Mean Kinetic Energy and Temperature

Spec mapping: OCR H556 Module 5.2 — Kinetic theory of gases (the relationship $\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$21​m⟨c2⟩=23​kT between mean translational kinetic energy and absolute temperature; the root-mean-square speed $c_{\text{rms}} = \sqrt{3kT/m}$crms​=3kT/m​; the temperature scale as a microscopic property of an ideal gas). Refer to the official OCR H556 specification document for exact wording.

Lesson 9 gave us the kinetic-theory pressure equation $pV = \tfrac{1}{3} N m \langle c^{2} \rangle$pV=31​Nm⟨c2⟩ from a microscopic argument about molecular bombardment. Lesson 7 gave us the macroscopic ideal-gas equation $pV = NkT$pV=NkT. Both describe the same gas, so the right-hand sides must be equal — and equating them gives the deepest result of classical statistical mechanics:

$\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$21​m⟨c2⟩=23​kT

This single line, derivable from two earlier results in three steps, redefines the concept of temperature. Where Lesson 1 treated temperature as "what a thermometer reads", Lesson 10 reveals that temperature is fundamentally a measure of the mean translational kinetic energy per molecule, with the Boltzmann constant $k$k as the universal conversion factor. Every molecule in an ideal monatomic gas at temperature $T$T has an average translational kinetic energy of $\tfrac{3}{2} k T$23​kT — about $6 \times 10^{-21}$6×10−21 J at room temperature, irrespective of which gas the molecule is.

The consequences are profound. Lighter molecules at a given temperature move faster than heavier ones ($c_{\text{rms}} \propto 1/\sqrt{m}$crms​∝1/m​). At absolute zero, classical molecular motion ceases. The internal energy of an ideal monatomic gas depends only on temperature ($U = \tfrac{3}{2} N k T$U=23​NkT). And the macroscopic gas laws of Modules 5.1 and 5.2 are recast as statistical statements about Newtonian mechanics applied to enormous numbers of molecules.

This lesson completes Module 5.2. It works the derivation in three lines, unpacks its physical meaning in four corollaries, works eight numerical examples spanning room temperatures, atmospheric escape and quantum onset, and ends with the contract sections (synoptic links, specimen question, mark-loss patterns, going further, A-Level misconceptions, summary). OCR H556 mark schemes specifically test the derivation, the formula $c_{\text{rms}} = \sqrt{3kT/m}$crms​=3kT/m​, and the equipartition-light statement that mean KE per molecule depends only on $T$T.

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Equating the Two Expressions for $pV$pV

From Lesson 9: $pV = \tfrac{1}{3} N m \langle c^{2} \rangle$pV=31​Nm⟨c2⟩. From Lesson 7: $pV = N k T$pV=NkT.

Both must be equal:

$$\tfrac{1}{3} N m \langle c^{2} \rangle = N k T$$31​Nm⟨c2⟩=NkT

The $N$N's cancel:

$$\tfrac{1}{3} m \langle c^{2} \rangle = k T$$31​m⟨c2⟩=kT

Multiply both sides by $\tfrac{3}{2}$23​:

$$\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$$21​m⟨c2⟩=23​kT

The left-hand side is the mean translational kinetic energy per molecule: $\tfrac{1}{2} m c^{2}$21​mc2 averaged over all molecules, with $\langle c^{2} \rangle$⟨c2⟩ the mean-square speed. The right-hand side is $\tfrac{3}{2} k T$23​kT, a simple multiple of the absolute temperature.

$$\langle \text{KE} \rangle = \tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$$⟨KE⟩=21​m⟨c2⟩=23​kT

This is arguably the most important equation in thermal physics:

The average translational kinetic energy of a molecule in an ideal gas depends only on the absolute temperature, and equals $\tfrac{3}{2} k T$23​kT.

