Microscopic Derivation of Pressure

Learn how molecular motion leads to the formula relating gas pressure and molecular speeds.

1. Idea Behind the Derivation

The goal is to connect the pressure of a gas with the motion of its molecules. If we assume gas molecules move freely and collide elastically with the container walls, each collision gives a tiny push. Adding up all these pushes gives the pressure.

The derivation shows how pressure depends on the number of molecules and the average of their squared speeds.

2. Assumptions Used

To derive the pressure formula, we assume:

  • Molecules behave like point masses.
  • Collisions with walls are perfectly elastic.
  • Motion is random and equally likely in all directions.
  • No intermolecular forces except during collisions.

2.1. Why These Assumptions Help

They allow us to treat molecular motion using simple mechanics without worrying about forces between molecules.

3. Momentum Change During Wall Collision

Consider one molecule of mass m moving with velocity cx in the x-direction. When it hits the wall and bounces back elastically, its velocity becomes -cx.

So the change in momentum is:

\( \Delta p = m(-c_x) - m(c_x) = -2mc_x \)

3.1. Force on the Wall

The molecule keeps hitting the same wall repeatedly. Time between two collisions with the same wall is:

\( \Delta t = \dfrac{2L}{c_x} \)

Thus, average force exerted by one molecule is:

\( F = \dfrac{\Delta p}{\Delta t} = \dfrac{2mc_x}{\dfrac{2L}{c_x}} = \dfrac{mc_x^2}{L} \)

4. Pressure Due to Many Molecules

If there are N molecules in the container, the total force is the sum of forces from all molecules. Pressure is force divided by area.

The resulting expression becomes:

\( P = \dfrac{1}{V} m (c_{x1}^2 + c_{x2}^2 + ... + c_{xN}^2) \)

4.1. Using Symmetry

Because motion is random in all directions:

\( \bar{c_x^2} = \bar{c_y^2} = \bar{c_z^2} = \dfrac{1}{3}\bar{c^2} \)

Thus:

\( P = \dfrac{1}{3} \rho \bar{c^2} \)

5. Final Pressure Relation

The derived formula connecting pressure to molecular motion is:

\( \boxed{ P = \dfrac{1}{3} \rho \bar{c^2} } \)

Where:

  • P = pressure of the gas
  • ρ = mass density of the gas
  • \( \bar{c^2} \) = mean square speed of molecules

5.1. Meaning of the Result

Pressure increases if:

  • molecules move faster (higher speed)
  • molecules are heavier
  • there are more molecules in a given volume