Thermodynamics: The Science of What Cannot Be Undone

Why a New Chapter? The Limits of Calorimetry

The Thermal Properties chapter gave you powerful empirical tools. You can compute the heat absorbed by a block of iron: \(Q = mc\Delta T\). You can calculate the energy released during a phase transition: \(Q = mL\). You know that heat and temperature are not the same thing, and that the Stefan–Boltzmann law governs how a hot object radiates energy into space.

Those tools answer the question: how much energy is exchanged?

They cannot answer a deeper class of questions:

Can all of that energy be recovered as useful work?
If I run the process in reverse, do I get the original state back?
Is there a theoretical maximum efficiency for any engine operating between two given temperatures?
Why does a scrambled egg never unscramble itself?

These questions require a new framework—thermodynamics—whose central concepts are work done by a system, internal energy, and entropy. The framework is built on two laws. The First Law is an energy-accounting statement: the total energy of a system plus its surroundings is conserved. The Second Law is more subtle and more profound: it tells us which energy conversions nature actually permits, even among those that would balance the books.

Core intuition: Calorimetry vs. Thermodynamics

Calorimetry tells you how much energy moves as heat. Thermodynamics tells you what you can do with it and what you cannot undo. The First Law counts; the Second Law constrains. Together they set the absolute boundaries of any engine, refrigerator, or spontaneous process.

Core intuition: Chapter Roadmap

We define the vocabulary of thermodynamic systems and state variables and state the Zeroth and First Laws. The First Law is given computational weight through work integrals for specific processes. The Second Law enters via entropy and irreversibility, and its most important consequence is the Carnot efficiency theorem. Refrigerators, heat pumps, and a deeper look at entropy complete the physics. The chapter closes with a summary that resolves every Perplexing Question.

Thermodynamic Systems and State Variables

Before writing a single equation, we need a precise vocabulary. Thermodynamics is exceptionally careful about the words it uses, and nearly every beginner error traces back to a vocabulary slip.

System and Surroundings

The system is the portion of the universe we choose to study—a specific quantity of gas in a cylinder, a chemical mixture in a beaker, the working fluid in a turbine. Everything outside the system is the surroundings. The boundary between them may be real (a cylinder wall) or imaginary (an invisible surface enclosing a parcel of air).

Systems are classified by how their boundary interacts with the surroundings:

Open system: both matter and energy can cross the boundary. (A boiling pot with the lid off.)
Closed system: energy can cross but matter cannot. (Gas in a sealed cylinder with a moveable piston. This is the default system in this chapter.)
Isolated system: neither matter nor energy crosses. (A perfectly insulated, rigid container—an idealisation of central importance in the Second Law.)

State Variables and Equilibrium

A thermodynamic state is a complete description of the system at one instant. For an ideal gas, the state is fully specified by three quantities: pressure \(p\), volume \(V\), and temperature \(T\). These are state variables: their values depend only on the current state, not on the history of how the system arrived there.

This last property is crucial and worth stating separately.

Core intuition: State vs. Path: the Central Distinction

A state variable has a definite value at each equilibrium state. Its change between two states, \(\Delta X = X_{\text{final}} - X_{\text{initial}}\), is the same regardless of the path taken. Heat \(Q\) and work \(W\), by contrast, are path functions: they depend on precisely how the process is carried out, not just on the endpoints. Saying “the gas has \(Q = 500\;\mathrm{J}\)” is as meaningless as saying “the gas has \(5\;\mathrm{km}\) of journey in it.” Heat and work are transfers, not possessions.

Other common state variables include internal energy \(U\), enthalpy \(H\), and entropy \(S\). We will introduce \(U\) and \(S\) in the sections that follow.

Thermodynamic Processes and Quasi-static Idealization

A thermodynamic process is a change from one equilibrium state to another. In practice, real processes happen at finite speed and pass through non-equilibrium intermediate states where \(p\), \(V\), \(T\) are not well-defined everywhere.

For the purpose of drawing \(p\)-\(V\) diagrams and computing work integrals, we invoke the quasi-static approximation: the process is assumed to occur infinitely slowly, so the system passes through a continuous sequence of equilibrium states, each characterised by well-defined \(p\) and \(V\). Only quasi-static processes can be drawn as smooth curves on a \(p\)-\(V\) diagram; a sudden irreversible process (like a gas rushing into vacuum) can only be drawn as two points—the initial and final states—with no connecting curve.

