Thermodynamics: Class 11 Physics · Chapter 11

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

100 min
1 simulation
16 practice questions
34 MCQs

What you will learn in Thermodynamics

Explain thermal equilibrium and use the zeroth law to justify the idea of temperature
Distinguish internal energy, a state variable, from heat and work, which are energy in transit
Apply the first law of thermodynamics with correct signs, including ΔW = PΔV for constant pressure
Define specific and molar heat capacities, explain C = 3R for solids and derive Cp − Cv = R for an ideal gas
Describe quasi-static isothermal, adiabatic, isobaric, isochoric and cyclic processes and calculate W, Q and ΔU for each
State the Kelvin–Planck and Clausius forms of the second law and tell reversible from irreversible processes
Analyse the Carnot cycle, derive η = 1 − T2/T1 and explain why no engine between two reservoirs can beat it

Original companion text aligned to NCERT Physics Class 11, Chapter 11, pp. 226–243. Not affiliated with or endorsed by NCERT.

11.1 Introduction

A street knife-grinder pedals his stone wheel, presses a dull kitchen knife against it, and a spray of sparks flies off. A minute later the blade is too hot to hold. Work done by his legs has ended up warming the steel. A pressure cooker runs the other way: steam pushes up the heavy weight on the vent and sets it whistling, so energy that came in from the flame ends up doing work. Thermodynamics (ऊष्मागतिकी) is the physics of such exchanges between heat, work and the energy stored inside matter.

2/2Heat from the flame makes steam, and the steam does work lifting the weight on the vent.

Fig. 11.1Energy moves both ways between work and the internal energy of matter.

On the left a grinding wheel turns while a knife blade is pressed against it, with sparks flying from the contact point; the label reads work to hot blade. On the right a pressure cooker sits on a flame. Step 1 highlights the grinding wheel. Step 2 highlights the cooker: its weight lifts off the vent and steam escapes; the label reads heat to lifted weight.

For a long time heat was pictured as a substance. Eighteenth-century scientists spoke of caloric, a weightless fluid that soaked into bodies and seeped from hot ones into cold ones until both were equally charged with it. The idea handled mixing hot and cold water quite well. What it could not handle was friction, which seemed able to produce heat without limit.

The decisive observation came from an arsenal in Munich. In 1798 Benjamin Thompson, better known as Count Rumford, had a brass cannon barrel bored by a deliberately blunt drill turned by horses. The barrel stayed hot for as long as the horses kept working, hot enough to bring a tank of water around it to the boil. A blunt drill removes very little metal, so on the fluid picture it should have released very little caloric, and any stored supply should eventually run out. Neither prediction held: the heat simply tracked the work done by the horses. Heat, Rumford concluded, is not a stored fluid but a form of motion, and here work was being turned into it.

Timeline

From caloric to the laws of thermodynamics

1798
Rumford's cannon boring
Heat produced by friction keeps pace with the work done, not with any stored fluid.
1824
Sadi Carnot's engines
Carnot asks how much work a heat engine can possibly deliver (Section 11.11).
1840s
Mayer and Joule
Work and heat are shown to be interchangeable forms of energy; Joule measures the conversion with falling weights and paddle wheels.
1850–1851
Clausius and Kelvin
The second law is stated in two equivalent forms (Section 11.9).
1930s
A law numbered zero
Ralph Fowler names the most basic law of all the zeroth law (Section 11.3).

The laws in this chapter were found by nineteenth-century scientists, many of whom doubted that atoms were real, and the laws never needed them. Thermodynamics works with a handful of bulk quantities that ordinary instruments can read: pressure, volume, temperature, mass and composition. Compare that with a molecule-by-molecule account of a single litre of air, which would have to follow about $2.5 \times 10^{22}$ particles. Kinetic theory, in the next chapter, sits in between: it uses molecules, but only their average behaviour.

The two subjects also ask different questions. Mechanics follows a body as a whole: where its centre of mass goes and how it turns under forces and torques. Thermodynamics looks inside: how hot the body is, how compressed, how much energy is tied up in the jiggling of its molecules. How fast a cricket ball leaves the bat is a question for mechanics. How much the ball and the wicketkeeper's gloves warm up when it is caught is a question for thermodynamics.

Quick check

1 of 2

What did Rumford's cannon-boring observations show?

the heat produced kept pace with the work donea blunt drill released the most stored caloricbrass stores more heat than iron doesheat flows from cold bodies to hot ones

Show answer

Answer (A) the heat produced kept pace with the work done

Heat kept appearing for as long as the horses worked, with no sign of a stored supply running out. That pointed to heat as a form of energy produced from work, not a fluid squeezed out of the metal.

11.2 Thermal equilibrium

A sealed bottle of water that has stood in a room since yesterday is in a very boring condition: check its pressure, volume or temperature at any moment and you read the same values. Thermodynamics calls this a state of thermodynamic equilibrium, one whose bulk variables (pressure, volume, temperature, mass and composition) no longer change with time. The word means something different from the balance of forces and torques in mechanics; a book lying on a table is in mechanical equilibrium whether it is warm or cold.

Note that this is a statement about averages only. Inside the flask the molecules still race about and collide billions of times a second; in the mechanical sense none of them is in equilibrium. What has stopped changing is the bulk description.

Whether a system stays put depends on its walls. A cold bottle packed in a foam cool box stays cold for hours; the same bottle standing in a steel tumbler of warm water is soon lukewarm. Foam is close to an adiabatic wall, one that blocks heat, while thin steel or copper is a diathermic wall, one that lets heat through. Fig. 11.2 shows the same contrast for two gas samples, A and B. Behind an adiabatic partition each gas can keep whatever pressure and volume it has, however different from its neighbour's. Through a diathermic partition energy flows as heat, both gases change, and they end in new states that stop changing. A and B are then in thermal equilibrium with each other.

2/3A copper (diathermic) wall lets heat flow from A to B: A cools, B warms.

Fig. 11.2Whether heat can cross the wall decides whether A and B reach a common state. (A and B are equal samples of the same gas.)NCERT Fig. 11.1 · p. 227

Two boxes of gas sit side by side, A on the left and B on the right, each with a thermometer. Step 1: a thick foam wall separates them; A reads 70 °C, B reads 20 °C, and the readings stay put. Step 2: the foam is replaced by a thin copper wall; a heat arrow points from A to B and the readings move to 58 °C and 32 °C. Step 3: both thermometers read 45 °C and the label says no more change.

What do A and B share once the flow has stopped? Experience suggests the answer at once: they are equally hot. Turning that intuition into a quantity you can measure is the job of the next law.

Quick check

1 of 2

In a gas that is in thermodynamic equilibrium, the molecules are at rest.

TrueFalse

Show answer

Answer False

Thermodynamic equilibrium means the bulk variables no longer change. The molecules keep moving and colliding; only their averages are steady.

11.3 Zeroth law of thermodynamics

How can you tell whether two bodies that have never touched would exchange heat if they did? Use a go-between. Let a third body C, a thermometer for instance, come to equilibrium first with A and then with B (Fig. 11.3). If C ends in the same state both times, experiment shows that A and B, when finally placed in contact, exchange no heat at all: neither of their states changes.

LawZeroth law of thermodynamics

If each of two systems is in thermal equilibrium with a third system, then the two are in thermal equilibrium with each other.

It sounds too obvious to need saying, and for a long time nobody said it. The first and second laws had already been numbered when physicists noticed that both quietly depend on this one. In the 1930s Ralph Fowler proposed calling it the zeroth law, because it comes logically before the others.

The zeroth law is what makes temperature a meaningful number. It makes 'is in equilibrium with' behave like an equals sign, which is transitive. Bodies therefore sort themselves into classes, each class holding all the bodies that would be in equilibrium with one another, and each class can be labelled by a number. That label is the temperature $T$ : bodies in thermal equilibrium share the same $T$ , and a body that would exchange heat with another has a different one. A thermometer is just a convenient body C that settles quickly and shows its label on a scale. Making those scales agree from one thermometer to another is thermometry, which you met in the previous chapter.

3/3Lower B into A: no net heat flows, just as the zeroth law predicts.

