Chapter 12 — Thermodynamics

Thermodynamics is the macroscopic theory of heat, work, and the limits of converting one into the other. From the Zeroth law (defining temperature) through the First (energy conservation) and Second (direction of spontaneous processes), we will end at Carnot's celebrated upper bound on heat-engine efficiency.

Concept Map

• 12.1 Systems, surroundings, types
• 12.2 Thermal equilibrium; Zeroth law
• 12.3 Heat, work, internal energy
• 12.4 First law — sign conventions & applications
• 12.5 Specific heats $C_p, C_v$; Mayer's relation
• 12.6 Processes — isothermal, adiabatic, isobaric, isochoric
• 12.7 Heat engines; refrigerators; COP
• 12.8 Second law — Kelvin-Planck & Clausius statements
• 12.9 Reversible & irreversible processes
• 12.10 Carnot cycle; Carnot theorem

12.1 Thermodynamic System, Surroundings, Types

Definitions

A thermodynamic system is the specific portion of matter (or radiation) under study. Everything outside is the surroundings. The boundary may be real or imaginary.

A system is described by state variables: $P, V, T, n$ (and $U, S$ for energy/entropy). Two are independent for a simple gas; equation of state $f(P,V,T)=0$ ties them.

Pitfalls

• The Earth is not isolated — receives solar radiation.
• "Closed" in thermodynamics means matter-tight, not energy-tight (different from everyday English).

12.2 Thermal Equilibrium; Zeroth Law

Two systems in thermal contact reach a common temperature — thermal equilibrium — after which no net heat flows.

Zeroth Law

If $A$ is in thermal equilibrium with $C$, and $B$ is in thermal equilibrium with $C$, then $A$ and $B$ are in thermal equilibrium with each other.

This is what justifies the very notion of temperature as a state variable: temperature is the property all bodies in mutual equilibrium share. Without the zeroth law, a thermometer ($C$) couldn't compare $A$ and $B$.

Pitfalls

• The label "zeroth" was tacked on after the First and Second laws were named, because logical priority belongs to it.
• Equilibrium is thermal equilibrium here — for a full thermodynamic equilibrium, also mechanical (no pressure difference) and chemical (no concentration difference).

12.3 Heat, Work, Internal Energy

Heat $Q$

Energy transferred because of a temperature difference. Path-dependent, not a state variable.

Work $W$

Energy transferred because of a generalised displacement. For a gas:

$\boxed{\ W = \int_{V_1}^{V_2} P\,dV\ }$

— area under the $P$-$V$ curve. Path-dependent.

Internal Energy $U$

Sum of kinetic and potential energies of all microscopic constituents. A state variable: depends only on $T$ (and $n$) for an ideal gas. For a monatomic ideal gas:

$U = \tfrac{3}{2} n R T.$

For a diatomic gas (rigid rotor) $U = \tfrac{5}{2}nRT$.

Pitfalls

• $U$ depends only on current state; $Q$ and $W$ depend on the path.
• For an ideal gas, $U$ depends only on $T$, not on $V$.

12.4 First Law of Thermodynamics

Statement

$\boxed{\ \Delta U = Q - W\ }$

The heat added to a system equals the increase in its internal energy plus the work done by the system.

Sign Conventions (NCERT)

• $Q > 0$: heat added to system.
• $W > 0$: work done by system on surroundings.
• $\Delta U > 0$: internal energy increases.

Applications

• Isolated system ($Q=0, W=0$): $\Delta U = 0$ → conservation of energy.
• Free expansion of an ideal gas ($Q=0$, $W=0$, expanding into vacuum): $\Delta U = 0$, so $\Delta T = 0$. Tricky: $W=0$ even though volume changes, because there's nothing to push against.
• Cyclic process: returns to start, $\Delta U = 0 \Rightarrow Q = W$. Engine output = heat input minus heat rejected.

