Explain temperature in terms of the average translational kinetic energy of particles ($\bar{E}_k = \frac{3}{2} k_B T$), distinguishing absolute (kelvin) and celsius temperature scales

What this dot point is asking

VCAA wants you to connect temperature to the microscopic motion of particles. The key relation is that temperature in kelvin is proportional to the average translational kinetic energy of particles.

Kinetic theory result

For an ideal gas:

where $k_B = 1.38 \times 10^{-23}$ J K$^{-1}$ is Boltzmann's constant, and $T$ is in kelvin.

The factor of $3/2$ comes from three translational degrees of freedom (motion in $x$, $y$ and $z$). Rotational and vibrational motion contribute additional kinetic energy at higher temperatures, but only translational motion contributes to temperature in this definition.

Absolute (kelvin) and celsius scales

Kelvin is the SI temperature scale. It is anchored to absolute zero (the point at which classical kinetic theory predicts particle motion would stop). Celsius offsets the same temperature interval to put $0$°C at the freezing point of water at standard pressure.

The size of a degree is identical in K and °C, so temperature differences ($\Delta T$) have the same numerical value in both scales.

Absolute zero is $0$ K $= -273.15$°C. Room temperature is approximately $293$ K $= 20$°C.

What this means in practice

• Temperature is a measure of particle motion. Heat (energy in transit) is not the same thing.
• Doubling the absolute temperature doubles the average translational kinetic energy of particles.
• The rms speed is $v_{\rm rms} = \sqrt{3 k_B T / m}$, so doubling $T$ multiplies $v_{\rm rms}$ by $\sqrt{2}$, not by $2$.

VCAA exam style

Year 11 SACs and Unit 1 assessments commonly ask:

• Compare microscopic motion at two temperatures.
• Convert celsius to kelvin before substituting.
• Distinguish heat from temperature in a written-answer question.

Common traps

Using celsius in the kinetic-theory formula

The formula requires kelvin. Using $T = 30°$C instead of $T = 303$ K gives wildly wrong answers.

Confusing temperature with internal energy

A bathtub of warm water has more internal energy than a cup of boiling water. Temperature compares the average per-particle kinetic energy; internal energy depends on both temperature and amount.

Treating absolute zero as achievable

Classical kinetic theory predicts particle motion stops at $0$ K. Quantum mechanics shows zero-point motion persists, but the third law of thermodynamics still makes $0$ K unattainable.

In one sentence

Temperature in kelvin is proportional to the average translational kinetic energy of particles via $\bar{E}_k = \frac{3}{2} k_B T$, with $T (\text{K}) = T (\text{°C}) + 273.15$, so doubling the absolute temperature doubles the average kinetic energy and multiplies the rms speed by $\sqrt{2}$.

Examples in context

Example 1. Argon atoms in an Australian Synchrotron beam-line cryogenic cooler. The Australian Synchrotron at Clayton uses cryogenically cooled beryllium windows and liquid-nitrogen-cooled monochromators to control thermal motion. At $77$ K (liquid nitrogen), the average translational kinetic energy of an atom is $\bar{E}_k = (3/2) k_B T = 1.5 \times 1.38 \times 10^{-23} \times 77 = 1.59 \times 10^{-21}$ J. For argon ($m = 6.63 \times 10^{-26}$ kg), the root-mean-square speed is $v_{\rm rms} = \sqrt{2\bar{E}_k/m} = 219$ m s$^{-1}$. Cooling to $77$ K from room temperature reduces the rms speed by $\sqrt{293/77} = 1.95$, sharpening diffraction peaks and reducing thermal blur in protein crystallography.

Example 2. Bass Strait natural gas at the wellhead. Methane at a Bass Strait wellhead emerges at about $360$ K and $20$ MPa. Average translational kinetic energy per molecule is $(3/2) \times 1.38 \times 10^{-23} \times 360 = 7.45 \times 10^{-21}$ J. For CH$_4$ ($m = 2.66 \times 10^{-26}$ kg), $v_{\rm rms} = 750$ m s$^{-1}$. When the gas expands through the choke into a $5$ MPa pipeline, Joule-Thomson cooling drops the temperature to about $310$ K, and rms speed falls to $695$ m s$^{-1}$. This temperature shift is purely a kinetic-energy redistribution, confirming the kinetic-theory link between $T$ in kelvin and average molecular motion.

Try this

Q1. State the relationship between absolute temperature and the average translational kinetic energy of gas particles. [2 marks]

• Cue. $\bar{E}_k = (3/2) k_B T$, with $T$ in kelvin and $k_B = 1.38 \times 10^{-23}$ J K$^{-1}$.

Q2. Calculate the average translational kinetic energy of a nitrogen molecule at (a) $300$ K and (b) $1200$ K, and find the ratio of root-mean-square speeds. [4 marks]

• Cue. (a) $6.21 \times 10^{-21}$ J. (b) $2.48 \times 10^{-20}$ J. Ratio $v_{\rm rms}$(b)/(a) $= \sqrt{4} = 2$.

Q3. Refer to argon at $77$ K. (a) Calculate the average translational KE per atom. (b) Determine the rms speed. (c) Explain why cooling reduces thermal motion noise in X-ray diffraction. [2+2+2 marks]

• Cue. (a) $1.59 \times 10^{-21}$ J. (b) $219$ m s$^{-1}$. (c) Lower thermal vibration sharpens diffraction peaks because atoms stay closer to ideal lattice positions.