Fig. 11.3The thermometer C plays the third system of the zeroth law.NCERT Fig. 11.2 · p. 228

A bucket of water A on the left, a copper block B on the right and a thermometer C. Step 1: C stands in the water and reads 31 °C. Step 2: C is moved onto the copper block and again reads 31 °C. Step 3: the block is lowered into the water, the thermometer is set aside, and the label says there is no net heat flow; T A equals T B.

Try this

One thermometer, two liquids

You’ll need

a thermometer
a steel tumbler of tap water
a closed bottle of cooking oil
a steel spoon and a wooden spoon

Predict first

The water and the oil have stood in the same room overnight. What will the thermometer show?

the oil reads warmerthe water reads warmerboth read the sameit depends on the amounts

Leave a steel tumbler of tap water and a closed bottle of cooking oil on the same table overnight.
Hold the thermometer in the water for a minute and note the reading. (A digital cooking thermometer is the safest choice.)
Wipe it, open the oil and read it in the oil.
Close the bottle, stand it in the water for ten minutes, then read the water again.
Finally touch a steel spoon and a wooden spoon from the same drawer. Which feels colder?

Table 11.1Your readings

Measurement	Prediction	Reading (°C)
Water
Oil
Water after the bottle stood in it

Your entries are saved automatically.

11.4 Heat, internal energy and work

Temperature tells us which way heat flows when two bodies touch: from the hotter to the colder, until the temperatures match. To go further we need three more ideas: internal energy, heat and work.

Zoom in on any object, a brick, a glass of water, the air in a balloon, and you find molecules darting about at random and tugging on one another. The energy of all that hidden activity, kinetic energy of the random motion plus potential energy of the molecular forces, is the internal energy $U$ (आंतरिक ऊर्जा). It is counted in the frame where the object as a whole is at rest: kinetic energy of the whole object moving belongs to mechanics, not to $U$ (Fig. 11.4).

Definition

Internal energy

The sum of the kinetic energies of the random motions of a system's molecules and the potential energies of their interactions, measured in the frame where the system's centre of mass is at rest.

ExampleA sealed can of gas on a moving train has the same $U$ as on the platform, provided its temperature is the same.

2/3Every molecule gains the same forward velocity. This ordered motion is mechanics: $U$ is unchanged.

Fig. 11.4Only random motion relative to the box counts in

U

; the box's own motion does not, until a collision scrambles it.NCERT Fig. 11.3 · p. 229

A box of gas molecules, each with a short arrow showing random motion, rides on a trolley. Step 1: the trolley is at rest; the label says the random motion of the molecules is the internal energy U. Step 2: the trolley has moved right and a large arrow shows the whole box moving; the molecular arrows are unchanged and the label says U is unchanged. Step 3: the trolley has hit a wall and stopped; the molecular arrows are longer, the gas is tinted warmer and the label says U increases.

We pictured $U$ through molecules, but thermodynamics treats it simply as another bulk property. What makes $U$ so useful is that the present condition of the system fixes it completely; the route by which the system got there does not matter. A litre of nitrogen at a given pressure and temperature has the same $U$ whether it was warmed from cold, squeezed from a larger volume or reached that state any other way. A quantity like this is called a state variable; pressure, volume and temperature are others (Section 11.7). In a gas the molecular forces are weak, so $U$ is almost all kinetic energy. As the next chapter shows, that includes the molecules' travel across the container, their rotation and, when hot enough, the vibration of their atoms.

SimulationInternal energy as molecular motion

A closed container of water molecules drawn as moving dots, with a thermometer and a temperature slider from 5 to 120 degrees Celsius. Raising the temperature makes the dots move faster; above 100 degrees they break away from one another and fill the container as a gas.

Fig. 11.5Drag the temperature slider and watch the molecules. Faster random motion means more internal energy.

Try this

Raise the temperature from 20 °C to 90 °C. What happens to the speed of the molecules, and so to U?
Go past 100 °C. The molecules separate from one another; pulling them apart also costs energy, which is why boiling needs so much heat (Example 11.1).

Suppose you want to raise the internal energy of the gas in a cylinder. You have exactly two options (Fig. 11.6): put something hotter against it, such as a hot plate, or shove the piston in. Either works, and either can be run backwards: cool surroundings drain energy away, and a gas allowed to push its piston out spends some of its own $U$ in doing so.

3/4Pushing the piston in does work on the gas, with no temperature difference needed: $U$ rises again.

Fig. 11.6Heat and work are two different routes by which energy enters or leaves a system.NCERT Fig. 11.4 · p. 229

A vertical cylinder of gas with a piston, and a gauge on the right showing the level of internal energy U. Step 1: the gas at rest with the gauge at a middle level. Step 2: the cylinder stands on a hot plate, heat arrows point up into the gas and the gauge rises. Step 3: an arrow pushes the piston down, the gas is compressed and the gauge rises further. Step 4: the piston moves up with an arrow labelled work out, heat arrows point out of the walls, and the gauge falls.

This gives two everyday words a precise meaning. Heat (ऊष्मा) is energy transferred between a system and its surroundings because of a temperature difference. Work is energy transferred in any other way: a piston moving under a force, a paddle stirring a liquid, a weight lifted by a rope.

Compare

Two ways energy crosses a boundary

Heat

Driven by a temperature difference
Flows by itself only from hotter to colder
Example: a pan of water on a flame

Work

Needs no temperature difference
A force acts through a displacement
Example: pushing a bicycle-pump handle

Both

Measured in joules
Describe a process, not a state
Change the internal energy $U$

In everyday speech we say a hot iron 'has a lot of heat'. Thermodynamics is stricter: the iron has a lot of internal energy, and heat is what flows out of it when it touches your cooler shirt. Keeping the two apart avoids most of the confusion in this chapter.

11.5 First law of thermodynamics

Heat and work are the only two ways energy crosses the boundary of a system, so conservation of energy gives a simple bookkeeping rule. Write

$Δ Q$ for the heat supplied to the system by its surroundings,
$Δ W$ for the work done by the system on its surroundings,
$Δ U$ for the change in the system's internal energy.

LawFirst law of thermodynamics

The heat supplied to a system equals the increase in its internal energy plus the work the system does on its surroundings.

First law of thermodynamics

Δ Q = Δ U + Δ W

(11.1)

Read it as a household budget: income arriving as heat is either saved, raising $U$ , or spent again as work. The first law is nothing more than conservation of energy for a system that can exchange heat as well as work. The signs do the bookkeeping, so it pays to fix them once (Table 11.2).

Table 11.2Sign convention for the first law (physics)

Quantity	Positive when	Negative when
$Δ Q$	heat flows into the system	heat flows out
$Δ W$	the system does work (it expands)	work is done on it (it is compressed)
$Δ U$	internal energy rises	internal energy falls

Your chemistry book counts work done on the system as positive and writes the same law as $Δ U = q + w$ . Nothing physical differs; only the sign attached to work does. Rearranged, our form reads

Δ Q - Δ W = Δ U

(11.2)

This version hides a surprise. Draw any two routes between the same pair of states on a $P$ – $V$ diagram (Fig. 11.7). The work, the area under each route, plainly differs, and so does the heat. Yet $Δ Q - Δ W$ comes out identical, because it equals $Δ U$ and $U$ is a state variable. Heat and work belong to the journey; their difference belongs to the two end points.

3/4Go straight from $i$ to $f$ , the pressure falling steadily as the gas expands: $W_{2}$ is smaller.

Fig. 11.7The work done is the area under the path on a

P

–

V

diagram, so it depends on the route.

A pressure against volume graph with an initial state i at high pressure and small volume and a final state f at low pressure and large volume. Step 1: only the two points. Step 2: route 1 goes right at high pressure and then straight down; the large area under it is shaded and labelled W1. Step 3: route 2 is a straight line from i down to f; the smaller area under it is shaded and labelled W2. Step 4: going out by route 1 and back along route 2 encloses a triangle, labelled W1 minus W2, the net work of the loop.

If $Δ U = 0$ , as in the slow isothermal expansion of an ideal gas (Section 11.8.2), then $Δ Q = Δ W$ : every joule of heat taken in is passed straight on as work.