Equation of State for an Ideal Gas

The state variables of an ideal gas are constrained by a single algebraic relationship. We will derive this from first principles in the Kinetic Theory chapter; for now we state it and use it: \[ pV = nRT, \] where \(n\) is the number of moles, \(R = 8.314\;\mathrm{J\,mol^{-1}K^{-1}}\) is the universal gas constant, and \(T\) is the absolute temperature in kelvin. (Recall that all thermodynamic formulas containing a standalone \(T\) require kelvin.)

This relation is the bridge between mechanics (where \(p\) and \(V\) appear naturally) and thermal physics (where \(T\) is central). Every standard process here is obtained by holding one combination of these variables constant.

Misconception alert

“Temperature is the amount of heat in a body.” Temperature is a state variable: it has a definite value at equilibrium and describes how energetically the microscopic constituents are moving. Heat is a process variable: it is energy crossing a boundary because of a temperature difference. A body at high temperature does not “contain” heat; it contains internal energy. It can transfer that energy as heat when placed in contact with a cooler body. Confusing the two leads directly to errors in applying the First Law.

✓ Checkpoint · Systems and State

A sealed, thermally insulated rigid box contains gas. After vigorous shaking, the gas is turbulent. Is this an equilibrium state? Once the turbulence dies out, which state variables change and which are restored to their original values?
Not an equilibrium state during shaking. After settling: \(p\), \(T\), and \(U\) are generally higher (turbulent KE converts to internal energy); \(V\) is unchanged; the state variables are now well-defined again.
Two gases are mixed irreversibly in a rigid, insulated container. Can you draw this mixing process as a curve on a \(p\)-\(V\) diagram? Explain why or why not.
No. Mixing is irreversible and passes through non-equilibrium states; only the initial and final equilibrium points can be plotted.
(Trap question) A student says: “I added \(200\;\mathrm{J}\) of heat to a gas, so the gas now contains \(200\;\mathrm{J}\) of heat.” Identify every error in this statement.
Two errors: (1) Heat is not a state function—it cannot be “contained.” (2) Some of the \(200\;\mathrm{J}\) may have left the gas as work; the internal energy increased by \(\Delta U = Q - W\), not by \(Q\).

The Zeroth Law and the First Law

◈ The first law — heat, work & internal energy

live bench

Add heat, let the gas do work, and watch the ledger balance: \(\Delta Q=\Delta U+\Delta W\). Internal energy is a state function; heat and work are not.

◈ loading the interactive bench…

↗ open in the Simulations page

Fig 14.1 The first law is energy bookkeeping. Heat \(Q\) added to the gas and work \(W\) done by it change the internal energy by \(\Delta U=Q-W\) — and \(U\) is a state function, so it forgets the path.

The Zeroth Law: What Temperature Really Means

The Zeroth Law is not a computational tool—it is the logical foundation of temperature measurement.

Zeroth Law of Thermodynamics: If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.

This sounds obvious. But its consequence is profound: it establishes that temperature is a property of a system that can be compared across systems. System C acts as a thermometer. Without the Zeroth Law, there would be no guarantee that two objects “at the same temperature” would exchange no heat when brought into contact. The Zeroth Law says they will not.

Internal Energy

Before stating the First Law we need the concept it is really about.

When a gas is in equilibrium, its molecules are in constant, random motion. They have kinetic energy of translation, rotation, and vibration. Between them, they have potential energy due to intermolecular forces. The total of all these microscopic energies is the internal energy \(U\) of the system.

Three facts about \(U\):

\(U\) is a state function: it depends only on the current equilibrium state.
For an ideal gas, intermolecular forces are zero, so \(U\) depends only on \(T\)—not on \(p\) or \(V\) separately. (We will derive \(U = \tfrac{f}{2}nRT\) in the Kinetic Theory chapter; for now, accept that \(\Delta U\) depends only on \(\Delta T\) for an ideal gas.)
\(U\) is extensive: double the amount of gas and \(U\) doubles.

