3.6.2.2 - Ideal Gases

Experimental gas laws and their controls

The gas laws are empirical: they summarise experiments. Each simple relationship keeps both the amount of gas and one other variable constant. A sealed sample fixes the amount; it does not necessarily fix the volume if its boundary can move.

Here $T$ is absolute temperature in kelvins and $p$ is absolute pressure, measured from vacuum. A gauge often reads pressure above atmospheric pressure: then $p_{\mathrm{absolute}}=p_{\mathrm{gauge}}+p_{\mathrm{atmospheric}}$. Using the gauge reading alone changes the proportionality.

For Boyle's law, doubling pressure halves volume. A plot of $p$ against $V$ is an inverse curve; a plot of $p$ against $1/V$ is a straight line through the origin. Its gradient is the constant $pV$. A straight line with a nonzero intercept is linear, but does not show direct proportionality.

[DIAGRAM: asset_slug: 6.2.2 - Ideal Gases - Diagram 1; caption: Two equivalent plots of Boyle’s law; description: Equivalent inverse-volume and reciprocal-volume plots for a fixed amount of gas at constant temperature.]

These exact illustrative curves both represent $pV=400\,\mathrm{Pa\,m^3}$. The reciprocal-volume axis has unit $\mathrm{m^{-3}}$.

These relationships describe equilibrium states. Rapid compression heats a gas, so its immediate pressure is not a Boyle's-law measurement at the original temperature. Compress slowly and wait for thermal equilibrium. For Charles's law a movable boundary allows expansion while pressure stays constant; a rigid sealed flask instead follows the pressure law when heated. Required practical 8 investigates Boyle's and Charles's laws with these controls.

Absolute temperature and extrapolation

The Celsius zero is a chosen reference, so doubling a Celsius temperature does not double the thermal state that controls gas behaviour. Absolute zero is $0\,\mathrm{K}$, equivalent to $-273.15\,{}^\circ\mathrm{C}$. For the ordinary gases and thermal processes in this course, absolute temperature cannot be negative.

A Celsius interval and a kelvin interval have the same size. Heating from $20$ to $40\,{}^\circ\mathrm{C}$ changes $T$ from about $293$ to $313\,\mathrm{K}$: the ratio is $313/293$, not 2. The temperature difference is $20\,\mathrm{K}$, not $293\,\mathrm{K}$. Use 273 instead of 273.15 when appropriate to the data precision.

For a dilute gas at fixed volume, pressure plotted against Celsius temperature is approximately linear. Extending the fitted trend to zero pressure estimates an intercept near $-273\,{}^\circ\mathrm{C}$. At fixed pressure, a volume-versus-Celsius plot gives the same ideal extrapolated intercept. Plotting either quantity against kelvin temperature instead gives direct proportionality.

This is an extrapolation, not a way of experimentally shrinking a real gas to zero volume. Real gases condense or depart from ideal behaviour before reaching the extrapolated point. In the classical ideal-gas model the random translational kinetic energy tends to zero as $T$ tends to zero; do not infer that all microscopic motion in all matter literally ceases. A line continued into negative kelvin temperatures is outside the ordinary gas model's physical domain.

Moles, molecules and the ideal gas equation

For a fixed amount of gas, the three experimental laws combine to make $pV/T$ constant. Increasing the amount at fixed $T$ and $V$ raises pressure in proportion. An ideal gas obeys

Use $p$ in $\mathrm{Pa}$, $V$ in $\mathrm{m^3}$ and $T$ in $\mathrm{K}$. Here $n$ is the amount in moles and $N$ the number of particles. Helium particles are atoms; nitrogen particles are molecules. One mole contains $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$ particles, so

Substitution gives $pV=NRT/N_A=NkT$, where $k=R/N_A$. With $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$, this gives $k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$. The two forms are the same model, expressed per mole or per particle.