B.4 Thermodynamics · IB DP Physics

      
      IB DP Physics · Theme B · The particulate nature of matter
Thermodynamics.B.4 · The first law, gas processes, entropy and the second law, heat engines
Higher level only: there is no standard level content in B.4

              Additional HL · 8 hours
            
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      The journey
Four parts.
Part 01
The first lawHeat, work and internal energy
Part 02
Gas processesFour special changes, and cycles
Part 03
EntropyDisorder, and the second law
Part 04
Heat enginesEfficiency and the Carnot limit
All calculations use a monatomic ideal gas, from B.3.
B.4 THERMODYNAMICS · HL ONLY


      
      Part 01 · HL only
The first law.Conservation of energy for a gas: heat in, work out, internal energy changed
B.4 THERMODYNAMICS · HL ONLY


      What are we talking about?
System and surroundings.The system is the thing studied, usually a fixed amount of gas. Everything else is the surroundings
Closed system
No mass in or outEnergy can cross the boundary, as heat or as work

Example: gas sealed in a cylinder with a piston
Isolated system
Neither mass nor energy in or outIts total energy is fixed

Example: a sealed, perfectly insulated flask. The whole universe
B.4 THERMODYNAMICS · HL ONLY
Why the distinction mattersThe first law is written for a closed system: it keeps account of the energy that crosses the boundary. The second law, in its simplest form, is about an isolated system: it says what the entropy inside can do when nothing crosses. Using a law with the wrong kind of system is the most common source of confusion in this topic.
Heat and work are transfers, not contentsA gas does not “contain” heat or work. It contains internal energy. Heat is energy crossing the boundary because of a temperature difference; work is energy crossing because the boundary moves against a force.
Draw a dotted line round the system before you start. Anything crossing that line is \(Q\) or \(W\). Anything that stays inside is \(U\).


      Energy inside the gas
Internal energy of a monatomic ideal gas.\(\Delta U = \dfrac{3}{2} n R \Delta T = \dfrac{3}{2} N k_\mathrm{B} \Delta T\)
The total random kinetic energy of the atoms
Depends only on temperature, for a fixed amount of gas
\(n\) moles, or \(N\) atoms

            2.0 mol of helium heated from 300 K to 450 K
            \(\Delta U = \dfrac{3}{2} n R \Delta T\)
            \(\Delta U = \dfrac{3}{2} \times 2.0 \times 8.31 \times 150\)
            \(\Delta U = \class{res}{3.7 \times 10^{3} \,\mathrm{J}}\)
          
B.4 THERMODYNAMICS · HL ONLY
Where the equation comes fromIn B.3 the average kinetic energy of one atom of an ideal gas is \(\frac{3}{2} k_\mathrm{B} T\). An ideal gas has no forces between its atoms, so no potential energy. Its internal energy is just the kinetic energy of all \(N\) atoms: \(U = \frac{3}{2} N k_\mathrm{B} T\). Since \(N k_\mathrm{B} = nR\), the two forms are the same.
The key consequenceIf the temperature of the gas does not change, its internal energy does not change, whatever happens to its pressure and volume. If the temperature returns to its starting value, so does the internal energy. This one fact settles half the questions in the topic.
Because \(PV = nRT\), you can also write \(\Delta U = \frac{3}{2} \Delta (PV)\). It saves working out the temperatures when a question gives pressures and volumes.
Using this for a gas of molecules such as oxygen or nitrogen. The factor \(\frac{3}{2}\) is for single atoms only.


      Moving the boundary
Work done by a gas.GAS IN A CYLINDERPRESS → TO LET IT EXPAND

              \(W = F \Delta x = P A \Delta x\)
              \(W = P \Delta V\)
            
At constant pressure
Expansion: work done by the gas, positive
Compression: work done on the gas, negative
B.4 THERMODYNAMICS · HL ONLY
The step in the middleThe piston has area \(A\) and moves a distance \(\Delta x\), so the volume of the gas grows by \(A \Delta x\). That is \(\Delta V\). Force times distance has become pressure times change in volume, and the shape of the container no longer matters: the same result holds for a balloon or a bubble.
Where the energy comes fromAtoms that hit a piston moving away from them bounce off more slowly, as a ball does off a bat that is being pulled back. So a gas that does work loses internal energy, and cools, unless heat is supplied at the same time to make up for it.
A numerical case: a gas at \(1.0 \times 10^{5} \,\mathrm{Pa}\) expands by \(2.0 \times 10^{-3} \,\mathrm{m}^{3}\), two liters. \(W = 1.0 \times 10^{5} \times 2.0 \times 10^{-3} = 200 \,\mathrm{J}\).