The First Law

A system can exchange energy with its surroundings in two ways: as heat \(Q\) (energy flowing because of a temperature difference) and as work \(W\) (energy flowing because of a force acting through a displacement—here, a pressure acting through a volume change).

The First Law of Thermodynamics: \[ \boxed{\Delta U = Q - W} \]

In words: the increase in internal energy equals the heat absorbed by the system minus the work done by the system.

Misconception alert

Physics convention vs. Chemistry convention. In this book (and in most physics curricula, including JEE and AP Physics): \[ W = \text{work done \emph{by} the gas}, \qquad \Delta U = Q - W. \] In chemistry texts: \[ W = \text{work done \emph{on} the gas}, \qquad \Delta U = Q + W. \] The underlying physics is identical; the sign of \(W\) is reversed. If the gas expands and does \(100\;\mathrm{J}\) of work on the piston, then \(W = +100\;\mathrm{J}\) in the physics convention and \(W = -100\;\mathrm{J}\) in the chemistry convention. Mixing the two conventions in a single calculation is a guaranteed source of sign errors. This book uses the physics convention throughout.

Birth of the formula: First Law from Energy Conservation

Consider a closed system (a gas in a cylinder) that absorbs a small quantity of heat \(\delta Q\) and simultaneously does a small amount of work \(\delta W\) on its surroundings. By conservation of energy, the energy entering the system as heat minus the energy leaving as work must appear as an increase in the system's internal energy: \[ dU = \delta Q - \delta W. \] Integrated over a finite process from state 1 to state 2: \[ \Delta U = U_2 - U_1 = Q - W. \] Three sign conventions follow immediately:

\(Q \gt 0\): heat flows into the system.
\(Q \lt 0\): heat flows out of the system.
\(W \gt 0\): the system does work on the surroundings (expansion).
\(W \lt 0\): the surroundings do work on the system (compression).

Note that \(dU\) is an exact differential (it integrates to a path-independent \(\Delta U\)), whereas \(\delta Q\) and \(\delta W\) are inexact differentials (path-dependent). This is the mathematical expression of the fact that \(U\) is a state function but \(Q\) and \(W\) are not.

Connection to Chapter 13: Latent Heat as the First Law

When a solid melts at constant temperature and pressure, no work is done (the volume change of a solid-to-liquid transition is negligible): \(W \approx 0\), so \(\Delta U = Q = mL_f\). All the heat absorbed goes directly into increasing the internal energy—breaking intermolecular bonds.

For a liquid vaporising at pressure \(p\), the volume change is substantial: the work done by the expanding vapour against atmospheric pressure is \(W = p\,\Delta V\), and the latent heat of vaporisation \(L_v\) must supply both this work and the increase in internal energy: \(Q = mL_v = \Delta U + p\,\Delta V\). Latent heat, in other words, is simply the First Law applied at a phase transition.

Sign Drill: A First Example

Example 1

A gas absorbs \(450\;\mathrm{J}\) of heat and simultaneously does \(180\;\mathrm{J}\) of work by expanding. Find \(\Delta U\). If the same amount of heat is added to the gas but the gas is kept in a rigid container so it cannot expand, find \(\Delta U\) in that case.

Solution. Case 1: \(Q = +450\;\mathrm{J}\), \(W = +180\;\mathrm{J}\). \[ \Delta U = Q - W = 450 - 180 = +270\;\mathrm{J}. \] Internal energy increases by \(270\;\mathrm{J}\). Case 2: \(Q = +450\;\mathrm{J}\), rigid container so \(W = 0\). \[ \Delta U = Q - W = 450 - 0 = +450\;\mathrm{J}. \] All the heat goes into internal energy. The temperature rise is larger in Case 2 than in Case 1 because no energy leaks out as work—this is the physical basis for the difference between \(C_P\) and \(C_V\), which we will quantify shortly.

Example 2

In one complete cycle of a heat engine, the working gas absorbs \(800\;\mathrm{J}\) of heat from a hot reservoir, does \(300\;\mathrm{J}\) of work, and rejects heat to a cold reservoir. (a) Find the heat rejected. (b) Find \(\Delta U\) for the full cycle.

