Unit 1

Thermodynamics

Everything in this unit is one idea wearing different hats: energy is conserved, and it spreads out. The first law is the conservation half. The second law is the spreading-out half. The formulas are bookkeeping for those two sentences.

0 · Foundations: the ideal gas

Almost every problem in this unit is secretly about a gas in a container, so get these three relationships solid before anything else.

The equation of state

The useful form in practice is the ratio form, because you almost never know n. For a fixed amount of gas going from state 1 to state 2:

What temperature actually is

Temperature is average translational kinetic energy per molecule, nothing more. That is the bridge between the microscopic and macroscopic pictures:

Internal energy of a monatomic ideal gas

Because U depends only on T, the single most useful sentence in this unit is: ΔT = 0  ⇒  ΔU = 0.

1 · Thermal expansion

Heat something and it gets bigger. The interesting part is why, and the exam-worthy part is that holes, gaps, and cavities get bigger too — they expand exactly as if they were filled with the surrounding material.

Why materials expand

Atoms in a solid sit in a potential energy well, vibrating about an equilibrium separation. If that well were a symmetric parabola, heating would just widen the vibration and the average separation would not change — solids would not expand at all.

Real interatomic potentials are asymmetric: steep on the close-in side (atoms repel hard when pushed together) and shallow on the far side. Add energy and the vibration widens more to the right than to the left, so the average separation creeps outward. That creep, summed over billions of bonds, is thermal expansion.

Linear expansion

For a rod, wire, rail, or any length L₀, the change in length is proportional to both the original length and the temperature change:

Two things to notice. First, ΔL scales with L₀ — a 1 km bridge moves a thousand times as far as a 1 m ruler for the same temperature swing, which is why long structures need expansion joints and short ones do not. Second, α values are around 10⁻⁵ K⁻¹, so fractional changes are tiny; keep your significant figures.

Area and volume expansion

Expansion is isotropic in an ordinary solid: every dimension grows by the same fractional amount. So a square of side L grows to L(1 + αΔT) on each side, and its area becomes L²(1 + αΔT)². Expanding that square and dropping the (utterly negligible) α² term:

The factors of 2 and 3 are just “how many dimensions are growing.” They are not extra physics, and they are worth being able to re-derive in one line rather than memorizing.

The hole rule

This is the single most-tested idea on the page.

The reason is that expansion is a uniform scaling. Imagine photocopying the plate at 100.2% — every feature, including the empty ones, scales up. Nothing about the material “fills in” the hole.

Practical consequence: a stuck metal jar lid loosens under hot water partly because the lid’s inner diameter grows. And to fit a tight ring onto a shaft, you heat the ring — or cool the shaft.

Bimetallic strips

Bond two metals with different α face to face. Heat the pair and the high-α metal wants to be longer than the low-α one, but they are glued together — so the strip bends, curving toward the metal that expands less. Cool it below the bonding temperature and it curves the other way.

That bend is a temperature-controlled switch, which is what an old thermostat, a toaster timer, and a car turn-signal flasher all are.

Water: the exception that matters

Between 0 °C and 4 °C, water contracts as you heat it. Its density peaks at about 4 °C and falls off in both directions. Above 4 °C it behaves normally.

Two consequences worth being able to explain in words:

• Lakes freeze from the top down. As surface water cools toward 4 °C it gets denser and sinks. Below 4 °C it gets less dense, so the coldest water stays on top and freezes there. The ice then insulates the water beneath, and fish survive the winter.
• Ice floats. Solid water is about 9% less dense than liquid water (917 vs 1000 kg/m³) — far off the bottom of the graph above. Water is one of very few substances whose solid floats on its liquid.

Worked examples

1. Expansion joint on a bridge

A steel bridge span is 1.20 km long. Local temperatures run from −15 °C in winter to +40 °C in summer. How much gap must the expansion joints absorb?

Setup. α_steel = 12×10⁻⁶ K⁻¹, L₀ = 1200 m, ΔT = 40 − (−15) = 55 K.

ΔL = (12×10⁻⁶)(1200)(55) = 0.79 m

Nearly 80 cm. Note that ΔT is a difference, so it is the same number in K or °C — no conversion needed here.

