Unit 7: Thermal Physics & Thermodynamics

Thermal Physics + Thermodynamics together contribute 8–12 % of the JEE-Main physics paper and is one of the highest-yield topics on JEE-Advanced because problems mix several sub-areas (e.g. calorimetry with conduction, or kinetic theory with a thermodynamic process). Expect:

• JEE-Main: 2–3 single-correct MCQs + 1 numerical, usually one on kinetic theory (rms / Cp,Cv / mean-free-path), one on a thermodynamic process (W, Q, $\Delta U$), one on heat transfer (Fourier / Stefan / Newton's cooling) and frequently one calorimetry mixture.
• JEE-Advanced: multi-concept problems — e.g. Carnot engine driving a refrigerator that melts ice; a gas inside a thermally-conducting cylinder; PV–T graphs requiring you to identify the process and compute $\eta$ or COP.

The chapter is "formula-heavy" but conceptually shallow if you internalise three skeletons: first law $Q = \Delta U + W$, kinetic gas $PV = \tfrac{1}{3} N m \bar{v^2}$, and Fourier $H = -kA\,dT/dx$. Everything else is a corollary.

Concept Map

THERMAL PHYSICS
│
├── Temperature & Expansion
│     ├── Linear (α)  →  ΔL = L₀ α ΔT
│     ├── Areal  (β = 2α)
│     └── Volume (γ = 3α)  + anomalous water
│
├── Calorimetry
│     ├── Q = mcΔT (sensible heat)
│     ├── Q = mL    (latent heat)
│     └── Principle of mixtures (Σ Q = 0)
│
├── Heat Transfer
│     ├── Conduction — Fourier  H = kA ΔT / L
│     │      • Thermal resistance R = L/(kA)
│     │      • Series/Parallel rules
│     ├── Convection — Newton's law of cooling
│     │      dT/dt = -k(T − T_s)
│     └── Radiation — Stefan-Boltzmann, Wien, Kirchhoff
│
KINETIC THEORY OF GASES
│
├── PV = ⅓ N m ⟨v²⟩    →  ⟨KE⟩ = (3/2) kT
├── v_rms : v_avg : v_mp  =  √3 : √(8/π) : √2
├── Equipartition  →  C_v = (f/2)R, C_p = C_v + R, γ = 1 + 2/f
└── Mean free path  λ = 1/(√2 n π d²)
│
THERMODYNAMICS
│
├── First law  Q = ΔU + W
├── Processes
│     ├── Isothermal     W = nRT ln(V₂/V₁)
│     ├── Adiabatic      PVᵞ = const, W = (P₁V₁ − P₂V₂)/(γ−1)
│     ├── Isobaric       W = PΔV
│     ├── Isochoric      W = 0
│     └── Polytropic     PVⁿ = const, W = (P₁V₁ − P₂V₂)/(n−1)
│
├── Cyclic processes — Net W = area on PV diagram
│
├── Engines  η = W/Q_h  ≤  1 − T_c/T_h  (Carnot bound)
├── Refrigerator  COP = Q_c/W
└── Second law (Kelvin–Planck ⇔ Clausius), Entropy ↑

Topic 1: Thermal Expansion

Sub-topic A: Linear, Areal & Volume Expansion

When a solid is heated, its dimensions change. For small temperature changes $\Delta T$, the relative change is linear:

where $\alpha$, $\beta$, $\gamma$ are the coefficients of linear, superficial, and cubical expansion respectively.

Derivation of $\beta = 2\alpha$ and $\gamma = 3\alpha$ for isotropic solids:

Take a square of side $L_0$. Its area is $A_0 = L_0^2$. After heating,

(neglecting $(\alpha \Delta T)^2$ since $\alpha \sim 10^{-5}\,\text{K}^{-1}$).

So $\Delta A / A_0 = 2\alpha \Delta T \Rightarrow \beta = 2\alpha$.