Solution. (b) In a complete cycle the system returns to its initial state, so all state variables are restored: \(\Delta U_{\text{cycle}} = 0\). (a) From the First Law over the cycle: \(0 = Q_{\text{net}} - W_{\text{net}}\), so \(Q_{\text{net}} = W_{\text{net}} = 300\;\mathrm{J}\). The gas absorbed \(Q_{\text{in}} = 800\;\mathrm{J}\) and rejected \(Q_{\text{out}}\), so: \[ Q_{\text{net}} = Q_{\text{in}} - Q_{\text{out}} = 800 - Q_{\text{out}} = 300\;\mathrm{J} \implies Q_{\text{out}} = 500\;\mathrm{J}. \] The engine rejects \(500\;\mathrm{J}\) to the cold reservoir. Efficiency \(\eta = W/Q_{\text{in}} = 300/800 = 37.5\%\).

✓ Checkpoint · Zeroth and First Laws

A gas is compressed, and \(200\;\mathrm{J}\) of work is done on it by the piston, while \(80\;\mathrm{J}\) of heat escapes to the surroundings. Using the physics convention, find \(\Delta U\).
\(W = -200\;\mathrm{J}\) (done by gas), \(Q = -80\;\mathrm{J}\) (heat out); \(\Delta U = Q - W = -80 - (-200) = +120\;\mathrm{J}\).
Over a complete thermodynamic cycle, a system does \(500\;\mathrm{J}\) of net work. What is the net heat absorbed by the system?
\(\Delta U = 0\) for any complete cycle, so \(Q_{\text{net}} = W_{\text{net}} = 500\;\mathrm{J}\).
(Trap question) A student argues: “If I insulate a gas perfectly and compress it, \(Q = 0\), so by \(\Delta U = Q - W\) the internal energy cannot change.” What is the error?
Compression means the surroundings do work on the gas, so \(W \lt 0\) in the physics convention. \(\Delta U = 0 - W = -W \gt 0\): internal energy (and temperature) increase. This is adiabatic heating, the mechanism behind diesel ignition.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · First law: bookkeeping and sign convention

A monatomic ideal gas absorbs \(600\,\mathrm{J}\) of heat while expanding and doing \(250\,\mathrm{J}\) of work on its surroundings. (a) Find the change in internal energy. (b) Does the gas temperature rise or fall?

Solution. Given: \(Q = +600\,\mathrm{J}\) (heat into gas); work done by the gas \(W = +250\,\mathrm{J}\).
Find: \(\Delta U\), and the sign of \(\Delta T\).
Setup: The first law in the convention “\(W\) done by the gas” is \(\Delta U = Q - W\). For an ideal gas \(U = U(T)\) only, so the sign of \(\Delta U\) fixes the sign of \(\Delta T\).
Solve: \[ \Delta U = Q - W = 600 - 250 = 350\,\mathrm{J}. \] Since \(\Delta U \gt 0\) and \(U = nC_V T\), the temperature rises.
Answer: \(\boxed{\Delta U = +350\,\mathrm{J};\ \text{temperature rises}}\)
Check: Energy accounting: the gas takes in \(600\,\mathrm{J}\), spends \(250\,\mathrm{J}\) pushing its surroundings, and must keep the remaining \(350\,\mathrm{J}\) as internal energy. ✓

Example 2 · Isobaric work

Two moles of an ideal gas are heated at a constant pressure of \(2.0\times10^{5}\,\mathrm{Pa}\), expanding from \(0.010\,\mathrm{m^3}\) to \(0.025\,\mathrm{m^3}\). Find the work done by the gas.

Solution. Given: \(n = 2\,\mathrm{mol}\), \(P = 2.0\times10^{5}\,\mathrm{Pa}\) (constant), \(V_1 = 0.010\,\mathrm{m^3}\), \(V_2 = 0.025\,\mathrm{m^3}\).
Find: \(W\) done by the gas.
Setup: At constant pressure the area under the isobar is a rectangle, so \(W = P\,\Delta V\).
Solve: \[ W = P\,\Delta V = (2.0\times10^{5})(0.025 - 0.010) = (2.0\times10^{5})(0.015) = 3000\,\mathrm{J}. \] Answer: \(\boxed{W = 3.0\times10^{3}\,\mathrm{J}}\)
Check: Via temperatures: at constant \(P\), \(W = nR\,\Delta T\), and \(\Delta T = P\,\Delta V/(nR) = 3000/(2\times8.314) = 180.4\,\mathrm{K}\), so \(W = nR\,\Delta T = 2(8.314)(180.4) = 3000\,\mathrm{J}\). ✓

