Hydrogen and oxygen samples are at the same temperature. Which statement is correct?

Their molecules have the same average kinetic energy

Their molecules have the same average speed

Their molecules have the same average kinetic energy

Both the speed and the kinetic energy are the same

Oxygen molecules move faster, being heavier

A refrigerator has a coefficient of performance of 4. This means:

It removes 4 J of heat from the cabinet per joule of work input

It violates the first law of thermodynamics

It removes 4 J of heat from the cabinet per joule of work input

It is 400% efficient at converting heat into work

Its efficiency as an engine would also be 4

On a PV diagram, the net work done by a gas over a complete cycle is:

The area enclosed by the cycle, positive if clockwise

Zero, since the gas returns to its initial state

The area enclosed by the cycle, positive if clockwise

Equal to the change in internal energy

Thermal Physics — Kinetic Theory, the Laws of Thermodynamics and Heat Engines

Thermodynamics is the chapter where sign conventions decide your marks. The physics is a single equation, ΔU = Q − W, and almost every wrong answer in the topic comes from getting a sign backwards or from forgetting which quantity is a state function and which is path-dependent. So this chapter puts the conventions and that distinction first, then works through the four standard processes, and closes with engines and entropy. One structural idea is worth carrying throughout: internal energy depends only on where you are, while heat and work depend on how you got there.

Kinetic Theory — Temperature Is Kinetic Energy

The kinetic theory replaces pressure and temperature with statements about molecules: P = (1/3)ρv²_rms and, most importantly, average KE per molecule = (3/2)k_BT. That second relation is the conceptual heart of the chapter, because it says temperature is nothing but the mean translational kinetic energy of the molecules.

Molecular speeds and degrees of freedom
Item	Expression	Note
RMS speed	v_rms = √(3RT/M)	The largest of the three; ∝ √T and ∝ 1/√M
Average speed	v_av = √(8RT/πM)	Middle value
Most probable speed	v_mp = √(2RT/M)	The smallest: v_mp < v_av < v_rms
Monatomic gas (He, Ar)	f = 3 · C_v = 3R/2 · γ = 1.67	Translation only
Diatomic gas (O₂, N₂)	f = 5 · C_v = 5R/2 · γ = 1.40	Adds two rotational modes
Polyatomic (non-linear)	f = 6 · C_v = 3R · γ = 1.33	Three rotational modes

🧠 Equipartition generates C_v, C_p and γ from one number

The equipartition theorem gives each degree of freedom an average energy of ½k_BT per molecule, so U = (f/2)nRT and therefore C_v = fR/2. Then Mayer's relation C_p − C_v = R gives C_p, and γ = C_p/C_v = 1 + 2/f. So you never need to memorise nine numbers — count the degrees of freedom and everything else follows in two steps. Two checks worth having: γ is always greater than 1, and it decreases as the molecule gets more complicated, because more storage modes make the gas harder to heat per degree. Note also that C_v and γ depend only on atomicity, not on which particular gas it is, so helium and argon behave identically here.

⚠️ At the same temperature all gases have the same kinetic energy, not the same speed

Since average KE = (3/2)k_BT depends only on T, hydrogen and oxygen molecules at 300 K carry identical average kinetic energies. But ½mv² being equal with different m means the speeds differ: v_rms ∝ 1/√M, so hydrogen (M = 2) moves about four times faster than oxygen (M = 32), because √(32/2) = 4. This pair of facts is asked constantly, usually as a statement question offering "same speed" and "same kinetic energy" as separate options. The same reasoning explains why hydrogen and helium escape a planet's atmosphere while nitrogen does not — the light molecules reach escape velocity far more often.

The First Law and the Four Processes

The first law is conservation of energy for a gas: ΔU = Q − W, with the convention that Q is positive when heat is supplied to the gas and W is positive when the gas does work by expanding. Getting these two signs right is most of the topic.

The four standard processes
Process	Held constant	Key results
Isothermal	Temperature	ΔU = 0, so Q = W = nRT ln(V₂/V₁). PV = constant
Adiabatic	No heat exchange	Q = 0, so ΔU = −W. PVᵞ = constant, TVᵞ⁻¹ = constant
Isobaric	Pressure	W = PΔV = nRΔT; Q = nC_pΔT
Isochoric	Volume	W = 0, so ΔU = Q = nC_vΔT

🎯 Why internal energy depends only on temperature, and what follows

For an ideal gas U = (f/2)nRT, so U is a function of temperature alone — it does not care about the pressure, the volume, or the route taken. Three consequences are examined repeatedly. In any isothermal process ΔU = 0 exactly, so all the heat supplied becomes work. In any cyclic process ΔU = 0 over the whole cycle, whatever happens in between, so Q_net = W_net. And in a free expansion into a vacuum the gas does no work and exchanges no heat, so ΔU = 0 and the temperature does not change at all — which surprises students who expect an expanding gas to cool. Internal energy is a state function; heat and work are not, and that asymmetry is the single most useful idea in the chapter.