Similarly for a cube of volume $V_0 = L_0^3$:

Hence the famous ratio:

Sub-topic B: Apparent vs Real Expansion of Liquids

Liquids do not have shape — only $\gamma$ matters. But if the container also expands,

This is why a flask appears to "lose" liquid when first heated (container expands faster) then overflows once the liquid catches up.

Sub-topic C: Anomalous Expansion of Water

Water has a density maximum at 4 °C. From 0 to 4 °C it contracts, then expands above 4 °C. This is why lakes freeze top-down — denser 4 °C water sinks, ice (less dense) floats and insulates, allowing aquatic life to survive.

Sub-topic D: Bimetallic Strip & Thermal Stress

A bimetallic strip is two metals (different $\alpha$) bonded together. On heating, the metal with higher $\alpha$ expands more — the strip bends towards the metal with lower $\alpha$. Used in thermostats.

Thermal stress in a clamped rod (cannot expand):

derived from $\Delta L / L = \alpha \Delta T$ would be the expansion if free, but it's prevented; so an equivalent compressive strain $\alpha \Delta T$ is imposed, giving stress $Y \alpha \Delta T$.

Worked Problem 1 (JEE-Main level)

A steel rod of length $1\,\text{m}$ and area $1\,\text{cm}^2$ is clamped between two rigid walls at $20\,^\circ \text{C}$. Find the force on the walls when heated to $120\,^\circ \text{C}$. Take $Y = 2\times 10^{11}$ Pa and $\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}$.

Solution. $\sigma = Y \alpha \Delta T = 2\times 10^{11} \times 1.2\times 10^{-5}\times 100 = 2.4 \times 10^{8}$ Pa.

$F = \sigma A = 2.4 \times 10^8 \times 10^{-4} = 2.4 \times 10^{4}$ N $= 24$ kN.

Worked Problem 2 (JEE-Advanced trap)

A circular hole of diameter $2\,\text{cm}$ is drilled in a steel plate at $20\,^\circ\text{C}$. To what diameter does the hole grow at $100\,^\circ\text{C}$? ($\alpha_{\text{steel}} = 1.2 \times 10^{-5}\,\text{K}^{-1}$.)

Trap: Beginners think the hole shrinks. It doesn't — every linear dimension (including a void) scales by $(1 + \alpha \Delta T)$.

$d = 2\,\text{cm} \times (1 + 1.2\times 10^{-5}\times 80) = 2\,\text{cm}\,(1 + 9.6\times 10^{-4}) \approx 2.00192\,\text{cm}$.

Topic 2: Calorimetry

Sub-topic A: Specific Heat and Latent Heat

Sensible heat (raises temperature):

Latent heat (causes phase change at constant temperature):

Standard JEE values to memorise:

Sub-topic B: Principle of Mixtures

In an isolated system, $\sum Q = 0$ — heat lost by hot bodies equals heat gained by cold bodies, including any phase changes.

Worked Problem 3 (Ice-Steam)

$10$ g of ice at $0\,^\circ\text{C}$ is added to $50$ g of water at $40\,^\circ\text{C}$ in a thermally-insulated calorimeter (mass effects ignored). Find the final temperature.

Solution. Heat available from cooling the water from 40 °C to 0 °C: $Q_{\text{available}} = 50 \times 1 \times 40 = 2000$ cal.

Heat to melt all ice: $Q_{\text{melt}} = 10 \times 80 = 800$ cal.

Excess = $2000 - 800 = 1200$ cal — used to warm the $60$ g of water (now all liquid) from $0\,^\circ$C.

$T_f = 1200 / (60 \times 1) = 20\,^\circ\text{C}$.

Worked Problem 4 (JEE-Advanced: partial phase change)

$1$ kg of steam at $100\,^\circ\text{C}$ is mixed with $1$ kg of ice at $0\,^\circ\text{C}$. Find the final state.

Solution. Maximum heat steam can release while condensing: $Q_1 = 1000 \times 540 = 5.4\times 10^5$ cal.