      When the pressure changes
Work on a pressure–volume diagram.Work done = the area under the path
Path to the right: expansion, work done by the gas
Path to the left: compression, work done on the gas
Vertical path: no work
AN EXPANSION
B.4 THERMODYNAMICS · HL ONLY
Why the areaSplit the change into thin vertical strips. Across each strip the pressure is nearly constant, so the work is \(P \Delta V\): the height of the strip times its width. Adding the strips gives the area under the curve. It is the same argument as for a force–displacement graph in A.3.
Work depends on the routeThere are many paths between the same two states on the diagram, and each has a different area under it. So the work done, and the heat needed, depend on how the change is carried out. The change in internal energy does not: it depends only on the two end states. That is why \(U\) is called a property of the state, and \(Q\) and \(W\) are not.
Estimate an awkward area by counting squares, after working out how many joules one square is worth: pascals times cubic meters.


      Conservation of energy
The first law of thermodynamics.\(\Delta U = Q - W\)
Internal energy rises by the thermal energy supplied, minus the work done by the gas
For a closed system

            PositiveNegative

            \(Q\)Thermal energy supplied to the systemThermal energy lost by it
\(W\)Work done by the systemWork done on it
\(\Delta U\)Temperature risesTemperature falls

          
B.4 THERMODYNAMICS · HL ONLY
Reading it as a budget\(Q\) is income. \(W\) is what is spent. \(\Delta U = U_\mathrm{f} - U_\mathrm{i}\) is the change in savings: income minus spending. If the gas spends more in work than it receives as heat, its internal energy falls. The gas exchanges energy with its surroundings: \(\Delta U\) need not be zero. For the isolated gas plus surroundings, \(\Delta E_\mathrm{total} = 0\).
The sign conventionThis is the Clausius convention, and it is the one used in every exam question: \(Q\) is counted positive when it goes into the gas, \(W\) is positive when it is done by the gas. Some books write \(\Delta U = Q + W\) with \(W\) as the work done on the gas. The physics is identical, but mixing the two gives wrong signs.
The data bookletIt prints the same law rearranged, as \(Q = \Delta U + W\). Nothing is different: add \(W\) to both sides of the form on the slide.
Before substituting, write beside each quantity whether it is into or out of the gas, and give it its sign. Most mistakes are sign mistakes.


      Worked example
Using the first law.
            A gas is supplied with 500 J of thermal energy
            and does 200 J of work as it expands
            \(\Delta U = Q - W\)
            \(\Delta U = 500 - 200 = \class{res}{+300 \,\mathrm{J}}\)
            Its temperature rises
          

            150 J of work is done on a gas
            while it loses 400 J of thermal energy
            \(Q = -400 \,\mathrm{J}\)  ·  \(W = -150 \,\mathrm{J}\)
            \(\Delta U = Q - W = -400 - (-150) = \class{res}{-250 \,\mathrm{J}}\)
            Its temperature falls
          
B.4 THERMODYNAMICS · HL ONLY
Checking the second one by common senseCompressing the gas puts 150 J in. Cooling it takes 400 J out. The gas ends 250 J down. The signs in the equation only keep track of that.
From internal energy to temperatureFor 0.50 mol of a monatomic gas, a change of −250 J means \(\Delta T = \frac{\Delta U}{\frac{3}{2} n R} = \frac{-250}{1.5 \times 0.50 \times 8.31} = -40 \,\mathrm{K}\).
Writing \(W = +150 \,\mathrm{J}\) because “work is done”. Work done on the system is negative in this convention.


      Quick check
The first law.01What can cross the boundary of a closed system?
Energy, but not mass
02On what does the internal energy of an ideal gas depend?
Its temperature (and the amount of gas)
03A gas at \(2.0 \times 10^{5} \,\mathrm{Pa}\) expands by \(3.0 \times 10^{-3} \,\mathrm{m}^{3}\). Work done?
\(600 \,\mathrm{J}\), by the gas
04What is the sign of \(W\) when a gas is compressed?
Negative
05\(Q = +800 \,\mathrm{J}\) and \(W = +300 \,\mathrm{J}\). Change in internal energy?
\(+500 \,\mathrm{J}\)
06What does the area under a P–V graph give?
The work done
PRESS → TO REVEAL EACH ANSWER


      
      Part 02 · HL only
Gas processes.Four changes, each with one thing held fixed, and how they join into a cycle
B.4 THERMODYNAMICS · HL ONLY


      Keeping one variable fixed
Four processes.Isobaric: constant pressure
Isovolumetric: constant volume
Isothermal: constant temperature
Adiabatic: no thermal energy in or out
FOUR LINES THROUGH ONE STATEEITHER DIRECTION
B.4 THERMODYNAMICS · HL ONLY
The names“Iso” means the same. Bar is pressure, as in barometer; therm is temperature, as in thermometer. Adiabatic comes from Greek words meaning “not passing through”: no heat passes through the walls.
Telling the two curves apartAn isothermal and an adiabatic expansion both curve downwards, but the adiabatic curve is steeper. In an isothermal expansion heat flows in and keeps the temperature up. In an adiabatic expansion nothing replaces the energy the gas spends on work, so it cools, and its pressure falls for two reasons at once: larger volume and lower temperature.
Each point on a P–V diagram is one state of the gas, with one temperature: \(T = \frac{PV}{nR}\). The larger the product \(PV\), the hotter the gas.
The lines have no arrows because each change can go either way: to the right is an expansion, to the left a compression. An arrow belongs on a diagram only when a question says which way the gas went.


