B.3 Gas laws

Use the normal component

Only the component of force perpendicular to the surface contributes to pressure on that surface. A tangential component produces shear rather than normal pressure.

Read the proportionality

At fixed force, doubling area halves pressure. At fixed area, doubling the perpendicular force doubles pressure. Pressure is a scalar even though the force producing it has direction.

Worked example from the mapped local textbook

A $51\,\mathrm{kg}$ student stands on one foot with contact area $62\,\mathrm{cm^2}=62\times10^{-4}\,\mathrm{m^2}$. The perpendicular force is the weight, $F=mg=(51)(9.8)=5.0\times10^2\,\mathrm{N}$.

$$P=\frac{F}{A}=\frac{5.0\times10^2}{62\times10^{-4}}=8.1\times10^4\,\mathrm{Pa}$$

The result is large because the same weight acts over a small area.

Connect mass to moles

If molar mass is M, then

$$n=\frac{m}{M}$$

Use matching units for m and M. Once n is known, the number of particles is $$N=nN_A$$.

Use ratios efficiently

For equal numbers of particles, the samples contain equal amounts in moles even if their masses differ. For isotope or element comparisons, calculate moles before comparing particle counts.

Worked comparison from local Question Bank row 28984

For $40\,\mathrm{g}$ of argon-40, $n_{Ar}=40/40=1.0\,\mathrm{mol}$. For $8\,\mathrm{g}$ of helium-4, $n_{He}=8/4=2.0\,\mathrm{mol}$. Since $N=nN_A$,

$$\frac{N_{Ar}}{N_{He}}=\frac{1.0N_A}{2.0N_A}=\frac12$$

The larger argon mass does not mean more atoms; its molar mass is also larger.

Connect microscopic and macroscopic quantities

Temperature is related to average translational kinetic energy. Pressure comes from momentum transfer when particles collide with the container walls. More energetic or more frequent collisions can increase pressure.

It is an approximation

Real gases have finite-size particles and intermolecular forces. The ideal model is most reliable when particles are far apart and interactions are relatively unimportant.

Compare two states

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Temperatures must be absolute. If pressure and temperature stay fixed, volume is fixed; if volume and temperature stay fixed, pressure is fixed.

Use the constraint first

Name what is constant before choosing a simplified law: Boyle-type behaviour uses constant T, Charles-type behaviour uses constant P, and pressure–temperature behaviour uses constant V.

Worked example from the mapped local textbook

A fixed gas changes from $P_1=1.1\times10^5\,\mathrm{Pa}$, $V_1=0.27\,\mathrm{m^3}$ and $T_1=289\,\mathrm{K}$ to $V_2=0.35\,\mathrm{m^3}$ and $T_2=423\,\mathrm{K}$.

$$P_2=\frac{P_1V_1T_2}{T_1V_2}=\frac{(1.1\times10^5)(0.27)(423)}{(289)(0.35)}=1.2\times10^5\,\mathrm{Pa}$$

The amount of gas must remain fixed for this two-state relation.

Particle form

Using the number of particles N, the same law is

$$PV=Nk_BT$$

where $$k_B$$ is the Boltzmann constant. The forms are equivalent because $$R=N_Ak_B$$ and $$N=nN_A$$.

Choose the form from the data

Use the molar form when amount of substance is given. Use the particle form when a question asks for the number of molecules or gives microscopic quantities. Rearrange before substituting, for example $$N=PV/(k_BT)$$.

Worked example from local Question Bank row 22708

$N=2.7\times10^{15}$ helium atoms occupy $V=1.3\times10^{-5}\,\mathrm{m^3}$ at $18\,^{\circ}\mathrm{C}=291\,\mathrm{K}$. The particle form matches the data:

$$P=\frac{Nk_BT}{V}=\frac{(2.7\times10^{15})(1.38\times10^{-23})(291)}{1.3\times10^{-5}}=0.83\,\mathrm{Pa}$$

The very small pressure is consistent with the tiny number of atoms compared with a macroscopic mole.

Kinetic-theory relation

For an ideal gas,

$$P=\frac13\rho\overline{v^2}$$

where ρ is gas density and $$\overline{v^2}$$ is the mean square molecular speed. The speed in this equation is not simply the square of the average speed.

Read the trends

Greater molecular speed increases momentum change per collision and collision rate, increasing pressure. At fixed speed, greater density means more mass per unit volume and therefore greater pressure.

Worked example from the mapped local textbook

Nitrogen at pressure $1.0\times10^5\,\mathrm{Pa}$ has density $1.17\,\mathrm{kg\,m^{-3}}$. Rearranging the kinetic-theory relation gives

$$v_{\mathrm{rms}}=\sqrt{\overline{v^2}}=\sqrt{\frac{3P}{\rho}}=\sqrt{\frac{3(1.0\times10^5)}{1.17}}=5.1\times10^2\,\mathrm{m\,s^{-1}}$$

This is the root-mean-square speed, not the ordinary arithmetic mean speed.

What is included

The model includes translational kinetic energy only. It neglects intermolecular potential energy and does not include rotational or vibrational molecular energy.

Read the dependence

At fixed amount of gas, U is proportional to T. At fixed temperature, U is proportional to N or n. Particle mass does not appear directly in $$U=\frac32Nk_BT$$.

Worked example from the mapped local textbook

For $1.0\,\mathrm{mol}$ of an ideal monatomic gas at $300\,\mathrm{K}$,

$$U=\frac32nRT=\frac32(1.0)(8.31)(300)=3.7\times10^3\,\mathrm{J}$$

This is the total random translational kinetic energy in the model. At the same temperature, doubling the amount of gas doubles $U$.

Temperature matters

Higher temperature gives particles more kinetic energy, making attractive interactions less important. Low temperature increases the importance of intermolecular attractions and can bring the gas closer to condensation.

Where it fails

At high pressure or high density, particles are crowded: their finite size and interactions matter. Near phase changes, the ideal model is especially unreliable.