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In Lesson 9 we derived the pressure equation of the kinetic theory of gases,
pV = (1/3) N m <c²>
In Lesson 7 we derived the Boltzmann form of the ideal gas equation,
pV = N k T
Both give the same pV for the same gas. If we equate them, we get a profound result: a microscopic relationship between the average kinetic energy of a molecule and the macroscopic temperature. This is the final and most important destination of Module 5.1.4 of the OCR A-Level Physics A specification (H556), and it is one of the most beautiful results in classical physics. Temperature, which began this chapter as "what a thermometer reads", is revealed to be nothing more than a measure of molecular kinetic energy.
pVFrom Lesson 9: pV = (1/3) N m <c²>.
From Lesson 7: pV = N k T.
Both must be equal:
(1/3) N m <c²> = N k T
The N's cancel:
(1/3) m <c²> = k T
Now multiply both sides by 3/2:
(1/2) m <c²> = (3/2) k T
And there it is. The left-hand side is the average kinetic energy per molecule ((1/2) m v² averaged over all molecules, where <c²> is the mean-square speed). The right-hand side is (3/2) k T, a simple multiple of the absolute temperature.
⟨KE⟩ = (1/2) m <c²> = (3/2) k T
This is arguably the most important equation in thermal physics. It says:
The average translational kinetic energy of a molecule in an ideal gas depends only on the absolute temperature, and is equal to
(3/2)kT.
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.
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,
<c²> = 3kT/m
c_rms = √(3kT/m)
m ≈ 3.3 × 10⁻²⁷ kg): c_rms ≈ 1920 m s⁻¹m ≈ 4.65 × 10⁻²⁶ kg): c_rms ≈ 511 m s⁻¹m ≈ 5.31 × 10⁻²⁶ kg): c_rms ≈ 478 m s⁻¹m ≈ 7.31 × 10⁻²⁶ kg): c_rms ≈ 407 m s⁻¹Both 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.
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.
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 × (1/2) m <c²> = N × (3/2) k T = (3/2) N k T
or, using N k = n R,
U = (3/2) n R T
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".
Find the mean translational kinetic energy of a molecule at T = 293 K.
<KE> = (3/2) k T
= (3/2)(1.38 × 10⁻²³)(293)
= (3/2)(4.04 × 10⁻²¹)
≈ 6.07 × 10⁻²¹ J
That is about 6 × 10⁻²¹ joules per molecule — a tiny amount. But there are about 2.7 × 10²⁵ molecules per cubic metre at STP, so the total translational kinetic energy per cubic metre is
U_per_m³ ≈ (2.7 × 10²⁵)(6.07 × 10⁻²¹) ≈ 1.64 × 10⁵ J
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.
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:
<c²> = 3 k T / m
= (3)(1.38 × 10⁻²³)(293) / (4.65 × 10⁻²⁶)
= 1.213 × 10⁻²⁰ / (4.65 × 10⁻²⁶)
≈ 2.61 × 10⁵ m² s⁻²
c_rms = √<c²> ≈ 511 m 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.
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
c_rms(600 K) = c_rms(293 K) × √(600/293)
= 511 × √(2.048)
= 511 × 1.431
≈ 731 m 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.
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?
T = m <c²> / (3k)
= (4.82 × 10⁻²⁶)(3.0 × 10⁵) / ((3)(1.38 × 10⁻²³))
= 1.446 × 10⁻²⁰ / (4.14 × 10⁻²³)
≈ 349 K
So about 76 °C — warm air, but not unreasonably so.
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
c_rms = √(3kT/m)
= √((3)(1.38 × 10⁻²³)(1500)/(3.3 × 10⁻²⁷))
= √((6.21 × 10⁻²⁰)/(3.3 × 10⁻²⁷))
= √(1.88 × 10⁷)
≈ 4340 m 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.
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.
c_rms(He) / c_rms(Ar) = √(m_Ar / m_He) = √(6.63 × 10⁻²⁶ / 6.64 × 10⁻²⁷) ≈ √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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