Worked Example

A gas absorbs $200$ J of heat and does $50$ J of work. Change in internal energy?

$\Delta U = Q - W = 200 - 50 = 150\ \text{J}.$

Pitfalls

• Chemistry/IUPAC convention writes $\Delta U = Q + W$ with $W =$ work done on system. Watch the sign.
• "Heat of a system" is meaningless — heat is a transfer, not a property.

12.5 Specific Heats of Gases; Mayer's Relation

Definitions

For one mole of gas:

• $C_V$: molar heat capacity at constant volume — heat per mole per K to raise $T$ at fixed $V$.
• $C_P$: molar heat capacity at constant pressure.

$Q_V = nC_V\Delta T,\qquad Q_P = nC_P\Delta T.$

Derivation — Mayer's Relation

At constant volume, $W = 0$, so first law gives $\Delta U = Q_V = nC_V\Delta T$. Therefore for an ideal gas

$\left(\frac{\partial U}{\partial T}\right) = nC_V \quad\text{(independent of process)}.$

At constant pressure, $W = P\Delta V = nR\Delta T$ (ideal gas), and $\Delta U$ still equals $nC_V\Delta T$. First law:

$Q_P = \Delta U + W = nC_V\Delta T + nR\Delta T = nC_P\Delta T,$ $\boxed{\ C_P - C_V = R\ }$

Define the adiabatic index (ratio of specific heats):

$\gamma = \frac{C_P}{C_V}.$

Values from equipartition (each degree of freedom contributes $\tfrac{1}{2}R$ to $C_V$):

Worked Example

For helium gas, $C_V = (3/2)R = 12.5$ J/(mol·K). Heat $300$ J added at constant volume to $0.1$ mol. Find $\Delta T$.

$\Delta T = \frac{Q}{nC_V} = \frac{300}{(0.1)(12.5)} = 240\ \text{K}.$

Pitfalls

• Mayer's relation $C_P - C_V = R$ holds per mole; in specific (per kg) form: $c_P - c_V = R/M$.
• $C_V, C_P$ are essentially temperature-independent only in the rigid-rotor model; vibrations excite at high $T$ and bump up $C_V$.

12.6 Thermodynamic Processes

Isothermal ($T =$ const)

Slow process in contact with a heat reservoir; $\Delta U = 0$ for ideal gas $\Rightarrow Q = W$.

Work derivation:

$W = \int_{V_1}^{V_2} P\,dV = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT\ln\frac{V_2}{V_1}.$

$\boxed{\ W_\text{iso} = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{P_1}{P_2}\ }$

Since $\Delta U = 0$: $Q_\text{iso} = W_\text{iso}$.

PV curve: hyperbola.

Adiabatic ($Q = 0$)

Fast process (no heat exchange) or perfectly insulated.

Derivation of $PV^\gamma =$ const: first law gives $dU = -dW$, i.e. $nC_V dT = -P\,dV$. Use $PV = nRT$ to write $dT$:

$d(PV) = nR\,dT \Rightarrow P\,dV + V\,dP = nR\,dT.$ $nC_V dT = -P\,dV.$

Multiply: $V\,dP = nR\,dT - P\,dV = (nR/nC_V)(-P\,dV) - P\,dV - ...$

Cleaner: differentiate $T = PV/nR$, plug into $nC_V\,dT = -P\,dV$:

$C_V\frac{d(PV)}{R} = -P\,dV \Rightarrow C_V(P\,dV + V\,dP) = -RP\,dV,$ $(C_V + R) P\,dV + C_V V\,dP = 0,$ $C_P P\,dV + C_V V\,dP = 0,$ $\gamma\frac{dV}{V} + \frac{dP}{P} = 0,$ $\ln P + \gamma\ln V = \text{const},$ $\boxed{\ PV^\gamma = \text{const}\ }$

Other useful forms:

$TV^{\gamma - 1} = \text{const},\quad T^\gamma P^{1-\gamma} = \text{const}.$

Work done:

$W_\text{adi} = \int P\,dV = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}.$

Since $Q=0$: $W = -\Delta U = nC_V(T_1 - T_2)$.