      Constant volume, constant pressure
Isovolumetric and isobaric changes.Isovolumetric
\(W = 0\), so \(\Delta U = Q\)All the heat supplied raises the temperature
Isobaric
\(W = P \Delta V\), so \(\Delta U = Q - P \Delta V\)Some of the heat is spent on work

            Isobaric: \(P = 1.0 \times 10^{5} \,\mathrm{Pa}\)
            \(V\) from \(2.0 \times 10^{-3}\) to \(5.0 \times 10^{-3} \,\mathrm{m}^{3}\)
            \(W = P \Delta V = 1.0 \times 10^{5} \times 3.0 \times 10^{-3} = \class{res}{300 \,\mathrm{J}}\)
            \(\Delta U = \dfrac{3}{2} P \Delta V = \dfrac{3}{2} \times 300 = \class{res}{450 \,\mathrm{J}}\)
            \(\Delta U = Q - W\)  so  \(Q = \Delta U + W = \class{res}{750 \,\mathrm{J}}\)
          
B.4 THERMODYNAMICS · HL ONLY
Why heating at constant pressure costs moreTo raise the temperature of the same gas by the same amount, more heat is needed at constant pressure than at constant volume. At constant volume every joule goes into the internal energy. At constant pressure the gas must expand as it warms, and for a monatomic ideal gas only \(\frac{3}{5}\) of the heat raises the internal energy: here 450 J out of 750 J. The other \(\frac{2}{5}\) leaves as work.
Where \(\Delta U = \frac{3}{2} P \Delta V\) comes from\(\Delta U = \frac{3}{2} n R \Delta T\), and at constant pressure \(n R \Delta T = P \Delta V\) from the ideal gas equation.
In an isovolumetric change, \(\frac{P}{T}\) is constant. In an isobaric change, \(\frac{V}{T}\) is constant. Both follow from \(PV = nRT\).


      Constant temperature
Isothermal changes.\(T\) constant, so \(\Delta U = 0\)
First law: \(0 = Q - W\), so \(Q = W\)
\(PV\) = constant: Boyle’s law
All the heat supplied comes straight out as work.
AN ISOTHERMAL CHANGE
B.4 THERMODYNAMICS · HL ONLY
How to do it in practiceThe gas must be able to exchange heat freely with surroundings at a fixed temperature, and the change must be slow, so that heat has time to flow. Push a piston in slowly on a gas in a thin metal cylinder standing in a water bath and the compression is very nearly isothermal.
CompressionEverything reverses. Work is done on the gas, \(W\) is negative, so \(Q\) is negative by the same amount: the gas must give out heat equal to the work done on it, or its temperature would rise.
Assuming “no temperature change” means “no heat”. In an isothermal change heat does flow. It is the internal energy that does not change.


      No thermal energy transferred
Adiabatic changes.\(P V^{\frac{5}{3}} = \text{constant}\)
\(Q = 0\), so \(\Delta U = -W\)
Expansion: the gas does work and cools
Compression: work is done on it and it warms
ADIABATIC AND ISOTHERMAL COMPARED
B.4 THERMODYNAMICS · HL ONLY
How to do it in practiceEither insulate the gas well, or make the change so fast that heat has no time to flow. A bicycle pump gets hot because the air in it is compressed quickly. Gas escaping from a pressurized can expands quickly and comes out cold. Sound waves compress and expand air so rapidly that the changes are adiabatic.
Where the temperature goesWith no heat exchanged, the work done by the gas comes entirely out of its internal energy: \(\Delta U = -W\). Internal energy is temperature, for an ideal gas, so the temperature must fall.
The power \(\frac{5}{3}\) is for a monatomic ideal gas only. It is given in the data booklet.
Clouds form this way: moist air rises, expands adiabatically in the lower pressure, cools, and its water vapor condenses.