DerivationWork done against a constant pressure

Let a gas at pressure $P$ push a piston of area $A$ outward through a small distance $Δ x$ . The force on the piston is $P A$ , so the gas does work

\begin{aligned} Δ W & = P A Δ x \\ = P Δ V \end{aligned}

because $A Δ x$ is the increase $Δ V$ in the gas's volume. If the pressure stays fixed through a finite change of volume, the same expression covers the whole change, and the first law becomes

Δ Q = Δ U + P Δ V

(11.3)

In a compression $Δ V < 0$ , so $Δ W < 0$ : work is done on the gas.

Example 11.1NCERT 11.5 · p. 231

Where does the heat of boiling go?

In a laboratory, $1.00 m o l$ ( $46.1 g$ ) of ethanol is boiled at its normal boiling point, ${78.4}^{\circ} C$ , under atmospheric pressure, $1.013 \times 10^{5} P a$ . This takes $38.6 k J$ of heat. Treat the vapour as an ideal gas and the liquid's volume (about $0.06 L$ ) as negligible. Find the work done by the ethanol in pushing back the atmosphere, and the increase in its internal energy.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A gas absorbs

600 J

of heat and does

250 J

of work on its surroundings. By how much does its internal energy change?

Show answer

Answer 350 J

Δ U = Δ Q - Δ W = 600 - 250 = 350 J,

an increase.

11.6 Specific heat capacity

Give a body a small amount of heat $Δ Q$ and its temperature rises by $Δ T$ . The ratio is the body's heat capacity:

S = \frac{Δ Q}{Δ T}

(11.4)

A bucket of water has a far larger heat capacity than a cup of it, so $S$ describes an object, not a material. Dividing by the mass $m$ gives a property of the substance alone, its specific heat capacity (विशिष्ट ऊष्मा धारिता):

s = \frac{S}{m} = \frac{1}{m} \frac{Δ Q}{Δ T}

(11.5)

Definition

Specific heat capacity

The heat needed to raise the temperature of unit mass of a substance by one kelvin; SI unit $J k g^{- 1} K^{- 1}$ .

ExampleAbout $4.2 k J$ warms $1 k g$ of water by $1 K$ .

The value depends on the substance and slightly on its temperature. Counting matter in moles instead, dividing by the number of moles $μ$ , gives the molar specific heat capacity:

C = \frac{S}{μ} = \frac{1}{μ} \frac{Δ Q}{Δ T}

(11.6)

Its unit is $J m o l^{- 1} K^{- 1}$ , and like $s$ it does not depend on how much substance you have. Both $s$ and $C$ can also depend on the conditions under which the heat is supplied, as gases will show. The advantage of counting in moles is that it exposes a striking regularity (Table 11.3).

Table 11.3Heat capacities of six solids near

25^{\circ} C

(standard data)NCERT Table 11.1 · p. 231

Solid	$M$ ( $g m o l^{- 1}$ )	$s$ ( $J k g^{- 1} K^{- 1}$ )	$C = s M$ ( $J m o l^{- 1} K^{- 1}$ )
Iron	55.85	449	25.1
Zinc	65.38	388	25.4
Nickel	58.69	444	26.1
Gold	196.97	129	25.4
Silicon	28.09	705	19.8
Beryllium	9.01	1825	16.4

The specific heats in the table differ by a factor of fourteen, yet the molar values of the four metals all lie close to $25 J m o l^{- 1} K^{- 1} .$ Gold atoms are heavy, so a kilogram of gold contains few of them; atom for atom, gold needs about as much heat as iron. The light, stiffly bonded elements at the bottom fall short, and diamond falls shortest of all, at about a quarter of the common value. A simple model explains the common value.

DerivationMolar heat capacity of a solid

Model a solid as $N$ atoms, each vibrating about a fixed position. A one-dimensional oscillator stores energy in two forms, kinetic and potential, and the law of equipartition of energy (Chapter 12) gives each form an average of $\frac{1}{2} k_{B} T$ . So each direction of vibration holds $k_{B} T$ on average, and an atom vibrating in three directions holds $3 k_{B} T$ .

For one mole $N = N_{A}$ , and $k_{B} N_{A} = R$ , so

\begin{aligned} U & = 3 k_{B} T \times N_{A} \\ = 3 R T \end{aligned}

Heating a solid at atmospheric pressure barely changes its volume, so the work term $P Δ V$ in Eq. (11.3) is negligible and $Δ Q \approx Δ U$ . Therefore

\begin{aligned} C & = \frac{Δ Q}{Δ T} = \frac{Δ U}{Δ T} \\ = 3 R \end{aligned}

(11.7)

Numerically $3 R = 24.9 J m o l^{- 1} K^{- 1},$ matching the metals in Table 11.3. The rule fails at low temperatures, where every solid's $C$ falls towards zero, and for light, stiff solids such as diamond even at room temperature. Explaining those failures needed quantum physics.

Example 11.2

Identifying a metal by its heat capacity

A jeweller has a $0.250 k g$ bar of an unknown pure metal. Heating it from ${20.0}^{\circ} C$ to ${60.0}^{\circ} C$ takes $2.35 k J$ . (a) Find its specific heat capacity. (b) Taking $C = 3 R$ , estimate its molar mass and suggest which metal it is.

Try it yourself first. Then reveal the solution one step at a time.

Specific heat capacity of water

Why is the calorie, the old unit of heat, defined with such a fussy pair of temperatures? Because water does not need exactly the same heat for every degree. Its specific heat is highest near $0^{\circ} C$ , dips to a shallow minimum in the middle thirties and climbs again towards $100^{\circ} C$ , a total change of under 1% (Fig. 11.8). A unit meant as 'the heat for one gram and one degree' therefore had to say which degree. The calorie was fixed as the heat needed to warm one gram of water through the single degree between ${14.5}^{\circ} C$ and ${15.5}^{\circ} C$ , and that amount is $4.186 J$ .

Fig. 11.8Water's specific heat changes by less than 1% between

0^{\circ} C

and

100^{\circ} C

(values from steam tables).NCERT Fig. 11.5 · p. 232

A curve of specific heat capacity, in joules per gram per kelvin, against temperature from 0 to 100 degrees Celsius. It starts at about 4.22 at 0 °C, falls to a broad minimum of about 4.18 between 30 and 40 °C, and rises again to about 4.22 at 100 °C. A thin yellow band marks the interval from 14.5 to 15.5 °C used to define the calorie.

In SI units the specific heat of water is about $4186 J k g^{- 1} K^{- 1},$ unusually large: few common substances store as much energy per kilogram per kelvin. In Joule's day the number $4.186 J$ per calorie had a grand name, the 'mechanical equivalent of heat', and measuring it was a landmark. Today it is simply the exchange rate between two units of energy, like rupees and paise, and since SI uses the joule for heat and work alike, the old name has fallen out of use.

Specific heats of an ideal gas

For a solid or a liquid it hardly matters how you heat it. For a gas it matters a great deal (Fig. 11.9). Pin the piston and every joule of heat stays in the gas as internal energy. Let the piston rise against a steady load and part of the heat leaves again as work, so the temperature rises less. A gas therefore has two principal molar heat capacities, $C_{v}$ at constant volume and $C_{p}$ at constant pressure, with $C_{p} > C_{v}$ . For an ideal gas the gap is exactly $R$ :

2/2Pinned: all of $Q$ raises $U$ , so $Δ T$ is larger. Free: part of $Q$ lifts the load, so $Δ T$ is smaller.

Fig. 11.9At constant pressure part of the heat does work, so more heat is needed per kelvin:

C_{p} > C_{v}

Two cylinders of gas on identical burners, each with a thermometer. The left piston is pinned so the volume is fixed; the right piston carries a load and is free to move, so the pressure is fixed. Step 1: equal heat arrows labelled Q enter both. Step 2: the left thermometer has risen higher and the label says all heat goes to U; the right piston has risen with an arrow, its thermometer has risen less, and the label says heat goes to U plus work.