2. Does the hole get bigger?

An aluminum plate at 20 °C has a circular hole of diameter 2.000 cm. The plate is heated to 120 °C. Find the new diameter.

Key move. Treat the hole’s diameter as an ordinary length in aluminum. Do not try to reason about the metal “squeezing inward” — it does not.

Δd = (24×10⁻⁶)(2.000 cm)(100 K) = 0.0048 cm New diameter = 2.000 + 0.0048 = 2.005 cm. Bigger.

Follow-up they like to ask: a steel bolt of diameter 2.002 cm will not fit at 20 °C but slides through once the plate is hot.

3. Overflow — when the container expands too

A 500 mL ordinary-glass beaker is filled to the brim with water at 20 °C, then heated to 80 °C. How much water spills?

Key move. The container expands too, so the spill is the difference of the two volume expansions, not the water’s alone.

ΔV_water = (207×10⁻⁶)(500)(60) = 6.21 mL
ΔV_glass = 3(9×10⁻⁶)(500)(60) = 0.81 mL Glass is quoted with α, so use β = 3α = 27×10⁻⁶ K⁻¹.

Spill = 6.21 − 0.81 = 5.4 mL. If the question had used Pyrex (α = 3.3), the glass term would be 0.30 mL and the spill 5.9 mL — closer to the water-only answer, because low-expansion glass barely moves.

4. Conceptual: which way does the strip bend?

A strip is brass on top, invar on the bottom, bonded flat at 25 °C. Describe its shape at 5 °C.

Reason it out. Cooling, so both shrink — but brass (α = 19) shrinks much more than invar (α = 1.2). The top layer becomes the shorter one, so the strip curves upward: it bows with the brass on the inside of the curve.

The general rule: the layer on the outside of the curve is always the one that is currently longer. On heating, that is the high-α metal; on cooling, the low-α one.

Checklist

• I can state ΔL = αL₀ΔT and say what each symbol is.
• I can explain in one sentence why holes get bigger.
• I can get β = 3α from scratch instead of memorizing it.
• I remember that in an overflow problem the container expands too.
• I can explain why a lake freezes from the top.

2 · Heat, specific heat & calorimetry

How much energy does it take to change something’s temperature — and what happens when the temperature refuses to change at all? Two equations, Q = mcΔT and Q = mL, and the whole skill is knowing which one you are in.

Heat vs. temperature vs. internal energy

These three get used interchangeably in everyday speech and never in physics.

Specific heat

Specific heat c is the energy needed to raise one kilogram by one kelvin:

A big c means the substance is hard to heat up and slow to cool down. Water’s is enormous — the highest of any common substance — which is why coastal climates are mild, why radiators and engine coolant use water, and why the ocean is a planetary thermal flywheel.

Latent heat and phase change

During melting or boiling, energy pours in and the temperature does not move. The energy is going into breaking intermolecular bonds — raising potential energy, not kinetic energy. Since temperature only tracks kinetic energy, the thermometer sits still.

Notice L_v for water is about 6.8× its L_f. Melting ice only has to loosen the crystal; boiling has to rip molecules entirely free of each other. That ratio explains two everyday facts: a pot takes forever to boil dry after it reaches 100 °C, and steam at 100 °C burns far worse than water at 100 °C — condensing on your skin dumps 2.26×10⁶ J/kg before the temperature even starts to drop.

The heating curve

Put 1 kg of ice at −20 °C on a burner of constant power and plot temperature against energy added. You get the single most useful diagram in this topic:

How to read it:

• Sloped = one phase, temperature climbing. Use Q = mcΔT with that phase’s c.
• Flat = phase change in progress, two phases coexisting. Use Q = mL.
• Steeper slope = smaller c. The ice and steam segments rise faster than the liquid-water segment precisely because c_ice and c_steam are about half c_water. If a graph question asks you to compare specific heats, compare slopes.

Calorimetry

Put hot and cold things in an insulated container. No energy leaves, no work is done, so all the heat lost by some objects is gained by the others:

Method

1. Write Q for every object, each as mc(T_f − T_i), all with the same unknown T_f.
2. Add any phase-change terms as separate mL pieces (positive if the substance absorbs energy: melting or boiling).
3. Set the sum to zero and solve.
4. Sanity check: T_f must land between the starting temperatures. If it does not, you have a sign error — or a phase change you did not account for.