Example 3 · Isochoric heating

A rigid sealed vessel holds \(3\,\mathrm{mol}\) of a diatomic ideal gas \(\left(C_V = \tfrac{5}{2}R\right)\). It absorbs \(1500\,\mathrm{J}\) of heat at constant volume. Find (a) the work done, (b) the change in internal energy, and (c) the temperature rise.

Solution. Given: \(n = 3\,\mathrm{mol}\), diatomic \(C_V = \tfrac{5}{2}R = 20.79\,\mathrm{J\,mol^{-1}K^{-1}}\), \(Q = +1500\,\mathrm{J}\), rigid vessel.
Find: \(W\), \(\Delta U\), \(\Delta T\).
Setup: Rigid vessel \(\Rightarrow \Delta V = 0 \Rightarrow W = 0\); all heat raises internal energy, \(\Delta U = nC_V\,\Delta T\).
Solve: \[\begin{aligned} W &= 0, \qquad \Delta U = Q - W = 1500\,\mathrm{J},\\ \Delta T &= \frac{\Delta U}{nC_V} = \frac{1500}{3(20.79)} = \frac{1500}{62.36} = 24.1\,\mathrm{K}. \end{aligned}\] Answer: \(\boxed{W = 0,\quad \Delta U = 1500\,\mathrm{J},\quad \Delta T = 24.1\,\mathrm{K}}\)
Check: Direct route: for an isochoric process \(Q = nC_V\,\Delta T\), giving \(\Delta T = 1500/62.36 = 24.1\,\mathrm{K}\) with no separate appeal to the first law. ✓

Example 4 · Specific heats and \(\gamma\) from degrees of freedom

A diatomic ideal gas has \(f = 5\) active degrees of freedom (three translational, two rotational). Find \(C_V\), \(C_P\), \(\gamma\), and verify \(C_P - C_V = R\).

Solution. Given: diatomic gas, \(f = 5\).
Find: \(C_V\), \(C_P\), \(\gamma\), \(C_P - C_V\).
Setup: Equipartition gives \(C_V = \tfrac{f}{2}R\); Mayer's relation gives \(C_P = C_V + R\); and \(\gamma = C_P/C_V\).
Solve: \[\begin{aligned} C_V &= \tfrac{5}{2}R = 2.5(8.314) = 20.79\,\mathrm{J\,mol^{-1}K^{-1}},\\ C_P &= C_V + R = \tfrac{7}{2}R = 29.10\,\mathrm{J\,mol^{-1}K^{-1}},\\ \gamma &= \frac{C_P}{C_V} = \frac{7}{5} = 1.40, \qquad C_P - C_V = R = 8.314\,\mathrm{J\,mol^{-1}K^{-1}}. \end{aligned}\] Answer: \(\boxed{C_V = \tfrac{5}{2}R,\ C_P = \tfrac{7}{2}R,\ \gamma = 1.40,\ C_P - C_V = R}\)
Check: Independent formula: \(\gamma = 1 + 2/f = 1 + 2/5 = 1.40\), matching \(C_P/C_V\). ✓

Example 5 · Isothermal expansion

One mole of an ideal gas at \(T = 300\,\mathrm{K}\) expands isothermally and reversibly from \(2.0\,\mathrm{L}\) to \(6.0\,\mathrm{L}\). Find \(W\), \(Q\), and \(\Delta U\).