C_{p} - C_{v} = R

(11.8)

DerivationProof that

C_{p} - C_{v} = R

Take one mole of an ideal gas and apply Eq. (11.3), $Δ Q = Δ U + P Δ V$ . At constant volume $Δ V = 0$ , so

\begin{aligned} C_{v} & = {(\frac{Δ Q}{Δ T})}_{v} \\ = \frac{Δ U}{Δ T} \end{aligned}

(11.9)

A subscript records what is held fixed. On the right it is unnecessary: an ideal gas's $U$ is a function of $T$ only, so $Δ U / Δ T$ does not care whether $V$ or $P$ was held constant. At constant pressure,

\begin{aligned} C_{p} & = {(\frac{Δ Q}{Δ T})}_{p} \\ = \frac{Δ U}{Δ T} \\ + P {(\frac{Δ V}{Δ T})}_{p} \end{aligned}

(11.10)

The first term is the same as in Eq. (11.9), again because $U$ depends only on $T$ . For one mole $P V = R T$ , so at fixed $P$ a temperature change $Δ T$ changes the volume by $Δ V = R Δ T / P$ , and

P {(\frac{Δ V}{Δ T})}_{p} = R

(11.11)

Putting Eqs. (11.9) and (11.11) into Eq. (11.10) gives $C_{p} = C_{v} + R$ , which is Eq. (11.8).

This is the clearest case of a general rule: a heat capacity depends on the process as well as on the substance. Heat a gas isothermally and $Δ T = 0$ , so $Δ Q / Δ T$ is unbounded; compress it adiabatically and its temperature rises with $Δ Q = 0$ , so the ratio is zero. Only once the conditions are stated does a heat capacity have a definite value.

Example 11.3

Helium heated two ways

$0.50 m o l$ of helium, a monatomic ideal gas with $C_{v} = \frac{3}{2} R$ , is warmed by $40 K$ , (a) in a rigid steel cylinder and (b) in a cylinder whose piston moves freely so the pressure stays constant. Find the heat needed in each case and the work done in (b). Take $R = 8.314 J m o l^{- 1} K^{- 1} .$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Using equipartition, what is the molar heat capacity of a typical solid at room temperature? Take

R = 8.31 J m o l^{- 1} K^{- 1} .

J/(mol K)

Show answer

Answer 24.9 J/(mol K)

C = 3 R = 3 \times 8.31 = 24.9 J m o l^{- 1} K^{- 1},

the Dulong–Petit value.

11.7 Thermodynamic state variables and equation of state

To pin down an equilibrium state you need only a short list of numbers. For gas in a cylinder the list is its pressure, volume, temperature and mass, and for a mixture such as air, its composition. These are its state variables. Once they are known, so is everything else thermodynamics can say about the gas, its internal energy included.

Not every state can be described this way. Puncture a partition with gas on one side and vacuum on the other, and for a moment the gas rushes through as a jet: dense near the hole, thin elsewhere, with no single pressure or temperature for the whole sample (Fig. 11.10). The same is true of the fuel–air mixture in a motorcycle engine just after the spark fires. Only when the gas has calmed down to uniform pressure and temperature, in equilibrium with its surroundings, do state variables describe it again.

2/3The partition breaks: the gas surges into the vacuum, uneven and far from equilibrium.

Fig. 11.10State variables describe only equilibrium states; the rushing gas in the middle step has none.NCERT Fig. 11.6 · p. 233

An insulated box divided by a partition, with gas molecules on the left and vacuum on the right. Step 1: molecules spread evenly over the left half; the label says P, V and T are defined. Step 2: the partition has gone and the molecules crowd near the middle with arrows pointing right; the label says there is no single P or T. Step 3: the molecules are spread evenly through the whole box and the label says equilibrium again.

The state variables are not independent of one another. The relation between them is the system's equation of state. For an ideal gas it is the ideal gas equation:

Equation of state of an ideal gas

P V = μ R T

For a fixed amount of gas only two variables are free, say $P$ and $V$ ; the third follows. A graph of $P$ against $V$ at one fixed temperature is an isotherm. Real gases follow more complicated equations of state, especially when dense.

State variables come in two kinds. Pour a litre of milk at $30^{\circ} C$ equally into two glasses. Each glass holds half the volume, half the mass and half the internal energy, but the milk in each is still at $30^{\circ} C$ , at the same pressure and with the same density. Quantities that get shared out like this are extensive: they grow with the size of the system. Quantities that stay the same in each part, such as temperature, pressure and density, are intensive.

Compare

Cut the system in half

Intensive (unchanged)

pressure $P$
temperature $T$
density $ρ$

Extensive (halved)

volume $V$
mass $M$
internal energy $U$

The idea doubles as an error check, rather like checking units: in a correct equation every term is of the same kind. Test it on $Δ Q = Δ U + P Δ V$ . $Δ U$ is extensive; $P Δ V$ is intensive times extensive, which is extensive; and $Δ Q$ , though not a state variable, doubles when the system doubles. All three match.

11.8 Thermodynamic processes

A process carries a system from one state to another, and the path it follows decides how much heat and work are involved. A few standard kinds of process cover most situations. First, though, a condition on how fast things are allowed to happen.

11.8.1 Quasi-static process

Imagine a gas under a piston loaded with sand, sitting quietly in a room. Tip off all the sand at once (Fig. 11.11) and the piston leaps up. The gas next to the piston thins out before the gas lower down has 'heard the news', so for a moment there is no single pressure or temperature to speak of. Pressing a cold gas suddenly against a hot plate scrambles it in a similar way. States like these are messy to analyse.

The cure is patience. Remove the sand one grain at a time and wait after each grain: the piston creeps up, and at every moment the gas is uniform, its pressure only a grain's weight away from the load above it. Temperature can be changed just as gently, by warming the gas through a long line of reservoirs, each a hair hotter than the last. In the limit of infinitely small steps the gas passes through nothing but equilibrium states. Such an idealised process is called quasi-static (स्थैतिककल्प), meaning 'almost static'.

3/3Remove grains one at a time: the gas passes through a staircase of equilibrium states that hugs a smooth curve.

Fig. 11.11Only a process slow enough to stay in equilibrium can be drawn as a curve on a

P

–

V

diagram.NCERT Fig. 11.7 · p. 234

On the left a cylinder of gas whose piston carries a heap of sand; on the right a small pressure against volume graph with states A and B. Step 1: the heap presses the piston down and only state A is marked. Step 2: the whole heap has gone, the piston has jumped up, swirls show churning gas, and on the graph a dotted line with a question mark joins A to B. Step 3: only a little sand remains after removing grains one by one, and the graph shows a fine staircase from A to B that follows a smooth curve.

No real process is infinitely slow, so this is an idealisation, much like the frictionless surfaces of mechanics. Still, a piston that moves gently and heat that crosses only small temperature differences come very close, and everything that follows assumes this ideal unless we say otherwise.

Four special processes recur throughout physics and engineering (Table 11.4). In an isothermal process (समतापीय प्रक्रम) the temperature is held fixed, for instance by sinking a thin steel bottle of gas in a big tank of water: there is so much water that the heat it swaps with the gas leaves its temperature practically unchanged. In an isobaric process the pressure is fixed, and in an isochoric process the volume is fixed. In an adiabatic process (रुद्धोष्म प्रक्रम) the system is insulated, so no heat enters or leaves. The table collects the results derived in the rest of this section.

Table 11.4Four special processes for an ideal gasNCERT Table 11.2 · p. 234

Process	Held fixed	$Δ U$	$W$ (by gas)	$Q$
Isothermal	$T$	$0$	$μ R T \ln (V_{2} / V_{1})$	$Q = W$
Adiabatic	no heat flow	$- W$	$μ R (T_{1} - T_{2}) / (γ - 1)$	$0$
Isobaric	$P$	$μ C_{v} Δ T$	$P Δ V = μ R Δ T$	$μ C_{p} Δ T$
Isochoric	$V$	$μ C_{v} Δ T$	$0$	$μ C_{v} Δ T$

11.8.2 Isothermal process

At constant temperature the ideal gas equation reduces to $P V = constant$ : halve the volume and the pressure doubles. This is Boyle's law, now seen as one particular thermodynamic process.