How heat travels

For steady conduction through a slab:

Read the structure: transfer is faster through a wide, thin, conductive slab with a big temperature difference, and slower through a narrow, thick, insulating one. Copper’s k is about 400 W/(m·K); fiberglass insulation is about 0.04 — a factor of 10⁴. Most insulation works by trapping air (k ≈ 0.026) in pockets small enough to stop convection.

Worked examples

1. Identify a metal from a calorimetry run

A 150 g metal block is heated to 95.0 °C and dropped into 200 g of water at 21.0 °C in a calorimeter of negligible heat capacity. The mixture settles at 25.8 °C. What is the metal?

Step 1 — heat gained by the water.

Q_w = (0.200)(4186)(25.8 − 21.0) = +4.02×10³ J

Step 2 — the metal supplied it. ΣQ = 0, so Q_m = −4.02×10³ J.

−4.02×10³ = (0.150)c(25.8 − 95.0) = −10.38 c c = 4.02×10³ / 10.38 = 387 J/(kg·K)

That matches copper (385). Note the sign took care of itself: ΔT for the metal is negative because it cooled.

2. Ice into water — does it all melt?

50 g of ice at 0 °C is added to 300 g of water at 25 °C. Find the final temperature.

Step 1 — test the phase change first.

• Energy to melt all the ice: mL_f = (0.050)(3.34×10⁵) = 16 700 J
• Energy available from cooling the water to 0 °C: (0.300)(4186)(25) = 31 400 J

31 400 > 16 700, so all the ice melts and there is energy left over. Good — the final temperature will be above 0 °C.

Step 2 — spend the surplus warming the combined 350 g of water.

ΔT = 31 400 − 16 700(0.350)(4186) = 14 7001465 = 10.0 K T_f = 10.0 °C

Sanity check: between 0 and 25 °C. ✓   Had the ice been 200 g instead, melting it all would have cost 66 800 J — more than the 31 400 J available — and the answer would have been 0 °C with ice remaining.

3. Reading specific heat off a graph

A 0.500 kg sample is heated at a constant 200 W. Its temperature rises from 20 °C to 60 °C in 90 s, with no phase change. Find c.

Q = Pt = (200)(90) = 1.80×10⁴ J c = Q / (mΔT) = 1.80×10⁴ / [(0.500)(40)] = 900 J/(kg·K) — aluminum.

On a T-vs-Q graph this is the reciprocal of the slope divided by mass: c = 1/(m · slope). Steep line, small specific heat.

4. Conceptual: two burns

Why is a burn from 100 °C steam worse than one from 100 °C water, when both are at the same temperature?

Because of the phase change. Steam condensing on skin releases L_v = 2.26×10⁶ J/kg before its temperature drops at all. Only after condensing does it start cooling from 100 °C like ordinary water. Per gram, condensation alone delivers about 2260 J — roughly the same as cooling liquid water from 100 °C all the way down to 46 °C. The steam burn simply carries far more energy.

Checklist

• I can say what heat is without using the word “temperature.”
• I know when to use mcΔT and when to use mL, and I never use both for the same stage.
• I can sketch the heating curve from memory and label which equation applies to each piece.
• I check whether ice fully melts before solving for T_f.
• I can explain why metal feels colder than wood.

3 · The first law & work done on or by a gas

The first law is conservation of energy with the microscopic bookkeeping made explicit. There are only two ways to change a gas’s internal energy: heat it, or push on it. That is the whole law — the difficulty is entirely in the signs.

The law itself

In words: the energy inside goes up by whatever you put in. Energy can enter as disorganized molecular collisions (heat) or as organized macroscopic pushing (work), and once inside, the gas cannot tell the difference.

Signs — the part that costs points

The intuition for W: compressing a gas warms it. A bike pump gets hot at the barrel. You did work on the gas, energy went in, temperature rose. Conversely a can of compressed air gets cold as it sprays — the gas expands, does work on the outside world, loses energy, cools.

Where W = −PΔV comes from

Take a cylinder of cross-sectional area A sealed by a frictionless piston, gas at pressure P. Push the piston in a distance d, slowly enough that the pressure stays essentially uniform.

