Themes › Theme B The particulate nature of matter
B.4Thermodynamics HL only
Thermodynamics is the physics of engines, fridges and power stations, and of why some things only ever happen one way. The first law is energy conservation for gases that are heated and do work. The second law sets a limit on how efficient any engine can be, and introduces entropy, which explains why time seems to have a direction. It builds on B.1 and B.3. The whole of B.4 is HL only.
Knowledge and science
Nature of science
Global impact. The steam engine came first and the theory second. Engineers built engines during the industrial revolution, and Sadi Carnot (1824) then asked how efficient an engine could possibly be. The answer still guides the design of every power station and car engine.
Theories. The second law can be stated in three equivalent ways: Kelvin's (about engines), Clausius's (about heat flow) and in terms of entropy. Showing that different statements say the same thing is a powerful test of a theory.
Shared endeavour. Carnot in France, Joule and Kelvin in Britain, Clausius in Germany and Boltzmann in Austria all contributed, often debating fiercely in the 1800s.
Models. Boltzmann linked entropy to counting the ways particles can be arranged, $S = k_B\ln\Omega$. The equation is carved on his gravestone in Vienna.
ToK: questions to think about
- Can a law be based on probability? The second law says entropy increases, but only because a decrease is overwhelmingly unlikely, not impossible. Is a statistical law as certain as Newton's laws?
- Why does time have a direction? Newton's laws work the same forwards and backwards, yet a broken egg never reassembles. Is the "arrow of time" a feature of the world, or of how we experience it?
- Can we know the fate of the universe? The second law suggests the universe will end in a uniform, cold "heat death". What kind of knowledge is a prediction that can never be tested?
- Does technology drive science, or science drive technology? Thermodynamics was developed to understand engines that already existed. What does this suggest about how knowledge grows?
1. Systems and the sign convention
In thermodynamics we study a system (usually a gas in a cylinder) and its surroundings.
- A closed system can't exchange mass with its surroundings, but it can exchange energy, by heating (thermal energy transfer) or by doing work.
- An isolated system can exchange neither mass nor energy.
The IB uses the Clausius sign convention:
- $Q$ is the net thermal energy supplied to the system: positive when the gas is heated, negative when it loses thermal energy.
- $W$ is the net work done by the system: positive when the gas expands and pushes on its surroundings, negative when work is done on the gas (it is compressed).
2. The first law of thermodynamics
The first law is the conservation of energy applied to a closed system. Energy supplied by heating either raises the gas's internal energy or is used by the gas to do work:
$Q$ = thermal energy supplied to the system, $\Delta U$ = increase in its internal energy, $W$ = work done by the system. All in joules.
Worked example: using the signs
(a) A gas is heated with 500 J of thermal energy and does 200 J of work as it expands. (b) Later it is compressed, with 150 J of work done on it, while it loses 60 J of thermal energy to the surroundings. Find $\Delta U$ in each case.
(a) $Q = +500$ J, $W = +200$ J: $\Delta U = Q - W = 500 - 200 = +300$ J. The internal energy rises.
(b) $Q = -60$ J (energy lost) and $W = -150$ J (work done on the gas): $\Delta U = -60 - (-150) = +90$ J. The internal energy still rises, because more energy goes in as work than leaves as heat.
3. Work done by a gas
A gas at pressure $P$ pushes a piston of area $A$ with force $F = PA$. If the piston moves out a small distance $\Delta x$, the work done by the gas is $F\Delta x = PA\Delta x = P\Delta V$:
for a change at constant pressure. $W$ is positive when the gas expands ($\Delta V > 0$) and negative when it is compressed. When the pressure changes, the work done is the area under the p–V graph.
Worked example: constant and changing pressure
(a) A gas at a constant pressure of $2.0 \times 10^5$ Pa expands from $1.5 \times 10^{-3}\ \text{m}^3$ to $4.0 \times 10^{-3}\ \text{m}^3$. (b) Another gas expands from $1.0 \times 10^{-3}\ \text{m}^3$ to $4.0 \times 10^{-3}\ \text{m}^3$ while its pressure falls steadily (a straight line on the p–V graph) from $3.0 \times 10^5$ Pa to $1.0 \times 10^5$ Pa. Find the work done by each gas.
(a) $W = P\Delta V = 2.0 \times 10^5 \times 2.5 \times 10^{-3} = 500$ J.
(b) The area under a straight line is a trapezium: $W = \dfrac{(3.0 + 1.0) \times 10^5}{2} \times 3.0 \times 10^{-3} = 600$ J.