      Worked example
An adiabatic compression.
            \(P_{1} = 1.0 \times 10^{5} \,\mathrm{Pa}\) · \(V_{1} = 4.0 \times 10^{-3} \,\mathrm{m}^{3}\) · \(T_{1} = 300 \,\mathrm{K}\)
            Compressed to \(V_{2} = 1.0 \times 10^{-3} \,\mathrm{m}^{3}\)
            \(P_{1} V_{1}^{\frac{5}{3}} = P_{2} V_{2}^{\frac{5}{3}}\)  so  \(P_{2} = P_{1} \left(\dfrac{V_{1}}{V_{2}}\right)^{\frac{5}{3}}\)
            \(P_{2} = 1.0 \times 10^{5} \times 4.0^{\frac{5}{3}} = \class{res}{1.0 \times 10^{6} \,\mathrm{Pa}}\)
          

            The new temperature
            \(\dfrac{P_{1} V_{1}}{T_{1}} = \dfrac{P_{2} V_{2}}{T_{2}}\)
            \(T_{2} = T_{1} \dfrac{P_{2} V_{2}}{P_{1} V_{1}} = 300 \times \dfrac{10.1 \times 1.0}{1.0 \times 4.0}\)
            \(T_{2} = \class{res}{760 \,\mathrm{K}}\)
          
B.4 THERMODYNAMICS · HL ONLY
Compare with an isothermal compressionSqueezed slowly to a quarter of its volume at 300 K, the gas would reach only four times the pressure, \(4.0 \times 10^{5} \,\mathrm{Pa}\). Done adiabatically the pressure rises to ten times its initial value, because the gas has also become much hotter. A diesel engine compresses air so sharply that the fuel ignites without a spark.
The work doneWith \(Q = 0\), the work done on the gas equals the rise in its internal energy: \(\Delta U = \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) = \frac{3}{2} (1010 - 400) = 910 \,\mathrm{J}\).
On a calculator, \(4.0^{\frac{5}{3}}\) is 4 ^ (5 / 3) = 10.1. The brackets matter.
Using \(P_{1} V_{1} = P_{2} V_{2}\). That is the isothermal law. The ideal gas equation with temperatures still holds for any change.


      Side by side
The four processes compared.
          
            ProcessConstantFirst law becomesOn a P–V diagram

            Isovolumetric\(V\)\(\Delta U = Q\)Vertical line
Isobaric\(P\)\(\Delta U = Q - P \Delta V\)Horizontal line
Isothermal\(T\)\(\Delta U = 0\), so \(Q = W\)Curve, \(PV\) constant
Adiabaticnone: \(Q = 0\)\(\Delta U = -W\)Steeper curve

          
        
B.4 THERMODYNAMICS · HL ONLY
One idea, four timesIn each process one term of \(\Delta U = Q - W\) is zero, or one quantity is easy to find, and the law collapses to something simpler. Constant volume: no work. Constant temperature: no change in internal energy. Adiabatic: no heat. Constant pressure: all three terms survive, but the work is simply \(P \Delta V\).
A plan for any questionName the process. Write down which term vanishes. Find the others from \(W = P \Delta V\) or the area, from \(\Delta U = \frac{3}{2} n R \Delta T\), and from the first law.
The ideal gas equation, \(PV = nRT\), holds at every point of every process. Use it to find any temperature you need.


      Back to the start
Cyclic processes.The gas returns to its starting state: over a cycle \(\Delta U = 0\)
Net work done = the area enclosed
Clockwise: net work done by the gas

              Pressure in \(10^{5} \,\mathrm{Pa}\) · volume in \(10^{-3} \,\mathrm{m}^{3}\)
              \(W_\mathrm{net} = (P_\mathrm{high} - P_\mathrm{low})(V_\mathrm{max} - V_\mathrm{min})\)
              \(W = 2.0 \times 10^{5} \times 3.0 \times 10^{-3} = \class{res}{600 \,\mathrm{J}}\)
            
A RECTANGULAR CYCLE
B.4 THERMODYNAMICS · HL ONLY
Why the enclosed areaFrom A to B the gas expands at high pressure and does work equal to the whole area under the top edge: \(3.0 \times 10^{5} \times 3.0 \times 10^{-3} = 900 \,\mathrm{J}\). From C to D it is compressed at low pressure, and work is done on it equal to the area under the bottom edge, 300 J. The two vertical sides involve no work. The signed work during compression is \(-300 \,\mathrm{J}\), so \(W_\mathrm{net} = 900 + (-300) = 600 \,\mathrm{J}\). The differences in the area formula are the loop’s height and width; over the complete cycle, \(\Delta P = 0\) and \(\Delta V = 0\), because final equals initial.
Where it comes fromOver the cycle \(\Delta U = 0\), so the first law gives \(Q_\mathrm{net} = W_\mathrm{net}\). The gas takes in more heat than it gives out, and the difference appears as work. A gas taken round and round such a loop is an engine.
Go round the loop counterclockwise and the net work is done on the gas: that is a refrigerator or heat pump.


