IB Physics HL Thermodynamics Paper 1 & 2 Entropy Calculations ~10 min read

Entropy Change Calculations

Last page we said what entropy is and which way it points. Now we put a number on it. There are two ways to do that, and the beauty is they agree: a big-picture one built from heat (ΔS = ΔQ/T), and a molecule-counting one built from arrangements (S = k_B ln Ω). Let’s meet both.

📚 What you need to know

From heat (at constant temperature): ΔS = ΔQ / T, measured in J K⁻¹
Heat in → ΔQ > 0 → entropy rises; heat out → ΔQ < 0 → entropy falls
A perfect reversible round-trip has no net heat, so ΔS = 0
From counting arrangements: S = k_B ln Ω, where Ω is the number of microstates (ways to arrange the particles and their energy)
For a change: ΔS = k_B ln(Ω₂/Ω₁)
Just one arrangement (perfect order) means Ω = 1, so S = 0
Let a gas fill twice the space and its microstates jump from 1 to 2^N, giving ΔS = Nk_B ln 2

The big-picture formula: entropy from heat

At a steady temperature T, moving heat in or out of a system changes its entropy by a simple amount — the heat transferred divided by the temperature:

Entropy change from heat ΔS = ^ΔQ⁄_T

The units are joules per kelvin (J K⁻¹), and T must be in kelvin. The sign follows the heat: add heat (+ΔQ) and entropy climbs; take heat away (−ΔQ) and it drops. For a perfect reversible process that ends where it started, no net heat flows, so ΔS = 0.

Why divide by T? Because the same lump of heat causes more disorder in a cold, orderly system than in a hot, already-jumbled one. A spark in a quiet library is a bigger disturbance than the same spark at a noisy concert — so the colder the system, the bigger the entropy jump per joule.

This one idea explains the puzzle from last page: why heat always flows hot → cold. When a lump of heat leaves a hot body and enters a cold one, the cold body (small T) gains more entropy than the hot body (large T) loses — so the total goes up. That’s the allowed direction.

The cold reservoir’s smaller T makes its entropy gain bigger than the hot reservoir’s loss. Total entropy rises — which is exactly why heat flows this way and never the reverse.

WE 1

1500 J of heat is added to a large tank of water held at a steady 300 K. Find the entropy change of the water.

Step 1 — heat added, so ΔQ = +1500 J; use ΔS = ΔQ / T ΔS = 1500 ÷ 300 ΔS = +5.0 J K⁻¹ Positive, because adding heat spreads energy around and raises disorder.

WE 2

1200 J of heat flows from a hot reservoir at 600 K into a cold reservoir at 300 K. Find the entropy change of each, and of the two together.

Hot reservoir (loses heat): ΔQ = −1200 J ΔS_hot = −1200 ÷ 600 = −2.0 J K⁻¹ Cold reservoir (gains heat): ΔQ = +1200 J ΔS_cold = +1200 ÷ 300 = +4.0 J K⁻¹ Total ΔS_total = +2.0 J K⁻¹ The cold side gains more than the hot side loses, so the total is positive — the second law is happy, and this is why heat flows this way.

The molecule-counting formula: entropy from arrangements

The second route goes right down to the particles. Ludwig Boltzmann linked entropy directly to the number of microstates — the number of different ways you can arrange the particles and their energy while the system still “looks” the same overall. Call that number Ω (Greek “omega”):

Entropy from microstates S = k_B ln Ω

Here k_B is the Boltzmann constant (1.38 × 10⁻²³ J K⁻¹) and Ω is the count of microstates. Notice what happens when there’s only one way to arrange things (Ω = 1): ln 1 = 0, so S = 0. Perfect order means zero entropy. As Ω grows, entropy grows too — but through a logarithm, so it rises fast at first and then more gently.

Entropy climbs with the number of microstates, but as a log: Ω = 1 gives S = 0 (only one arrangement — perfect order), and each extra arrangement adds a little less.

