Thermodynamics

This chapter is a single continuous argument, and it helps to see the whole climb before starting.

• Thermal equilibrium and the zeroth law of thermodynamics build the very idea of temperature from scratch, using nothing but the notion of an insulating (adiabatic) wall versus a conducting (diathermic) wall.
• Heat, internal energy and work are carefully separated — a distinction the chapter insists is not mere pedantry but essential to getting problems right.
• The first law of thermodynamics assembles these three quantities into one conservation statement, $\Delta Q = \Delta U + \Delta W$.
• Specific heat capacity and molar specific heat capacity are defined, along with the relation $C_p - C_v = R$ for an ideal gas and the Dulong-Petit rule $C = 3R$ for solids.
• Thermodynamic state variables, the extensive-intensive distinction, and the equation of state $PV = \mu RT$ describe what "the state of a gas" actually means.
• Thermodynamic processes — quasi-static, isothermal, adiabatic, isochoric, isobaric and cyclic — are catalogued with the work done and heat exchanged in each.
• The second law of thermodynamics, stated by Kelvin-Planck and by Clausius, explains why heat engines can never be perfectly efficient and refrigerators never perfectly free.
• Reversible and irreversible processes are distinguished, leading to the Carnot engine and the universal ceiling on heat-engine efficiency, $\eta = 1 - T_2/T_1$.

Thermal equilibrium and the zeroth law

Equilibrium in mechanics means zero net force and torque. Thermodynamic equilibrium means something different: a system is in an equilibrium state if its macroscopic variables — pressure, volume, temperature, mass, composition — do not change with time. Whether a system reaches such a state depends on the wall separating it from its surroundings. An adiabatic wall is an insulating wall that blocks any flow of heat; a diathermic wall is a conducting wall that allows heat to pass. Put two gases A and B in contact through an adiabatic wall and any pair of states $(P_A, V_A)$ and $(P_B, V_B)$ can coexist unchanged. Replace that wall with a diathermic one, and the macroscopic variables of both systems drift spontaneously until they stop changing — the two systems have reached thermal equilibrium with each other. What single property is shared by two systems in thermal equilibrium? Experience says: their temperatures are equal. This is formalised by the zeroth law of thermodynamics: if two systems A and B are each separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. Symbolically, if $T_A = T_C$ and $T_B = T_C$, then $T_A = T_B$. R. H. Fowler stated this law in 1931, long after the first and second laws had already been numbered — hence the awkward "zeroth" — but logically it comes first, because it is what lets us define temperature as a single number that every thermometer, however built, can agree on.

Heat, internal energy and work — three quantities that must never be confused

Internal energy $U$ of a system is the sum of the kinetic and potential energies of all its molecules, measured in the frame where the centre of mass of the system is at rest — the ordered kinetic energy of the system moving as a whole (a bullet's speed, say) is never part of $U$; only the disordered, random molecular motion counts. Firing a bullet changes its mechanical (kinetic) energy, not its temperature; only when friction stops the bullet does that ordered energy degrade into disordered internal energy, raising the temperature of the bullet and the wood. Crucially, $U$ is a state variable: it depends only on the gas's current pressure, volume and temperature, never on the path by which that state was reached. Heat and work are different in kind. They are not properties a system possesses in a given state; they are the two modes of energy transfer by which $U$ changes. Heat is energy transferred because of a temperature difference between system and surroundings; work is energy transferred by any other means — pushing a piston down, or letting the gas push it out. It is meaningless to say "a gas has a certain amount of heat," exactly as it would be to say it has "a certain amount of work" — but it is perfectly meaningful to say a gas has a certain internal energy, or that heat was supplied to it, or that it did work. Confusing heat with internal energy is the single most common error in this chapter.

The first law of thermodynamics

Let $\Delta Q$ be the heat supplied to a system, $\Delta W$ the work done by the system on its surroundings, and $\Delta U$ the resulting change in internal energy. Energy conservation, applied to a thermodynamic system, reads:

where: $\Delta Q$ is positive when heat flows in, $\Delta W$ is positive when the system does work on its surroundings (as in expansion), and $\Delta U$ is the net change in internal energy. The physical reading is simple bookkeeping: heat fed into a system is spent two ways — part raises its internal energy, the rest leaves as work on the outside world. Because $U$ is a state variable, $\Delta U$ depends only on the initial and final states, whatever route the system takes between them; $\Delta Q$ and $\Delta W$ individually depend on the path, but the combination $\Delta Q - \Delta W$ never does. For a gas in a cylinder with a movable piston, pushing the piston out by $\Delta V$ against constant external pressure $P$ does work $\Delta W = P\Delta V$, so Eq. (11.1) becomes $\Delta Q = \Delta U + P\Delta V$. NCERT's own illustration: turning 1 g of water to steam at atmospheric pressure absorbs latent heat $\Delta Q = 2256\ \text{J}$; the volume swells from $1\ \text{cm}^3$ to $1671\ \text{cm}^3$, so the work done against $P = 1.013\times10^5\ \text{Pa}$ is $\Delta W = 1.013\times10^5\times1671\times10^{-6} = 169.2\ \text{J}$, leaving $\Delta U = 2256 - 169.2 = 2086.8\ \text{J}$ — most of the heat raises internal energy, separating the water molecules, and only a small share does mechanical work. This example is worked in full in the Worked examples section below.

