Energy Changes with Heating

In Methods of Heating and Heating you studied energy transfers for solids and liquids changing temperature and state. Here we study those same ideas — energy in, temperature and state changing — in much more depth, but for one state of matter only: the gas. Gases are the easiest state to model mathematically, and it is historically no accident that the science of thermodynamics itself grew out of trying to understand gases in steam engines. Before we can study how a gas's energy changes as it is heated, we need a model of a gas we can actually calculate with — the ideal gas. We won't derive that model from scratch; we'll take its assumptions and its key equation as given, and use them as the foundation for everything that follows: the first law of thermodynamics, the different ways a gas can be heated or compressed, heat engines, and finally entropy and the second law.

By the end of this workbook you should be able to:

State the assumptions of the ideal gas model, and use n = N/N_A to relate amount of substance, number of particles and the Avogadro constant
Use the ideal gas law PV = Nk_BT and PV = nRT, and explain how it follows from the empirical gas laws
Calculate the internal energy of an ideal monatomic gas from U = 3/2 Nk_BT = 3/2 nRT
Apply the first law of thermodynamics, Q = ΔU + W, with the correct sign convention
Identify and calculate work, heat and internal energy change for isothermal, isovolumetric, adiabatic and isobaric processes
Interpret PV diagrams for thermodynamic cycles, and calculate the efficiency of a heat engine and of a Carnot cycle
Explain entropy and the second law of thermodynamics, both in macroscopic and microscopic (microstate) terms

Name

1. A model for a gas

Before we can study how the energy of a gas changes as it is heated, cooled, compressed or expanded, we need a way to describe the physical state of a gas — and a model of what a gas actually is at the particle level. This section sets up both: the measurable quantities pressure and amount of substance, and the kinetic theory model of an ideal gas.

Pressure

A force acting on a surface has an effect that depends not just on its size, but on the area over which it acts. Pressure is defined as the force acting perpendicular (normal) to a surface, per unit area of that surface.

P = F / A

Key idea. Pressure, P = F/A, where F is the force exerted perpendicular to the surface. SI unit: the pascal, Pa (1 Pa = 1 N m⁻²).

Worked example 1.1

A student of mass 58 kg stands on one foot. The area of contact between her shoe and the floor is 130 cm². Calculate the pressure she exerts on the floor.

Answer:
P = F/A = mg/A = (58 × 9.81) / (130 × 10⁻⁴)

Amount of substance

A gas sample typically contains an enormous number of particles — far too many to count individually. Instead we measure the amount of substance, n, in moles.

n = N / N_A

Key idea. The amount of substance, n, as given by n = N/N_A, where N is the number of particles (atoms or molecules) and N_A is the Avogadro constant, N_A = 6.02 × 10²³ mol⁻¹.

Mole, mol: the SI unit of amount of substance. One mole of a substance contains exactly the Avogadro constant's worth of its defining particles (6.02 × 10²³ atoms or molecules).

If you know the mass of a sample and its molar mass (the mass of one mole, in g mol⁻¹), you can find n directly: n = mass / molar mass. Combined with n = N/N_A, this lets you move between mass, moles and number of particles for any sample of gas.

Worked example 1.2

A sample of neon gas (molar mass 20.2 g mol⁻¹) has a mass of 15.0 g. Calculate (a) the number of moles of neon in the sample, (b) the number of neon atoms in the sample.

Answer:
(a) n = mass / molar mass = 15.0 / 20.2
(b) N = n × N_A = (15.0 / 20.2) × 6.02 × 10²³

The ideal gas model

To predict how a gas behaves, physicists use a simplified, idealized model: the ideal gas. Real gases only approximate this model, but under most everyday conditions the approximation is excellent — good enough that the equations built on it (which we meet in Section 2) accurately predict how real gases behave.

Key idea. An ideal gas is described by the following assumptions:

it contains a very large number of identical particles;
the volume of the particles themselves is negligible compared with the volume occupied by the gas;
the particles move in random directions, with a wide range of speeds;
there are no forces between particles except during collisions, so an ideal gas has no (electrical) potential energy between its particles;
the particles obey Newton's laws of motion, and all collisions (between particles, and between particles and the container walls) are perfectly elastic.

Fig. 1.1 A gas exerts pressure because its particles are constantly colliding with the walls of their container and rebounding (amber arrows show the rebound direction). By Newton's third law, each collision also exerts an equal and opposite force on the wall itself, perpendicular to its surface; the pressure is the average effect of an enormous number of these collisions, spread over the wall's area. (This model can be developed mathematically to show that pressure is related to the average of (particle speed)², P = ⅓ρv² — see your presentation for that derivation. The essential idea for us is simpler: pressure is a large-scale, measurable effect of huge numbers of random particle collisions.)