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What This Equation Tells Us

1. Temperature is kinetic energy

The formal meaning of "temperature" is now made explicit: it is a measure of the average molecular kinetic energy, with k as the constant of proportionality. Doubling the absolute temperature doubles the average kinetic energy per molecule. Halving it halves the kinetic energy.

Temperature is not the total kinetic energy — that depends on how many molecules you have, which is an extensive property. It is the average kinetic energy per molecule, which is intensive: the same in a drop of water at 20 °C as in the entire Atlantic Ocean at 20 °C.

2. All molecules have the same average kinetic energy at the same temperature

The equation (1/2)m<c²> = (3/2)kT does not involve the molecular mass m on the right-hand side. So at a given temperature, the average kinetic energy per molecule is the same for every ideal gas, regardless of mass.

This has a direct consequence: heavier molecules move more slowly. At T = 293 K,

$$\begin{aligned} \langle c^{2} \rangle &= 3kT/m \\ c_{\text{rms}} &= \sqrt{3kT/m} \end{aligned}$$⟨c2⟩crms​​=3kT/m=3kT/m​​

• Hydrogen (m ≈ 3.3 × 10⁻²⁷ kg): c_rms ≈ 1920 m s⁻¹
• Nitrogen (m ≈ 4.65 × 10⁻²⁶ kg): c_rms ≈ 511 m s⁻¹
• Oxygen (m ≈ 5.31 × 10⁻²⁶ kg): c_rms ≈ 478 m s⁻¹
• Carbon dioxide (m ≈ 7.31 × 10⁻²⁶ kg): c_rms ≈ 407 m s⁻¹

Both air components (N_2 and O_2) move at similar speeds, but hydrogen (much lighter) is four times faster than nitrogen. This is why hydrogen escapes from Earth's gravitational field — its molecules are fast enough to reach escape velocity — while heavier gases stay in the atmosphere.

3. Absolute zero means zero molecular motion (classically)

If T → 0, then <c²> → 0 and all molecules come to rest. This is the classical picture of absolute zero: the temperature at which molecular motion ceases. In reality, quantum mechanics gives every molecule a small residual zero-point energy, and the absolute minimum energy of a gas is a little above zero — but the classical picture gets you the right answer to an excellent approximation for almost all purposes.

4. Internal energy of an ideal (monatomic) gas

For an ideal monatomic gas, each molecule has only translational kinetic energy — there are no rotational or vibrational modes because the molecule is a single atom. The total internal energy is therefore

$$U = N \times \tfrac{1}{2} m \langle c^{2} \rangle = N \times \tfrac{3}{2} k T = \tfrac{3}{2} N k T$$U=N×21​m⟨c2⟩=N×23​kT=23​NkT

or, using $Nk = nR$Nk=nR,

$$U = \tfrac{3}{2} n R T$$U=23​nRT

The internal energy of an ideal monatomic gas depends only on T and is directly proportional to it. Doubling the temperature doubles the internal energy.

For polyatomic gases (O_2, N_2, CO_2, ...), the molecules also store energy in rotational (and, at high temperatures, vibrational) modes. The internal energy is then larger than (3/2)nRT — a fact described by the equipartition theorem, which lies beyond the A-Level specification. At A-Level, the formula U = (3/2)NkT strictly applies only to monatomic ideal gases, but you may assume it in exam questions that specify "monatomic".

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Worked Example 1: Mean KE at Room Temperature

Find the mean translational kinetic energy of a molecule at T = 293 K.

$$\begin{aligned} \langle \text{KE} \rangle &= \tfrac{3}{2} k T \\ &= \tfrac{3}{2}(1.38 \times 10^{-23})(293) \\ &= \tfrac{3}{2}(4.04 \times 10^{-21}) \\ &\approx 6.07 \times 10^{-21}\,\text{J} \end{aligned}$$⟨KE⟩​=23​kT=23​(1.38×10−23)(293)=23​(4.04×10−21)≈6.07×10−21J​