DerivationWork done in an isothermal expansion

Let $μ$ moles of an ideal gas at temperature $T$ go quasi-statically from $(P_{1}, V_{1})$ to $(P_{2}, V_{2})$ . Over a tiny volume change $d V$ the pressure is practically constant, so the gas does work $d W = P d V$ . Adding all these pieces means integrating:

\begin{aligned} W & = \int_{V_{1}}^{V_{2}} P d V \\ = μ R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} \end{aligned}

Here we used $P = μ R T / V$ and took the constants outside the integral. The integral of $1 / V$ is $\ln V$ , so

Isothermal work

W = μ R T \ln \frac{V_{2}}{V_{1}}

(11.12)

For an ideal gas $U$ depends only on $T$ , so $Δ U = 0$ and the first law gives $Q = W$ . In an expansion ( $V_{2} > V_{1}$ ), $W > 0$ : the gas takes heat from the bath and passes all of it on as work. In a compression, $W < 0$ : the work done on the gas flows out into the bath as heat.

Example 11.4

Squeezing air slowly

A large glass syringe holds $0.020 m o l$ of air at $27^{\circ} C$ . Held in a big water bath, its plunger is pushed in very slowly from $500 m L$ to $200 m L$ . Treating the air as an ideal gas, find (a) the work done by the air, (b) the heat exchanged with the bath and (c) the final pressure.

Try it yourself first. Then reveal the solution one step at a time.

11.8.3 Adiabatic process

In an adiabatic process no heat crosses the boundary, so the first law reduces to $Δ U = - Δ W$ . When the gas does work it pays entirely out of its internal energy, and an ideal gas cools; when work is done on it, it warms. A short calculus argument, beyond this chapter, shows that in a quasi-static adiabatic change an ideal gas obeys

Adiabatic process

P V^{γ} = constant

(11.13)

where

$γ = C_{p} / C_{v}$: the ratio of the specific heats (molar or per unit mass)

Because $C_{p} > C_{v}$ , $γ$ is greater than 1: about $1.67$ for a monatomic gas such as helium and $1.40$ for air. Between any two states on the same adiabat,

P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}

(11.14)

Substituting $P = μ R T / V$ gives an equivalent form, $T V^{γ - 1} = constant$ , which we shall need for the Carnot engine. Since $γ > 1$ , pressure falls faster with volume along an adiabat than along an isotherm, so through any point the adiabat is the steeper curve (Fig. 11.12). A gas that expands adiabatically cools and ends up on a lower isotherm.

3/3C lies on a cooler isotherm (dotted): the gas paid for its work out of $U$ .

Fig. 11.12From the same starting point an adiabat falls more steeply than an isotherm (drawn here for

γ = 5 / 3

).NCERT Fig. 11.8 · p. 235

A pressure against volume graph with a starting state A at high pressure. Step 1: a solid blue isotherm, labelled T1, runs from A down to state B at four times the volume. Step 2: a dashed vermillion adiabat from A falls more steeply to state C at the same final volume, lower than B. Step 3: a dotted cooler isotherm, labelled T2 less than T1, passes through C.

DerivationWork done in an adiabatic change

Take an ideal gas adiabatically from $(P_{1}, V_{1}, T_{1})$ to $(P_{2}, V_{2}, T_{2})$ . Along the way $P = K / V^{γ}$ , where $K$ is the constant in Eq. (11.13), so

\begin{aligned} W & = \int_{V_{1}}^{V_{2}} P d V \\ = K \int_{V_{1}}^{V_{2}} \frac{d V}{V^{γ}} \\ = \frac{K}{1 - γ} [V^{1 - γ}]_{V_{1}}^{V_{2}} \end{aligned}

(11.15)

Evaluate at the two limits, writing $K = P_{2} V_{2}^{γ}$ at $V_{2}$ and $K = P_{1} V_{1}^{γ}$ at $V_{1}$ , so that $K V_{2}^{1 - γ} = P_{2} V_{2}$ and $K V_{1}^{1 - γ} = P_{1} V_{1}$ :

Adiabatic work

\begin{aligned} W & = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.16)

The last step uses $P V = μ R T$ . If the gas does work ( $W > 0$ ), then $T_{2} < T_{1}$ and it cools; if work is done on it ( $W < 0$ ), then $T_{2} > T_{1}$ and it warms. This agrees with $W = - Δ U = μ C_{v} (T_{1} - T_{2}),$ because $C_{v} = R / (γ - 1)$ follows from $C_{p} - C_{v} = R$ and $γ = C_{p} / C_{v}$ .

Example 11.5

Why a diesel engine needs no spark plug

In a diesel engine, air at $27^{\circ} C$ and $1.0 \times 10^{5} P a$ is compressed to one-sixteenth of its volume so quickly that hardly any heat escapes. Treat the compression as quasi-static and adiabatic with $γ = 1.40$ . Find (a) the final pressure, (b) the final temperature and (c) the work done on one mole of the air.

Try it yourself first. Then reveal the solution one step at a time.

Try this

Feel adiabatic heating

You’ll need

a bicycle pump

Predict first

With the nozzle blocked you push the handle in sharply several times. What happens to the barrel near the nozzle?

it becomes coolernothing you can feelit becomes warmerit becomes damp

Block the nozzle of a bicycle pump firmly with your thumb.
Push the handle in sharply and let it spring back, ten times.
Feel the barrel near the nozzle, then near the handle.
Repeat with the nozzle open and compare.

Table 11.5What you felt

Trial	Prediction	Barrel warmer?
Nozzle blocked
Nozzle open

Your entries are saved automatically.

With the nozzle blocked, the barrel near the nozzle warms noticeably. Each quick stroke compresses the trapped air faster than heat can leak away, so your work goes into the air's internal energy, nearly an adiabatic compression, and the hot air then warms the metal. With the nozzle open, most of the air escapes instead of being squeezed, and the barrel warms much less. Some warming in both trials comes from friction between the plunger seal and the barrel.

Quick check

1 of 3

Two identical gas samples start in the same state. One expands isothermally to twice its volume, the other adiabatically to twice its volume. Which ends at the lower pressure?

the adiabatic onethe isothermal oneboth end at the same pressureit depends on the gas

Show answer

Answer (A) the adiabatic one

Isothermally the pressure halves. Adiabatically it falls by the factor

2^{γ}

, which exceeds 2 because the gas also cools. 'It depends on the gas' is tempting since

γ

varies, but

2^{γ} > 2

for every gas.

11.8.4 Isochoric process

Lock the volume and the gas cannot push anything, so $Δ W = 0$ . Every joule of heat then lands in $U$ : $Δ Q = Δ U = μ C_{v} Δ T$ . That is exactly how $C_{v}$ was defined, so $C_{v}$ decides how hot the gas gets. The air in a closed steel tiffin box left in the sun is heated roughly this way, and its pressure rises in proportion to its kelvin temperature.

11.8.5 Isobaric process

Now hold the pressure fixed and let the gas expand as it warms. It does work

\begin{aligned} W & = P (V_{2} - V_{1}) \\ = μ R (T_{2} - T_{1}) \end{aligned}

(11.17)

while its internal energy rises by $μ C_{v} Δ T$ . The heat has both bills to pay, $μ (C_{v} + R) Δ T = μ C_{p} Δ T$ in all, so at constant pressure it is $C_{p}$ that sets the temperature rise (Example 11.3). Air warming in a room with an open window stays at atmospheric pressure, and so does the steam escaping from an open pan.

11.8.6 Cyclic process

Engines repeat themselves: after each cycle the working gas is back exactly where it began. Because $U$ depends only on the state, a complete cycle brings no change in $U$ , and the first law then forces the net heat taken in to equal the net work delivered. On a $P$ – $V$ diagram a cycle is a closed loop. The work on the expanding leg is the area under it; on the compressing leg, the area under that leg counts as negative. The net work is the area enclosed by the loop, positive when the loop runs clockwise, as in the out-and-back route of Fig. 11.7. Every engine you know, from a motorcycle to a thermal power station, runs through a cycle.

Example 11.6

A rectangular cycle

A gas in an engine cylinder is taken round a rectangle on the $P$ – $V$ diagram. It expands from $1.0 L$ to $3.0 L$ at $2.0 \times 10^{5} P a$ ; its pressure is lowered to $1.0 \times 10^{5} P a$ at constant volume; it is compressed back to $1.0 L$ at that pressure; and its pressure is raised at constant volume to the start. Find (a) the net work done by the gas per cycle, (b) the net heat it absorbs and (c) its power output at 25 cycles per second.