      Quick check
Gas processes.01In which process is no work done?
Isovolumetric
02In which process is \(\Delta U = 0\)?
Isothermal
03A gas expands adiabatically. What happens to its temperature?
It falls
04Which is steeper on a P–V diagram, an isothermal or an adiabatic curve?
The adiabatic curve
05What is \(\Delta U\) over one complete cycle?
Zero
06What does the area inside a cycle on a P–V diagram give?
The net work done
PRESS → TO REVEAL EACH ANSWER


      
      Part 03 · HL only
Entropy and the second law.Why some things happen by themselves and their reverse never does
B.4 THERMODYNAMICS · HL ONLY


      A direction for time
Processes that only go one way.GAS IN HALF A BOXPRESS → TO REMOVE THE PARTITION
The gas spreads out. It never un-spreads
Heat flows from hot to cold, never back
Energy would be conserved either way: the first law allows both
Real processes are irreversible
B.4 THERMODYNAMICS · HL ONLY
What needs explainingFilm a ball bouncing to rest, an ice cube melting in a warm drink or smoke spreading through a room, and run the film backwards. Nothing in the reversed film breaks conservation of energy. The floor could cool slightly and give its thermal energy back to the ball. Yet it never happens. Some other law must pick out the direction in which things really go.
The answer in one sentenceThere are overwhelmingly more ways for the gas to be spread through the whole box than for it to be in one half, so that is where chance takes it. The rest of this part makes that idea exact.
A reversible process is an ideal limit: one carried out so slowly and gently, with no friction and no temperature differences, that it could be run backwards leaving no trace.


      Counting the ways
Microstates and macrostates.
            Macrostate: what you can see, such as “2 heads”
Microstate: the exact arrangement, such as HTHT
Every microstate is equally likely
\(\Omega\): the number of microstates of a macrostate

          
FOUR COINS16 MICROSTATES
B.4 THERMODYNAMICS · HL ONLY
The six ways to get two headsHHTT, HTHT, HTTH, THHT, THTH, TTHH. Each is exactly as likely as HHHH, which can happen in only one way. So “two heads” turns up six times as often as “four heads”, not because any arrangement is favored but because more arrangements fit the description.
What happens with more coinsWith 100 coins there are \(2^{100} \approx 10^{30}\) microstates. Only one of them is all heads. The macrostates close to 50 heads account for nearly all the rest. Shake the coins and you will always find a number close to 50. For the \(10^{23}\) atoms of a gas the same effect is so extreme that the most likely macrostate is the only one ever seen.
For the gas: the macrostate is its pressure, volume and temperature. A microstate is the position and velocity of every atom.


      A measure of disorder
Entropy.\(S = k_\mathrm{B} \ln \Omega\)
More microstates: more disorder, more entropy
Unit: J K⁻¹
\(k_\mathrm{B} = 1.38 \times 10^{-23} \,\mathrm{J} \,\mathrm{K}^{-1}\)

            Four coins, all heads: \(\Omega = 1\)
            \(S = k_\mathrm{B} \ln 1 = \class{res}{0}\)
            Two heads and two tails: \(\Omega = 6\)
            \(S = 1.38 \times 10^{-23} \times \ln 6\)
            \(S = \class{res}{2.5 \times 10^{-23} \,\mathrm{J} \,\mathrm{K}^{-1}}\)
          
B.4 THERMODYNAMICS · HL ONLY
Why a logarithmPut two systems side by side. Every microstate of one can go with every microstate of the other, so the numbers of microstates multiply: \(\Omega = \Omega_{1} \Omega_{2}\). But we want entropy to add, like energy or volume: two identical blocks should have twice the entropy of one. The logarithm turns multiplying into adding: \(\ln (\Omega_{1} \Omega_{2}) = \ln \Omega_{1} + \ln \Omega_{2}\).
What “disorder” means hereNot untidiness, but the number of ways the atoms could be arranged without your noticing any difference. A crystal has few. A gas has vastly more. A gas in a large container has more than the same gas in a small one.
Entropy rises when a solid melts, a liquid boils, a gas expands, two substances mix, or anything is heated.


      The story · Vienna, 1877–1906
Background story · not assessed
The equation on a tombstone.
            Ludwig Boltzmann in 1898
            His grave in Vienna, with S = k. log W carved above the bust
          
In 1877 Boltzmann linked entropy to the number of arrangements of atoms
Many physicists then doubted that atoms existed at all
He defended the idea for thirty years, and died in 1906
Within a few years experiments on Brownian motion had settled the question
His W is our \(\Omega\). The constant \(k\) now carries his name.
Portrait: Rudolf Fenzl, 1898, public domain. Grave: Feldkurat Katz, CC BY-SA 4.0.
B.4 THERMODYNAMICS · HL ONLY


      The large-scale view
Entropy change from thermal energy.\(\Delta S = \dfrac{\Delta Q}{T}\)
\(\Delta Q\): thermal energy transferred at absolute temperature \(T\)
Energy in: entropy rises. Energy out: it falls
The same heat counts for more at a low temperature

