1The idea in one paragraph

A gas in a cylinder can gain or lose energy in two ways: by heat, thermal energy that flows because of a temperature difference, and by work, when its boundary moves against a force. The first law of thermodynamics is conservation of energy for the gas: heat in equals the rise in internal energy plus the work the gas does. Take the gas round a closed loop of changes and it returns to where it started, so any net heat absorbed has been turned into work: that is a heat engine. The second law says the conversion is never complete. Some energy must always be rejected to something colder, and the best any engine working between two temperatures can do is the Carnot efficiency, 1 − T_c/T_h. Behind the second law is entropy: a count of how many microscopic arrangements match what you see. Isolated systems drift towards the macrostates with the most arrangements, and that drift is why heat flows from hot to cold and never back on its own.

2Systems, and the sign convention

A system is whatever you choose to draw a boundary around: here, usually, a fixed amount of gas in a cylinder. Two kinds matter.

• A closed system exchanges no matter with its surroundings, but energy can cross the boundary in both directions, as heat or as work. A gas sealed in a cylinder with a movable piston is closed.
• An isolated system exchanges neither matter nor energy. Nothing crosses the boundary at all. A perfectly insulated, rigid, sealed box would be isolated; so, taken as a whole, is the universe.

The IB uses the Clausius sign convention, and every mark in this subtopic depends on it. Figure 1 shows it on a cylinder.

• Q is the resultant thermal energy supplied to the system. Heat in: Q is positive. Heat out: Q is negative.
• W is the resultant work done by the system. The gas expands and pushes the piston out: W is positive. The gas is compressed, so work is done on it: W is negative.
• ΔU is the change in the system's internal energy. For an ideal gas, ΔU is positive exactly when the temperature rises.

3The first law of thermodynamics

Energy supplied to the gas as heat has only two places to go: into the gas's internal energy, or out again as work done by the gas. So:

Q = ΔU + W

This is conservation of energy applied to a closed system, nothing more. What makes it tricky is the signs.

Worked example 1: the signs in action. (a) A gas absorbs 500 J of heat and does 200 J of work pushing a piston out. (b) A different gas is compressed, with 150 J of work done on it, while it loses 40 J of heat to the surroundings. Find ΔU in each case.

In (b), compression pushes 150 J in, heat leaks 40 J out, and 110 J stays. If your answer came out as −190 J, you added the work instead of subtracting a negative.

Internal energy of a monatomic ideal gas. From B.3, U = (3/2)nRT, so

ΔU = (3/2)Nk_BΔT = (3/2)nRΔT

and since PV = nRT, the same thing can be written ΔU = (3/2)Δ(PV). That last form is a gift in cycle questions: you can read ΔU straight off the pressure and volume at two points of a P–V diagram without ever finding n or T.

4Work done by a gas, and the area under the curve

Picture the gas pushing a piston of area A through a small distance Δx at constant pressure P. The force is PA, so the work is PA × Δx, and AΔx is the change in volume:

W = PΔV (constant pressure)

On a P–V diagram that is the area of a rectangle, as Figure 2(a) shows. If the pressure changes as the gas expands, split the change into thin strips; each strip is a small PΔV, and together they add up to the area under the line. That is true whatever the shape of the line, and it is how the guide expects you to handle "situations where pressure is not constant".

The area gives the size; the direction gives the sign. Moving to the right (expansion), W is positive. Moving to the left (compression), W is negative. A vertical line (constant volume) has no area under it, so W = 0.

Worked example 2: pressure not constant. A monatomic ideal gas goes from X (1.0 × 10⁻³ m³, 4.0 × 10⁵ Pa) to Y (4.0 × 10⁻³ m³, 1.0 × 10⁵ Pa) along the straight line of Figure 2(b). Determine W, ΔU and Q.

The gas does 750 J of work but only 300 J came in as heat; the other 450 J came out of its internal energy, so it cooled. The guide says quantitative problems use monatomic ideal gases, which is why (3/2) is safe here.

5Four processes, each keeping one thing fixed

The guide names four processes. Each is obtained by keeping one quantity fixed, and each makes one term of the first law vanish or simplifies it. Figure 3 draws all four from the same starting state S.