Try it yourself first. Then reveal the solution one step at a time.

Try all four processes in the lab below. Whatever the process and whichever gas you pick, the three bars always balance: $Q = Δ U + W$ .

Fig. 11.13Pick a process and a gas, set the final volume (or temperature) and press Play. The shaded area is the work; the bars check the first law.

Quick check

1 of 3

Match each process with what stays fixed.

Tap an item on the left, then its match on the right.

Show answer

Answer

isothermal : temperature
isobaric : pressure
isochoric : volume
adiabatic : no heat exchange

The Greek roots help: iso = same, therm = heat/temperature, baros = weight/pressure, choros = space. Adiabatic means 'impassable' to heat.

11.9 Second law of thermodynamics

The first law is energy conservation, and it rules out any machine that creates energy from nothing. Yet many events that would conserve energy perfectly are never seen. A cyclist brakes to a stop halfway up a slope, and the brake pads and rims warm up. The first law has no objection to the reverse: the rims could cool by themselves and spend that energy rolling the bicycle back up the slope. It never happens. Some principle beyond energy conservation must forbid it, and that principle is the second law of thermodynamics.

The second law shows itself most clearly in devices that turn heat into work. A heat engine runs in a cycle: it takes heat $Q_{1}$ from a hot reservoir, delivers work $W$ and rejects heat $Q_{2}$ to a cold reservoir. Its efficiency is

Efficiency of a heat engine

η = \frac{W}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}}

A refrigerator runs the other way. Work $W$ is done on it to pull heat $Q_{2}$ out of a cold space, and it dumps $Q_{1} = Q_{2} + W$ into the warmer surroundings. Its coefficient of performance is

Coefficient of performance

α = \frac{Q_{2}}{W}

The second law says that $η$ can never reach 1 and $α$ can never be infinite. Two classic statements put this precisely (Fig. 11.14).

LawSecond law: Kelvin–Planck statement

It is impossible for a device working in a cycle to take heat from a single reservoir and convert all of it into work, with no other effect.

LawSecond law: Clausius statement

It is impossible for heat to pass from a colder body to a hotter body with no other effect, such as work supplied from outside.

1/4A real engine: $Q_{1}$ in, work $W$ out, and some heat $Q_{2}$ always rejected to the cold reservoir.

Fig. 11.14Kelvin–Planck forbids the perfect engine; Clausius forbids the perfect refrigerator.

A hot reservoir at the top, a cold reservoir at the bottom and a device in the middle. Step 1, real engine, allowed: heat Q1 flows down into the engine, work W leaves to the right and heat Q2 flows down to the cold reservoir. Step 2, perfect engine: heat Q flows in and all of it leaves as work, with nothing to the cold reservoir; a large cross marks it forbidden. Step 3, refrigerator, allowed: heat Q2 rises from the cold reservoir, work W enters from the right and heat Q1 rises to the hot reservoir. Step 4, perfect refrigerator: heat Q rises straight from cold to hot with no work; a large cross marks it forbidden.

The two statements look different, one about engines and one about refrigerators, yet each implies the other: a perfect refrigerator paired with an ordinary engine would make a perfect engine, and a perfect engine could drive an ordinary refrigerator to make a perfect one. They are two faces of one law. Notice the words 'no other effect'. Turning heat completely into work is allowed if something else changes as well. A gas expanding isothermally converts all the heat it absorbs into work, but it ends with a larger volume, so that process is not a cycle.

Quick check

1 of 3

Which proposed device would violate the Kelvin–Planck statement?

an electric heater that turns all its electrical energy into heata refrigerator powered by electricitya ship engine that takes heat from the sea and turns all of it into worka car engine that rejects hot exhaust gas

Show answer

Answer (C) a ship engine that takes heat from the sea and turns all of it into work

Drawing heat from a single reservoir (the sea) and converting all of it into work, cycle after cycle, is exactly what Kelvin–Planck forbids. Turning work entirely into heat, as a heater does, is perfectly allowed.

11.10 Reversible and irreversible processes

Film almost any everyday event and play it backwards, and you can tell at once that the film is reversed (Fig. 11.15). The backward version would break no law of mechanics and would conserve energy perfectly, yet it never happens. Processes that occur by themselves run one way only; physicists call them irreversible. A few examples:

A drop of ink spreads through a glass of water until the colour is even; the ink never gathers itself back into a drop.
Smoke from an agarbatti drifts through the whole room; it never streams back into the stick.
A hot iron set down on its stand cools to room temperature; the cooler room never warms it up again.
A burnt matchstick never reassembles itself from smoke and ash.
Brakes that have warmed up while stopping a bicycle never push it forward again. Turning their extra internal energy wholly back into motion would break the Kelvin–Planck statement.

2/4The ink spreads out by itself.

Fig. 11.15Mixing runs one way: a film of it played backwards looks impossible.

A glass of water seen from the side. Step 1: a small dense cluster of purple ink dots near the surface. Step 2: the dots have spread into a loose cloud in the upper middle. Step 3: the dots are scattered evenly through all the water. Step 4: the dots have crowded back into the drop at the top, with a large cross over them and the label film reversed: never seen.

Irreversibility has two main sources:

Non-equilibrium states. Free expansion, explosions, mixing and anything driven by a finite pressure or temperature difference carry the system through states far from equilibrium.
Dissipation. Friction, viscosity and electrical resistance turn ordered energy into internal energy: brake blocks grow hot as a train slows down, and a charging cable warms as current flows through it.

No machine is entirely free of friction or resistance, so in practice every process is at least a little irreversible.

Definition

Reversible process

A process that can be run backwards so that both the system and its surroundings return to their initial states, with no change anywhere else.

ExampleAn ideal gas expanding isothermally, infinitely slowly, against a frictionless piston (an idealisation).

Two conditions must both hold for a process to be reversible (उत्क्रमणीय): it must never leave equilibrium (it is quasi-static), and nothing may be lost to friction or other dissipation. Picture an ideal gas pushing a perfectly smooth piston outward, infinitely slowly, while sitting in a bath: nudge the outside pressure up by a whisker and the gas retraces every state, handing back exactly the heat it took. Why care about an ideal that never quite happens? Because it sets the benchmark. The second law forbids an engine that turns all its heat into work, so engineers want to know the best that is possible between two temperatures. That best belongs to an engine made only of reversible steps; every irreversibility pulls it lower.

Quick check

1 of 2

Which of these processes are irreversible?

ink spreading through watera bicycle braking to a stopa quasi-static isothermal expansion against a frictionless pistona matchstick burning

Select all that apply

Show answer

Answer

(A) ink spreading through water
(B) a bicycle braking to a stop
(D) a matchstick burning

Mixing, friction and combustion are all irreversible. The ideal quasi-static, frictionless expansion is the textbook example of a reversible process.

11.11 Carnot engine

In 1824 a young French engineer, Sadi Carnot, noticed that steam engines were being improved largely by trial and error. He asked a sharper question: given a hot reservoir at $T_{1}$ and a cold one at $T_{2}$ , what is the best that any engine could possibly do, and with what cycle? Remarkably, he reached the right answer while still thinking of heat as caloric.

The ideal engine should be reversible, since irreversibility wastes work. So heat must never cross a finite temperature difference. That fixes two steps at once: heat $Q_{1}$ must be taken in isothermally at $T_{1}$ , and heat $Q_{2}$ given out isothermally at $T_{2}$ . To close the cycle the working substance must also travel from $T_{1}$ down to $T_{2}$ and back up. Any leg on which it swapped heat, an isochoric one say, would need a ladder of reservoirs at every in-between temperature, and our engine is allowed only two. So these legs must involve no heat at all: they must be reversible adiabatic steps. A reversible engine working between two temperatures is a Carnot engine, and its cycle of two isotherms joined by two adiabats is the Carnot cycle (Fig. 11.16).

1/5Two isotherms (solid) joined by two adiabats (dashed): $1 \to 2 \to 3 \to 4 \to 1$ .