That is about $6 \times 10^{-21}$6×10−21 joules per molecule — a tiny amount. But there are about $2.7 \times 10^{25}$2.7×1025 molecules per cubic metre at STP, so the total translational kinetic energy per cubic metre is

$$U_{\text{per m}^{3}} \approx (2.7 \times 10^{25})(6.07 \times 10^{-21}) \approx 1.64 \times 10^{5}\,\text{J}$$Uper m3​≈(2.7×1025)(6.07×10−21)≈1.64×105J

About 164 kJ per cubic metre. This energy is completely invisible in everyday life, because all the molecular motions are random and cancel out on average. But if you could somehow capture and redirect it all, 164 kJ would be enough to boil about 0.07 kg of water — not trivial.

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Worked Example 2: Rms Speed of Nitrogen at Room Temperature

Find the rms speed of nitrogen molecules at T = 293 K. Take the mass of one nitrogen molecule as m = 4.65 × 10⁻²⁶ kg.

From (1/2) m <c²> = (3/2) k T:

$$\begin{aligned} \langle c^{2} \rangle &= 3 k T / m \\ &= (3)(1.38 \times 10^{-23})(293) / (4.65 \times 10^{-26}) \\ &= 1.213 \times 10^{-20} / (4.65 \times 10^{-26}) \\ &\approx 2.61 \times 10^{5}\,\text{m}^{2}\,\text{s}^{-2} \\ c_{\text{rms}} &= \sqrt{\langle c^{2} \rangle} \approx 511\,\text{m s}^{-1} \end{aligned}$$⟨c2⟩crms​​=3kT/m=(3)(1.38×10−23)(293)/(4.65×10−26)=1.213×10−20/(4.65×10−26)≈2.61×105m2s−2=⟨c2⟩​≈511m s−1​

Consistent with the earlier result from the pressure equation. Nitrogen molecules move at about 511 m s⁻¹ at room temperature — faster than the speed of sound, about 343 m s⁻¹.

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Worked Example 3: Rms Speed at a Higher Temperature

The same nitrogen is heated to T = 600 K. What is the new rms speed?

Since <c²> ∝ T, and since c_rms = √<c²>, we have c_rms ∝ √T. So

$$\begin{aligned} c_{\text{rms}}(600\,\text{K}) &= c_{\text{rms}}(293\,\text{K}) \times \sqrt{600/293} \\ &= 511 \times \sqrt{2.048} \\ &= 511 \times 1.431 \\ &\approx 731\,\text{m s}^{-1} \end{aligned}$$crms​(600K)​=crms​(293K)×600/293​=511×2.048​=511×1.431≈731m s−1​

The ratio √(600/293) ≈ 1.43 tells us that doubling the temperature (roughly) increases the rms speed by a factor of √2 ≈ 1.41. So the rms speed grows slowly with temperature — you have to quadruple T to double the speed.

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Worked Example 4: Finding the Temperature

Air molecules of average mass m = 4.82 × 10⁻²⁶ kg have a mean-square speed of <c²> = 3.0 × 10⁵ m² s⁻². What is the temperature of the gas?

$$\begin{aligned} T &= m \langle c^{2} \rangle / (3k) \\ &= (4.82 \times 10^{-26})(3.0 \times 10^{5}) / ((3)(1.38 \times 10^{-23})) \\ &= 1.446 \times 10^{-20} / (4.14 \times 10^{-23}) \\ &\approx 349\,\text{K} \end{aligned}$$T​=m⟨c2⟩/(3k)=(4.82×10−26)(3.0×105)/((3)(1.38×10−23))=1.446×10−20/(4.14×10−23)≈349K​

So about 76 °C — warm air, but not unreasonably so.

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Worked Example 5: Hydrogen at Very High Altitude

Near the top of the Earth's atmosphere (the exosphere), the temperature can reach T ≈ 1500 K (even hotter during solar activity). The escape velocity from Earth at that altitude is about 11.2 km s⁻¹ = 11 200 m s⁻¹.