The work done by the external force on the gas is force times displacement:

The minus sign is not a convention someone invented to annoy you. It is there because volume goes down when you do positive work.

“On the gas” vs. “by the gas”

Exam questions switch between these two phrasings constantly, sometimes inside a single problem. They are the same number with opposite signs:

State functions vs. path functions

This distinction quietly runs the entire unit.

• U, P, V, T are state functions. They depend only on where the gas is, not how it got there. Two different processes ending at the same point produce the same ΔU.
• Q and W are path functions. They describe what happened along the way. Different routes between the same two states give different Q and different W — but their sum always comes out the same, because that sum is ΔU.

Two immediate payoffs, both heavily tested:

1. Any complete cycle has ΔU = 0, because the gas returns to the same state. So for a cycle, Q_net = −W_net,on = W_net,by.
2. ΔT = 0 ⇒ ΔU = 0 for an ideal gas, no matter what route was taken.

The four processes

Every AP thermodynamics problem is one of these four, or a sequence of them. This table is worth reproducing from memory.

Adiabatic is the one worth a second look. Q = 0 happens either because the container is insulated or because the process is fast enough that no appreciable heat has time to flow. So an adiabatic compression puts all of the work straight into internal energy — the gas heats up sharply. That is a diesel engine: compress air fast enough and it ignites fuel with no spark plug.

Worked examples

1. Isobaric compression with heat leaving

A gas is compressed at a constant pressure of 2.0×10⁵ Pa from 3.0×10⁻³ m³ to 1.0×10⁻³ m³. During the process 250 J of heat leaves the gas. Find ΔU, and say whether the gas warmed or cooled.

Work. ΔV = 1.0×10⁻³ − 3.0×10⁻³ = −2.0×10⁻³ m³.

W = −PΔV = −(2.0×10⁵)(−2.0×10⁻³) = +400 J Positive, as it must be — the gas was compressed.

Heat. Leaving the gas, so Q = −250 J.

ΔU = Q + W = −250 + 400 = +150 J

ΔU > 0, so the temperature rose. Even though heat was flowing out, the compression put in more energy than the heat carried away.

2. Isothermal expansion

0.50 mol of ideal gas expands isothermally at 300 K from 0.010 m³ to 0.030 m³. Find W, Q, and ΔU.

Start with the zero. Isothermal ⇒ ΔT = 0 ⇒ ΔU = 0.

W = −nRT ln(V_f/V_i) = −(0.50)(8.31)(300) ln(3.0) = −1.4×10³ J Negative because the gas expanded — it did work on its surroundings.

Then from ΔU = Q + W = 0: Q = −W = +1.4×10³ J.

Physically: to hold the temperature fixed while the gas does 1.4 kJ of work, the reservoir must feed in exactly 1.4 kJ of heat. Energy passes straight through the gas without accumulating.

3. Adiabatic compression

0.20 mol of a monatomic ideal gas is compressed adiabatically; 500 J of work is done on it. Find ΔU and the temperature change.

Start with the zero. Adiabatic ⇒ Q = 0.

ΔU = 0 + 500 = +500 J

Now convert to a temperature change using the monatomic internal-energy formula:

ΔT = 2ΔU3nR = 2(500)3(0.20)(8.31) = +201 K

A 200-kelvin jump from half a kilojoule. Adiabatic compressions heat gases dramatically, which is exactly why diesel engines work and why a fire piston can ignite tinder.

4. Path dependence — same endpoints, different work

A gas goes from state A (1.0×10⁵ Pa, 2.0×10⁻³ m³) to state B (0.50×10⁵ Pa, 4.0×10⁻³ m³) two ways:

• Path 1: expand at constant pressure to 4.0×10⁻³ m³, then drop the pressure at constant volume.
• Path 2: drop the pressure at constant volume first, then expand at constant pressure.

Path 1. Only the isobaric leg does work, at the high pressure:

W₁ = −(1.0×10⁵)(2.0×10⁻³) = −200 J

Path 2. Only the isobaric leg does work, at the low pressure:

W₂ = −(0.50×10⁵)(2.0×10⁻³) = −100 J

Different work for the same start and end. But ΔU is identical for both (same endpoints, and P_AV_A = P_BV_B, so in fact ΔU = 0 here). The heat must therefore differ too — by exactly 100 J — so that Q + W comes out the same both times.

