DPST1021 Chap.10 Kinetic Theory and the Ideal Gas Law
Kinetic Theory and the Ideal Gas Law
Lectures 20 and 21 open the gas part of the thermal physics block, arriving in week 8 straight after temperature, specific heat and latent heat. The material links two views of a gas. On the large scale you measure pressure, volume and temperature with gauges and thermometers; on the small scale you picture an enormous number of molecules flying about and bouncing off the container.
Kinetic theory shows that both pictures describe the same gas, a link the first law relies on the following week.
Three results do most of the work. Pressure is force per unit area, and the force on a wall is the rate at which molecules deliver momentum to it. The ideal gas law, PV = nRT = NkT, ties the large-scale quantities together and can be written per mole or per molecule.
Temperature, measured in kelvin, fixes the average translational kinetic energy of a molecule at three halves of kT.
Lecture 21 then extends the picture: molecules that can rotate or vibrate store extra energy, equipartition hands each active degree of freedom an equal share, and the internal energy of the gas follows from counting those degrees of freedom.
The lecture also treats two features of a real gas sample: the spread of molecular speeds described by the Maxwell distribution, and the mean free path between collisions.
Lab 6 tests the gas law directly. Test 2 questions here reward careful unit conversion and a correct count of degrees of freedom.
What this chapter covers
- 01
Pressure from collisions: treat each impact as a normal momentum transfer of 2mv and relate the average force on a wall to pressure through P = F/A.
- 02
Ideal gas law: apply PV = nRT with moles or PV = NkT with molecules, always using absolute pressure, cubic metres and kelvin.
- 03
Kinetic temperature: connect the average translational kinetic energy of one molecule to temperature through three halves of kT, and find rms speeds from it.
- 04
Degrees of freedom: count f = 3 for monatomic gases and f = 3, 5 or 7 for diatomic gases according to the temperature ranges on the formula sheet.
- 05
Equipartition and internal energy: give each active degree of freedom half of kT per molecule and compute internal energy or its change for n moles.
- 06
Speed distribution and mean free path: describe how the spread of speeds shifts with temperature and how collision spacing depends on density and molecular size.
Worked example · free
Doubling the rms speed of nitrogen
- 1Convert the mass: m = 28.0 × 1.661 × 10−27 = 4.651 × 10−26 kg.
- 1From ½mvrms² = 1.5kT, vrms = √(3kT/m) = √(3 × 1.381 × 10−23 × 300 / 4.651 × 10−26) = 517 m/s.
- 1Since vrms grows as the square root of T, doubling the speed needs four times the temperature: 1200 K, giving 1034 m/s.
- 1At 300 K nitrogen sits in the 100 K to 1000 K band, so f = 5 and Eint = 2.5 × 1.00 × 8.314 × 300 = 6.24 kJ.
- 1At 1200 K vibration is active, so f = 7 and Eint = 3.5 × 1.00 × 8.314 × 1200 = 34.9 kJ.
Key terms
- Ideal gas
- A model gas of point-like molecules that interact only through brief elastic collisions, obeying PV = nRT exactly.
- Boltzmann constant
- The gas constant per molecule, k = R/NA = 1.381 × 10−23 J/K, used whenever a formula counts particles rather than moles.
- Root-mean-square speed
- The square root of the average of the squared molecular speeds, equal to √(3kT/m) for an ideal gas.
- Degree of freedom
- An independent way a molecule can hold energy, such as motion along one axis, rotation about one axis, or a vibration.
- Equipartition theorem
- The principle that every active degree of freedom holds an average energy of ½kT per molecule at equilibrium.
- Mean free path
- The average distance a molecule travels between successive collisions with other molecules.
Kinetic Theory and the Ideal Gas Law FAQ
Why does gauge pressure give wrong answers in the gas law?
A tyre gauge reads the excess above the surrounding atmosphere, so a reading of 220 kPa corresponds to roughly 321 kPa absolute. The gas law relates the total force per area the molecules exert, which includes the atmospheric share. Add 1.013 × 10⁵ Pa to any gauge reading before substituting.
Do heavier gas molecules carry more kinetic energy at the same temperature?
No. At a given temperature every ideal gas has the same average translational energy per molecule, 1.5kT. What changes with mass is the speed: oxygen molecules, sixteen times heavier than hydrogen, move at one quarter of the rms speed. Rotational and vibrational energy can differ between species, which is why internal energy per mole depends on f.
Why do diatomic gases have different values of f at different temperatures?
Rotation and vibration are quantised, so each mode needs a minimum energy before it can absorb any share. Below about 100 K collisions are too gentle to start rotation, between roughly 100 K and 1000 K two rotational modes are active, and above about 1000 K vibration adds two more. Measured heat capacities step up to match.
Exam move
Begin each gas problem by writing a units line: pressure in pascals absolute, volume in cubic metres, temperature in kelvin, and whether you are counting moles or molecules. Next, practise the collision argument once on paper until you can reproduce the chain from 2mv per impact to three halves of kT without notes, since lecture outcomes expect you to explain the link, not just quote it.
Then drill calculations both ways: temperature to rms speed and back, and internal energy from f, n and T. Keep the diatomic temperature bands from the formula sheet beside you and decide f before any arithmetic. Finish with qualitative questions on the speed distribution and mean free path, sketching how the curve shifts when the gas is heated. Round only at the end, because the quiz checks significant figures.
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