Isobaric ($P =$ const)

$W = P(V_2 - V_1) = nR(T_2 - T_1)$. $Q = nC_P\Delta T$. $\Delta U = nC_V\Delta T$.

Isochoric ($V =$ const)

$W = 0$. $Q = nC_V\Delta T = \Delta U$.

Comparison Table

Adiabatic vs isothermal: on a PV diagram an adiabat is steeper than an isotherm through the same point — for $V_2 > V_1$ the adiabatic drop is larger.

Worked Example — adiabatic

$2$ mol of an ideal monatomic gas at $300$ K, $1$ atm, is compressed adiabatically to half its volume. Final $T$? ($\gamma = 5/3$.)

$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300\,(2)^{2/3} = 300(1.587) = 476\ \text{K}.$

Worked Example — work in isothermal

1 mol ideal gas at $300$ K expands isothermally from $1$ L to $10$ L. Work?

$W = (1)(8.314)(300)\ln 10 = (2494)(2.303) = 5743\ \text{J}.$

Pitfalls

• "Adiabatic" doesn't mean "fast" automatically — it means no heat exchange. A perfectly insulated slow process is still adiabatic.
• The $\gamma$ in $PV^\gamma$ is $C_P/C_V$, not a fudge factor.
• For an ideal gas, $\Delta U = nC_V\Delta T$ always, regardless of process — not just isochoric.

12.7 Heat Engines; Refrigerators

Heat Engine

A device that converts heat (from a hot reservoir at $T_1$) partially into work, dumping the rest into a cold reservoir at $T_2$.

         T1 (hot)
           |
           v  Q1
        [ENGINE] ----> W
           |
           v  Q2
         T2 (cold)

Energy conservation: $W = Q_1 - Q_2$. Efficiency:

$\boxed{\ \eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}\ }$

Refrigerator / Heat Pump

Reverses the engine: input work pumps heat from cold reservoir to hot.

         T1 (hot)
           ^  Q1
           |
        [FRIDGE] <---- W
           ^
           |  Q2
         T2 (cold)

$Q_1 = Q_2 + W$.

Coefficient of Performance (COP):

• Refrigerator (we care about $Q_2$ extracted from cold side):

$\text{COP}_\text{R} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}.$

• Heat pump (we care about $Q_1$ delivered to warm side):

$\text{COP}_\text{HP} = \frac{Q_1}{W} = \text{COP}_\text{R} + 1.$

COP can be greater than 1 — that does not violate the first law; we move heat, not create energy.

Worked Example

An engine absorbs $1000$ J of heat per cycle and delivers $250$ J of work. Efficiency?

$\eta = \frac{250}{1000} = 25\%.$

Pitfalls

• Engine efficiency is dimensionless and always $< 1$ for real engines.
• COP > 1 is normal — typical household fridge has $\text{COP}\sim 3-4$.

12.8 Second Law of Thermodynamics

Two Statements

Kelvin-Planck: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work.

(Equivalently: no engine of $100\%$ efficiency.)

Clausius: No process is possible whose sole result is the transfer of heat from a cold body to a hot body.

(Equivalently: refrigerator needs external work.)

Equivalence

Suppose Kelvin-Planck violated → an engine $E$ converts all heat from $T_1$ into work, with no rejection. Couple it to an ordinary refrigerator using that work to pump heat from $T_2$ to $T_1$. Net effect: heat flows from cold to hot with no other change — violates Clausius. Reverse argument also holds. Hence the two statements are logically equivalent.

Worked Example

Why can't a ship engine extract heat from sea water and convert it entirely to motion, leaving the ocean slightly cooler? — Violates Kelvin-Planck: this would be a "single-reservoir engine" with 100% efficiency.