Worth knowing Real gases vs the ideal gas model. No real gas is perfectly ideal, but most real gases behave very close to an ideal gas under normal conditions. The model breaks down at high pressure or high density (particles are pushed close enough together that their own volume, and the forces between them, are no longer negligible) and at low temperature (particles move slowly enough that intermolecular forces start to matter — this is exactly what causes a real gas to condense into a liquid).

Check your understanding

1State four assumptions of the kinetic theory model of an ideal gas.

Any four of: the gas contains a very large number of identical particles; the volume of the particles is negligible compared with the volume of the gas; the particles move randomly with a wide range of speeds; there are no forces between particles except during collisions; the particles obey Newton's laws of motion; all collisions are perfectly elastic.

2Using the model in Fig. 1.1, explain why the pressure of a fixed mass of gas increases if it is compressed into a smaller volume at constant temperature.

At constant temperature the particles' average speed is unchanged. In a smaller volume, the particles have less distance to travel between collisions with the walls, so they collide with a given area of wall more frequently. More collisions per second per unit area means a greater average force per unit area — that is, a greater pressure.

3A gas cylinder contains 3.2 kg of oxygen gas (O₂, molar mass 32.0 g mol⁻¹). Calculate (a) the number of moles of oxygen in the cylinder, (b) the number of oxygen molecules in the cylinder.

(a) n = mass / molar mass = 3200 / 32.0
(b) N = n × N_A = (3200 / 32.0) × 6.02 × 10²³

4Explain why the ideal gas model is a good approximation to a real gas at low density, but a poor approximation at high density or low temperature.

At low density the particles are, on average, far apart, so the volume of the particles themselves really is negligible compared with the volume of the container, and the particles spend almost all their time far enough apart that intermolecular forces are negligible — matching the assumptions of the model closely. At high density the particles are forced close together, so their own volume and the forces between them are no longer negligible. At low temperature the particles move more slowly, so intermolecular forces (which the model assumes are zero) have more time to act between collisions — this is what eventually causes a real gas to condense into a liquid, a state the ideal gas model cannot describe at all.

2. The ideal gas law

A fixed amount of gas, n, sealed in a container, can be fully described by three measurable quantities: its pressure P, volume V and temperature T. Experiments carried out over the last few centuries — keeping one of these quantities fixed at a time — found simple, direct relationships between the other two. These are the empirical gas laws: "empirical" because they come purely from observation and experiment, not from a theoretical model.

Law	Held constant	Relationship
Boyle's law	temperature, T	P ∝ 1/V, i.e. P₁V₁ = P₂V₂
Pressure law	volume, V	P ∝ T (K), i.e. P₁/T₁ = P₂/T₂
Charles' law	pressure, P	V ∝ T (K), i.e. V₁/T₁ = V₂/T₂

Key idea. In every empirical gas law, temperature must be measured in kelvin, not degrees Celsius — the proportionality only holds on the absolute scale.

Explore each gas law

Boyle's law: change the volume of the gas and watch how the pressure responds at constant temperature.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by Dr Jones Physics.

The pressure law: heat the gas at constant volume and watch how the pressure responds.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by Dr Jones Physics.

Charles' law: heat the gas at constant pressure and watch how the volume responds.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by Dr Jones Physics.

Worked example 2.1

A sample of gas has a volume of 480 cm³ at a pressure of 1.0 × 10⁵ Pa. The gas is compressed at constant temperature until its pressure is 2.4 × 10⁵ Pa. Calculate its new volume.

Answer:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂ = (1.0 × 10⁵ × 480) / (2.4 × 10⁵)

Combining the gas laws

The three empirical laws are all special cases of one more general relationship: for a fixed amount of gas, PV/T is constant.

PV / T = constant, or P₁V₁/T₁ = P₂V₂/T₂

Key idea. The ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature: PV/T = constant.

This tells us how P, V and T are related for a fixed amount of gas — but it says nothing about how much gas there is. Experiments show that, at constant temperature and pressure, volume is directly proportional to the amount of gas, n. Combining this with PV/T = constant gives the full ideal gas law (also called the equation of state for an ideal gas):

PV = nRT

Key idea. The equations governing the behaviour of ideal gases: PV = Nk_BT and PV = nRT, where R is the universal (molar) gas constant, R = 8.31 J K⁻¹ mol⁻¹, and k_B is the Boltzmann constant, k_B = 1.38 × 10⁻²³ J K⁻¹. Since n = N/N_A, the two forms are linked by k_B = R/N_A — R is simply the macroscopic (per mole) version of the same constant that k_B expresses per particle.