4. Change in internal energy
For an ideal monatomic gas, $U = \frac{3}{2}nRT$ (B.3). So a change in internal energy depends only on the change in temperature:
Because $PV = nRT$, you can also write $\Delta U = \frac{3}{2}\Delta(PV)$, which is handy when a question gives pressures and volumes instead of temperatures.
Worked example: heating at constant pressure or constant volume
0.20 mol of helium is heated from 300 K to 400 K, (a) at constant volume and (b) at constant pressure. Find the thermal energy needed in each case.
In both cases $\Delta U = \tfrac{3}{2}nR\Delta T = 1.5 \times 0.20 \times 8.31 \times 100 = 249$ J.
(a) Constant volume: no work is done, so $Q = \Delta U = 249$ J.
(b) Constant pressure: the gas expands, doing work $W = P\Delta V = nR\Delta T = 0.20 \times 8.31 \times 100 = 166$ J. So $Q = 249 + 166 = 415$ J.
Heating at constant pressure needs more energy, because some of it goes into pushing the surroundings back.
5. Four types of process
Each type of process keeps one quantity fixed, and that tells you which term in the first law disappears:
- Isovolumetric (isochoric): constant volume, so $W = 0$ and $Q = \Delta U$.
- Isobaric: constant pressure, so $Q = \Delta U + P\Delta V$.
- Isothermal: constant temperature, so $\Delta U = 0$ and $Q = W$.
- Adiabatic: no thermal energy transfer, so $Q = 0$ and $\Delta U = -W$.
- In an isothermal expansion, all the thermal energy supplied is turned into work, and the temperature stays the same. It must happen slowly, with the gas in good thermal contact with its surroundings.
- In an adiabatic expansion, the gas does work using its own internal energy, so it cools. In an adiabatic compression, the work done on the gas raises its internal energy, so it heats up. Adiabatic changes happen when the gas is insulated, or when the change is so fast that there is no time for heat to flow.
6. Adiabatic processes
For an adiabatic change of an ideal monatomic gas:
On a p–V diagram, an adiabatic curve (an adiabat) is steeper than an isotherm through the same point. When you compress a gas adiabatically, its pressure rises for two reasons: the smaller volume, and the rise in temperature.
Worked example: compressing a gas quickly
A monatomic ideal gas at 300 K is compressed adiabatically to a quarter of its original volume. Find the factor by which its pressure increases, and its new temperature.
$P_1V_1^{5/3} = P_2V_2^{5/3}$, so $\dfrac{P_2}{P_1} = \left(\dfrac{V_1}{V_2}\right)^{5/3} = 4^{5/3} = 10.1$.
Then use $\dfrac{PV}{T} = \text{constant}$: $\dfrac{T_2}{T_1} = \dfrac{P_2V_2}{P_1V_1} = 10.1 \times \dfrac{1}{4} = 2.52$.
$T_2 = 2.52 \times 300 = 760$ K (about 490 °C). An isothermal compression would only have quadrupled the pressure. Diesel engines use this effect: the air is compressed so fast and so much that it becomes hot enough to ignite the fuel without a spark.
7. Cycles and heat engines
A heat engine turns thermal energy into mechanical work, over and over again. To keep going, its gas must follow a cycle: a sequence of processes that returns it to its starting state. After a full cycle, the gas has the same temperature, so $\Delta U = 0$ for the whole cycle.
- The engine takes in thermal energy $Q_h$ from a hot reservoir (for example burning fuel).
- It does useful work $W$.
- It must give out the remaining thermal energy $Q_c$ to a cold reservoir (for example the air or a river). By the first law, $W = Q_h - Q_c$.
On a p–V diagram, a cycle is a closed loop. The net work done in one cycle equals the area enclosed by the loop. A loop traced clockwise is an engine (net work done by the gas); anticlockwise, net work is done on the gas, as in a fridge or heat pump.
Worked example: a rectangular cycle
A monatomic ideal gas is taken round the cycle A → B → C → D → A shown. Find the net work done per cycle, the thermal energy taken in, and the efficiency.
Net work = enclosed area = $(3.0 - 1.0) \times 10^5 \times (3.0 - 1.0) \times 10^{-3} = 400$ J.
Use $\Delta U = \frac{3}{2}\Delta(PV)$ and $Q = \Delta U + W$ for each stage:
- A → B (constant volume): $\Delta U = 1.5 \times 1.0 \times 10^{-3} \times 2.0 \times 10^5 = 300$ J, $W = 0$, so $Q = +300$ J (in).