Fig. 11.16The Carnot cycle: two isotherms joined by two adiabats (drawn for a monatomic gas).NCERT Fig. 11.9 · p. 237

On the left a pressure against volume graph with four states, 1 to 4: a solid vermillion isotherm at T1 from 1 to 2, a dashed adiabat from 2 to 3, a solid sky-blue isotherm at T2 from 3 to 4 and a dashed adiabat from 4 back to 1. On the right a cylinder with a piston stands on a base. Step 1 shows the whole loop. Step 2 highlights 1 to 2: the cylinder stands on the hot reservoir, heat arrows labelled Q1 go in and the piston rises. Step 3 highlights 2 to 3: on an insulating stand the piston rises further. Step 4 highlights 3 to 4: on the cold reservoir heat Q2 flows out and the piston moves down. Step 5 highlights 4 to 1: on the insulating stand the piston returns to its starting height.

DerivationEfficiency of a Carnot engine

Take $μ$ moles of an ideal gas as the working substance. Step 1 → 2, isothermal expansion at $T_{1}$ from $V_{1}$ to $V_{2}$ : by Eq. (11.12) the gas does work equal to the heat it absorbs,

\begin{aligned} Q_{1} & = W_{1 \to 2} \\ = μ R T_{1} \ln \frac{V_{2}}{V_{1}} \end{aligned}

(11.18)

Step 2 → 3, adiabatic expansion from $T_{1}$ to $T_{2}$ : by Eq. (11.16) the gas does work

\begin{aligned} W_{2 \to 3} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.19)

Step 3 → 4, isothermal compression at $T_{2}$ from $V_{3}$ to $V_{4}$ : work $W_{3 \to 4}$ is done on the gas, and the same amount leaves it as heat $Q_{2}$ ,

\begin{aligned} Q_{2} & = W_{3 \to 4} \\ = μ R T_{2} \ln \frac{V_{3}}{V_{4}} \end{aligned}

(11.20)

Step 4 → 1, adiabatic compression from $T_{2}$ back to $T_{1}$ : the work done on the gas is

\begin{aligned} W_{4 \to 1} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.21)

The net work done by the gas in one cycle is $W_{1 \to 2} + W_{2 \to 3} - W_{3 \to 4} - W_{4 \to 1} .$ The two adiabatic works are equal, so they cancel, leaving

\begin{aligned} W & = W_{1 \to 2} - W_{3 \to 4} \\ = μ R T_{1} \ln \frac{V_{2}}{V_{1}} \\ - μ R T_{2} \ln \frac{V_{3}}{V_{4}} \end{aligned}

(11.22)

and the efficiency is

\begin{aligned} η & = \frac{W}{Q_{1}} \\ = 1 - \frac{Q_{2}}{Q_{1}} \\ = 1 - \frac{T_{2} \ln \frac{V_{3}}{V_{4}}}{T_{1} \ln \frac{V_{2}}{V_{1}}} \end{aligned}

(11.23)

The adiabats link the volumes. Applying $T V^{γ - 1} = constant$ to step 2 → 3,

\begin{array}{c} T_{1} V_{2}^{γ - 1} = T_{2} V_{3}^{γ - 1} \\ {(\frac{V_{2}}{V_{3}})}^{γ - 1} = \frac{T_{2}}{T_{1}} \end{array}

(11.24)

and to step 4 → 1,

\begin{array}{c} T_{2} V_{4}^{γ - 1} = T_{1} V_{1}^{γ - 1} \\ {(\frac{V_{1}}{V_{4}})}^{γ - 1} = \frac{T_{2}}{T_{1}} \end{array}

(11.25)

The right-hand sides are equal, so $V_{2} / V_{3} = V_{1} / V_{4}$ , which rearranges to

\frac{V_{3}}{V_{4}} = \frac{V_{2}}{V_{1}}

(11.26)

The two logarithms in Eq. (11.23) are therefore equal and cancel, leaving

Efficiency of a Carnot engine

η = 1 - \frac{T_{2}}{T_{1}}

(11.27)

The efficiency depends only on the two reservoir temperatures, which must be in kelvin. It would reach 1 only if $T_{2}$ were absolute zero, and it is zero when $T_{1} = T_{2}$ : an engine needs a temperature difference to run at all.

Example 11.7

The limit for a thermal power station

A coal-fired power station raises steam at $540^{\circ} C$ and condenses it with river water at $35^{\circ} C$ . (a) What is the Carnot efficiency for these temperatures? (b) The station delivers $500 M W$ of electrical power at an overall efficiency of 38%. At what rate does it reject heat? (c) What would that rate be for a Carnot engine with the same output?

Try it yourself first. Then reveal the solution one step at a time.

Because nothing in the Carnot cycle is irreversible, it can also be run the other way round, $1 \to 4 \to 3 \to 2 \to 1$ . The gas then soaks up $Q_{2}$ at the cold end, needs work $W$ from outside and dumps $Q_{1}$ at the hot end: the engine has become a refrigerator, and no refrigerator between the same two temperatures can do better. In fact the Carnot engine is the only reversible engine that can work between just two reservoirs.

3/3Together: no change at the source, and heat $W^{'} - W$ from the sink turned entirely into work. Forbidden.

Fig. 11.17If any engine beat the Carnot engine, this pair would turn heat from one reservoir entirely into work.NCERT Fig. 11.10 · p. 239

A hot reservoir along the top and a cold reservoir along the bottom. Step 1: engine I on the left takes heat Q1 from the hot reservoir, sends work W prime to the right and rejects Q1 minus W prime to the cold reservoir. Step 2: engine R on the right, run backwards, takes Q1 minus W from the cold reservoir and returns Q1 to the hot reservoir, driven by work W from I; the leftover W prime minus W points upward. Step 3: a dashed box encloses both; the only flows are heat W prime minus W from the cold reservoir into the box and the same amount of work out, and the hot reservoir label says no net change.

TheoremCarnot's theorem

(a) No engine working between two given reservoirs can be more efficient than a Carnot engine working between the same reservoirs. (b) The efficiency of a Carnot engine does not depend on the working substance.

Suppose some engine I, perhaps irreversible, were more efficient than a Carnot engine R working between the same source at $T_{1}$ and sink at $T_{2}$ . Run I forwards: it takes heat $Q_{1}$ from the source, delivers work $W^{'}$ and rejects $Q_{1} - W^{'}$ to the sink (Fig. 11.17).

Now run R backwards as a refrigerator, sized so that it returns exactly $Q_{1}$ to the source. As an engine taking in $Q_{1}$ it would deliver work $W$ ; reversed, it needs the same work $W$ and takes $Q_{1} - W$ from the sink. If I is more efficient, $W^{'} > W$ , so I can drive R and still have $W^{'} - W$ to spare.

Treat the pair as one machine. The source gives $Q_{1}$ to I and gets $Q_{1}$ back from R: no net change. The sink receives $Q_{1} - W^{'}$ but gives up $Q_{1} - W$ , a net loss of $W^{'} - W$ . The combination does nothing except take heat $W^{'} - W$ from a single reservoir and turn all of it into work, which the Kelvin–Planck statement forbids. So no engine can beat R. Part (b) follows in the same way: if a Carnot engine using one gas beat a Carnot engine using another, coupling the two would again break the second law.

Since every Carnot engine between $T_{1}$ and $T_{2}$ has the same efficiency, $1 - Q_{2} / Q_{1} = 1 - T_{2} / T_{1}$ whatever the working substance. Hence

Carnot cycle

\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}

(11.28)

Here $Q_{1}$ and $Q_{2}$ are the heats exchanged isothermally with the hot and cold reservoirs. Because Eq. (11.28) mentions no substance at all, it can serve as the definition of a temperature scale: the ratio of two temperatures is the ratio of the heats a Carnot engine exchanges at them. This thermodynamic temperature scale does not depend on the properties of mercury, of a gas or of any other thermometric material, and for an ideal gas it agrees with the ideal gas temperature.

Example 11.8

An ideal refrigerator

A refrigerator keeps its inside at $5^{\circ} C$ in a kitchen at $35^{\circ} C$ . Suppose it worked as a reversed Carnot engine. (a) What would its coefficient of performance be? (b) How much work would it need to remove $1.0 k J$ of heat from the food, and how much heat would it release into the kitchen?