That is path dependence in one problem. It is drawn out in Section 4 (PV diagrams).

Checklist

• I can state the first law and say what each symbol means, including which direction W counts.
• I can derive W = −PΔV from F = PA in two lines.
• Given “work done by the gas,” I know to flip the sign before using the law.
• I know which quantity is zero for each of the four processes without looking.
• I know ΔU = 0 for any complete cycle and for any isothermal process.

4 · PV diagrams

A PV diagram turns a thermodynamics problem into a geometry problem. Work becomes an area, internal energy becomes a position, and a whole engine cycle becomes a loop you can measure with a ruler. Learn to read these and most of the unit stops requiring algebra.

Anatomy of the diagram

Pressure on the vertical axis, volume on the horizontal. Every point is a complete state of the gas: fix P and V and the ideal gas law fixes T too. Every curve is a process, and the arrow on it tells you which way time runs.

Work is the area underneath

For a thin slice of the process where the volume changes by dV, the gas does P dV of work on its surroundings. Add up the slices:

The graph gallery

Four shapes. Learn to recognize them instantly.

Isobaric — constant pressure

Isochoric — constant volume

Isothermal — constant temperature

Adiabatic — no heat exchanged

All four on one plot

Start every process from the same state and the differences become a matter of slope:

Path dependence

Same start, same finish, two routes. The endpoints fix ΔU, but the areas underneath are visibly different — so the work differs, and the heat must differ by exactly the same amount to compensate.

Cycles

Close the loop and the gas returns to its original state, so ΔU = 0 over the whole cycle. The first law then gives Q_net = W_net,by: over one cycle, whatever net heat went in came out as net work. And that net work is just the enclosed area.

How to read any PV problem

1. Label every state point with its P and V, and compute PV for each. That product is proportional to T, and for a monatomic gas U = ³⁄₂PV.
2. Identify each leg by its shape (horizontal, vertical, hyperbola, other).
3. Work, leg by leg, from the area. Vertical legs contribute nothing. Sign from the direction of travel.
4. ΔU, leg by leg, from ³⁄₂Δ(PV) — no path reasoning required.
5. Q = ΔU − W for each leg. Never compute Q directly if you can avoid it.
6. Check the totals. Around a full cycle, ΣΔU must be 0, and ΣW must equal the enclosed area with the right sign. If it does not, you have a sign error.

Worked examples

1. Work from a slanted line (trapezoid)

A gas expands along a straight line on the PV diagram from (1.0×10⁻³ m³, 4.0×10⁵ Pa) to (5.0×10⁻³ m³, 2.0×10⁵ Pa). Find the work done by the gas.

Do not use PΔV — the pressure is not constant. The region under a straight line is a trapezoid, so use the average height:

W_by = P_avgΔV = 4.0×10⁵ + 2.0×10⁵2 (4.0×10⁻³) = +1200 J And W_on = −1200 J. Expanding, so the gas did work on the world.

Units check: Pa · m³ = (N/m²)(m³) = N·m = J. Areas on a PV diagram are always energies — that is worth verifying once so you trust it forever.

2. Full cycle analysis (the big one)

A monatomic ideal gas is taken clockwise around the rectangular cycle A→B→C→D→A:

State	P (Pa)	V (m³)	PV (J)	U = 3⁄2PV (J)
A	1.0×105	2.0×10−3	200	300
B	3.0×105	2.0×10−3	600	900
C	3.0×105	6.0×10−3	1800	2700
D	1.0×105	6.0×10−3	600	900

Computing U at every corner first is the whole trick — every ΔU below is then just a subtraction.

Leg	Type	Won (J)	ΔU (J)	Q = ΔU − W (J)
A→B	Isochoric (up)	0	+600	+600
B→C	Isobaric (right)	−1200	+1800	+3000
C→D	Isochoric (down)	0	−1800	−1800
D→A	Isobaric (left)	+400	−600	−1000
Cycle		−800	0	+800

Checks, all three of which must pass:

• ΣΔU = 600 + 1800 − 1800 − 600 = 0 ✓ (it is a closed cycle)
• Enclosed area = ΔP × ΔV = (2.0×10⁵)(4.0×10⁻³) = 800 J, and W_by = +800 J ✓ (clockwise, so positive)
• Q_net = +800 J = W_by ✓

Bonus — efficiency. Heat entered on A→B and B→C: Q_H = 600 + 3000 = 3600 J. So

e = |W_net||Q_H| = 8003600 = 0.22 = 22% Only heat absorbed goes in the denominator — the 1800 J and 1000 J expelled do not.