Heat to melt ice + warm to $100\,^\circ\text{C}$: $Q_2 = 1000 \times 80 + 1000 \times 1 \times 100 = 1.8\times 10^5$ cal.

Since $Q_1 > Q_2$, steam over-supplies. Heat left after $Q_2$ is used: $5.4\times 10^5 - 1.8\times 10^5 = 3.6\times 10^5$ cal.

This $3.6\times 10^5$ cal further condenses steam at $100\,^\circ\text{C}$: mass condensed $= 3.6\times 10^5 / 540 = 667$ g.

So $667$ g of steam condenses to water, and the final mixture at $100\,^\circ\text{C}$ contains: $667 + 1000 = 1667$ g of water + $1000 - 667 = 333$ g of steam.

Trap: Don't assume the system ends as a single phase — sometimes it equilibrates partway through a phase change.

Topic 3: Heat Transfer

Sub-topic A: Conduction — Fourier's Law

Heat current (rate) through a slab of cross-section $A$, length $L$, with hot face at $T_1$ and cold face at $T_2$ in steady state:

For a rod with constant cross-section in steady state, $H$ is uniform, so $dT/dx$ is constant, and

Thermal resistance (analogous to electrical):

Derivation of $T(x)$ profile (linear):

Steady state $\Rightarrow$ $H$ independent of $x$:

So the temperature drops linearly along a uniform rod in steady state. (Not so if $k$ or $A$ varies with $x$.)

Sub-topic B: Series and Parallel Thermal Resistance

Series (rods end-to-end, same heat current):

For two rods of equal area:

Parallel (rods side-by-side, same $\Delta T$):

For two rods of equal length:

Worked Problem 5 (Compound rod)

Two rods of equal length and area but conductivities $k$ and $3k$ are joined end-to-end. Hot end at $100\,^\circ$C, cold end at $0\,^\circ$C. Find junction temperature.

Solution. Same $H$ through both:

$\Rightarrow 100 - T = 3T \Rightarrow T = 25\,^\circ\text{C}$.

Sub-topic C: Convection — Newton's Law of Cooling

For a body at temperature $T$ in surroundings at $T_s$, with $T - T_s$ small,

Derivation (from Stefan's law):

Net radiative loss is $P = e\sigma A (T^4 - T_s^4)$. Writing $T = T_s + \theta$ with $\theta \ll T_s$:

So $P \propto \theta$. Energy balance $mc \, d\theta/dt = -P \Rightarrow d\theta/dt = -k\theta$ where $k = 4e\sigma A T_s^3/(mc)$.

Solution: $T(t) = T_s + (T_0 - T_s) e^{-kt}$.

Sub-topic D: Radiation

Stefan-Boltzmann Law (total radiant exitance of a black body):

For a body with emissivity $e \in [0,1]$: $P = e\sigma A T^4$.

Net rate of radiation between body at $T$ and surroundings at $T_s$:

Wien's Displacement Law: peak emission wavelength

(So the sun at $\sim 5800$ K peaks at $\sim 500$ nm — visible.)

Kirchhoff's Law: at thermal equilibrium, emissivity = absorptivity at every wavelength. Good absorbers are good emitters — a black body is the perfect absorber and emitter.

Worked Problem 6 (Stefan)

A body at $327\,^\circ$C is placed in surroundings at $27\,^\circ$C. Power radiated is $80$ W. Find the power radiated when the body's temperature is $727\,^\circ$C, surroundings unchanged.

Solution. $T_1 = 600$ K, $T_2 = 1000$ K, $T_s = 300$ K. Emissivity, area same:

$P_2 \approx 8.16 \times 80 \approx 653$ W.

Worked Problem 7 (Newton's cooling)

A liquid cools from $80\,^\circ$C to $60\,^\circ$C in $5$ min in surroundings at $20\,^\circ$C. Find time to cool from $60\,^\circ$C to $40\,^\circ$C.