Check whether a hydrogen molecule (m ≈ 3.3 × 10⁻²⁷ kg) can escape. The rms speed at 1500 K is

$$\begin{aligned} c_{\text{rms}} &= \sqrt{3kT/m} \\ &= \sqrt{(3)(1.38 \times 10^{-23})(1500)/(3.3 \times 10^{-27})} \\ &= \sqrt{(6.21 \times 10^{-20})/(3.3 \times 10^{-27})} \\ &= \sqrt{1.88 \times 10^{7}} \\ &\approx 4340\,\text{m s}^{-1} \end{aligned}$$crms​​=3kT/m​=(3)(1.38×10−23)(1500)/(3.3×10−27)​=(6.21×10−20)/(3.3×10−27)​=1.88×107​≈4340m s−1​

So the average hydrogen molecule at 1500 K has a speed of only about 4340 m s⁻¹ — well below escape velocity. But the Maxwell-Boltzmann distribution has a long high-speed tail: some fraction of molecules move considerably faster than c_rms. Over geological time, the tail of the distribution is enough to let Earth lose its entire hydrogen inventory to space. This is why Earth has very little hydrogen in its atmosphere today, despite hydrogen being the most abundant element in the universe. Heavier gases like nitrogen and oxygen have rms speeds one order of magnitude below escape velocity, and essentially none of them ever escape — Earth keeps its air.

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Worked Example 6: Comparing Gases at the Same Temperature

Two gases, helium (m_He = 6.64 × 10⁻²⁷ kg) and argon (m_Ar = 6.63 × 10⁻²⁶ kg), are at the same temperature. Compare their rms speeds.

$$\frac{c_{\text{rms}}(\text{He})}{c_{\text{rms}}(\text{Ar})} = \sqrt{\frac{m_{\text{Ar}}}{m_{\text{He}}}} = \sqrt{\frac{6.63 \times 10^{-26}}{6.64 \times 10^{-27}}} \approx \sqrt{9.98} \approx 3.16$$crms​(Ar)crms​(He)​=mHe​mAr​​​=6.64×10−276.63×10−26​​≈9.98​≈3.16

So helium moves about 3.16 times faster than argon at the same temperature. Again, this is because argon is about 10 times heavier.

But note: since <KE> = (3/2)kT depends only on temperature, each helium molecule carries the same kinetic energy as each argon molecule. Argon molecules are heavier but slower; helium molecules are lighter but faster. The product (1/2)m v² is the same in both cases.

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The Maxwell-Boltzmann Distribution (brief mention)

Real molecules do not all move at exactly c_rms. Different molecules have different speeds, and the distribution of speeds follows the Maxwell-Boltzmann distribution:

$$f(c) = 4 \pi N \left(\frac{m}{2 \pi k T}\right)^{3/2} c^{2} \exp\!\left(-\frac{mc^{2}}{2kT}\right)$$f(c)=4πN(2πkTm​)3/2c2exp(−2kTmc2​)

This is not on the OCR A-Level specification, but you may meet it in further reading. The distribution is a skewed bell curve, with most molecules moving at or near a "most probable" speed

$$c_p = \sqrt{2kT/m}$$cp​=2kT/m​

and a long tail extending to very high speeds. Three important speeds can be extracted from the distribution:

• Most probable speed: c_p = √(2kT/m)
• Mean speed: <c> = √(8kT/(πm)) ≈ 1.13 c_p
• Rms speed: c_rms = √(3kT/m) ≈ 1.22 c_p

All three are related by constants of order 1, but they are not equal. For energy calculations, always use c_rms.

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The Power of the Result

(1/2) m <c²> = (3/2) k T is, in a certain sense, the bridge between two pictures of thermodynamics:

• Macroscopic thermodynamics, as it developed from Sadi Carnot and Clausius, treats heat, temperature, and entropy as primitive quantities defined operationally.
• Kinetic theory, as developed by Maxwell and Boltzmann, derives all of these from the Newtonian mechanics of atoms and molecules.