Pitfalls

• The second law forbids certain directions of spontaneous change; it does not forbid heat flowing cold→hot when work is supplied (that's a refrigerator).
• "Spontaneous" = no external work required.

12.9 Reversible & Irreversible Processes

Definitions

A reversible process can be reversed without any net change in the system and surroundings. Conditions:

1. Quasi-static — proceeds through a continuous sequence of equilibrium states (infinitely slow).
2. No dissipation — friction, viscosity, electrical resistance, etc. must be absent.

All real-world processes are irreversible to some extent: friction, finite temperature differences for heat flow, free expansion, inelastic collisions.

Examples

Pitfalls

• Quasi-static alone is not enough — frictionless is also required.
• Reversibility is a theoretical limit; real engines approach but never attain it.

12.10 Carnot Engine

The Cycle

A reversible engine operating between two temperatures $T_1$ (hot) and $T_2$ (cold) executes four reversible strokes:

1. A → B: Isothermal expansion at $T_1$ — gas absorbs $Q_1$ from hot reservoir, does work.
2. B → C: Adiabatic expansion — gas cools from $T_1$ to $T_2$.
3. C → D: Isothermal compression at $T_2$ — gas dumps $Q_2$ to cold reservoir.
4. D → A: Adiabatic compression — gas warms back to $T_1$.

         P
         ^
       A *  isotherm T1
         |\
         | \ adiabat
         |  \
       B *   \
         |    *  D
   adiab.|  / |
         | /  *  C
       (isotherm T2)
         |
         +---------------> V

Derivation of efficiency

Along the two isotherms:

$Q_1 = nRT_1\ln\frac{V_B}{V_A},\qquad Q_2 = nRT_2\ln\frac{V_C}{V_D}.$

Along the two adiabats:

$T_1 V_B^{\gamma-1} = T_2 V_C^{\gamma-1},\qquad T_1 V_A^{\gamma-1} = T_2 V_D^{\gamma-1}.$

Dividing: $V_B/V_A = V_C/V_D$. Hence

$\frac{Q_2}{Q_1} = \frac{T_2}{T_1},$

and

$\boxed{\ \eta_\text{Carnot} = 1 - \frac{T_2}{T_1}\ }$

(temperatures in kelvin).

Carnot's Theorem

(a) No heat engine working between two reservoirs can be more efficient than a Carnot engine between the same two reservoirs.

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

Proof of (a): if a hypothetical super-Carnot engine existed, coupling it to a Carnot refrigerator would create a perpetual-motion machine of the second kind — violating Kelvin-Planck.

Worked Example

A Carnot engine operates between $500$ K and $300$ K. Efficiency?

$\eta = 1 - \frac{300}{500} = 0.4 = 40\%.$

If $1000$ J of heat is absorbed at $T_1$, work done = $400$ J, heat rejected = $600$ J.

Worked Example — Carnot fridge COP

Domestic fridge at $T_2 = 273$ K, kitchen at $T_1 = 300$ K. Carnot COP?

$\text{COP} = \frac{T_2}{T_1 - T_2} = \frac{273}{27} = 10.1.$

(Real fridges achieve $\sim 3$, well below this ideal limit.)

Pitfalls

• Carnot efficiency is always $< 100\%$ because $T_2 > 0$ K.
• Even if $T_2 = 0$ K, the third law forbids reaching it — so 100% efficiency is unattainable, not just rare.
• Temperatures must be in kelvin; using Celsius gives nonsense.

Solved Problems

1. $1$ mol of ideal gas at $27^\circ$C is compressed adiabatically until its temperature rises to $127^\circ$C. Work done on the gas? ($\gamma=7/5$.)

$W_\text{by gas} = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{(1)(8.314)(300 - 400)}{0.4} = -2079\ \text{J}.$

Work on the gas = $+2079$ J.