            0.10 kg of ice melts at 273 K
            Specific latent heat \(3.34 \times 10^{5} \,\mathrm{J} \,\mathrm{kg}^{-1}\)
            \(\Delta Q = mL = 0.10 \times 3.34 \times 10^{5} = 3.34 \times 10^{4} \,\mathrm{J}\)
            \(\Delta S = \dfrac{\Delta Q}{T} = \dfrac{3.34 \times 10^{4}}{273}\)
            \(\Delta S = \class{res}{+122 \,\mathrm{J} \,\mathrm{K}^{-1}}\)
          
B.4 THERMODYNAMICS · HL ONLY
Two definitions, one quantityClausius defined entropy this way in 1865, from heat and temperature, with no mention of atoms. Boltzmann’s counting of microstates came later and gives the same numbers. Adding thermal energy to a body gives its atoms more energy to share out, and so more ways to share it: more microstates.
Why temperature is underneathA given amount of heat makes a big difference to a cold body, whose atoms have little energy, and a small difference to a hot one, which already has a huge number of microstates. A whisper is noticed in a library and lost in a crowded street.
Use this form when the temperature stays constant during the transfer: a change of state, an isothermal process, or a reservoir so large that its temperature hardly changes. The transfer must also be gentle and reversible. In a free expansion no heat flows at all, yet the entropy of the gas rises.
Using °C. The temperature must be in kelvin.


      Three ways of saying it
The second law of thermodynamics.Entropy form
The entropy of an isolated system never decreasesReversible process: it stays the same
Irreversible process: it increases
Clausius form
Heat does not flow by itself from cold to hotMoving thermal energy from a colder body to a hotter one needs work
Kelvin form
No engine can turn heat entirely into workWorking in a cycle, it must reject some thermal energy to a colder body
B.4 THERMODYNAMICS · HL ONLY
They are the same lawSuppose heat could flow by itself from cold to hot, against Clausius. Then the waste heat of any engine could be sent back to its hot source for nothing, and the engine would in effect turn heat entirely into work, against Kelvin. Each statement implies the other, and both follow from the entropy form, as the next slide shows for the first.
Real processesFriction, heat flow across a temperature difference, mixing and sudden expansion all increase entropy, and one or more of them is present in every real process. So in practice the entropy of an isolated system always goes up. It can never be brought back down, which is exactly what “irreversible” means.
The universe as a whole is an isolated system. Its entropy has been rising since the beginning, and this is what gives time its direction.


      Worked example
Why heat flows from hot to cold.
            600 J flows from a body at 400 K to one at 300 K
            \(\Delta S = \dfrac{\Delta Q}{T}\) for each body
            Hot body: \(\Delta S = \dfrac{-600}{400} = -1.5 \,\mathrm{J} \,\mathrm{K}^{-1}\)
            Cold body: \(\Delta S = \dfrac{+600}{300} = +2.0 \,\mathrm{J} \,\mathrm{K}^{-1}\)
            Total: \(\class{res}{+0.5 \,\mathrm{J} \,\mathrm{K}^{-1}}\)
          

            The total entropy rises: this happens by itself
Reversed, the total would be \(-0.5 \,\mathrm{J} \,\mathrm{K}^{-1}\): forbidden
The hot body’s entropy fell, but the cold body’s rose by more

          
B.4 THERMODYNAMICS · HL ONLY
The Clausius form, derivedThe same 600 J is divided by a smaller temperature for the cold body than for the hot one, so the gain is always larger than the loss. Heat flowing from hot to cold always raises the total entropy. Heat flowing the other way would always lower it. The entropy form of the second law therefore forbids it.
Equal temperaturesIf both bodies were at 350 K, the two changes would cancel and the total entropy would not change. That is the reversible limit, and also the reason heat stops flowing once two bodies reach the same temperature.
Always add up the entropy changes of everything involved. One part can go down as long as the total does not.


      Systems that are not isolated
Entropy can fall locally.The entropy of a system that is not isolated can decrease
The entropy of its surroundings then rises by at least as much
Water freezing
Liquid to ordered crystalThe latent heat it gives out raises the entropy of the colder air
A refrigerator
Cools its contentsIt rejects more thermal energy to the room than it removes
A living thing
Builds order from foodIt gives out heat and waste of higher entropy
B.4 THERMODYNAMICS · HL ONLY
A freezing pond, with numbers1.0 kg of water freezes at 273 K, giving out \(3.34 \times 10^{5} \,\mathrm{J}\) to air at 263 K. The water’s entropy changes by \(\frac{-3.34 \times 10^{5}}{273} = -1223 \,\mathrm{J} \,\mathrm{K}^{-1}\). The air’s changes by \(\frac{+3.34 \times 10^{5}}{263} = +1270 \,\mathrm{J} \,\mathrm{K}^{-1}\). The total is \(+47 \,\mathrm{J} \,\mathrm{K}^{-1}\): the water has become more ordered, and the universe less.
Life does not break the second lawA growing plant or animal is a highly ordered arrangement of atoms. It is not isolated. It takes in low-entropy energy, sunlight or food, and gives back heat at a low temperature. The entropy it exports is far greater than the entropy it loses.
When a question says entropy has decreased, look for what is outside the system. The answer is always there.