An ideal gas is defined as a gas that obeys PV = nRT perfectly, for any P, V, T and n. As you saw in Section 1, real gases obey this equation very closely under most everyday conditions.

Worked example 2.2

A cylinder of volume 8.0 × 10⁻³ m³ contains an ideal gas at a pressure of 3.0 × 10⁵ Pa and a temperature of 290 K. Calculate the amount of gas, in moles, in the cylinder.

Answer:
PV = nRT
n = PV / RT = (3.0 × 10⁵ × 8.0 × 10⁻³) / (8.31 × 290)

Representing the state of a gas: PV diagrams

Because P, V and T are all linked by PV = nRT, a fixed amount of gas can be represented as a single point on a graph of pressure against volume — a PV diagram. A curve of constant temperature on a PV diagram is called an isotherm; every point on it obeys Boyle's law, PV = constant. PV diagrams will become an essential tool from Section 5 onwards, for tracking exactly how a gas's state changes as it is heated, compressed or allowed to expand.

Explore how pressure, volume, temperature and the number of particles are all linked for an ideal gas, using PhET's Gas Properties simulation. Try holding volume constant and heating the gas, or pushing the wall in at constant temperature, and watch the pressure gauge respond.

Launch simulation ↗

Simulation: Gas Properties by PhET Interactive Simulations, University of Colorado Boulder.

Check your understanding

5State Boyle's law, in words, and give its equation.

For a fixed amount of gas at constant temperature, the pressure is inversely proportional to the volume: P ∝ 1/V, or P₁V₁ = P₂V₂.

6A fixed mass of gas at a pressure of 1.4 × 10⁵ Pa and a temperature of 15 °C occupies a volume of 620 cm³. The gas is heated at constant volume until its pressure is 1.9 × 10⁵ Pa. Calculate its final temperature in °C.

Convert to kelvin first: T₁ = 15 + 273 = 288 K.
P₁/T₁ = P₂/T₂
T₂ = P₂T₁ / P₁ = (1.9 × 10⁵ × 288) / (1.4 × 10⁵)
Then subtract 273 to convert the answer back to °C.

7A weather balloon contains 4.5 mol of helium at a pressure of 1.0 × 10⁵ Pa and a temperature of 293 K. Calculate the volume of the balloon.

PV = nRT
V = nRT / P = (4.5 × 8.31 × 293) / (1.0 × 10⁵)

8Explain why PV = nRT can be written as PV = Nk_BT, and state how k_B and R are related.

The amount of substance n is related to the number of particles N by n = N/N_A. Substituting into PV = nRT gives PV = (N/N_A)RT = N(R/N_A)T. Defining k_B = R/N_A gives PV = Nk_BT. k_B is simply R "per particle" rather than "per mole": k_B = R/N_A = 8.31/(6.02 × 10²³).

3. Internal energy of an ideal gas

Recall from Heating that the internal energy of a substance is the sum of the total random kinetic energy and total random potential energy of all its particles. For an ideal gas this simplifies dramatically: because the model assumes there are no forces between particles (Section 1), an ideal gas has no potential energy at all. Its internal energy is purely the total random translational kinetic energy of its particles.

Key idea. The internal energy, U, of an ideal monatomic gas is related to the number of molecules (or amount of substance) by:

U = 3/2 Nk_BT = 3/2 nRT

This result follows from combining the ideal gas law PV = Nk_BT with the kinetic theory relationship between pressure and the particles' average (speed)² (Section 1) — on average, each particle in an ideal gas has translational kinetic energy 3/2 k_BT, so N particles together have total internal energy 3/2 Nk_BT.

Key idea. Because it has no potential energy, the internal energy of an ideal gas depends only on its temperature — not on its pressure or volume individually. Two samples of the same ideal gas at the same temperature have the same internal energy, however different their pressure and volume happen to be. This single fact will be essential when we look at different ways of heating a gas in Sections 4 and 5.

Worth knowing Monatomic vs molecular gases. U = 3/2 Nk_BT applies to a monatomic ideal gas (single atoms, such as helium or argon), whose particles can only have translational kinetic energy. Most real gases are molecular (O₂, N₂, CO₂...), and their molecules can also rotate and vibrate, storing extra internal energy in those motions. This makes their internal energy larger, for the same n and T, than the monatomic formula predicts — but the monatomic case is the one this course focuses on, since it is the simplest to calculate exactly.