2. A Carnot engine between $400$ K and $300$ K absorbs $400$ J per cycle. Work done?

$\eta = 1 - 300/400 = 0.25, \quad W = (0.25)(400) = 100\ \text{J}.$

3. $5$ mol of ideal monatomic gas isothermally expand at $300$ K from $V$ to $3V$. Heat absorbed?

$Q = W = nRT\ln 3 = (5)(8.314)(300)(1.0986) = 13700\ \text{J} = 13.7\ \text{kJ}.$

4. A diatomic ideal gas ($\gamma=1.4$) is suddenly compressed to $1/5$ of its initial volume. Ratio $T_2/T_1$ and $P_2/P_1$?

$T_2/T_1 = (V_1/V_2)^{\gamma-1} = 5^{0.4} = 1.903,$ $P_2/P_1 = (V_1/V_2)^\gamma = 5^{1.4} = 9.518.$

5. An engine of efficiency $40\%$ rejects $300$ J of heat per cycle to the sink. Heat absorbed from source?

$Q_2/Q_1 = 0.60 \Rightarrow Q_1 = 300/0.60 = 500\ \text{J}.$

Work $= 200$ J.

6. A refrigerator extracts $400$ J from inside per cycle, requiring $100$ J of input work. COP?

$\text{COP} = 400/100 = 4.$

7. Show free expansion of ideal gas leaves $T$ unchanged. — Free expansion: into vacuum, so $W=0$. Walls insulated: $Q=0$. First law: $\Delta U = 0$. For ideal gas $U$ depends only on $T$, so $\Delta T = 0$. $\checkmark$

JEE/NEET Edge Cases

• PV diagram area = work (cyclic). Clockwise loop = positive work (engine); counterclockwise = work on gas (refrigerator).
• Two processes between same end-points — work and heat differ; $\Delta U$ is the same. Classic JEE problem.
• Polytropic process: $PV^n =$ const generalises everything. $n=0$ isobaric; $n=1$ isothermal; $n=\gamma$ adiabatic; $n=\infty$ isochoric.
• Slope of adiabat vs isotherm: on PV plot, $\vert \text{slope}_\text{adi}\vert /\vert \text{slope}_\text{iso}\vert = \gamma$.
• Cyclic process through Q ≠ 0 and W ≠ 0; only $\Delta U = 0$.
• Sign of $\Delta U$ in adiabatic expansion: gas does work → loses internal energy → $\Delta U < 0$, $\Delta T < 0$. (Cooling.)
• Carnot is not the only reversible cycle — Stirling and Ericsson also achieve Carnot efficiency with appropriate regeneration.
• Refrigerator that "freezes" by extracting heat from inside: input work is added to the heat extracted, both delivered to outside — that's why the back of a fridge feels warm.
• Heat is a path function: writing "heat of the gas" is meaningless; "heat absorbed during process A→B" is meaningful.
• First-law sign sign-flip: NCERT $\Delta U = Q - W$ (W done by gas). IUPAC/chemistry $\Delta U = Q + W$ (W done on gas). Both correct in context.

Quick Recap

• Zeroth law $\to$ defines temperature.
• First law: $\Delta U = Q - W$ — energy conservation.
• $U$ is a state function; $Q$ and $W$ are path functions.
• $C_P - C_V = R$ (Mayer); $\gamma = C_P/C_V$.
• Isothermal: $W = nRT\ln(V_2/V_1)$, $\Delta U = 0$.
• Adiabatic: $PV^\gamma =$ const, $Q = 0$.
• Engine efficiency $\eta = 1 - Q_2/Q_1$; refrigerator COP $= Q_2/W$.
• Second law forbids 100%-efficient engine (Kelvin-Planck) and spontaneous cold-to-hot flow (Clausius).
• Carnot is the upper bound: $\eta_\text{max} = 1 - T_2/T_1$.
• All reversible engines between the same two reservoirs have identical efficiency, independent of working fluid.

Formula Sheet