      Quick check
Entropy and the second law.01What is a microstate?
One exact arrangement of all the particles
02How many microstates have three heads from four coins?
4
03900 J is supplied to a body at 300 K. Entropy change?
\(\frac{900}{300} = +3.0 \,\mathrm{J} \,\mathrm{K}^{-1}\)
04What happens to the entropy of a real isolated system?
It always increases
05State the Clausius form of the second law.
Heat does not flow by itself from cold to hot
06Water freezes: its entropy falls. Why is that allowed?
The surroundings gain more
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      Part 04 · HL only
Heat engines.Turning thermal energy into work, and the limit that no design can beat
B.4 THERMODYNAMICS · HL ONLY


      The story · Paris, 1824
Background story · not assessed
Carnot’s question.
            Sadi Carnot at 17, painted by Louis-Léopold Boilly
            A Newcomen engine, from an engraving of 1717
          
Steam engines had been pumping mines since 1712, improved by trial and error
Carnot, a young army engineer, asked: is there a limit to how good an engine can be?
His answer, in 1824: yes, and it depends only on the two temperatures
He paid to print 600 copies. He died of cholera at 36
The limit came before the first and second laws were known.
Portrait: Louis-Léopold Boilly, 1813, public domain. Engine: Digital Public Library of America, public domain.
B.4 THERMODYNAMICS · HL ONLY


      Work from thermal energy
Heat engines.\(\eta = \dfrac{\text{useful work}}{\text{input energy}} = \dfrac{W}{Q_\mathrm{h}}\)
Takes in \(Q_\mathrm{h}\) from a hot reservoir
Does work \(W\), using a gas taken round a cycle
Rejects \(Q_\mathrm{c}\) to a cold reservoir
\(W = Q_\mathrm{h} - Q_\mathrm{c}\)
ENERGY FLOW IN ONE CYCLE
B.4 THERMODYNAMICS · HL ONLY
Why some heat must be thrown awayOver a cycle the gas returns to its starting state, so its entropy returns to its starting value. But it gained entropy when it took in \(Q_\mathrm{h}\). It must get rid of that entropy before the next cycle, and the only way to remove entropy is to give out thermal energy. That is \(Q_\mathrm{c}\). An engine with no cold reservoir would break the Kelvin form of the second law.
A worked caseAn engine takes in 2000 J and rejects 1400 J each cycle. \(W = 2000 - 1400 = 600 \,\mathrm{J}\), and \(\eta = \frac{600}{2000} = 0.30\). Equivalently, \(\eta = 1 - \frac{Q_\mathrm{c}}{Q_\mathrm{h}}\).
A reservoir is a body so large that its temperature does not change when it gives out or takes in the heat: the furnace gases, the atmosphere, a river.


      The best possible engine
The Carnot cycle.A → B: isothermal expansion at \(T_\mathrm{h}\), taking in \(Q_\mathrm{h}\)
B → C: adiabatic expansion, cooling to \(T_\mathrm{c}\)
C → D: isothermal compression at \(T_\mathrm{c}\), rejecting \(Q_\mathrm{c}\)
D → A: adiabatic compression, warming to \(T_\mathrm{h}\)
TWO ISOTHERMALS, TWO ADIABATICS
B.4 THERMODYNAMICS · HL ONLY
Why these four stepsEntropy is created whenever heat flows across a temperature difference. To waste nothing, heat must enter the gas only while the gas is at the temperature of the hot reservoir, and leave only while it is at the temperature of the cold one: two isothermal steps. To get the gas from one temperature to the other without any heat flow at all, the other two steps must be adiabatic.
Why nobody builds oneA reversible isothermal step would have to be infinitely slow, so a true Carnot engine would deliver its work at zero power. It is a yardstick, not a design: every real engine is compared with it.
Heat is exchanged only on the two apricot curves. On the two terracotta curves \(Q = 0\).


