Thermodynamics

The first law of thermodynamics is the principle of energy conservation applied to a gas, given by:

$$Q = \Delta U + W$$

where $Q$ is the thermal energy added to the gas, $\Delta U$ is the change in its internal energy and $W$ is the work done by the gas.

The signs carry the physics:

• $Q > 0$ when heat flows in
• $W > 0$ when the gas expands and pushes on its surroundings
• $\Delta U > 0$ when the temperature rises

For an ideal gas the internal energy is purely the random kinetic energy of the molecules, so $\Delta U$ depends only on the temperature change. For a fixed amount of monatomic ideal gas:

$$\Delta U = \frac{3}{2} n R \Delta T = \frac{3}{2} N k_B \Delta T$$

A bicycle pump shows the law in action: pushing the piston in quickly does work on the trapped air ($W < 0$) before much heat can escape, so $\Delta U$ is positive and the barrel becomes noticeably warm.

You must be able to assign the correct sign to each term from a written description, rearrange the law to find any one quantity, and link the sign of $\Delta U$ to whether the gas heats up or cools down.

IB problems use four standard processes, each defined by what stays constant and each drawn as a different curve on a $P\text{-}V$ diagram. In every case the work done equals the area under the curve.

• Isobaric (constant pressure): a horizontal line, with $W = P\Delta V$.
• Isovolumetric (constant volume): a vertical line; zero area means $W = 0$, so $Q = \Delta U$.
• Isothermal (constant temperature): the curve follows $PV = \text{constant}$; since $\Delta U = 0$ for an ideal gas, all heat absorbed becomes work, $Q = W$.
• Adiabatic ($Q = 0$): no heat is exchanged, so $\Delta U = -W$; for a monatomic ideal gas the curve is steeper than any isotherm through the same point.

In an adiabatic compression the gas temperature rises even though no heat is added. Microscopically, gas molecules collide elastically with the inward-moving piston and rebound with a higher speed, gaining kinetic energy. Because the internal energy of an ideal gas is solely the sum of its molecular kinetic energies, this work done on the gas appears directly as a rise in temperature.

You must be able to identify each process from its curve, apply the first law to it, and carry out adiabatic calculations using the curve a monatomic ideal gas obeys:

$$PV^{5/3} = \text{constant}$$

Entropy measures how spread out the energy and particles of a system are. Counting arrangements makes this precise: if a macrostate can be produced by $\Omega$ different microstates, its entropy is:

$$S = k_B \ln \Omega$$

States that can occur in more ways have higher entropy, which is why gases mix and hot objects share their energy: the evened-out arrangements vastly outnumber the ordered ones.

For thermal energy $\Delta Q$ transferred at constant absolute temperature $T$, the entropy change is:

$$\Delta S = \frac{\Delta Q}{T}$$

The same energy transfer counts for more at a low temperature, and this is the key to the second law of thermodynamics: in any real (irreversible) process the total entropy of a system plus its surroundings increases. Heat therefore flows spontaneously from hot to cold, because the cold body gains more entropy than the hot body loses, and never the reverse on its own.

Local decreases are allowed. A freezer turns water into highly ordered ice ($\Delta S < 0$ for the water), but the heat it pumps into the kitchen raises the entropy of the surroundings by more, so $\Delta S_{\text{total}} > 0$.

You must be able to calculate entropy changes for both a system and its surroundings, predict the sign of $\Delta S$, and explain in entropy terms why a process is irreversible.

A heat engine is any device that runs a gas around a repeating cycle to convert thermal energy into useful mechanical work. In each cycle it absorbs heat $Q_H$ from a hot reservoir, delivers net work $W$, and rejects waste heat $Q_C$ to a cold reservoir.

Because the gas returns to its starting state, $\Delta U = 0$ over a complete cycle and conservation of energy gives:

$$Q_H = W + Q_C$$

Efficiency compares useful output with energy input:

$$\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}$$

The second law forbids $Q_C = 0$: no cyclic engine can convert all of its heat input into work, so $\eta < 1$ always. A petrol car engine, with hot combustion gases as its hot reservoir and the exhaust and radiator as its cold one, typically manages $\eta \approx 0.3$; the remaining $70\%$ of the fuel's energy leaves as waste heat.

On a $P\text{-}V$ diagram a cycle is a closed loop, and the area enclosed equals the net work per cycle. A clockwise loop is an engine (net work done by the gas); a counter-clockwise loop is a refrigerator or heat pump (net work done on the gas).

You must be able to track $Q_H$, $W$ and $Q_C$ through a cycle, calculate efficiencies, and read net work from loop areas.

The Carnot cycle is the idealized engine cycle that sets the efficiency ceiling for any engine working between the same two temperatures. It has four reversible steps:

• an isothermal expansion at $T_h$ during which the gas absorbs $Q_H$
• an adiabatic expansion that cools the gas to $T_c$
• an isothermal compression at $T_c$ that rejects $Q_C$
• an adiabatic compression that returns the gas to its starting state

Its efficiency depends only on the reservoir temperatures in kelvin:

$$\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}$$

A power station running between $T_h = 750\ \text{K}$ and $T_c = 300\ \text{K}$ can therefore never exceed $\eta = 1 - \frac{300}{750} = 0.60$, however well it is engineered.

The Carnot value is unachievable in practice because every step would have to be truly reversible. Reversible heat transfer in the isothermal stages requires an infinitesimally small temperature difference between the gas and the reservoir, meaning the transfer would be infinitely slow, and the adiabatic stages would need zero friction and no turbulence.

Real engines must operate in finite time and always experience friction and heat leaks, so they inevitably generate entropy and their efficiencies fall well below the Carnot limit.

You must be able to sketch and label the four steps on a $P\text{-}V$ diagram, calculate $\eta_{\text{Carnot}}$, and explain why real engines cannot reach it.