Why a logarithm? Because microstates multiply but entropy should add. Put two systems together and their arrangement-counts multiply (Ω₁ × Ω₂), yet we’d like their entropies to simply add. A logarithm does exactly that magic — it turns “times” into “plus”.

For a change, we compare the before and after counts:

Entropy change from microstates ΔS = k_B ln(Ω₂ / Ω₁)

Counting microstates: a gas in a box

Here’s the classic example that ties it all together. Take a box split in two, holding N labelled gas molecules. To start, they’re all trapped in the left half. There’s only one way to do that — every molecule on the left — so Ω₁ = 1^N = 1.

With the partition in, there’s just one arrangement (Ω₁ = 1). Remove it and each molecule independently picks left or right — 2 choices each — so Ω₂ = 2^N.

Pull out the partition and each molecule can now be on the left or right — 2 choices each. With N molecules that’s 2 × 2 × … = 2^N arrangements. Feed the two counts into the change formula:

ΔS = k_B ln(2^N / 1) = k_B ln(2^N) ΔS = Nk_B ln 2

And here’s the payoff: letting the gas fill twice the space is a volume doubling, V → 2V. Work it out the big-picture way and you get the same answer, Nk_B ln 2. The heat formula and the counting formula agree perfectly — two different windows onto the same idea.

WE 3

3 labelled gas molecules sit in one half of a box. The partition is removed. (a) How many microstates before and after? (b) Find the entropy change. (k_B = 1.38 × 10⁻²³ J K⁻¹.)

(a) Before: Ω₁ = 1. After: Ω₂ = 2³ = 8 (b) Use ΔS = k_B ln(Ω₂/Ω₁) ΔS = (1.38×10⁻²³) × ln(8 ÷ 1) ΔS = (1.38×10⁻²³) × 2.08 ΔS ≈ 2.9 × 10⁻²³ J K⁻¹ Same as Nk_B ln 2 with N = 3. Tiny for 3 molecules — but a real gas has ~10²³ of them.

All on one side
Ω₁ = 1

remove
partition

Free either side
Ω₂ = 2^N

S = k_B ln Ω

ΔS = Nk_B ln 2

💡 Top tips

ΔS = ΔQ/T needs T in kelvin and a steady temperature — that’s why we use big reservoirs whose T barely moves.
Sign of the heat sets the sign of ΔS: heat in is +, heat out is −.
Ω = 1 means S = 0 — only one arrangement is perfect order.
It’s ln Ω, not Ω: always take the natural log. For a doubling of space, Ω₂/Ω₁ = 2^N, so ΔS = Nk_B ln 2.
Units of entropy are J K⁻¹, and k_B = 1.38 × 10⁻²³ J K⁻¹.

⚠ Common mistakes

Using temperature in °C instead of kelvin in ΔS = ΔQ/T
Forgetting the sign of ΔQ — heat leaving gives a negative entropy change
Writing S = k_BΩ instead of S = k_B ln Ω
Thinking Ω₁ is more than 1 — when all particles are on one side, there’s exactly one arrangement
Forgetting that ln(2^N) = N ln 2 when simplifying

Quick recap: Entropy change comes two equivalent ways. From heat: ΔS = ΔQ/T (kelvin), positive for heat in, negative for heat out, zero for a reversible round-trip. From counting: S = k_B ln Ω and ΔS = k_B ln(Ω₂/Ω₁), with Ω = 1 giving S = 0. Doubling a gas’s space gives ΔS = Nk_B ln 2 either way.

Superb — you can now calculate entropy changes from both ends, heat and arrangements. With that in hand, we’re ready to state the rule that ties the whole topic together: the Second Law of Thermodynamics, which says the total entropy of the Universe can only ever increase. That’s next.

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Entropy Change Calculations

📚 What you need to know

The big-picture formula: entropy from heat

The molecule-counting formula: entropy from arrangements

Counting microstates: a gas in a box

💡 Top tips

⚠ Common mistakes

Want this to actually click before the exam?

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