state the basic assumptions of the kinetic theory of gases

Kinetic Theory of Gases – Cambridge A‑Level (9702)

Learning Objective

State the basic assumptions of the kinetic theory of gases, derive the kinetic‑theory pressure equation \[ pV=\frac13\,Nm\langle c^{2}\rangle , \] and compare it with the ideal‑gas law \(pV=Nk_{\!B}T\) to obtain the relationship \(\langle E_{\text{kin}}\rangle=\tfrac32k_{\!B}T\).

Fundamental Assumptions

The kinetic theory models a gas as a very large collection of identical point particles that move randomly and

interact only through very short, elastic collisions. All six assumptions below are required in the syllabus and are used explicitly in the derivation.

Assumption	What it means
Large number of particles	The gas contains a huge number \(N\) of identical atoms or molecules, so statistical averages are meaningful.
Negligible particle volume	Each particle is treated as a point; the total volume occupied by the particles is negligible compared with the container volume \(V\).
No intermolecular forces (except during collisions)	While separated, particles exert no attractive or repulsive forces on one another; forces act only during the instantaneous collisions.
Elastic collisions	Both particle–particle and particle–wall collisions are perfectly elastic, so kinetic energy is conserved in each collision.
Collision time \(\ll\) mean free time	The duration of a collision is extremely short compared with the average time between successive collisions, allowing particles to travel freely between impacts on the walls.
Kinetic energy ↔ temperature	The average translational kinetic energy of a molecule is proportional to the absolute temperature: \[ \langle E_{\text{kin}}\rangle=\frac12m\langle c^{2}\rangle=\frac32k_{\!B}T . \]

Molecular Motion and Pressure – Derivation

Consider a rectangular container of dimensions \(L\times A\) (length \(L\) along the \(x\)-axis, wall area \(A\) perpendicular to \(x\)). The volume is \(V=AL\). The derivation proceeds in four logical steps, each explicitly using the assumptions above.

1. Impulse from a single molecule.

   Take one molecule of mass \(m\) with velocity component \(c_x\) towards the wall at \(x=0\). Because the collision with the wall is elastic (Assumption 4), the \(x\)-component reverses sign:

   \[ \Delta p_x = m(-c_x)-m(c_x) = -2mc_x . \]

   The molecule travels a distance \(2L\) between successive impacts on the same wall, so the time between impacts is

   \[ \Delta t = \frac{2L}{|c_x|}. \]

   Hence the average force exerted by this molecule on the wall is

   \[ F_x = \frac{\Delta p_x}{\Delta t}= \frac{-2mc_x}{2L/|c_x|}= \frac{m c_x^{2}}{L}. \]

   Note that the free‑flight distance \(2L\) is justified by Assumption 5 (collision time ≪ mean free time).

2. Summation over all molecules.

   If the gas contains \(N\) molecules, the total force on the wall is the sum of the individual forces:

   \[ F = \frac{m}{L}\sum_{i=1}^{N}c_{x,i}^{2}. \]

   Define the mean‑square speed in the \(x\)-direction:

   \[ \langle c_x^{2}\rangle = \frac{1}{N}\sum_{i=1}^{N}c_{x,i}^{2}. \]

   Then

   \[ F = \frac{Nm\langle c_x^{2}\rangle}{L}. \]
3. From force to pressure.

   Pressure is force per unit area. Using the geometry \(A = V/L\) (since \(V=AL\)) gives

   \[ p = \frac{F}{A}= \frac{Nm\langle c_x^{2}\rangle}{L}\,\frac{L}{V}= \frac{Nm\langle c_x^{2}\rangle}{V}. \]
4. Isotropy of the gas.

   Because the motion is random (Assumption 1), the three Cartesian components have the same mean‑square value:

   \[ \langle c^{2}\rangle = \langle c_x^{2}\rangle+\langle c_y^{2}\rangle+\langle c_z^{2}\rangle = 3\langle c_x^{2}\rangle . \]

   Substituting \(\langle c_x^{2}\rangle = \langle c^{2}\rangle/3\) into the expression for \(p\) yields the fundamental kinetic‑theory result

   \[ \boxed{\,pV = \frac13\,Nm\langle c^{2}\rangle\,}. \]

Connection with the Ideal‑Gas Equation

The ideal‑gas law written for individual molecules is

\[ pV = Nk_{\!B}T . \]

Equating this with the kinetic‑theory expression gives

\[ \frac13\,Nm\langle c^{2}\rangle = Nk_{\!B}T \quad\Longrightarrow\quad \frac12\,m\langle c^{2}\rangle = \frac32\,k_{\!B}T . \]

Since \(\langle E_{\text{kin}}\rangle = \tfrac12 m\langle c^{2}\rangle\), we obtain the textbook relationship

\[ \boxed{\;\langle E_{\text{kin}}\rangle = \frac32\,k_{\!B}T\;}. \]

Key Formulae

• Pressure from molecular motion: \(pV = \dfrac13\,Nm\langle c^{2}\rangle\)
• Ideal‑gas law (molecular form): \(pV = Nk_{\!B}T\)
• Average translational kinetic energy: \(\langle E_{\text{kin}}\rangle = \dfrac32\,k_{\!B}T\)
• Root‑mean‑square speed: \(v_{\text{rms}} = \sqrt{\langle c^{2}\rangle}= \sqrt{\dfrac{3k_{\!B}T}{m}}\)

Worked Example (A‑Level style)

Question: A sample contains \(2.0\times10^{23}\) molecules of nitrogen (\(M = 28\ \text{g mol}^{-1}\)) at \(300\ \text{K}\). Find the pressure if the gas occupies a volume of \(5.0\times10^{-3}\ \text{m}^{3}\).

Solution:

1. Mass of one nitrogen molecule \[ m = \frac{M}{N_{\!A}} = \frac{28\times10^{-3}\ \text{kg mol}^{-1}}{6.022\times10^{23}\ \text{mol}^{-1}} = 4.65\times10^{-26}\ \text{kg}. \]
2. Mean‑square speed from \(\tfrac12 m\langle c^{2}\rangle = \tfrac32 k_{\!B}T\): \[ \langle c^{2}\rangle = \frac{3k_{\!B}T}{m} = \frac{3(1.38\times10^{-23}\,\text{J K}^{-1})(300\ \text{K})}{4.65\times10^{-26}\,\text{kg}} = 2.68\times10^{5}\ \text{m}^{2}\!\!/\text{s}^{2}. \]
3. Pressure from the kinetic‑theory expression \[ p = \frac13\frac{Nm\langle c^{2}\rangle}{V} = \frac13\frac{(2.0\times10^{23})(4.65\times10^{-26})(2.68\times10^{5})}{5.0\times10^{-3}} \approx 1.7\times10^{5}\ \text{Pa}. \]
4. Check with the ideal‑gas law: \[ p = \frac{Nk_{\!B}T}{V} = \frac{(2.0\times10^{23})(1.38\times10^{-23})(300)}{5.0\times10^{-3}} \approx 1.7\times10^{5}\ \text{Pa}, \] confirming consistency.

Suggested Diagram