Specific heat capacity

If heat $\Delta Q$ raises a substance's temperature by $\Delta T$, its heat capacity is $S = \Delta Q/\Delta T$. Dividing by the amount present gives quantities independent of quantity: the specific heat capacity

per unit mass ($\text{J kg}^{-1}\text{K}^{-1}$), and the molar specific heat capacity

per mole ($\text{J mol}^{-1}\text{K}^{-1}$), where $m$ is mass in kg and $\mu$ is the number of moles. Both depend on the substance, its temperature, and — for gases — the condition under which heat is supplied. For a solid of $N$ atoms, the equipartition of energy predicts average vibrational energy $3k_BT$ per atom, giving molar heat capacity $C = 3R$, the Dulong-Petit rule, which agrees well with measurement at ordinary temperature (failing at low temperature, and for carbon). Water's specific heat is $s = 4186\ \text{J kg}^{-1}\text{K}^{-1}$, and the calorie is defined as the heat needed to raise 1 g of water from $14.5\,^\circ\text{C}$ to $15.5\,^\circ\text{C}$, with $1\ \text{cal} = 4.186\ \text{J}$. For gases, $C_v$ (constant volume) and $C_p$ (constant pressure) differ because heating at constant pressure must also supply the work of expansion. The relation $C_p - C_v = R$ is one of the chapter's key derivations, given in full below.

State variables, extensive and intensive quantities, and the equation of state

An equilibrium state of a gas is fully specified by macroscopic state variables: pressure, volume, temperature, mass and composition. Not every state a system passes through is an equilibrium state — free expansion into a vacuum, or an explosive reaction, passes through states with no well-defined uniform pressure or temperature, states that cannot be plotted on a $P$-$V$-$T$ surface at all. The relation linking the state variables of an equilibrium state is its equation of state; for an ideal gas, $PV = \mu RT$, with $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$. For a fixed amount of gas this leaves only two independent variables, and a plot of $P$ against $V$ at fixed $T$ is an isotherm. State variables are of two kinds. Extensive variables scale with system size — $U$, $V$, total mass $M$ all halve if the system is split into equal parts. Intensive variables do not — $P$, $T$, density $\rho$ stay the same in each half. This is a useful consistency check: in $\Delta Q = \Delta U + P\Delta V$ every term is extensive, since intensive $P$ times extensive $\Delta V$ is extensive.

Thermodynamic processes

A process that changes a gas's state slowly enough that the system stays, at every instant, arbitrarily close to thermal and mechanical equilibrium is called quasi-static — meaning nearly static; pressure and temperature differ from the surroundings only infinitesimally at every stage. Strictly it takes infinite time, but sufficiently slow real processes approximate it well. All named processes below are quasi-static unless stated otherwise.

• Isothermal — $T$ fixed throughout, usually via contact with a large reservoir. $PV = \text{constant}$ (Boyle's law), and since $U$ of an ideal gas depends only on $T$, $\Delta U = 0$, so $\Delta Q = \Delta W$: heat absorbed converts entirely into work (or, in compression, work done on the gas releases entirely as heat).
• Adiabatic — thermally insulated, $\Delta Q = 0$. Work done by the gas comes entirely at the expense of internal energy, so adiabatic expansion cools the gas and compression heats it. Governed by $PV^\gamma = \text{constant}$, with $\gamma = C_p/C_v$.
• Isochoric — $V$ fixed, so $\Delta W = 0$ and all heat supplied changes internal energy (and temperature), governed by $C_v$.
• Isobaric — $P$ fixed. $\Delta W = P(V_2 - V_1) = \mu R(T_2 - T_1)$; the heat absorbed splits between raising $U$ and doing work, governed by $C_p$.
• Cyclic — the system returns to its initial state, so $\Delta U = 0$ over the cycle, and total heat absorbed equals total work done.

