The gaseous state: ideal and real gases and pV = nRT

In a gas, molecules are in constant, rapid, random motion. They continually collide with the walls of the container, and each collision exerts a tiny force on the wall.

Pressure is the total force of all these collisions per unit area of the container wall.

• More frequent or more forceful collisions → higher pressure.
• This is why heating a fixed volume of gas raises its pressure (faster molecules hit harder and more often), and why compressing a gas raises its pressure (collisions become more frequent).

An ideal gas is a model that obeys $pV = nRT$ exactly. It makes two key assumptions:

1. The gas molecules have zero (negligible) volume themselves.
2. There are no intermolecular forces of attraction between the molecules.

(Other kinetic assumptions — random motion, elastic collisions — are developed in Physics; chemistry focuses on these two.)

Real gases are not ideal. They deviate most under conditions where the two assumptions break down:

• High pressure — molecules are pushed close together, so their own volume becomes significant compared with the container.
• Low temperature — molecules move slowly, so intermolecular attractions become significant (and eventually the gas condenses).

Real gases behave most ideally at low pressure and high temperature, where molecules are far apart and fast-moving.

The ideal gas equation links pressure, volume, amount and temperature: $pV = nRT$ with $R = 8.31\,\text{J K}^{-1}\text{mol}^{-1}$ (given in the exam).

Units are where most marks are lost — convert first:

Quantity	SI unit	Common conversions
$p$	pascals (Pa)	$1\,\text{kPa} = 10^3\,\text{Pa}$; $1\,\text{atm} \approx 1.01 \times 10^5\,\text{Pa}$
$V$	cubic metres (m³)	$1\,\text{dm}^3 = 10^{-3}\,\text{m}^3$; $1\,\text{cm}^3 = 10^{-6}\,\text{m}^3$
$T$	kelvin (K)	$T(\text{K}) = \theta(°\text{C}) + 273$
$n$	moles	—

Worked. Find the moles of gas in $0.500\,\text{m}^3$ at $1.00 \times 10^5\,\text{Pa}$ and $300\,\text{K}$: $n = \frac{pV}{RT} = \frac{(1.00 \times 10^5)(0.500)}{(8.31)(300)} = 20.1\,\text{mol}$

Since $n = \dfrac{m}{M}$, substitute into $pV = nRT$: $pV = \frac{m}{M}RT \quad\Rightarrow\quad M = \frac{mRT}{pV}$

This lets you find the molar mass (and hence $M_r$) of a volatile liquid or gas from its mass, pressure, volume and temperature.

Worked. $0.240\,\text{g}$ of a gas occupies $200\,\text{cm}^3$ at $1.00 \times 10^5\,\text{Pa}$ and $373\,\text{K}$. Find $M_r$.

• Convert: $V = 200 \times 10^{-6} = 2.00 \times 10^{-4}\,\text{m}^3$. $M = \frac{mRT}{pV} = \frac{(0.240)(8.31)(373)}{(1.00 \times 10^5)(2.00 \times 10^{-4})} = 37.2\,\text{g mol}^{-1}$ So $M_r \approx 37$.