Solution. Use approximate form $\frac{T_1-T_2}{t} = k\left(\frac{T_1+T_2}{2}-T_s\right)$.

For 80→60: $\frac{20}{5} = k(70-20)\Rightarrow k = 4/50 = 0.08\,\text{min}^{-1}$.

For 60→40: $\frac{20}{t} = 0.08(50-20)=2.4\Rightarrow t \approx 8.33$ min.

Topic 4: Kinetic Theory of Gases

Sub-topic A: Derivation of Gas Pressure

Consider $N$ molecules of mass $m$ in a cube of side $L$. Molecules move randomly with mean square speed $\overline{v^2}$.

Take a molecule with velocity components $(v_x, v_y, v_z)$. On hitting the wall normal to $x$-axis it reverses $v_x \to -v_x$, $\Delta p = 2 m v_x$. It returns to the same wall after time $\Delta t = 2L/v_x$.

Force from one molecule on this wall:

Summing over $N$ molecules and averaging:

By isotropy, $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \overline{v^2}/3$. Pressure:

Hence the fundamental result:

Sub-topic B: Temperature & Average Kinetic Energy

Combining with $PV = N k_B T$:

So average translational KE per molecule is $\tfrac{3}{2} k_B T$, independent of mass — the deep physical meaning of temperature.

Sub-topic C: Maxwell Speed Distribution — Three Speeds

Maxwell-Boltzmann:

You only need three derived speeds:

Sub-topic D: Equipartition & Degrees of Freedom

Equipartition theorem: every quadratic degree of freedom contributes $\tfrac{1}{2}k_B T$ to the average energy.

Mayer's relation (proof for ideal gas):

$Q_p = nC_p \Delta T$, $Q_v = nC_v \Delta T$. At constant pressure $W = nR\Delta T$, $\Delta U = nC_v \Delta T$. First law $\Rightarrow nC_p\Delta T = nC_v\Delta T + nR\Delta T$:

Also $\gamma = C_p/C_v = 1 + 2/f$.

Sub-topic E: Mean Free Path

Average distance between collisions:

where $n = N/V$ is number density and $d$ is the molecular diameter. The $\sqrt{2}$ comes from relative velocity of two moving molecules.

Worked Problem 8 (mixing gases)

$1$ mol of $\text{He}$ ($\gamma_1 = 5/3$) and $2$ mol of $\text{O}_2$ ($\gamma_2 = 7/5$) are mixed. Find $\gamma_{\text{mix}}$.

Solution. Equivalent $C_v$:

$C_{p,\text{mix}} = C_{v,\text{mix}} + R = 19R/6$. So

Worked Problem 9 (rms vs avg)

Find the temperature at which the rms speed of $\text{O}_2$ equals the rms speed of $\text{N}_2$ at $300$ K. ($M_{\text{O}_2}=32$, $M_{\text{N}_2}=28$.)

Solution. $v_{rms} = \sqrt{3RT/M}$. Equating:

Topic 5: First Law of Thermodynamics & Processes

Sub-topic A: First Law

with the IUPAC/physics convention: $Q > 0$ if heat is added to the system, $W > 0$ if work is done by the system. ($W_{\text{on}} = -W_{\text{by}}$, careful!).

For an ideal gas, $\Delta U = nC_v \Delta T$ — always, regardless of the process — because $U$ depends only on $T$.

Sub-topic B: The Four Basic Processes

For an ideal gas (n moles):

Isothermal ($T$ const). $\Delta U = 0$, so $Q = W$.

Isobaric ($P$ const).

Isochoric ($V$ const).

Adiabatic ($Q = 0$). $\Delta U = -W$ and $PV^\gamma = \text{const}$.

Derivation of $PV^\gamma = \text{const}$: First law adiabatic $\Rightarrow nC_v dT + P\, dV = 0$. Use $PV = nRT \Rightarrow nR\, dT = P\, dV + V\, dP$:

Integrate: $\gamma \ln V + \ln P = \text{const}\Rightarrow PV^\gamma = \text{const}$.