🧠 Reading work off a PV diagram

The work done by a gas is the area under the curve on a PV diagram, which makes several questions purely visual. Expansion moves rightward and the work is positive; compression moves leftward and it is negative. For a closed cycle the net work is the area enclosed, positive if the cycle runs clockwise (an engine) and negative if anticlockwise (a refrigerator). Two further readings are useful: a vertical line is isochoric so it encloses no area and does no work, and an adiabatic curve is always steeper than an isothermal one through the same point, by the factor γ. That steepness is why an adiabatic compression raises the temperature and is asked as a graph-identification question.

The Second Law, Engines and Entropy

The first law permits any energy-conserving process; the second law says which of them actually happen. A heat engine takes Q₁ from a hot reservoir at T₁, does work W, and dumps Q₂ into a cold reservoir at T₂, with efficiency η = W/Q₁ = 1 − Q₂/Q₁. For a reversible (Carnot) engine this becomes η = 1 − T₂/T₁ with the temperatures in kelvin.

🎯 Why no engine can be 100% efficient, and where the temperatures must be in kelvin

η = 1 − T₂/T₁ reaches 1 only if T₂ = 0 K, which is unattainable, so some heat must always be rejected — that is the Kelvin–Planck statement of the second law in one line. Two practical readings follow. Raising T₁ helps, but lowering T₂ helps more per degree, which is why power stations care so much about cooling water. And the temperatures must be in kelvin: using °C is the single most common numerical error here, and it does not merely shift the answer, it can produce an efficiency above 1 or a negative one, which should be caught immediately as absurd. Note too that no real engine reaches the Carnot value, because Carnot assumes a reversible cycle with no friction and no finite temperature differences — it is an upper bound, not a target.

Engine, refrigerator and heat pump — the same cycle, different figure of merit
Device	What you want	Measure
Heat engine	Work out of heat in	η = W/Q₁ = 1 − T₂/T₁, always less than 1
Refrigerator	Heat removed from the cold space	COP = Q₂/W = T₂/(T₁ − T₂), can be greater than 1
Heat pump	Heat delivered to the warm space	COP = Q₁/W = T₁/(T₁ − T₂) = refrigerator COP + 1

⚠️ A coefficient of performance above 1 does not break energy conservation

A refrigerator with COP = 4 moves four joules of heat out of the cabinet for every joule of electrical work. Students read that as efficiency above 100% and assume a mistake. It is not: the device is not creating energy, it is pumping it from one place to another, and the joule of work is added to the four and dumped at the back as five. Energy is conserved exactly. The name is chosen deliberately — "coefficient of performance", not efficiency — because it is a ratio of what you want to what you pay, and nothing forbids that being greater than one. The same logic makes a heat pump an efficient way to warm a house: it delivers more heat than the electricity it consumes, by taking the difference from the cold outside air.

ℹ️ Entropy, and the direction of time

Entropy change is ΔS = Q/T for a reversible transfer, and the second law in its most general form says the entropy of an isolated system never decreases. That is what makes processes irreversible: heat flows from hot to cold and never spontaneously back, a gas expands to fill a container and never spontaneously gathers into a corner. Both directions conserve energy, so the first law permits both — only entropy distinguishes them, which is why the second law is sometimes called the arrow of time. Note that entropy can decrease locally, as it does inside a refrigerator; it is the total for the system plus its surroundings that cannot.