- B → C (constant pressure): $\Delta U = 1.5 \times 3.0 \times 10^5 \times 2.0 \times 10^{-3} = 900$ J, $W = 600$ J, so $Q = +1500$ J (in).
- C → D (constant volume): $Q = \Delta U = -900$ J (out).
- D → A (constant pressure): $\Delta U = -300$ J, $W = -200$ J, so $Q = -500$ J (out).
Energy input $Q_h = 300 + 1500 = 1800$ J; energy rejected $Q_c = 900 + 500 = 1400$ J. Check: $1800 - 1400 = 400$ J ✓.
$\eta = \dfrac{W}{Q_h} = \dfrac{400}{1800} = 0.22$, or 22%.
8. The Carnot cycle
The Carnot cycle is an ideal, reversible cycle made of two isothermal and two adiabatic processes:
- Isothermal expansion at $T_h$: the gas takes in $Q_h$ from the hot reservoir and does work. ($\Delta U = 0$, so $Q_h = W$ for this stage.)
- Adiabatic expansion: the gas keeps doing work and cools from $T_h$ to $T_c$.
- Isothermal compression at $T_c$: work is done on the gas, and it gives out $Q_c$ to the cold reservoir.
- Adiabatic compression: work done on the gas warms it from $T_c$ back to $T_h$.
No engine working between two reservoirs can be more efficient than a Carnot engine working between the same two temperatures:
with both temperatures in kelvin. Efficiency is higher when the hot reservoir is hotter and the cold reservoir is colder. It could only reach 100% if $T_c = 0$ K, which is impossible.
Worked example: limits on a power station
(a) A power station's steam enters the turbines at 560 °C, and the cooling water is at 30 °C. Find the maximum possible efficiency. (b) Compare the rectangular cycle above with a Carnot engine working between its highest and lowest temperatures.
(a) $T_h = 833$ K, $T_c = 303$ K: $\eta_{\text{Carnot}} = 1 - \dfrac{303}{833} = 0.64$, or 64%. Real power stations manage about 40%, because their processes are not reversible (friction, turbulence and heat leaks).
(b) $T \propto PV$, so the hottest state (C, $PV = 900$ J) is 9 times hotter than the coldest (A, $PV = 100$ J). $\eta_{\text{Carnot}} = 1 - \dfrac{1}{9} = 0.89$, far more than the 22% the rectangular cycle achieves.
9. The second law of thermodynamics
The first law says energy is conserved. The second law says which way processes go. The IB expects three equivalent forms:
- Kelvin form: it is impossible for an engine working in a cycle to take thermal energy from a hot reservoir and convert it completely into work. Some energy must always be given out to a cold reservoir, so $\eta < 1$.
- Clausius form: thermal energy cannot flow spontaneously from a colder body to a hotter body. A fridge can move energy from cold to hot, but only because work is done on it.
- Entropy form: the total entropy of an isolated system never decreases. It stays constant for a reversible process and increases for an irreversible one.
Real processes always involve friction, turbulence or energy flowing across a temperature difference, so they are irreversible. The entropy of a real isolated system therefore always increases. The universe as a whole is an isolated system, so its entropy is always increasing.
10. Entropy from heat and temperature
Entropy $S$ is a thermodynamic quantity that measures the disorder of the particles in a system: how spread out their energy and positions are. It is measured in $\text{J K}^{-1}$. Gases have more entropy than liquids, and liquids more than solids; hotter systems and bigger volumes have more entropy.
When thermal energy $\Delta Q$ is transferred to a system at constant temperature $T$, its entropy changes by:
$T$ in kelvin. $\Delta S$ is positive when energy flows in, and negative when it flows out.
Worked example: melting ice and heat flow
(a) Find the entropy change when 0.50 kg of ice melts at 0 °C ($L_f = 3.34 \times 10^5\ \text{J kg}^{-1}$). (b) 1000 J of thermal energy flows from a large body at 400 K to a large body at 300 K. Find the total entropy change.
(a) $\Delta Q = mL = 0.50 \times 3.34 \times 10^5 = 1.67 \times 10^5$ J, so $\Delta S = \dfrac{1.67 \times 10^5}{273} = 612\ \text{J K}^{-1}$. The water is more disordered than the ice.
(b) Hot body: $\Delta S = -\dfrac{1000}{400} = -2.50\ \text{J K}^{-1}$. Cold body: $\Delta S = +\dfrac{1000}{300} = +3.33\ \text{J K}^{-1}$.