3. Identify the process from the graph

A gas moves along a curve on which PV is the same at both endpoints, and the curve bulges toward the origin. What kind of process is it, and what are Q, W, and ΔU if the gas expanded?

Equal PV at both ends means equal temperature at both ends, so ΔU = 0. The bulge-toward-the-origin shape is a hyperbola — an isothermal process. Expanding means the gas did work on the world, so W_on < 0, and the first law forces Q = −W_on > 0: heat flowed in.

Careful, though: equal PV at the endpoints alone only guarantees ΔU = 0. The path could still have wandered off the isotherm in between, in which case W would be a different number — but ΔU would still be zero.

4. Which path takes more heat?

Using the two paths from the path-dependence figure, both running A→B: which requires more heat input?

ΔU is the same for both (same endpoints). From Q = ΔU − W_on, and with W_on = −(area) for a rightward path,

Q = ΔU + area Bigger area ⇒ more heat required.

So Path 1 (expand at high pressure first) needs more heat. It also delivers more work. That is not a coincidence — the extra heat is precisely what pays for the extra work, since the internal energy change is fixed.

Checklist

• I can name the process from the shape of the curve, instantly.
• I get the sign of W from the direction of travel, not from memorized cases.
• I compute U = ³⁄₂PV at every corner before doing anything else.
• I know clockwise = engine and counterclockwise = refrigerator.
• I check that ΣΔU = 0 around every closed cycle.

5 · The second law & entropy

The first law says energy is never lost. The second law says that is not enough — energy also has a quality, and every real process degrades it. That is why you cannot un-scramble an egg, why no engine is 100% efficient, and why time appears to run one direction.

Three statements, one law

The second law gets phrased three different ways depending on what is being discussed. They are logically equivalent — each implies the other two.

Heat engines

A heat engine runs a working substance around a cycle, taking heat Q_H from a hot reservoir, converting part of it to work W, and dumping the rest as Q_C into a cold reservoir. Because it is a cycle, ΔU = 0 and the first law gives an exact energy balance:

Efficiency and the Carnot limit

Efficiency is “what you want divided by what you paid for”:

Now the ceiling. No engine operating between two reservoirs can beat a fully reversible one, and every reversible engine between the same two reservoirs has the same efficiency:

Two consequences that are worth internalizing rather than deriving:

• 100% efficiency requires T_C = 0 K, which is unreachable. That is the Kelvin–Planck statement written as an equation.
• Efficiency depends on the ratio, not the difference. An engine running between 600 K and 300 K has the same Carnot limit (50%) as one between 400 K and 200 K, even though the second has half the temperature difference.

Refrigerators and heat pumps

Run the engine diagram backwards. Now work goes in, and heat is pulled from the cold side and dumped on the hot side. On a PV diagram this is the counterclockwise cycle.

Entropy

Entropy S measures how spread out energy is — equivalently, how many microscopic arrangements are consistent with what you can see macroscopically. For heat transferred reversibly at a temperature that does not appreciably change:

The T in the denominator is the interesting part. The same joule of heat produces a bigger entropy change when it lands somewhere cold. That single fact is why heat flows hot → cold and never the reverse:

Why entropy increases

Nothing forces gas molecules to spread out. There is no repulsive law pushing them apart. Spreading is simply overwhelmingly more likely.

Boltzmann made this exact: S = k_B ln Ω, where Ω is the number of microstates. You will not be asked to compute with this in AP Physics 2, but it is the reason the second law is true, and it explains why the law is statistical — entropy decreases are not strictly impossible, merely so improbable that they never happen.

Worked examples

1. Engine efficiency and waste heat

An engine absorbs 3600 J per cycle from a hot reservoir and expels 2800 J to a cold one. Find the work per cycle and the efficiency.

|W| = |Q_H| − |Q_C| = 3600 − 2800 = 800 J e = 800/3600 = 0.22 = 22%

This is the same engine as the cycle worked out in Section 4 (PV diagrams) — there the 800 J came from the enclosed area of the loop, here from the energy balance. Same number, two routes.

