149–150, 152ThermodynamicsStretch

Kinetic Theory

The gas laws were discovered by experiment. Kinetic theory\textbf{Kinetic theory} explains them from first principles: treat a gas as a huge number of tiny particles obeying Newton’s laws, work out the force of their collisions on the walls, and the ideal gas equation appears.

40 min Topic 9 · Thermodynamics

What you'll be able to do

  • State the assumptions of the kinetic theory model of an ideal gas
  • Use pV=13Nm⟨c2⟩pV = \tfrac13 Nm\langle c^{2}\rangle
  • Derive and use 12m⟨c2⟩=32kT\tfrac12 m\langle c^{2}\rangle = \tfrac32 kT
  • Calculate root-mean-square speeds of gas molecules
1

Assumptions of the model

The kinetic theory of an ideal gas assumes: the gas contains a very large number of molecules moving randomly; the volume of the molecules is negligible compared with the volume of the container; there are no forces between molecules except during collisions; collisions are perfectly elastic; and the duration of a collision is negligible compared with the time between collisions.

These assumptions are why ideal behaviour is approached at low pressure and high temperature. At high pressure molecules take up a significant fraction of the volume; at low temperature intermolecular attractions become important.

Tip — Learn at least four assumptions word for word. They are commonly asked for directly.

2

From collisions to pressure

Consider one molecule of mass mm moving with velocity component uu towards a wall. It rebounds elastically with −u-u, so its momentum change is 2mu2mu.

It travels to the opposite wall and back, a distance 2l2l, before hitting the same wall again, so it collides every 2l/u2l/u seconds. The average force it exerts is 2mu2l/u=mu2l\dfrac{2mu}{2l/u} = \dfrac{mu^{2}}{l}.

Summing over NN molecules, and noting that on average the motion is shared equally between three dimensions, gives ⟨u2⟩=13⟨c2⟩\langle u^{2}\rangle = \tfrac13\langle c^{2}\rangle. Dividing force by wall area l2l^{2} gives the result.

pV=13Nm⟨c2⟩pV = \tfrac13 Nm\langle c^{2}\rangle
Kinetic theory equation.
3

Mean square and r.m.s. speed

Molecules have a range of speeds, so the model uses the mean square speed\textbf{mean square speed} ⟨c2⟩\langle c^{2}\rangle: square every speed, then take the average.

Its square root is the root-mean-square\textbf{root-mean-square} speed, crms=⟨c2⟩c_{\mathrm{rms}} = \sqrt{\langle c^{2}\rangle}. It is not the same as the average speed, and is always slightly larger.

Since NmNm is the total mass and Nm/VNm/V is density, the equation can also be written p=13ρ⟨c2⟩p = \tfrac13\rho\langle c^{2}\rangle.

Example 1Four molecules have speeds 200, 400, 500 and 700 m s−1700\ \mathrm{m\,s^{-1}}. Find the mean speed and the r.m.s. speed.
1Mean: (200+400+500+700)/4=450 m s−1(200 + 400 + 500 + 700)/4 = 450\ \mathrm{m\,s^{-1}}.
2Mean square: (40 000+160 000+250 000+490 000)/4=235 000(40\,000 + 160\,000 + 250\,000 + 490\,000)/4 = 235\,000.
3r.m.s.: 235 000\sqrt{235\,000}.
AnswerMean 450 m s−1450\ \mathrm{m\,s^{-1}}; r.m.s. 485 m s−1485\ \mathrm{m\,s^{-1}}.
4

Temperature and molecular kinetic energy

Equating 13Nm⟨c2⟩\tfrac13 Nm\langle c^{2}\rangle with NkTNkT gives 13m⟨c2⟩=kT\tfrac13 m\langle c^{2}\rangle = kT. Multiplying by 32\tfrac32 gives the average kinetic energy of one molecule.

So the average molecular kinetic energy is proportional to absolute temperature and independent of the type of gas\textbf{independent of the type of gas}. At the same temperature, hydrogen and oxygen molecules have the same average kinetic energy — but hydrogen molecules, being lighter, move faster.

At absolute zero the average kinetic energy would be zero, which matches its definition. For an ideal gas, with no potential energy between molecules, the internal energy is entirely this kinetic energy: 32NkT\tfrac32 NkT.

12m⟨c2⟩=32kT\tfrac12 m\langle c^{2}\rangle = \tfrac32 kT
Average kinetic energy per molecule.
crms=3kTmc_{\mathrm{rms}} = \sqrt{\dfrac{3kT}{m}}
r.m.s. speed.
Example 1Find the r.m.s. speed of nitrogen molecules (m=4.65×10−26m = 4.65\times10^{-26} kg) at 300 K.
1crms=3×1.38×10−23×3004.65×10−26c_{\mathrm{rms}} = \sqrt{\dfrac{3 \times 1.38\times10^{-23} \times 300}{4.65\times10^{-26}}}.
2=2.67×105= \sqrt{2.67\times10^{5}}.
Answer517 m s−1517\ \mathrm{m\,s^{-1}}

Tip — Doubling the absolute temperature doubles the kinetic energy but increases r.m.s. speed only by 2\sqrt{2}.

Equation recap

pV=13Nm⟨c2⟩pV = \tfrac13 Nm\langle c^{2}\rangle
Pressure from motion.
p=13ρ⟨c2⟩p = \tfrac13\rho\langle c^{2}\rangle
Density form.
12m⟨c2⟩=32kT\tfrac12 m\langle c^{2}\rangle = \tfrac32 kT
Energy and temperature.

Common mistakes to avoid

Taking the r.m.s. speed as the average of the speeds.
Square, average, then square-root.
Saying heavier molecules have more kinetic energy at the same temperature.
Average kinetic energy depends only on TT; heavier molecules move more slowly.
Doubling crmsc_{\mathrm{rms}} when temperature doubles.
crms∝Tc_{\mathrm{rms}} \propto \sqrt{T}.
Using the mass of a mole instead of a molecule.
In 32kT\tfrac32 kT and 13Nm⟨c2⟩\tfrac13 Nm\langle c^{2}\rangle, mm is one molecule.

Key takeaways

  • Kinetic theory models a gas as many randomly moving molecules colliding elastically.
  • pV=13Nm⟨c2⟩pV = \tfrac13 Nm\langle c^{2}\rangle.
  • r.m.s. speed is the square root of the mean square speed.
  • Average molecular kinetic energy 12m⟨c2⟩=32kT\tfrac12 m\langle c^{2}\rangle = \tfrac32 kT.
  • At equal temperature, all gases have equal average molecular kinetic energy.

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