The second law of thermodynamics

The first law permits many processes that are never actually seen — a hot table could, in principle, cool itself and convert that lost internal energy into the mechanical energy of a book that hops off it, with energy balancing perfectly, yet nobody has ever watched this happen. Something beyond energy conservation forbids it: the second law of thermodynamics, which sets a fundamental ceiling on heat-engine efficiency and refrigerator performance. Two equivalent statements summarise it.

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

Clausius statement: no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Neither statement forbids converting heat into work, or moving heat from cold to hot — engines and refrigerators do exactly this every day. What is forbidden is doing so as the sole result, with nothing else in the universe changing: a refrigerator moves heat from cold to hot only by also consuming external work; an engine converts heat to work only by also dumping some leftover heat into a cold sink. The two statements can be proved logically equivalent — violating one lets you construct a device that violates the other.

Reversible and irreversible processes, and the Carnot engine

A process from state $i$ to state $f$ is reversible if it can be run backwards so that both the system and its surroundings return exactly to their original states, with no other change anywhere. Nearly every real process is irreversible: a hot vessel's base cools by spreading heat to its cooler rim, never the reverse; gas released into a vacuum diffuses to fill the container and never un-diffuses; a stirred liquid converts work into heat via friction, and that heat never spontaneously turns back into motion. Irreversibility has two roots — passing through non-equilibrium states, and dissipative effects such as friction and viscosity, present to some degree in essentially every real process. A process is reversible only if it is both quasi-static and free of dissipation, such as the slow, frictionless isothermal expansion of a gas. Reversibility matters because the second law rules out a perfect (100%-efficient) heat engine but leaves open the best possible efficiency between two temperatures. Sadi Carnot answered this in 1824. A reversible engine operating between a hot reservoir at $T_1$ and cold reservoir at $T_2$ is a Carnot engine, and its cycle consists of four reversible steps: isothermal expansion at $T_1$ absorbing heat $Q_1$, adiabatic expansion cooling the gas to $T_2$, isothermal compression at $T_2$ releasing heat $Q_2$, and adiabatic compression restoring the gas to $T_1$. The full derivation of its efficiency, $\eta = 1 - T_2/T_1$, is below; no engine between the same two temperatures can ever beat it, and no reversible engine does better than another regardless of working substance — the formula depends only on the two temperatures.

Board level — work and change in internal energy when water turns to steam

One gram of water is converted to steam at atmospheric pressure. The measured latent heat of vaporisation is $2256\ \text{J/g}$. At atmospheric pressure, $1\ \text{g}$ of water occupies $1\ \text{cm}^3$ as a liquid and $1671\ \text{cm}^3$ as vapour, and atmospheric pressure is $P = 1.013\times10^5\ \text{Pa}$. Find the work done by the expanding steam and the resulting change in internal energy.

Step 1. Identify the heat supplied. The latent heat is the heat absorbed at constant pressure, so $\Delta Q = 2256\ \text{J}$ for this 1 g sample.

Step 2. Compute the work done against the atmosphere. The volume change is $\Delta V = V_g - V_l = 1671\ \text{cm}^3 - 1\ \text{cm}^3 = 1670\ \text{cm}^3 \approx 1671\times10^{-6}\ \text{m}^3$ (NCERT's own calculation keeps the full vapour volume, since the liquid volume is negligible in comparison). Then $\Delta W = P\Delta V = 1.013\times10^5 \times 1671\times10^{-6} = 169.2\ \text{J}$.

Step 3. Apply the first law to find $\Delta U$. From $\Delta Q = \Delta U + \Delta W$, $\Delta U = \Delta Q - \Delta W = 2256 - 169.2 = 2086.8\ \text{J}$. Answer: the steam does $169.2\ \text{J}$ of work pushing back the atmosphere, and the internal energy of the water rises by $2086.8\ \text{J}$ — the great majority of the latent heat goes into separating the water molecules from each other, not into mechanical expansion work.

Board level — pressure rise when hydrogen gas is compressed adiabatically

A cylinder with a movable, thermally insulated piston contains 3 moles of hydrogen ($\gamma = 1.4$ for a diatomic gas) at standard temperature and pressure. The gas is compressed adiabatically to half its original volume. By what factor does the pressure increase?

Step 1. Identify the process: the cylinder and piston are insulators, so no heat enters or leaves — adiabatic, governed by $PV^\gamma = \text{constant}$, so $P_1V_1^\gamma = P_2V_2^\gamma$.

Step 2. Rearrange for the pressure ratio: $\dfrac{P_2}{P_1} = \left(\dfrac{V_1}{V_2}\right)^\gamma$. Here $V_2 = V_1/2$, so $\dfrac{P_2}{P_1} = 2^{1.4}$.