Work done in adiabatic:

Sub-topic C: Master Table

The polytropic molar heat capacity is a deep JEE-Advanced result, derivable from first law with $W = nR\Delta T/(1-n)$ and $\Delta U = nC_v\Delta T$.

Sub-topic D: Slopes on PV Diagram

• Isobar: horizontal
• Isochor: vertical
• Isotherm: $dP/dV = -P/V$
• Adiabat: $dP/dV = -\gamma P/V$ — steeper than isotherm at any common point.

Worked Problem 10 (Adiabatic compression)

A monatomic ideal gas at $300$ K, $1$ atm is compressed adiabatically to one-eighth its initial volume. Find the final temperature and pressure.

Solution. $\gamma = 5/3$. $TV^{\gamma-1} = \text{const}$:

$P_2 = P_1 (V_1/V_2)^\gamma = 1\cdot 8^{5/3} = 32$ atm.

Worked Problem 11 (Multi-process)

$1$ mole of ideal gas undergoes a cycle ABCA where AB is isobaric ($P=2P_0$, $V_0 \to 2V_0$), BC is isochoric ($P \to P_0$), CA is isothermal ($V \to V_0$). Find net work over the cycle.

Solution. At A: $(P,V) = (2P_0, V_0) \Rightarrow T_A = 2P_0 V_0/R$. At B: $(2P_0, 2V_0)\Rightarrow T_B = 4P_0 V_0/R$. At C: $(P_0, 2V_0) \Rightarrow T_C = 2P_0V_0/R = T_A$. Good, CA is isothermal.

$W_{AB} = P\Delta V = 2P_0 \cdot V_0 = 2P_0V_0$. $W_{BC} = 0$. $W_{CA} = nRT_A \ln(V_0/2V_0) = 2P_0V_0 \ln(1/2) = -2P_0V_0\ln 2$.

$W_{\text{net}} = 2P_0V_0(1 - \ln 2) \approx 2P_0V_0 (1 - 0.693) \approx 0.614\,P_0V_0$.

Topic 6: Cyclic Processes, Heat Engines & Carnot Cycle

Sub-topic A: Cyclic Processes — Net Work = Area

In a cycle, $\oint dU = 0$ since $U$ is a state function. Hence $\oint dQ = \oint dW$. The net work equals the enclosed area on a P-V diagram.

• Clockwise loop $\Rightarrow$ engine (net $W > 0$, heat absorbed from hot reservoir);
• Counter-clockwise $\Rightarrow$ refrigerator (net $W < 0$, heat dumped uphill).

Sub-topic B: Heat Engines & Efficiency

A heat engine takes $Q_h$ from a hot reservoir, dumps $Q_c$ to a cold reservoir, produces work $W = Q_h - Q_c$.

For a refrigerator:

For a heat pump:

Sub-topic C: Carnot Cycle — Full Derivation

Four reversible processes: A→B isothermal expansion at $T_h$, B→C adiabatic expansion ($T_h\to T_c$), C→D isothermal compression at $T_c$, D→A adiabatic compression ($T_c \to T_h$).

Heat absorbed in A→B (isothermal):

Heat released in C→D (isothermal):

Adiabatic legs give:

Dividing $\Rightarrow V_B/V_A = V_C/V_D$. So $\ln$ terms cancel and

Sub-topic D: Carnot Theorem

(i) No engine working between two reservoirs can be more efficient than a reversible (Carnot) engine operating between them.

(ii) All reversible engines between the same two reservoirs have the same efficiency, independent of working substance.

Proof is a classic reductio ad absurdum: suppose another engine $X$ with $\eta_X > \eta_C$. Use $X$ to drive a reversed Carnot engine — net effect would transfer heat from cold to hot reservoir with no work done, violating Clausius statement.

