems.
Absolute zero from a graph
If you plot p against temperature in °C for a gas at constant volume and extend the straight line back, it meets p = 0 at about -273 °C. This is absolute zero, which is why kelvin temperatures must be used in the gas equations. Real gases behave most like ideal gases at low pressure and high temperature.
Worked example
How many molecules are in 0.025 m3 of gas at 1.0 x 105 Pa and 300 K?
- N = pV / kT
- = 1.0 x 105 x 0.025 / (1.38 x 10-23 x 300)
- = 2500 / 4.14 x 10-21
Answer: N = 6.0 x 1023 molecules
Check yourself
- How many molecules are in 1.0 x 10-3 m3 of air at 1.01 x 105 Pa and 290 K? Give your answer in units of 1022, to 3 s.f.
Show answer
2.52 x 1022. N = pV / kT = 101 / (1.38 x 10-23 x 290) = 2.52 x 1022.
- 3.0 x 1023 molecules of gas occupy 0.010 m3 at 350 K. Find the pressure, in units of 105 Pa, to 3 s.f.
Show answer
1.45 x 105 Pa. p = NkT / V = 3.0 x 1023 x 1.38 x 10-23 x 350 / 0.010 = 144900 Pa = 1.45 x 105 Pa.
- A sealed can of gas at 1.00 x 105 Pa is heated from 20 °C to 80 °C. Find the new pressure, in units of 105 Pa, to 3 s.f.
Show answer
1.20 x 105 Pa. p_2 = p_1 x T_2 / T_1 = 1.00 x 105 x 353 / 293 = 1.20 x 105 Pa.
- Gas at 2.0 x 105 Pa occupies 3.0 x 10-3 m3. It is compressed at constant temperature to 1.2 x 10-3 m3. Find the new pressure, in units of 105 Pa.
Show answer
5 x 105 Pa. p_2 = p_1V_1 / V_2 = 2.0 x 105 x 3.0 x 10-3 / 1.2 x 10-3 = 5.0 x 105 Pa.
- A gas occupies 2.0 litres at 300 K. It is heated at constant pressure to 450 K. Find its new volume in litres.
Show answer
3 litres. V_2 = V_1 x T_2 / T_1 = 2.0 x 450 / 300 = 3.0 litres.
- Why must temperatures be in kelvin in pV = NkT?
- Pressure and volume are proportional to absolute temperature, which is zero at absolute zero
- Kelvin numbers are larger
- Celsius cannot be used for gases
- It makes k equal to 1
Show answer
Pressure and volume are proportional to absolute temperature, which is zero at absolute zero. Only on the absolute scale does T = 0 correspond to zero pressure.
- A graph of pressure against temperature in °C for a gas at constant volume is extended back. Where does it cross the temperature axis?
- At 0 °C
- At -100 °C
- At about -273 °C
- It never crosses
Show answer
At about -273 °C. This is absolute zero.
Sources
- Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
- Pearson qualifications: International Advanced Levels Physics (2018)
- Pearson Edexcel IAL Physics Unit 5 (WPH15) question paper, June 2023: list of data, formulae and relationships
Last updated 7 November 2026. Spotted a mistake? Tell us.