2. Is this engine possible?

An inventor claims an engine operating between 550 K and 300 K with 50% efficiency. Evaluate the claim.

e_Carnot = 1 − 300550 = 0.4545 = 45.5%

The claimed 50% exceeds the Carnot limit, so the engine is impossible. No design refinement can help: the bound depends only on the reservoir temperatures, not on the working substance, the cycle shape, or the engineering.

Follow-up they like: what is the most work per cycle this engine could produce from 3600 J of input? (0.4545)(3600) = 1636 J.

3. Entropy change of melting ice

1.0 kg of ice melts at 0 °C. Find the entropy change of the ice.

The temperature is constant during melting, so ΔS = Q/T applies directly with Q = mL_f.

ΔS = (1.0)(3.34×10⁵)273 = +1.2×10³ J/K

Positive — liquid water has vastly more available arrangements than a rigid ice crystal. Note the kelvin conversion: using 0 in the denominator would divide by zero, which is a useful reminder that this formula is kelvin-only.

4. Why heat only flows one way

1200 J of heat flows from a 600 K reservoir to a 300 K reservoir. Find ΔS_universe. Then check the reverse.

ΔS = −1200600 + 1200300 = −2.0 + 4.0 = +2.0 J/K Allowed — the total went up.

Now reverse it, 1200 J from cold to hot:

ΔS = −1200300 + 1200600 = −4.0 + 2.0 = −2.0 J/K Forbidden — a spontaneous decrease in total entropy.

Both directions conserve energy perfectly. The first law cannot tell them apart; only the second law rules one out. That is the cleanest demonstration of why we need a second law at all.

5. Free expansion — entropy without heat

An insulated container is divided by a partition, gas on one side and vacuum on the other. The partition is removed. What are Q, W, ΔU, ΔT, and ΔS?

• Q = 0 — the container is insulated.
• W = 0 — the gas expands into vacuum, pushing against nothing.
• ΔU = Q + W = 0, so ΔT = 0 for an ideal gas.
• ΔS = nR ln(V_f/V_i) > 0.

The point: entropy increased even though no heat flowed. The formula ΔS = Q/T only applies to reversible transfers, and free expansion is violently irreversible — the gas will never spontaneously crowd back into half the box. For an irreversible process you compute ΔS along an imagined reversible path between the same two states, which is where the logarithm comes from.

This is also a good check on the difference between the two laws: the first law is entirely satisfied by the gas returning to one side. Only the second law says it will not.

Checklist

• I can state the second law in all three forms and say what each one forbids.
• I use only absorbed heat in the denominator of the efficiency formula.
• I convert to kelvin before touching the Carnot formula.
• I can test a claimed engine against the Carnot limit in one line.
• I can explain, using Q/T, why heat flows hot to cold.
• I know that local entropy decreases are fine as long as the universe’s total rises.

6 · Formula sheet & constants

Everything the unit uses, with the condition each one needs to be legal.

Constants worth memorizing

7 · Unit-wide traps

• Celsius in a gas law. PV = nRT needs kelvin. Every time. A ΔT is the same in K and °C, but a bare T is not.
• Gauge vs absolute pressure. A tire gauge reading 200 kPa means 301 kPa absolute. Gas laws want absolute.
• Heat is not temperature. Heat is energy in transit because of a temperature difference. An object contains internal energy, never “heat.”
• Sign of W. This unit uses W = work done on the gas, so compression is positive. See the sign section.
• ³⁄₂nRT is monatomic-only. Diatomic gases store energy in rotation too. AP problems say “monatomic” when they want that formula — if they do not, avoid it.

8 · Labs for this unit

Every lab for this unit lives in Interactive Laboratories, each one paired with the section above that explains it.

Interactive lab Ideal gas & PV work Open it in Interactive Laboratories → Interactive lab Thermal conductivity Open it in Interactive Laboratories → Interactive lab Entropy Open it in Interactive Laboratories → Interactive lab Heat transfer & calorimetry Open it in Interactive Laboratories → Bench lab Boyle’s law Open it in Interactive Laboratories →