The formula tells us that these two pictures are not rivals but complements. Temperature is mean kinetic energy per molecule, up to a constant of proportionality (3k/2), and the macroscopic gas laws follow from Newton's laws applied statistically to vast numbers of molecules.

This unification, achieved by the end of the nineteenth century, was one of the great triumphs of classical physics. It demonstrated that thermodynamics — which seemed at first to be a world of its own, with its own laws (the zeroth, first, second, third laws) and its own irreversible arrow of time — was actually a statistical consequence of ordinary mechanics. The apparent irreversibility emerges from the enormous number of particles involved and the statistical preference for high-entropy states. Individual molecules obey reversible Newtonian dynamics, but vast collections of them display irreversible macroscopic behaviour.

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Connection Diagram: Micro to Macro

graph LR
    MK["Microscopic<br/>kinetic theory"] --> EQ1[pV = 1/3 N m c²]
    GL["Macroscopic<br/>ideal gas law"] --> EQ2[pV = N k T]
    EQ1 --> EQ3[equate]
    EQ2 --> EQ3
    EQ3 --> RES[1/2 m c² = 3/2 k T]
    RES --> CONS1["Temperature ∝<br/>mean KE per<br/>molecule"]
    RES --> CONS2["Internal energy<br/>U = 3/2 N k T<br/>monatomic"]
    RES --> CONS3["Lighter molecules<br/>move faster at<br/>same T"]

This diagram summarises the entire chain of argument from Lessons 7 to 10.

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Worked Example 7: Total Internal Energy

Find the total translational kinetic energy of the air in a 200 m³ room at 293 K. Treat air as an ideal gas.

Total translational KE = (3/2) N k T. The number of molecules from pV = NkT:

$$\begin{aligned} N &= pV/(kT) \\ &= (1.01 \times 10^{5})(200)/((1.38 \times 10^{-23})(293)) \\ &= 2.02 \times 10^{7} / (4.04 \times 10^{-21}) \\ &\approx 5.0 \times 10^{27} \end{aligned}$$N​=pV/(kT)=(1.01×105)(200)/((1.38×10−23)(293))=2.02×107/(4.04×10−21)≈5.0×1027​

Then

$$\begin{aligned} U &= \tfrac{3}{2} N k T \\ &= \tfrac{3}{2}(5.0 \times 10^{27})(1.38 \times 10^{-23})(293) \\ &\approx 3.03 \times 10^{7}\,\text{J} \\ &\approx 30\,\text{MJ} \end{aligned}$$U​=23​NkT=23​(5.0×1027)(1.38×10−23)(293)≈3.03×107J≈30MJ​

(For a diatomic gas like air, the true internal energy is higher because of rotational modes — typically a factor of 5/3 larger, so around 50 MJ. But the monatomic formula gives the order of magnitude.)

That is a lot of thermal energy — enough to run a 1 kW heater for over 8 hours, or to lift a 1-tonne object through 3 km. It is simply not accessible to us as work, because it is randomly distributed: the second law of thermodynamics forbids us from converting all of it back into ordered motion.

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Worked Example 8: The Temperature Below Which a Gas Behaves Quantum Mechanically

This is not on the OCR specification, but is worth mentioning for context. Quantum effects become important in a gas when the thermal de Broglie wavelength λ = h/(2πmkT)^(1/2) becomes comparable to the average separation between molecules.

For helium at standard density (about 2.7 × 10²⁵ molecules/m³, corresponding to STP), the mean separation is (2.7 × 10²⁵)^(-1/3) ≈ 3.3 nm. Setting λ ≈ 3.3 nm gives T ≈ 4 K — roughly the condensation temperature of helium.

This is why helium liquefies and then becomes a superfluid at temperatures of a few K: its atoms are light enough, and its interactions weak enough, that quantum effects dominate well above absolute zero. Heavy atoms like argon crystallise (rather than becoming superfluids) before quantum effects kick in.

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Synoptic Links