Try it yourself first. Then reveal the solution one step at a time.

Explore a Carnot engine yourself. Change the gas and the expansion ratio and watch $Q_{1}$ , $Q_{2}$ and $W$ change, while the efficiency stays pinned to $1 - T_{2} / T_{1}$ .

InteractiveCarnot engine lab

One mole of ideal gas runs a Carnot cycle starting at 10 litres. Sliders set the hot reservoir temperature from 400 to 800 kelvin, the cold reservoir temperature from 250 to 390 kelvin and the isothermal expansion ratio; buttons choose a monatomic or diatomic gas. Play moves the state round the loop on a pressure against volume graph while a cylinder sits on the hot reservoir, an insulating stand or the cold reservoir. Cards show Q1, Q2, the work, the efficiency and one minus T2 over T1, which always agree.

Fig. 11.18Set the two reservoir temperatures, the expansion ratio and the gas, then press Play one cycle.

Quick check

1 of 4

What is the efficiency of a Carnot engine working between

527^{\circ} C

and

127^{\circ} C

, as a percentage?

Show answer

Answer 50 %

T_{1} = 800 K

T_{2} = 400 K

, so

η = 1 - 400 / 800 = 0.50.

Using Celsius values (

1 - 127 / 527 = 0.76

) is the classic mistake.

Recap

Thermodynamics: key points and summary

In thermal equilibrium the bulk variables stop changing. The zeroth law (two systems each in equilibrium with a third are in equilibrium with each other) defines temperature (Section 11.3).
Internal energy $U$ is the molecular kinetic and potential energy in the centre-of-mass frame, and it is a state variable. Heat, driven by a temperature difference, and work are energy in transit (Section 11.4).
First law: $Δ Q = Δ U + Δ W$ , with $Δ Q$ the heat supplied to the system and $Δ W$ the work done by it; at constant pressure $Δ W = P Δ V$ (Section 11.5).
$s = Δ Q / (m Δ T)$ and $C = Δ Q / (μ Δ T)$ . Equipartition gives $C = 3 R$ for solids, which fits most at ordinary temperatures (Section 11.6).
The calorie is defined with water warmed through the degree between ${14.5}^{\circ} C$ and ${15.5}^{\circ} C$ ; $1 c a l = 4.186 J$ (Specific heat capacity of water).
For an ideal gas $C_{p} - C_{v} = R$ (Specific heats of an ideal gas).
State variables describe equilibrium states and are linked by an equation of state such as $P V = μ R T$ . Intensive variables stay the same when a system is halved; extensive ones halve (Section 11.7).
A quasi-static process is infinitely slow and passes only through equilibrium states (Section 11.8.1).
Isothermal ideal gas: $Q = W = μ R T \ln (V_{2} / V_{1})$ (Section 11.8.2). Adiabatic: $P V^{γ}$ constant and $W = μ R (T_{1} - T_{2}) / (γ - 1)$ (Section 11.8.3).
Isochoric: $W = 0$ . Isobaric: $W = P Δ V = μ R Δ T$ . Cyclic: $Δ U = 0$ , and the net work is the area of the loop (Section 11.8.6).
Second law: no engine can turn all its heat into work (Kelvin–Planck), and heat cannot flow from cold to hot by itself (Clausius); so $η < 1$ and $α$ is finite (Section 11.9).
Spontaneous processes are irreversible. A reversible process must be quasi-static and free of dissipation (Section 11.10).
A Carnot engine (two isotherms, two adiabats) has $η = 1 - T_{2} / T_{1}$ , the highest possible between two temperatures for any working substance; also $Q_{2} / Q_{1} = T_{2} / T_{1}$ (Section 11.11).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 11.1	Practice Q1
NCERT Exercise 11.2	Practice Q2
NCERT Exercise 11.3	Practice Q3
NCERT Exercise 11.4	Practice Q4
NCERT Exercise 11.5	Practice Q5
NCERT Exercise 11.6	Practice Q6
NCERT Exercise 11.7	Practice Q7
NCERT Exercise 11.8	Practice Q8

Quick check: MCQs on Thermodynamics

11.1 Introduction2 questions
11.2 Thermal equilibrium2 questions
11.3 Zeroth law of thermodynamics3 questions
11.4 Heat, internal energy and work3 questions
11.5 First law of thermodynamics3 questions
Specific heats of an ideal gas3 questions
11.7 Thermodynamic state variables and equation of state3 questions
11.8.3 Adiabatic process3 questions
11.8.6 Cyclic process3 questions
11.9 Second law of thermodynamics3 questions
11.10 Reversible and irreversible processes2 questions
11.11 Carnot engine4 questions

Glossary · 15

Key terms (English–Hindi)

adiabatic processरुद्धोष्म प्रक्रमruddhoshm prakram: A process with no heat exchange; for a quasi-static ideal gas $P V^{γ}$ stays constant.First seen in the chapter →

Carnot engineकार्नो इंजनkaarno injan: A reversible engine working between two reservoirs through two isothermal and two adiabatic steps; its efficiency is $1 - T_{2} / T_{1}$ .First seen in the chapter →

diathermic wall: A wall that lets heat pass, such as a thin copper sheet. A wall that blocks heat is called adiabatic.First seen in the chapter →

first law of thermodynamics: Energy conservation for a system: heat supplied equals the rise in internal energy plus the work done by the system, $Δ Q = Δ U + Δ W$ .First seen in the chapter →

heatऊष्माushma: Energy that flows between a system and its surroundings because of a temperature difference.First seen in the chapter →

internal energyआंतरिक ऊर्जाaantarik oorja: The total kinetic and potential energy of the molecules of a system, measured in the frame where the system as a whole is at rest; symbol $U$ .First seen in the chapter →
isothermal processसमतापीय प्रक्रमsamtaapiya prakram: A process at constant temperature; for an ideal gas $P V$ stays constant and $Δ U = 0$ .First seen in the chapter →

molar specific heat capacity: Heat needed per mole per kelvin rise in temperature, $C = Δ Q / (μ Δ T)$ , in $J m o l^{- 1} K^{- 1}$ .First seen in the chapter →

quasi-static processस्थैतिककल्प प्रक्रमsthaitikkalp prakram: An idealised, infinitely slow process in which the system passes only through equilibrium states.First seen in the chapter →

reversible processउत्क्रमणीय प्रक्रमutkramaniya prakram: A process that can be run backwards so that the system and surroundings both return to their starting states with no change elsewhere.First seen in the chapter →

specific heat capacityविशिष्ट ऊष्मा धारिताvishisht ushma dhaarita: Heat needed per unit mass per kelvin rise in temperature, $s = Δ Q / (m Δ T)$ , in $J k g^{- 1} K^{- 1}$ .First seen in the chapter →
state variableअवस्था चरavastha char: A quantity such as $P$ , $V$ , $T$ or $U$ whose value is fixed by the equilibrium state alone, not by how the state was reached.First seen in the chapter →

thermal equilibriumतापीय साम्यtaapiya saamya: The state of two systems in thermal contact when no net heat flows between them, so their macroscopic variables stop changing.First seen in the chapter →
thermodynamicsऊष्मागतिकीushmagatiki: The branch of physics that relates heat, work, temperature and energy using a few measurable bulk variables.First seen in the chapter →

zeroth law of thermodynamics: If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.First seen in the chapter →

Thermodynamics: Class 11 Physics · Chapter 11

What you will learn in Thermodynamics

11.1 Introduction

11.2 Thermal equilibrium

11.3 Zeroth law of thermodynamics

11.4 Heat, internal energy and work

11.5 First law of thermodynamics

11.6 Specific heat capacity

Specific heat capacity of water

Specific heats of an ideal gas

11.7 Thermodynamic state variables and equation of state

11.8 Thermodynamic processes

11.8.1 Quasi-static process

11.8.2 Isothermal process

11.8.3 Adiabatic process

11.8.4 Isochoric process

11.8.5 Isobaric process

11.8.6 Cyclic process

11.9 Second law of thermodynamics

11.10 Reversible and irreversible processes

11.11 Carnot engine

Thermodynamics: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Thermodynamics

Key terms (English–Hindi)

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