Solution. Given: \(n = 1\,\mathrm{mol}\), \(T = 300\,\mathrm{K}\) (constant), \(V_1 = 2.0\,\mathrm{L}\), \(V_2 = 6.0\,\mathrm{L}\).
Find: \(W\), \(Q\), \(\Delta U\).
Setup: Isothermal, ideal gas \(\Rightarrow \Delta U = 0\). Reversible isothermal work is \(W = nRT\ln(V_2/V_1)\); the first law then gives \(Q = W\).
Solve: \[ W = nRT\ln\frac{V_2}{V_1} = (1)(8.314)(300)\ln 3 = (2494.2)(1.0986) = 2740\,\mathrm{J}. \] \(\Delta U = 0\), so \(Q = W = 2740\,\mathrm{J}\) (heat absorbed).
Answer: \(\boxed{W = Q = 2.74\times10^{3}\,\mathrm{J},\quad \Delta U = 0}\)
Check: Consistency / magnitude: with \(\Delta U = 0\) the first law forces \(Q = W\); numerically \(nRT \approx 2500\,\mathrm{J}\) and \(\ln 3 \approx 1.1\), so \(W \approx 2.7\,\mathrm{kJ}\), the right order. ✓

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

First-Law Signs
A gas absorbs \(350\;\mathrm{J}\) of heat and does \(120\;\mathrm{J}\) of work on its surroundings. Find \(\Delta U\). In a second process, \(280\;\mathrm{J}\) of work is done on the gas while \(90\;\mathrm{J}\) of heat escapes. Find \(\Delta U\) for the second process.
Convention \(\Delta U = Q - W\) (\(W\) done by gas). Process 1: \(\Delta U = 350 - 120 = +230\;\mathrm{J}\). Process 2: \(W = -280\;\mathrm{J}\), \(Q = -90\;\mathrm{J}\), so \(\Delta U = -90 - (-280) = +190\;\mathrm{J}\).
Isochoric Heating
Two moles of a monatomic ideal gas are heated at constant volume from \(300\;\mathrm{K}\) to \(500\;\mathrm{K}\). Find \(Q\), \(W\), and \(\Delta U\).
\(W = 0\); \(\Delta U = nC_V\Delta T = 2\times\tfrac{3}{2}R\times200 = 600R = 4988\;\mathrm{J}\); \(Q = \Delta U = 4988\;\mathrm{J}\).
Isobaric Expansion
One mole of a diatomic ideal gas expands isobarically from \(250\;\mathrm{K}\) to \(450\;\mathrm{K}\). Find \(Q\), \(W\), \(\Delta U\), and the ratio \(W/Q\).
\(\Delta U = \tfrac{5}{2}R\times200 = 500R \approx 4157\;\mathrm{J}\); \(W = R\times200 = 200R \approx 1663\;\mathrm{J}\); \(Q = \tfrac{7}{2}R\times200 = 700R \approx 5820\;\mathrm{J}\); \(W/Q = 2/7\).
Rectangular Cycle on a \(p\)–\(V\) Diagram
One mole of a monatomic ideal gas runs the cycle A\((p_0, V_0) \to\) B\((p_0, 3V_0) \to\) C\((2p_0, 3V_0) \to\) D\((2p_0, V_0) \to\) A, with the A\(\to\)B and C\(\to\)D legs isobaric and the B\(\to\)C and D\(\to\)A legs isochoric. Find \(Q\), \(W\), \(\Delta U\) for each leg and the net work, and verify \(\Delta U_{\text{cycle}} = 0\).
\(T_A = p_0V_0/R\), \(T_B = 3p_0V_0/R\), \(T_C = 6p_0V_0/R\), \(T_D = 2p_0V_0/R\). AB: \(W = 2p_0V_0\), \(\Delta U = 3p_0V_0\), \(Q = 5p_0V_0\). BC: \(W = 0\), \(\Delta U = \tfrac{9}{2}p_0V_0 = Q\). CD: \(W = -4p_0V_0\), \(\Delta U = -6p_0V_0\), \(Q = -10p_0V_0\). DA: \(W = 0\), \(\Delta U = -\tfrac{3}{2}p_0V_0 = Q\). \(W_{\text{net}} = 2p_0V_0 - 4p_0V_0 = -2p_0V_0\) (net work done on the gas: this is an anticlockwise cycle). \(\sum\Delta U = 3 + \tfrac{9}{2} - 6 - \tfrac{3}{2} = 0\). ✓
Entropy Is a State Function: Two Routes
One mole of a monatomic ideal gas goes from \((300\;\mathrm{K}, V)\) to \((400\;\mathrm{K}, 2V)\). Compute \(\Delta S\) by (a) isothermal-then-isochoric and (b) isochoric-then-isothermal, and show they agree.
(a) Isothermal (\(V\to2V\)): \(R\ln2\); isochoric (\(300\to400\)): \(\tfrac32 R\ln\tfrac{400}{300}\). (b) Same two pieces in the other order. Both give \(\Delta S = R\ln2 + \tfrac32 R\ln\tfrac43 = 5.76 + 3.59 = 9.35\;\mathrm{J\,K^{-1}}\), independent of path—as a state function must be.