Step 3. Evaluate: $2^{1.4} = e^{1.4\ln 2} \approx 2.64$. Answer: the pressure increases by a factor of about $2.64$. A quick check with $TV^{\gamma-1} = \text{constant}$ gives a temperature rise by a factor $2^{0.4} \approx 1.32$ — consistent with the first law, since work done on the gas ($\Delta Q = 0$) must raise its internal energy.

Competitive level — efficiency, work and heat rejected by a Carnot engine

A Carnot engine operates between a hot reservoir at $T_1 = 500\ \text{K}$ and a cold reservoir at $T_2 = 300\ \text{K}$, absorbing $Q_1 = 840\ \text{J}$ of heat from the source in each cycle. Find (a) the efficiency, (b) the work done per cycle, and (c) the heat rejected to the sink per cycle.

Step 1. Efficiency from the reservoir temperatures alone: $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.4$, i.e. $40\%$. Note that only the ratio of absolute temperatures matters — reservoir temperatures must always be used in kelvin, never in Celsius, or the ratio (and hence the efficiency) comes out wrong.

Step 2. Work done per cycle: $W = \eta Q_1 = 0.4 \times 840 = 336\ \text{J}$.

Step 3. Heat rejected per cycle, from energy conservation over the cycle: $Q_2 = Q_1 - W = 840 - 336 = 504\ \text{J}$. As a cross-check, the universal Carnot relation $Q_2/Q_1 = T_2/T_1$ gives $Q_2 = 840\times(300/500) = 504\ \text{J}$, matching exactly. Answer: $\eta = 40\%$, $W = 336\ \text{J}$ per cycle, and $Q_2 = 504\ \text{J}$ rejected to the sink per cycle. Timing note: this should take well under two minutes. The chain is always the same — convert both temperatures to kelvin, compute $\eta = 1 - T_2/T_1$ first, then $W = \eta Q_1$, then $Q_2 = Q_1 - W$ (or cross-check with $Q_2/Q_1 = T_2/T_1$). The one trap under exam pressure is plugging Celsius values straight into the ratio; always glance at the units before dividing.

Treating heat as something a gas "has." Only internal energy is a state variable; heat and work are always transfers, described with phrases like "heat supplied to" or "work done by," never "heat contained in."

Getting the sign of work wrong. NCERT's convention takes $\Delta W$ as work done by the system, positive when the gas expands. Flipping this sign silently — a common slip when switching textbooks — turns $\Delta Q = \Delta U + \Delta W$ into the wrong equation.

Assuming $\Delta U = 0$ whenever a process looks "steady." That holds only for an isothermal process of an ideal gas ($U$ depends on $T$ alone) and for a full cyclic process ($U$ is a state variable) — not for isobaric, isochoric or adiabatic processes, where $\Delta U$ is generally nonzero.

Mixing up $PV = \text{constant}$ and $PV^\gamma = \text{constant}$. The first describes an isothermal process, the second an adiabatic one. Confusing them gives a P-V curve of the wrong steepness and a wrong final pressure or temperature.

Using Celsius temperatures in the Carnot formula. $\eta = 1 - T_2/T_1$ is a ratio of absolute temperatures. Using Celsius values changes the ratio and gives a badly wrong efficiency; always convert to kelvin first.

Thinking a refrigerator or a real engine violates the second law. Refrigerators do move heat from cold to hot, and engines do convert heat to work — the second law forbids only doing either as the sole result, with no other change (extra work input for the refrigerator; leftover heat rejected for the engine).

• State the zeroth law of thermodynamics and explain how it leads to the concept of temperature.
• Explain, in your own words, why heat and internal energy must never be treated as the same thing, and give an example (such as the bullet) that makes the distinction concrete.
• Write and apply the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$, correctly tracking signs for heat supplied, work done by the system, and change in internal energy.
• Distinguish specific heat capacity from molar specific heat capacity, and derive $C_p - C_v = R$ for an ideal gas.
• Identify whether a given variable (pressure, volume, temperature, internal energy, heat, work) is a state variable, and whether it is extensive or intensive.
• Recognise the five named thermodynamic processes from a description or a P-V diagram, and state what is zero (or constant) in each.
• Derive the work done by an ideal gas in an isothermal expansion and in an adiabatic process, and use $PV^\gamma = \text{constant}$ confidently.
• State the Kelvin-Planck and Clausius statements of the second law, and explain why they are equivalent.
• Explain what makes a process reversible, and why real processes are essentially always irreversible.
• Derive the Carnot engine's efficiency $\eta = 1 - T_2/T_1$ from the four steps of the Carnot cycle, and use it to compute work done and heat rejected.

Ask the tutor: to see how the P-V diagram for any of these five processes changes for a diatomic versus a monatomic gas, or to work through another Carnot-cycle numeric step by step.