Total: $+0.83\ \text{J K}^{-1}$. The cold body gains more entropy than the hot one loses, so the total increases. Flow the other way would decrease the total entropy, which is why it never happens on its own: this is the Clausius form of the second law.
11. Entropy and microstates
Entropy can also be found by counting. A macrostate describes a system as a whole (for example, "5 heads out of 10 coins"). A microstate is one particular arrangement that gives that macrostate (exactly which coins are heads). All microstates are equally likely.
Mixed-up macrostates have far more microstates than tidy ones, so a random system is overwhelmingly likely to be found in a disordered state. With $10^{23}$ particles instead of 10 coins, the odds become so extreme that the "unlikely" never happens. Boltzmann's equation links entropy to the number of microstates $\Omega$:
$k_B$ is the Boltzmann constant. A system with only one possible microstate ($\Omega = 1$) has zero entropy.
Worked example: coins
Find the entropy of the "5 heads" and "10 heads" macrostates of 10 coins, and the difference between them.
$S_{5} = k_B\ln 252 = 1.38 \times 10^{-23} \times 5.53 = 7.6 \times 10^{-24}\ \text{J K}^{-1}$.
$S_{10} = k_B\ln 1 = 0$.
The difference is tiny for 10 coins, but entropy grows with the number of particles. For a mole of gas it is measured in joules per kelvin.
12. Local decreases in entropy
The entropy of a non-isolated system can decrease, as long as the entropy of its surroundings increases by at least as much.
- Water freezing in a freezer becomes more ordered, so its entropy falls. But the freezer pumps thermal energy out into the warmer kitchen, and the extra work done by the motor also ends up as heat there. The kitchen's entropy rises by more than the water's falls.
- Living things build ordered structures from simple molecules, decreasing their own entropy. They do it by using energy from food or sunlight and releasing heat to their surroundings, increasing the entropy of the universe overall.
13. Common mistakes
- Getting the signs wrong. $W$ is work done by the gas: negative for a compression. $Q$ is energy supplied to the gas: negative when it loses energy.
- Saying no heat flows in an isothermal process. Heat does flow: $Q = W$. It is the temperature (and so $U$) that stays constant. In an adiabatic process, $Q = 0$ but the temperature changes.
- Using $W = P\Delta V$ when the pressure changes. Use the area under the p–V graph instead.
- Forgetting that $\Delta U = 0$ over a complete cycle. So for a whole cycle, net $Q$ = net $W$.
- Using Celsius temperatures in $\eta_{\text{Carnot}}$ or $\Delta S = \frac{\Delta Q}{T}$.
- Calculating efficiency as $\frac{W}{Q_c}$ or $\frac{Q_c}{Q_h}$. It is $\frac{W}{Q_h}$.
- Saying entropy can never decrease. It can decrease locally; the total entropy of an isolated system can't.
- Calling $k_B$ the Stefan–Boltzmann constant. $k_B$ is the Boltzmann constant ($1.38 \times 10^{-23}\ \text{J K}^{-1}$); $\sigma$ is the Stefan–Boltzmann constant.
14. Check your understanding
A gas expands adiabatically. What happens to its temperature, and why?
It falls. $Q = 0$, so $\Delta U = -W$. The gas does positive work, so its internal energy, and therefore its temperature, decreases.
An engine takes in 2.5 kJ of thermal energy each cycle and rejects 1.6 kJ. Find the work done per cycle and the efficiency.
$W = Q_h - Q_c = 0.9$ kJ, and $\eta = \dfrac{0.9}{2.5} = 0.36$, or 36%.
A Carnot engine works between 600 K and 300 K. Which raises its efficiency more: raising $T_h$ by 100 K, or lowering $T_c$ by 100 K?
Now: $\eta = 1 - \frac{300}{600} = 0.50$. Raising $T_h$: $1 - \frac{300}{700} = 0.57$. Lowering $T_c$: $1 - \frac{200}{600} = 0.67$. Lowering the cold temperature helps more.
Why must a heat engine have a cold reservoir?
By the Kelvin form of the second law, no cyclic engine can turn all the thermal energy it takes in into work. To return the gas to its starting state, some energy must be rejected to a colder body.
Four coins are tossed. How many microstates does the macrostate "2 heads, 2 tails" have, and what is its entropy?
$\Omega = {}^4C_2 = 6$ (HHTT, HTHT, HTTH, THHT, THTH, TTHH). $S = k_B\ln 6 = 2.5 \times 10^{-23}\ \text{J K}^{-1}$.