148, 150–151ThermodynamicsCore

The Ideal Gas Laws

A bicycle pump gets warm, a sealed crisp packet swells at altitude, and a tyre’s pressure rises on a hot day. All three follow from a single equation linking the pressure, volume, temperature and amount of a gas: the ideal gas equation\textbf{ideal gas equation} pV=NkTpV = NkT.

35 min Topic 9 · Thermodynamics

What you'll be able to do

  • State Boyle’s law, Charles’s law and the pressure law
  • Use pV=NkTpV = NkT and pV/T=pV/T = constant for a fixed mass of gas
  • Explain how the gas laws lead to the idea of absolute zero
  • Describe experiments to investigate the gas laws
1

Pressure

Pressure is force per unit area, p=F/Ap = F/A, measured in pascals, where 1 Pa=1 N m−21\ \mathrm{Pa} = 1\ \mathrm{N\,m^{-2}}. Atmospheric pressure is about 1.0×1051.0\times10^{5} Pa.

A gas exerts pressure because its particles collide with the container walls. Each collision reverses a component of a particle’s momentum, and the rate of change of momentum over the whole wall is the force.

Every gas law can be explained by thinking about how often particles hit the walls and how hard.

p=FAp = \dfrac{F}{A}
Pressure.
2

The three gas laws

Boyle’s law\textbf{Boyle’s law}: for a fixed mass of gas at constant temperature, pVpV is constant. Halve the volume and particles hit the walls twice as often, doubling the pressure.

Charles’s law\textbf{Charles’s law}: at constant pressure, V∝TV \propto T. Hotter particles move faster, so the gas must expand to keep the collision rate per area unchanged.

The pressure law\textbf{The pressure law}: at constant volume, p∝Tp \propto T. Faster particles hit harder and more often.

In each case TT must be in kelvin. A graph of VV or pp against Celsius temperature is a straight line, but it only passes through the origin when plotted in kelvin.

pV=constantpV = \text{constant}
Boyle’s law, constant TT.
p1V1T1=p2V2T2\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}
Combined law, fixed mass.
Example 1A tyre holds air at 2.2×1052.2\times10^{5} Pa at 12∘12^{\circ}C. After a long drive the air is at 42∘42^{\circ}C. Assuming constant volume, find the new pressure.
1T1=285T_{1} = 285 K, T2=315T_{2} = 315 K.
2p2=2.2×105×315285p_{2} = 2.2\times10^{5} \times \dfrac{315}{285}.
Answer2.4×105 Pa2.4\times10^{5}\ \mathrm{Pa}

Tip — Before using any gas law, convert every temperature to kelvin. Using Celsius is the single most common error.

3

Absolute zero from the gas laws

Plot pressure against Celsius temperature for a fixed volume of gas and you get a straight line. Extrapolate it backwards and it meets zero pressure at about −273∘-273^{\circ}C.

The same intercept appears for any gas and any starting pressure. That temperature is absolute zero\textbf{absolute zero}, where the particles would have minimum kinetic energy and exert no pressure.

In practice real gases liquefy long before this, so the extrapolation describes ideal behaviour, not a measurement.

4

The ideal gas equation

Combining the three laws and allowing for the amount of gas gives pV=NkTpV = NkT, where NN is the number of particles and k=1.38×10−23 J K−1k = 1.38\times10^{-23}\ \mathrm{J\,K^{-1}} is the Boltzmann constant\textbf{Boltzmann constant}.

An ideal gas\textbf{ideal gas} obeys this equation exactly at all pressures and temperatures. Real gases approach ideal behaviour at low pressure and high temperature, when particles are far apart and move fast.

To investigate Boyle’s law, trap air in a sealed tube, change the pressure with a pump, and read volume and pressure after waiting for the temperature to settle. A graph of pp against 1/V1/V is a straight line through the origin.

pV=NkTpV = NkT
Ideal gas equation.
Example 1How many molecules are in 2.0×10−3 m32.0\times10^{-3}\ \mathrm{m^{3}} of gas at 1.0×1051.0\times10^{5} Pa and 300 K?
1N=pVkT=1.0×105×2.0×10−31.38×10−23×300N = \dfrac{pV}{kT} = \dfrac{1.0\times10^{5} \times 2.0\times10^{-3}}{1.38\times10^{-23} \times 300}.
Answer4.8×10224.8\times10^{22} molecules

Tip — When compressing slowly in the Boyle’s law practical, wait between readings so the gas returns to room temperature — rapid compression warms it.

Equation recap

pV=NkTpV = NkT
Ideal gas equation.
p1V1/T1=p2V2/T2p_{1}V_{1}/T_{1} = p_{2}V_{2}/T_{2}
Fixed mass.
k=1.38×10−23 J K−1k = 1.38\times10^{-23}\ \mathrm{J\,K^{-1}}
Boltzmann constant.

Common mistakes to avoid

Using Celsius in pV/TpV/T.
Always convert to kelvin: T=θ+273T = \theta + 273.
Saying doubling temperature from 20°C to 40°C doubles pressure.
In kelvin, 293 K to 313 K is only a 7% increase.
Omitting "fixed mass".
The laws apply only if no gas enters or leaves.
Plotting pp against VV and expecting a straight line.
Boyle’s law gives a straight line for pp against 1/V1/V.

Key takeaways

  • Boyle: pVpV constant at constant temperature.
  • Charles: V∝TV \propto T at constant pressure; pressure law: p∝Tp \propto T at constant volume.
  • For a fixed mass, pV/TpV/T is constant, with TT in kelvin.
  • Ideal gas equation: pV=NkTpV = NkT.
  • Extrapolating the gas laws to zero pressure locates absolute zero.

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