Sub-topic E: Second Law — Equivalent Statements

Kelvin–Planck: It is impossible to construct a device which extracts heat from a single reservoir and converts all of it into work in a cycle.

Clausius: It is impossible to construct a device which transfers heat from a colder body to a hotter body without any external work.

Equivalence: Violation of K-P $\Rightarrow$ violation of Clausius (and vice versa), shown by chaining engines.

Sub-topic F: Entropy (Qualitative)

For a reversible process,

$S$ is a state function. For an isolated system $\Delta S \ge 0$ (= 0 only if reversible). This is the second law in its most powerful form.

For an ideal gas reversibly between two states:

Worked Problem 12 (Carnot)

A Carnot engine takes $1000$ J from a reservoir at $500$ K. The sink is at $300$ K. Find $W$ done and $Q_c$ rejected.

Solution. $\eta = 1 - 300/500 = 0.4$. $W = 0.4 \times 1000 = 400$ J; $Q_c = 600$ J.

Worked Problem 13 (Refrigerator + freezing water)

A refrigerator with COP = $5$ removes heat from water at $0\,^\circ$C and dumps it into a room at $27\,^\circ$C. How much water at $0\,^\circ$C is frozen per minute if the compressor draws $200$ W of electrical power?

Solution. Heat removed per second $Q_c = \text{COP}\cdot W = 5\cdot 200 = 1000$ J/s. In 1 minute $= 60$ kJ.

Mass frozen $m = Q_c / L_f = 60000\,\text{J} / (3.34\times 10^5\,\text{J/kg}) \approx 0.18$ kg = $180$ g.

Worked Problem 14 (JEE-Advanced: irreversible expansion)

$2$ mol of ideal monatomic gas at $T_0=300$ K, $V_0 = 10$ L expands irreversibly against a constant external pressure $P_{\text{ext}} = P_0/2$ where $P_0$ is initial pressure. Find the final temperature given the process is adiabatic.

Solution. Adiabatic $\Rightarrow Q = 0$, so $\Delta U = -W_{\text{by gas}}$. But for an irreversible expansion against constant $P_{\text{ext}}$, $W = P_{\text{ext}} \Delta V$, not the reversible formula.

$P_0 = nRT_0/V_0 = 2(8.314)(300)/(10\times10^{-3}) = 4.99\times 10^5$ Pa. $P_{\text{ext}} = P_0/2 = 2.49\times 10^5$ Pa.

Final state must have $P_f = P_{\text{ext}}$ (mechanical equilibrium): $V_f = nRT_f/P_f$.

$\Delta U = n C_v (T_f - T_0) = -P_{\text{ext}}(V_f - V_0)$:

With $C_v = \tfrac{3}{2}R = 12.47$, $C_p = \tfrac{5}{2}R = 20.78$:

Numerator $= 2(12.47)(300) + (2.49\times 10^5)(10\times 10^{-3}) = 7482 + 2495 = 9977$ J. Denominator $= 2(20.78) = 41.56$. $T_f \approx 240$ K.

(So the irreversible expansion is less cool than the reversible $T_f = T_0(P_f/P_0)^{(\gamma-1)/\gamma}= 300\cdot (0.5)^{0.4}=227$ K. A textbook JEE-Advanced trap.)

