Thermodynamics (HL)

The **total energy of all the particles**: their random **kinetic energy** + the **potential energy** of the forces between them.

**Temperature only** — an ideal gas has no inter-particle PE, so U is fixed by the random KE of the particles.

$Q = \Delta U + W$ — the heat **added** equals the rise in **internal energy** plus the **work done by** the gas.

$\Delta U = Q - W$ (heat in **minus** work done by the gas).

**Negative** — Q is the heat **added** to the gas, so heat leaving makes Q < 0.

**Negative** — W is the work done **by** the gas; on compression the surroundings do work on it, so W < 0.

$W = P\,\Delta V$ — pressure times the change in volume.

P in **pascals (Pa)**, ΔV in **cubic metres (m³)**, giving W in **joules (J)**.

**Internal energy** is energy a gas **already has** inside; **heat** is energy **flowing** in or out due to a temperature difference.

**ΔU = 0** — U depends on temperature alone, so no temperature change means no change in internal energy.

$\Delta U = Q - W = 500 - 200 = 300$ J (the gas warms).

Write each sign in **words** first ('heat removed → Q negative', 'gas compressed → W negative'), then plug into $\Delta U = Q - W$.

The **disorder** of a system — the **number of microstates** (microscopic arrangements) available. Unit: **J K⁻¹**.

One specific microscopic arrangement of the particles that gives the same overall (macroscopic) state. **More microstates ⇒ higher entropy**.

$\Delta S = \dfrac{\Delta Q}{T}$, with **T in kelvin**.

$\Delta S$ in **J K⁻¹**, $\Delta Q$ in **J**, $T$ in **K**.

Heat **in** ⇒ ΔQ is **positive** (entropy rises); heat **out** ⇒ ΔQ is **negative** (entropy falls).

The **entropy of an isolated system never decreases** — it increases for any irreversible (real) process.

Yes — but only if another part gains **more**, so the **total** entropy of the isolated system still does not decrease.

Because it **increases the total entropy** of the universe ($\Delta S_{total} > 0$); the reverse would decrease it, so it never happens unaided.

The **direction** of time set by the second law: real processes always run the way that **increases total entropy**.

Calculate $\Delta S_{total}$ for the isolated system. If it is **positive**, the process can occur (and is irreversible).

$\Delta S = \Delta Q/T$, and the **cold** body has the **smaller T**, so for the same ΔQ it gains **more** entropy than the hot body loses.

Entropy is the **joule per kelvin (J K⁻¹)**; energy is the **joule (J)** — do not confuse them.

$\Delta U = Q - W$, where **W** is the work done **by** the gas. Internal energy U depends only on temperature.

**T** is constant, so $\Delta U = 0$ and therefore $Q = W$.

**P** is constant; the work done by the gas is $W = P\,\Delta V$.

**V** is constant, so $W = 0$ and therefore $Q = \Delta U$.

**Q = 0** (no heat flows), so $\Delta U = -W$.

The **area under the curve** between the start and end volumes.

Takes in **Q_in** from the hot reservoir, does useful **work W**, and rejects **Q_out** to the cold reservoir. $W = Q_{in} - Q_{out}$.

$\eta = \dfrac{\text{useful work}}{\text{energy input}} = 1 - \dfrac{Q_{out}}{Q_{in}}$.

$\eta_{Carnot} = 1 - \dfrac{T_{cold}}{T_{hot}}$, with both temperatures in **kelvin**.

Friction, turbulence and unwanted heat loss waste energy, so the real efficiency is always **lower** than the Carnot ceiling.

$\eta = 1 - \dfrac{600}{800} = 0.25$, i.e. **25%**.

$\eta_{Carnot} = 1 - \dfrac{300}{500} = 0.40$, i.e. **40%**.