      The limit
The Carnot efficiency.\(\eta_\mathrm{Carnot} = 1 - \dfrac{T_\mathrm{c}}{T_\mathrm{h}}\)
The greatest efficiency possible between two reservoirs
Depends only on their temperatures, in kelvin
Real engines always do worse

            Steam at 800 K · cooling water at 300 K
            \(\eta_\mathrm{Carnot} = 1 - \dfrac{T_\mathrm{c}}{T_\mathrm{h}} = 1 - \dfrac{300}{800} = \class{res}{0.63}\)
            A real engine: 2000 J in, 600 J of work
            \(\eta = \dfrac{W}{Q_\mathrm{h}} = \dfrac{600}{2000} = \class{res}{0.30}\)
          
B.4 THERMODYNAMICS · HL ONLY
Where the limit comes fromIn the reversible Carnot cycle the total entropy does not change. The hot reservoir loses \(\frac{Q_\mathrm{h}}{T_\mathrm{h}}\) and the cold one gains \(\frac{Q_\mathrm{c}}{T_\mathrm{c}}\), so these are equal: \(\frac{Q_\mathrm{c}}{Q_\mathrm{h}} = \frac{T_\mathrm{c}}{T_\mathrm{h}}\). Put that into \(\eta = 1 - \frac{Q_\mathrm{c}}{Q_\mathrm{h}}\). Any real engine creates extra entropy, has to reject more heat to carry it away, and so does less work.
How to make an engine betterRaise \(T_\mathrm{h}\) or lower \(T_\mathrm{c}\). The cold reservoir is the air or a river, near 300 K, and cannot be changed much. So engineers push the hot temperature as high as materials allow. This is why power stations use steam at very high temperature and pressure, and why even the best turn well under two thirds of their heat into electricity.
Using °C. With 527 °C and 27 °C the formula gives 0.95: quite wrong. Convert to kelvin first.
An efficiency of 1 would need \(T_\mathrm{c} = 0 \,\mathrm{K}\), which cannot be reached.


      The story · Zürich, 1865
Background story · not assessed
Clausius names entropy.
            Rudolf Clausius
          
In 1850 Clausius put Carnot’s result together with Joule’s: energy is conserved, and heat will not flow uphill
In 1865 he found the quantity that measures this, \(\frac{\Delta Q}{T}\), and named it entropy
He built the word from the Greek for “transformation”, to sound like “energy”
His summary: the energy of the universe is constant; its entropy tends to a maximum.
Portrait: AIP Emilio Segrè Visual Archives, CC0.
B.4 THERMODYNAMICS · HL ONLY


      Quick check
Heat engines.01An engine takes in 500 J and rejects 350 J. Work done?
150 J
02Its efficiency?
\(\frac{150}{500} = 0.30\)
03Which four processes make up a Carnot cycle?
Two isothermal and two adiabatic
04Reservoirs at 600 K and 300 K. Greatest possible efficiency?
\(1 - \frac{300}{600} = 0.50\)
05Why must an engine reject heat?
To remove the entropy it took in
06How can the Carnot efficiency be raised?
Hotter source or colder sink
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      Summary
Thermodynamics on one slide.First law
\(\Delta U = Q - W\)\(Q\) into the gas, \(W\) done by it
\(\Delta U = \frac{3}{2} n R \Delta T\)
Work
\(W = P \Delta V\)Area under a P–V path
Area inside a cycle: net work
Processes
One thing fixedIsovolumetric, isobaric, isothermal
Adiabatic: \(P V^{\frac{5}{3}}\) constant
Entropy
\(S = k_\mathrm{B} \ln \Omega\)\(\Delta S = \frac{\Delta Q}{T}\)
A measure of disorder
Second law
Entropy never decreasesIn an isolated system
Clausius and Kelvin forms
Heat engines
\(\eta_\mathrm{Carnot} = 1 - \frac{T_\mathrm{c}}{T_\mathrm{h}}\)\(\eta = \frac{W}{Q_\mathrm{h}}\), always below the Carnot limit
B.4 THERMODYNAMICS · HL ONLY


      Summary
Key words.Closed systemEnergy can cross, mass cannot
Isolated systemNothing crosses its boundary
Internal energyTotal random energy of the particles
IsothermalAt constant temperature
AdiabaticNo thermal energy transferred
IsobaricAt constant pressure
IsovolumetricAt constant volume
MicrostateOne exact arrangement
EntropyA measure of disorder, in J K⁻¹
IrreversibleRaises the total entropy
Heat engineTurns thermal energy into work
Carnot cycleThe most efficient cycle possible
B.4 THERMODYNAMICS · HL ONLY


    
	Positive	Negative
\(Q\)	Thermal energy supplied to the system	Thermal energy lost by it
\(W\)	Work done by the system	Work done on it
\(\Delta U\)	Temperature rises	Temperature falls
Process	Constant	First law becomes	On a P–V diagram
Isovolumetric	\(V\)	\(\Delta U = Q\)	Vertical line
Isobaric	\(P\)	\(\Delta U = Q - P \Delta V\)	Horizontal line
Isothermal	\(T\)	\(\Delta U = 0\), so \(Q = W\)	Curve, \(PV\) constant
Adiabatic	none: \(Q = 0\)	\(\Delta U = -W\)	Steeper curve