Problem-Solving Heuristics

1. Choose the right $C$. Heat at constant $V$ uses $C_v$, at constant $P$ uses $C_p$. For a polytropic, derive $C = C_v - R/(n-1)$.
2. Sign of $W$. Always state clearly whether the convention is $W_{\text{by gas}}$ (most physics texts, the one used here) or $W_{\text{on gas}}$ (chemistry).
3. $\Delta U$ depends only on $\Delta T$ for an ideal gas. Use the easiest path (often isochoric) to compute $\Delta U$.
4. In cycles, jump straight to areas. Compute $Q$ for each leg only if asked; net work is the enclosed area.
5. Adiabat vs isotherm crossings. Adiabat is steeper by a factor of $\gamma$ — this distinguishes which is which on a P-V diagram.
6. Conduction in steady state $\Rightarrow$ same H everywhere. Use the series-resistance picture; for parallel rods use parallel-resistance.
7. Stefan involves $T^4$. If $T \to 2T$, power $\times 16$ — these factor-of-many MCQs come up.
8. Wien gives "the colour". $\lambda_{\max} \propto 1/T$ — hotter is bluer.
9. For irreversible expansion vs reversible expansion, compute $W$ from $W = P_{\text{ext}} \Delta V$ if pressure is constant external, not $\int P\,dV$.
10. Carnot bound $\eta \le 1 - T_c/T_h$ — any larger answer is wrong by the second law.
11. Mean free path drops with $P$ at fixed $T$: $\lambda \propto 1/n \propto T/P$.
12. Don't confuse $T_{\text{rms}}$ with $T$ itself. $v_{rms}\propto \sqrt T$, not $T$.

Common Traps & Mistakes

• Hole expands too. Anything you draw on a uniformly heated body — solid disk, hole, ring of dots — scales by the same $(1+\alpha\Delta T)$.
• Latent heat reservoirs. Don't assume mixing ends in a single phase; check whether the heat budget exceeds latent heat.
• Polytropic mol. heat. The formula $C = C_v + R/(1-n)$ has a sign that flips depending on which way you write it; double-check whose convention.
• Adiabatic process formulas apply only to reversible (quasi-static) adiabats. For sudden compression against constant external pressure, use first law directly.
• Stefan vs Newton's cooling. Newton's law is the small-$\Delta T$ linearisation of Stefan. They are not independent laws.
• Wien vs Stefan. Wien tells where the peak is; Stefan tells the total radiated power. Don't confuse them.
• Equipartition only at sufficiently high $T$. At low temperatures rotational/vibrational modes "freeze out" — that's why $\text{H}_2$ behaves like a monatomic gas below $\sim 100$ K.
• $Q_h = Q_c + W$ only for an engine in steady cycle. For a single process it does not hold.
• Sign of work in compression. If gas is compressed, $\Delta V < 0$, so $W_{\text{by}} < 0$. Plug signs algebraically, don't try to remember separate "expansion" and "compression" formulas.
• Cyclic $\Delta U = 0$ but not $\Delta S = 0$ for the universe — only for the gas.

Quick Revision Card

• Linear expansion: $\Delta L = L_0 \alpha \Delta T$; $\alpha:\beta:\gamma = 1:2:3$.
• Thermal stress: $\sigma = Y\alpha\Delta T$.
• Calorimetry: $Q = mc\Delta T$, phase change $Q = mL$.
• Fourier: $H = kA\Delta T/L$; $R = L/(kA)$; series $R$'s add, parallel $R$'s combine.
• Newton's cooling: $dT/dt = -k(T-T_s)$, exponential decay.
• Stefan: $P = e\sigma A T^4$; Wien: $\lambda_{\max}T = b$; Kirchhoff: $e = a$.
• Kinetic: $\langle\text{KE}\rangle = \tfrac{3}{2}k_BT$; $v_{rms} = \sqrt{3RT/M}$.
• $C_v = (f/2)R$, $C_p = C_v + R$, $\gamma = 1 + 2/f$.
• Mean free path $\lambda = 1/(\sqrt 2 n\pi d^2)$.
• First law: $Q = \Delta U + W$, $\Delta U = nC_v\Delta T$.
• Isothermal $W = nRT\ln(V_2/V_1)$; adiabatic $PV^\gamma =$ const, $W = (P_1V_1-P_2V_2)/(\gamma-1)$.
• Cycle: net $W$ = enclosed area; $\eta = W/Q_h = 1 - Q_c/Q_h \le 1 - T_c/T_h$.
• COP refrigerator $= Q_c/W$; COP heat pump $= Q_h/W = \text{COP}_{\text{ref}} + 1$.

Formula Sheet