Gas Laws

A gas exerts pressure because its molecules constantly bombard every surface they meet. Pressure is the normal force per unit area, given by:

$$P = \frac{F}{A}$$

It is measured in pascals ($1\ \text{Pa} = 1\ \text{N m}^{-2}$); typical atmospheric pressure is about $1.0 \times 10^5\ \text{Pa}$.

Because the number of molecules in even a small gas sample is enormous, we count them in moles: one mole contains $N_A = 6.02 \times 10^{23}$ particles (the Avogadro constant). The amount of substance $n$ links the particle count $N$ and the sample mass $m$ through:

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

where $M$ is the molar mass. For example, $0.032\ \text{kg}$ of oxygen gas ($M = 0.032\ \text{kg mol}^{-1}$) is exactly one mole, i.e. $6.02 \times 10^{23}$ molecules.

You must be able to convert fluently between $m$, $n$ and $N$, quote pressures in pascals, and explain in words that gas pressure arises from a vast number of tiny molecular collisions with the container walls rather than from a single steady push.

Experiments on a fixed mass of gas reveal three simple patterns.

• Boyle's law: at constant temperature, pressure is inversely proportional to volume, so $PV$ is constant; halving the volume of a sealed syringe doubles the pressure.
• Charles's law: at constant pressure, volume is directly proportional to absolute temperature, so $\frac{V}{T}$ is constant.
• The pressure law: at constant volume, pressure is directly proportional to absolute temperature, so $\frac{P}{T}$ is constant; this is why a sealed aerosol can must not be heated.

The three combine, for a fixed amount of gas changing between two states, into:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Every form only works with temperature in kelvin; extrapolating the straight-line $V$–$T$ or $P$–$T$ graphs back to zero volume or pressure points to $-273\ ^{\circ}\text{C}$, absolute zero.

Know the graph shapes:

• $P$ against $V$ is a hyperbola (an isotherm) that never touches the axes.
• $P$ against $\frac{1}{V}$ is a straight line through the origin.
• $V$–$T$ and $P$–$T$ plots in kelvin are straight lines through the origin.

In a problem, first identify which variable is held constant, then select the matching proportionality.

The empirical laws are special cases of a single equation of state:

$$PV = nRT$$

where $R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$ is the universal gas constant. Written per molecule it becomes:

$$PV = N k_B T$$

with $k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}$; the two constants are linked by $R = N_A k_B$. An ideal gas is the theoretical gas that obeys this equation exactly at all temperatures and pressures.

Real gases approximate ideal behaviour when the pressure is low and the temperature is well above the boiling point, because then the molecules' own volume is negligible compared with the container and intermolecular forces barely act. Near condensation (high pressure, low temperature) real gases deviate strongly.

A typical task is finding the amount of gas in a container: for air at $1.0 \times 10^5\ \text{Pa}$ filling $0.025\ \text{m}^3$ at $293\ \text{K}$, $n = \frac{PV}{RT} \approx 1.0\ \text{mol}$. Be ready to use either form of the law, to justify the assumptions of ideality, and to state the conditions under which they fail.

Kinetic theory explains the macroscopic gas laws from molecular motion. Its assumptions:

• a gas consists of a very large number of identical point molecules in continuous random motion.
• all collisions (with each other and with the walls) are perfectly elastic.
• intermolecular forces act only during collisions.
• the total volume of the molecules is negligible compared with the container.
• the duration of a collision is negligible compared with the time between collisions.

The macroscopic pressure of an ideal gas originates from the collective elastic collisions of its molecules with the boundary walls. When a gas molecule of mass $m$ with velocity component $v_x$ rebounds elastically from a wall perpendicular to the $x$-axis, its momentum changes by $\Delta p = 2mv_x$.

The rate of these collisions depends on the molecular speed and number density, translating into a continuous normal force on the wall. Averaging this microscopic force over all $N$ molecules moving in three dimensions leads directly to:

$$P = \frac{1}{3}\rho v^2$$

This shows that pressure depends on mass density and the mean-square molecular speed $v^2$. Combining this result with $PV = N k_B T$ gives the central link between the micro and macro worlds:

$$\bar{E}_k = \frac{3}{2} k_B T$$

Absolute temperature therefore measures the average random translational kinetic energy per molecule. You should be able to list the assumptions, outline the momentum argument in words, and use both equations numerically.

For a real substance, internal energy consists of both the random kinetic energy of its molecules and the potential energy due to intermolecular forces. An ideal gas, however, assumes zero intermolecular potential energy ($E_p = 0$) because molecules interact only during instantaneous collisions. Consequently, the internal energy $U$ of an ideal gas is entirely kinetic.

For a monatomic ideal gas, $U$ equals the total random translational kinetic energy of its molecules:

$$U = N \bar{E}_k = \frac{3}{2} N k_B T = \frac{3}{2} n R T$$

Internal energy therefore depends only on the amount of gas and its absolute temperature, not on pressure or volume separately. Doubling the kelvin temperature of a sealed sample doubles its internal energy, while compressing a gas at constant temperature changes $P$ and $V$ but leaves $U$ unchanged.

Since $PV = nRT$, you can also write $U = \frac{3}{2} PV$, a quick route in calculations: helium at $1.0 \times 10^5\ \text{Pa}$ filling $0.030\ \text{m}^3$ stores $U = 1.5 \times 1.0 \times 10^5 \times 0.030 = 4.5 \times 10^3\ \text{J}$.

Be able to explain why $U$ is purely kinetic for an ideal gas and to compute it from $T$ and $n$, from $N$, or directly from $PV$.