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

Work in a \(p\propto V^2\) Process
An ideal gas follows \(p = p_0(V/V_0)^2\) from \((p_0, V_0)\) to \((8p_0, 2V_0)\). For one mole of monatomic gas, find (a) \(W\), (b) \(\Delta U\), (c) \(Q\).
(a) \(W = \displaystyle\int_{V_0}^{2V_0}\!p_0(V/V_0)^2\,dV = \tfrac{p_0}{V_0^2}\cdot\tfrac{(2V_0)^3-V_0^3}{3} = \tfrac{7}{3}p_0V_0\). (b) \(\Delta T = (p_fV_f - p_0V_0)/(nR) = (16-1)p_0V_0/R = 15p_0V_0/R\), so \(\Delta U = \tfrac{3}{2}R\cdot\tfrac{15p_0V_0}{R} = \tfrac{45}{2}p_0V_0\). (c) \(Q = \Delta U + W = \tfrac{45}{2}p_0V_0 + \tfrac{7}{3}p_0V_0 = \tfrac{149}{6}p_0V_0 \approx 24.8\,p_0V_0\).
Engine with a Sloped Return Leg
One mole of diatomic gas runs A\((p_0,V_0)\to\)B\((2p_0,V_0)\) isochorically, B\(\to\)C\((2p_0,2V_0)\) isobarically, then C\(\to\)A along a straight line on the \(p\)–\(V\) diagram. Find the net work and the efficiency \(\eta = W_{\text{net}}/Q_{\text{in}}\).
\(W_{AB}=0\); \(W_{BC}=2p_0V_0\); \(W_{CA}=\tfrac12(2p_0+p_0)(V_0-2V_0)=-\tfrac32p_0V_0\); \(W_{\text{net}}=\tfrac12p_0V_0\). The line C\(\to\)A is \(p=(p_0/V_0)V\), along which \(dQ = 6(p_0/V_0)V\,dV \lt 0\) throughout (heat is rejected on the whole leg), so heat enters only on AB and BC. \(Q_{AB}=\tfrac52 R(T_B-T_A)=\tfrac52 p_0V_0\); \(Q_{BC}=\tfrac72 R(T_C-T_B)=7p_0V_0\); \(Q_{\text{in}}=\tfrac{19}{2}p_0V_0\). \(\eta = \tfrac12 / \tfrac{19}{2} = 1/19 \approx 5.3\%\).
Entropy of Mixing and the Gibbs Paradox
One mole of gas A and one mole of gas B, each at temperature \(T\) and volume \(V\), are separated by a partition that is then removed. Find the entropy of mixing if the gases are different, and if they are identical.
Different gases: each expands into \(2V\) at constant \(T\), so \(\Delta S = R\ln2 + R\ln2 = 2R\ln2 \approx 11.5\;\mathrm{J\,K^{-1}}\). Identical gases: removing the partition produces no new macrostate, so \(\Delta S = 0\) (Gibbs paradox); the discontinuity is resolved by the indistinguishability of identical particles.
Testing an Efficiency Claim
An inventor claims an engine between a furnace at \(527\,{}^{\circ}\mathrm{C}\) and a cooler at \(27\,{}^{\circ}\mathrm{C}\) producing \(900\;\mathrm{J}\) from \(1500\;\mathrm{J}\) of heat input. Evaluate the claim.
\(T_H = 800\;\mathrm{K}\), \(T_C = 300\;\mathrm{K}\), so \(\eta_{\text{Carnot}} = 1 - 300/800 = 62.5\%\). Claimed \(\eta = 900/1500 = 60\% \lt 62.5\%\): thermodynamically permissible (though optimistic for a real engine). Above \(62.5\%\) it would violate the second law.