Key takeaways

Average KE per molecule = (3/2)k_BT, so temperature IS mean translational kinetic energy.
v_mp < v_av < v_rms, all ∝ √T and ∝ 1/√M — so hydrogen is 4× faster than oxygen at the same T.
Count degrees of freedom: C_v = fR/2, C_p − C_v = R, γ = 1 + 2/f. Monatomic 1.67, diatomic 1.40.
ΔU = Q − W, with Q positive when heat enters and W positive when the gas expands.
U depends only on T, so ΔU = 0 for isothermal, cyclic and free-expansion processes alike.
Isothermal PV = constant with Q = W; adiabatic PVᵞ = constant with ΔU = −W; isochoric W = 0.
Work is the area under a PV curve; net work in a cycle is the area enclosed, clockwise for an engine.
Carnot η = 1 − T₂/T₁ in KELVIN, always below 1; a refrigerator's COP can exceed 1 without breaking anything.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Hydrogen and oxygen samples are at the same temperature. Which statement is correct?
1. Their molecules have the same average speed
2. Their molecules have the same average kinetic energy
3. Both the speed and the kinetic energy are the same
4. Oxygen molecules move faster, being heavier
Show answer
Answer: B — Their molecules have the same average kinetic energy
Average kinetic energy is (3/2)k_BT and depends only on temperature, so it is identical. But v_rms ∝ 1/√M, so hydrogen (M = 2) moves √(32/2) = 4 times faster than oxygen (M = 32). Equal kinetic energy with unequal mass necessarily means unequal speed, which is exactly the distinction being tested.
For a diatomic ideal gas, the ratio γ = C_p/C_v is:
1. 1.67
2. 1.40
3. 1.33
4. 1.00
Show answer
Answer: B — 1.40
A diatomic molecule has f = 5 (three translational plus two rotational), and γ = 1 + 2/f = 1 + 2/5 = 1.40. The value 1.67 is monatomic (f = 3) and 1.33 polyatomic (f = 6). Note that γ falls as the molecule gets more complex, because extra storage modes make the gas harder to heat per degree.
An ideal gas undergoes a free expansion into a vacuum. Its temperature:
1. Falls, since the gas expands
2. Remains unchanged
3. Rises
4. Falls to absolute zero
Show answer
Answer: B — Remains unchanged
In a free expansion the gas pushes against nothing, so W = 0, and the container is insulated, so Q = 0. Hence ΔU = 0, and since U depends only on temperature for an ideal gas, T is unchanged. Expecting cooling comes from confusing this with an adiabatic expansion, where the gas does work and therefore does cool.
A Carnot engine works between 500 K and 300 K. Its efficiency is:
1. 40%
2. 60%
3. 20%
4. 100%
Show answer
Answer: A — 40%
η = 1 − T₂/T₁ = 1 − 300/500 = 0.4, or 40%. Both temperatures must be in kelvin — using Celsius here is the commonest numerical error in the topic and can even produce an efficiency above 1, which should immediately be recognised as impossible.
In an isothermal expansion of an ideal gas:
1. Q = 0 and ΔU = −W
2. ΔU = 0 and Q = W
3. W = 0 and Q = ΔU
4. Both Q and W are zero
Show answer
Answer: B — ΔU = 0 and Q = W
Temperature is constant, and U depends only on temperature, so ΔU = 0 and the first law gives Q = W — every joule of heat supplied leaves as work. Option 1 describes an adiabatic process and option 3 an isochoric one, so this question is really testing whether the four processes are distinguished.
A refrigerator has a coefficient of performance of 4. This means:
1. It violates the first law of thermodynamics
2. It removes 4 J of heat from the cabinet per joule of work input
3. It is 400% efficient at converting heat into work
4. Its efficiency as an engine would also be 4
Show answer
Answer: B — It removes 4 J of heat from the cabinet per joule of work input
COP = Q₂/W, so four joules are pumped out of the cabinet per joule of work. Nothing is violated: the device moves energy rather than creating it, and the work input is added to the heat removed and rejected at the back as five joules. It is called a coefficient of performance rather than an efficiency precisely because it may exceed one.
On a PV diagram, the net work done by a gas over a complete cycle is:
1. Zero, since the gas returns to its initial state
2. The area enclosed by the cycle, positive if clockwise
3. Equal to the change in internal energy
4. Always negative
Show answer
Answer: B — The area enclosed by the cycle, positive if clockwise
Work is path-dependent, so it does not vanish over a cycle even though ΔU does. The net work is the enclosed area, positive for a clockwise cycle (an engine) and negative for an anticlockwise one (a refrigerator). Since ΔU = 0 over the cycle, the first law then gives Q_net = W_net.
Through the same point on a PV diagram, an adiabatic curve compared with an isothermal curve is:
1. Less steep
2. Steeper, by the factor γ
3. Identical in slope
4. Horizontal
Show answer
Answer: B — Steeper, by the factor γ
Isothermal gives PV = constant with slope −P/V, while adiabatic gives PVᵞ = constant with slope −γP/V, and γ > 1 always. That extra steepness is why an adiabatic compression raises the temperature while an isothermal one does not, and identifying the two curves on a graph is a standard question.