9478 / 2027
Temperature and Ideal Gases overview

Lesson 5 of 5

Temperature and particle energy

  • I can recall and apply mean translational kinetic energy = m mean(c^2)/2 = 3kT/2.Syllabus

    Syllabus 9478, 12(g). Recall and apply mean translational kinetic energy = m mean(c^2)/2 = 3kT/2.

Make a guess. You are not marked.

Helium and neon atoms are mixed in a container at one temperature. Which atoms have the greater root-mean-square speed?

  1. Neon, because heavier atoms carry more momentum and so move faster on average.
  2. Helium, because equal mean kinetic energy needs a smaller mass to move faster.
  3. They are equal, because the gas is at one temperature.
Show the answer

Helium, because equal mean kinetic energy needs a smaller mass to move faster.

At the same temperature the mean kinetic energy, 3/2 kT, is the same for both. 1/2 m<c2> is equal, so the lighter helium atoms have the larger rms speed.

Absolute temperature determines the mean translational kinetic energy per ideal-gas particle. Equal temperature gives equal mean energy, even when different particle masses lead to different characteristic speeds.

Connect the two pressure equations

For N particles of mass m, the ideal-gas relation is pV = NkT. The collision model gives pV = Nm mean(c2)/3. Equate them and cancel N:

NkT = ⅓Nm mean(c2)
kT = ⅓m mean(c2)
Mean translational kinetic energy
= ½m mean(c2) = (3/2)kT

This is the average kinetic energy associated with translation of each gas particle. Particles have a range of velocities; the equation does not assign the same energy to every particle. It also does not by itself give every contribution to the total internal energy of a molecular gas or another phase.

Distinguish mean speed from rms speed

Root-mean-square speed means square the speeds, find their mean, then take the square root:

crms = √[mean(c2)]
crms = √(3kT/m)

It is generally different from the mean speed and from the magnitude of the mean velocity. A zero mean velocity can result from opposite directions cancelling; speeds and their squares are non-negative.

Worked average

Three supplied speeds

For the illustrative speeds 100, 200 and 300 m/s:

Mean speed = (100 + 200 + 300)/3
= 200 m/s
Mean-square speed
= (1002 + 2002 + 3002)/3
= 140000/3 m2/s2
Rms speed = √(140000/3)
≈ 216 m/s

This small list illustrates how averaging operations differ. It is not a three-particle model of an isotropic equilibrium gas.

Estimate microscopic scales

With supplied k of order 10-23 J/K and T of order 300 K, mean translational energy is of order a few 10-21 J per particle. For a supplied light-particle mass of order 10-26 kg, √(3kT/m) is of order 103 m/s. These are rough scale checks, not precise measurements.

Worked equal-temperature comparison

Two different particle masses at 300 K

Use k = 1.38 × 10-23 J/K and a first particle mass m = 6.64 × 10-27 kg.

Mean translational energy
= (3/2)(1.38 × 10-23)(300)
= 6.21 × 10-21 J
crms = √[3(1.38 × 10-23)(300)/(6.64 × 10-27)]
≈ 1.37 × 103 m/s

At the same temperature, a second gas with particle mass 4m has the same mean translational energy. Its rms speed is 1/√4 = 1/2 of the first value: about 684 m/s.

Equal temperature gives equal mean energy, not equal speed

Compare two ideal gases at 300 K. Their particle masses are m = 6.64 × 10-27 kg and 4m. These are distribution averages, not a claim about every particle in either gas.

Four times the particle mass gives half the rms speed at the same temperatureTwo vertically separated comparisons share the same temperature, 300 kelvin. The first particle mass is m, 6.64 times ten to the minus 27 kilograms; the second is four m. Equal-length brown bars represent equal mean translational kinetic energies, 6.21 times ten to the minus 21 joules per particle. The blue rms-speed bars share a different common scale: the first is 240 drawing units long and represents about 1368 metres per second, while the second is 120 units and represents about 684 metres per second. Both kinds of bar start at zero. Energy and speed bars do not share units or one cross-quantity scale. These are scalar averages, with no assigned motion direction or claim that every particle has the shown speed. No particle radius is used to represent mass.Particle mass mT = 300 KMean translational energy06.21 × 10-21 J per particleRoot-mean-square speed0About 1368 m/sParticle mass 4mT = 300 KMean translational energy06.21 × 10-21 J per particleRoot-mean-square speed0About 684 m/s

The energy bars share one scale and are equal. The speed bars share a separate scale and have a 2:1 ratio: crms is proportional to 1/√m at fixed T. The bars compare averages rather than particle sizes or individual velocities.

Both gases are at 300 K. The equal energies are means per particle, and the speed comparison uses rms values. Neither value is assigned to every individual particle. Energy and rms-speed bars use separate scales; compare bar lengths only within the same quantity.

Check your understanding

Two ideal gases are at the same absolute temperature. A particle of the second gas has four times the mass of a particle of the first. Compare their mean translational kinetic energies and rms speeds.

  1. Their mean energies are equal and the heavier particles have one quarter of the rms speed.
  2. The heavier particles have four times the mean energy and the same rms speed.
  3. Their mean kinetic energies are equal and the heavier particles have one half of the rms speed.
Show the answer

Their mean kinetic energies are equal and the heavier particles have one half of the rms speed.

At equal T, both have mean translational energy 3kT/2. Since c_rms is proportional to 1/sqrt(m), multiplying m by four halves c_rms.

For the same gas, doubling the absolute temperature doubles mean translational energy but multiplies rms speed by √2, not two. A Celsius reading doubling does not establish either of these changes.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Calculate the mean translational kinetic energy of a nitrogen molecule at 300 K, and the root-mean-square speed of the molecules. The molar mass of nitrogen is 0.028 kg mol-1. [3]

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