Students swap the last two rows. Isothermal: T is fixed, so ΔU = 0; if the gas expands and does work, exactly that much heat must flow in. It needs a slow change with good thermal contact. Adiabatic: no heat crosses, so the work done by an expanding gas comes out of its internal energy and it cools; compress it and it heats up. It needs good insulation or a change too fast for heat to flow, like the compression stroke of a diesel engine, which heats the air enough to ignite the fuel.

That cooling is why the adiabat in Figure 3 is steeper than the isotherm. Expanding from S, both lose pressure because the volume grows, but the adiabatic gas also cools, so its pressure falls further. For a monatomic ideal gas the adiabat obeys:

PV^(5/3) = constant (adiabatic, monatomic ideal gas)

Worked example 3: an adiabatic compression. Helium at 1.0 × 10⁵ Pa and 300 K occupies 1.20 × 10⁻³ m³. It is compressed adiabatically to 0.30 × 10⁻³ m³. Calculate the final pressure, the final temperature and the work done.

The volume went down by a factor of 4 and the pressure up by more than 10, because the temperature rose as well. An isothermal compression by the same factor would only have multiplied the pressure by 4.

6Cycles and heat engines

A cyclic process takes a gas through a series of changes and back to its starting state. Because it ends where it began, its temperature is the same, so over a whole cycle ΔU = 0. The first law then says:

Over a cycle: net Q = net W

On a P–V diagram, the net work in a cycle is the area enclosed by the loop. Run clockwise, the gas does more work expanding (at higher pressure) than is done on it while it is compressed, so net work is positive: the cycle is an engine.

A heat engine is any device that runs a cycle like this to turn thermal energy into work. Figure 4 shows the energy flows. The engine takes in energy Q_h from a hot reservoir (burning fuel, steam from a boiler), does useful work W, and rejects the remainder Q_c to a cold reservoir (the air, a river, a cooling tower). Energy is conserved, so Q_h = W + Q_c. Its efficiency is

η = useful work ÷ input energy = W ÷ Q_h

Worked example 4: an engine cycle. A monatomic ideal gas is taken round the rectangular cycle of Figure 5. Find Q for each stage, the net work and the efficiency.

Use ΔU = (3/2)Δ(PV), with P in Pa and V in m³, so PV is in joules. PV is 100 J at A, 300 J at B, 900 J at C and 300 J at D.

The net work, 400 J, is the area of the rectangle, 2.0 × 10⁵ Pa × 2.0 × 10⁻³ m³, which is a good check. The input energy is only the heat that comes in: 300 + 1500 = 1800 J. The 1400 J that leaves on C → D and D → A is Q_c.

The classic mistake is to divide by the net heat, 400 J, and get an efficiency of 1.

7The Carnot cycle and the limit on efficiency

The Carnot cycle is an ideal, reversible cycle made of two isotherms and two adiabats, shown in Figure 6.

• A → B: isothermal expansion at the hot temperature T_h. Heat Q_h flows in.
• B → C: adiabatic expansion. No heat; the gas does work and cools from T_h to T_c.
• C → D: isothermal compression at the cold temperature T_c. Heat Q_c flows out.
• D → A: adiabatic compression. No heat; the gas warms from T_c back to T_h.

All the heat enters at the single highest temperature and leaves at the single lowest one, with no heat flowing across a temperature difference on the way. That is what makes it the most efficient cycle possible between two reservoirs. Its efficiency depends only on those two temperatures:

η_Carnot = 1 − T_c ÷ T_h (temperatures in kelvin)

No real engine working between the same two temperatures can beat this. Real engines have friction, heat leaks, and heat flowing across finite temperature differences, all of which are irreversible and push the efficiency below the Carnot value.

Worked example 5: a power station. Steam enters a turbine at 540 °C and the condenser is at 30 °C. Find the maximum possible efficiency.

Real power stations manage far less, and even the ideal limit leaves over a third of the input to be rejected, which is why they need cooling towers or a river. Plug in degrees Celsius instead (1 − 30/540 = 0.94) and the answer is wildly wrong.

