Pressure, Amount of Substance, and the Ideal Gas Law
Pressure: Force Distributed Over Area
Pressure
Pressure is the force applied per unit area.
Mathematically, it is expressed as:
$$
P = \frac{F}{A}
$$
where:
- $P$ is the pressure (in pascals, Pa)
- $F$ is the force applied (in newtons, N)
- $A$ is the area over which the force is distributed (in square meters, m²)
Pressure is a scalar quantity.
The unit of pressure, the pascal (Pa), is equivalent to one newton per square meter ($\text{N m}^{-2}$).
Consider a block weighing 100 N resting on a surface with an area of 0.5 m². To find the pressure it exerts:
$$P = \frac{F}{A} $$
$$=\frac{100 \, \text{N}}{0.5 \, \text{m}^2} = 200 \, \text{Pa}$$
A cylinder with a weight of 500 N and a base area of 0.2 m² exerts a pressure of:
$$
P = \frac{500 \, \text{N}}{0.2 \, \text{m}^2} = 2500 \, \text{Pa}
$$
Always ensure the force is perpendicular to the surface when calculating pressure.
Amount of Substance: The Mole and Avogadro’s Constant
The amount of substance is measured in moles, a fundamental concept in chemistry and physics.
Mole
A mole is defined as the amount of substance containing as many particles (atoms, molecules, etc.) as there are atoms in 12 grams of carbon-12.
This number is the Avogadro constant,$N_A$, approximately $6.022 \times 10^{23}$ particles per mole.
Calculating Moles
If a substance contains $N$ particles, the number of moles $n$ is given by:
$$
n = \frac{N}{N_A}
$$
where:
- $n$ is the number of moles
- $N$ is the total number of particles
- $N_A$ is the Avogadro constant ($6.022 \times 10^{23}$ particles/mol)
- The molar mass of a substance (in grams per mole) tells you the mass of one mole of that substance.
- For example, water (H₂O) has a molar mass of $18 \text{.g mol}^{-1}$l.
How many moles are in $1.2 \times 10^{24}$ molecules of water?
Solution
Using the formula:
$$n = \frac{N}{N_A} $$
$$= \frac{1.2 \times 10^{24}}{6.022 \times 10^{23}} $$
$$\approx 2 \, \text{mol}$$
The Ideal Gas Law: A Universal Equation
Ideal gas law
The ideal gas law is the equation of state of a hypothetical ideal gas which relates the pressure, volume, temperature, and amount of substance in a gas.
It is expressed as:
$$
PV = nRT
$$
where:
- $P$ is the pressure (in pascals, Pa)
- $V$ is the volume (in cubic meters, m³)
- $n$ is the number of moles
- $R$ is the universal gas constant ($8.31 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}$)
- $T$ is the absolute temperature (in kelvin, K)
Calculate the pressure of 0.5 moles of an ideal gas in a 0.02 m³ container at 300 K.
Solution
Using the ideal gas law:
$$
PV = nRT
$$
$$
P = \frac{nRT}{V} = \frac{0.5 \times 8.31 \times 300}{0.02} = 62,325 \, \text{Pa}
$$
Always convert temperature to kelvin when using the ideal gas law.
Alternative Form: Using the Boltzmann Constant
The ideal gas law can also be expressed in terms of the number of molecules $N$ and the Boltzmann constant$k_B$:
$$
PV = Nk_BT
$$
where:
- $N$ is the number of molecules
- $k_B$ is the Boltzmann constant ($1.38 \times 10^{-23} \, \text{J} \, \text{K}^{-1}$)
The Boltzmann constant $k_B$ is related to the universal gas constant $R$ by the equation $R = N_A k_B$.
Observing the Ideal Gas Law: Empirical Observations
The ideal gas law is derived from three fundamental gas laws:
Boyle’s Law
At constant temperature, the pressure of a gas is inversely proportional to its volume ($P \propto \frac{1}{V}$).
$$P_1 V_1 = P_2 V_2$$
Charles’s Law
At constant pressure, the volume of a gas is directly proportional to its absolute temperature ($V \propto T$).
$$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$
Gay-Lussac’s Law
At constant volume, the pressure of a gas is directly proportional to its absolute temperature ($P \propto T$).
$$\frac{P_1}{T_1}=\frac{P_2}{T_2}$$
- Combining these relationships, we find: $$
\frac{PV}{T} = \text{constant}
$$ - This constant is proportional to the number of moles $n$, leading to the ideal gas law: $$
PV = nRT
$$
Actual Derivation of the Ideal Gas Law (in case you are really curious)
Although the IB syllabus does not require you to derive the ideal gas law, understanding where it comes from helps connect macroscopic properties (like pressure and temperature) to microscopic particle motion.
- Consider a box filled with gas particles, each of mass $m$, moving randomly in all directions.
- The gas has $N$ particles contained in a cubic box of volume $V=L^3$.
- Each particle moves in straight lines until it collides with a wall.
- When a particle hits a wall and bounces back elastically, its momentum changes by $\Delta p = 2mv_x$, where $v_x$ is the component of velocity perpendicular to the wall.
- The time between two successive collisions of this particle with the same wall is $\Delta t = \dfrac{2L}{v_x}$.
- Hence, the average force that this single particle exerts on the wall is
\[
F = \frac{\Delta p}{\Delta t} = \frac{m v_x^2}{L}
\] - If we sum this effect for all $N$ particles and divide by the wall’s area $A=L^2$, we obtain the pressure:
\[
p = \frac{\sum F}{A} = \frac{N m \langle v_x^2 \rangle}{V}
\] - Because the motion of particles is random, the average squared speed is distributed equally among the three directions:
\[
\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle = \frac{1}{3}\langle v^2 \rangle
\] - Substituting this relation gives
\[
pV = \frac{1}{3} N m \langle v^2 \rangle = \frac{1}{3} N m v_{\text{rms}}^2,
\]
where $v_{\text{rms}}$ is the root-mean-square speed of the particles. - Now, from kinetic theory, the average translational kinetic energy of one particle is related to temperature:
\[
\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T,
\]
where $k_B$ is the Boltzmann constant. - Combining both results:
\[
pV = N k_B T
\] - If we write $N = n N_A$, where $n$ is the number of moles and $N_A$ is Avogadro’s number, and recall that $R = N_A k_B$, we finally obtain the ideal gas law:
\[
pV = nRT.
\] - Note that this derivation is based on the assumptions that:
- gas particles have negligible volume
- collisions are perfectly elastic
- there are no intermolecular forces
- These assumptions are not fully true for real gases, but they work remarkably well for gases at low pressure and high temperature.
Applications and Limitations of the Ideal Gas Law
- The ideal gas law is a powerful tool for understanding gas behavior, but it has limitations.
- It assumes:
- No intermolecular forces between gas particles.
- Negligible volume of gas particles compared to the container.
The ideal gas law breaks down under conditions of high pressure or low temperature, where intermolecular forces and particle volume become significant.
Gas Laws