For the cycle of Figure 5, the hottest point is C and the coldest A. Since T ∝ PV, T_C ÷ T_A = 900 ÷ 100 = 9, so a Carnot engine between those temperatures would reach 1 − 1/9 = 0.89. The rectangular cycle's 0.22 is far below it, because it takes in heat at every temperature from T_A up to T_C, not just at the top.

The history is a nature-of-science point: the steam engine came before the theory, and the question of why engines could not be made more efficient is what led to the second law.

8Entropy, two ways

Entropy, S, is a thermodynamic quantity that relates to the disorder of the particles in a system: roughly, how spread out the energy is and how many ways the particles could be arranged without the system looking any different. The guide gives two ways to calculate it.

From heat and temperature. When a quantity of heat ΔQ is transferred to a system at temperature T, its entropy changes by

ΔS = ΔQ ÷ T

with ΔQ positive for heat in and T in kelvin. The unit of entropy is J K⁻¹. The formula applies directly when T stays constant, as in melting or boiling, or when ΔQ is small enough that T barely moves.

Worked example 6: melting ice. 0.50 kg of ice at 0 °C melts in a room at 20 °C. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. Find the entropy change of the ice, of the room, and of the two together.

The ice gains more entropy than the room loses, because the same Q is divided by a smaller temperature. That is the whole second law in a single example.

From counting microstates. A macrostate is what you can measure: pressure, volume, temperature, or, with coins, how many heads. A microstate is one exact arrangement of every particle, or every coin, that gives that macrostate. If Ω is the number of microstates for a macrostate, then

S = k_B ln Ω

where k_B is the Boltzmann constant. The guide suggests a coin model, and Figure 7 uses it. Throw four coins. There is only one way to get four heads (HHHH), but six ways to get two heads (HHTT, HTHT, HTTH, THHT, THTH, TTHH). Every microstate is equally likely, so two heads turns up six times as often as four heads.

The entropy of the two-heads macrostate is k_B ln 6 = 1.38 × 10⁻²³ × 1.79 = 2.5 × 10⁻²³ J K⁻¹; that of four heads is k_B ln 1 = 0. Scale up: 100 coins have about 1.0 × 10²⁹ microstates with 50 heads and one with 100 heads. With 10²³ molecules, the evenly spread macrostates hold so many of the microstates that a system left alone is practically certain to be found in them. That is what "entropy increases" means: ordered arrangements are so outnumbered they are never chosen by chance.

9The second law of thermodynamics

The guide wants the second law in three forms. They say the same thing from different sides.

Clausius form. Thermal energy cannot flow spontaneously from a colder body to a hotter one. A refrigerator moves heat from cold to hot, but only because work is done on it; with no work, heat only flows from hot to cold.

Kelvin form. No cyclic process can take thermal energy from a single reservoir and convert it completely into work. Every heat engine must reject some energy to a colder reservoir, so η < 1 always. This is the form behind the Carnot limit and behind the guide's linking question on why every engine and energy source has an upper limit on efficiency.

Entropy form. The entropy of an isolated system never decreases. In a reversible process (an idealisation, like the Carnot cycle) it stays constant; in an irreversible process it increases. Processes in real isolated systems are almost always irreversible (friction, heat flowing across a temperature difference, gases mixing), so the entropy of a real isolated system always increases.

Figure 8 shows why the Clausius form and the entropy form are the same law. Let 1000 J flow from a body at 400 K to a body at 300 K, both inside an isolated system.

The total entropy rises by 0.83 J K⁻¹, so the flow is allowed. Run it backwards, from cold to hot, and every sign flips: the total would fall by 0.83 J K⁻¹, which the entropy form forbids. That is the Clausius statement.

Entropy can fall locally. The second law is about isolated systems. A system that is not isolated can lose entropy, as long as its surroundings gain at least as much. Water freezing in a freezer becomes more ordered and its entropy falls; but the freezer pumps that heat, plus the work its motor does, out into the kitchen, and the kitchen's entropy rises by more. A growing plant builds ordered structures, paid for by sunlight arriving from a very hot Sun and heat leaving to cold space. Whenever you meet a local decrease, find the surroundings where the larger increase happens.

The universe. As the ultimate isolated system, the universe's entropy can only rise. That gives time a direction (eggs break, never un-break) and points to a far future where energy is spread so evenly that no work can be extracted anywhere.