Worked example 3.1

Calculate the internal energy of 2.0 mol of a monatomic ideal gas at a temperature of 310 K.

Answer:
U = 3/2 nRT = 1.5 × 2.0 × 8.31 × 310

Worked example 3.2

The temperature of 0.40 mol of a monatomic ideal gas increases from 288 K to 320 K. Calculate the change in internal energy of the gas.

Answer:
ΔU = 3/2 nRΔT = 1.5 × 0.40 × 8.31 × (320 − 288)

Check your understanding

9Calculate the total internal energy of 1.0 mol of a monatomic ideal gas at 0 °C.

U = 3/2 nRT = 1.5 × 1.0 × 8.31 × 273

10Explain why the internal energy of an ideal gas depends only on its temperature, and not on its pressure or volume individually.

An ideal gas has no potential energy, because the model assumes there are no forces between its particles. So its internal energy is entirely kinetic. The average translational kinetic energy of a particle, 3/2 k_BT, is set only by the temperature — it doesn't matter what volume the gas occupies or what pressure it happens to be at. So for a fixed n, U = 3/2 nRT depends only on T.

11A sample of 2.5 × 10²⁴ atoms of a monatomic ideal gas has a total internal energy of 5200 J. Calculate the temperature of the gas.

U = 3/2 Nk_BT
T = 2U / (3Nk_B) = (2 × 5200) / (3 × 2.5 × 10²⁴ × 1.38 × 10⁻²³)

120.80 mol of a monatomic ideal gas is cooled from 400 K to 250 K. Calculate the change in internal energy of the gas, and state whether internal energy was transferred into or out of the gas.

ΔU = 3/2 nRΔT = 1.5 × 0.80 × 8.31 × (250 − 400)
Since ΔT is negative, ΔU is negative — the internal energy decreased, so energy was transferred out of the gas.

4. The first law of thermodynamics

We now have everything we need to describe how a gas's energy changes as it is heated. There are exactly two ways energy can be transferred into or out of a gas: as heating (thermal energy transfer, Q) and as work (mechanical energy transfer, W, done as the gas expands or is compressed).

Work done when a gas changes volume

Picture a gas trapped in a cylinder by a frictionless, gas-tight piston of area A. If the gas expands, pushing the piston out a small distance Δs at constant pressure P, the gas does work on the piston: W = (force) × (distance) = (PA) × Δs. Since the change in volume ΔV = AΔs, this becomes:

W = PΔV

Key idea. The work done by or on a closed system, W = PΔV, when its boundaries are changed, can be described in terms of pressure and change of volume of the system. If the pressure is not constant, the work done is equal to the area under the graph of the process on a PV diagram.

Fig. 4.1 A gas expanding at constant pressure P pushes its piston (area A, right edge shown in navy) from its initial position (dashed line) to its final position (solid grey), a distance Δs. The work done by the gas on the piston is W = (PA)Δs = PΔV, where ΔV is the extra volume — shaded blue — the gas now occupies.

Worked example 4.1

A gas expands from a volume of 0.020 m³ to 0.045 m³ at a constant pressure of 1.4 × 10⁵ Pa. Calculate the work done by the gas.

Answer:
W = PΔV = 1.4 × 10⁵ × (0.045 − 0.020)

The first law of thermodynamics

If an amount of thermal energy Q is transferred into a gas, the principle of conservation of energy tells us that this energy can only do one of two things: increase the gas's internal energy, and/or let the gas do work on its surroundings as it expands. This is the first law of thermodynamics:

Q = ΔU + W

Key idea. The first law of thermodynamics, Q = ΔU + W, results from applying conservation of energy to a closed system, and relates the internal energy of a system to the transfer of energy as heat and as work.

Common mistake Watch your signs. Thermal energy transferred into the gas: +Q; transferred out of the gas: −Q. An increase in internal energy: +ΔU; a decrease: −ΔU. Work done by the gas as it expands: +W; work done on the gas as it is compressed: −W. Get in the habit of writing every term with its sign in brackets, as in the worked examples below.

Worked example 4.2

80 J of work is done by a gas as it expands, while 150 J of thermal energy is transferred into it. Calculate the change in internal energy of the gas.

Answer:
Q = ΔU + W
(+150) = ΔU + (+80)
ΔU = 150 − 80

Combining the first law with the internal energy equation from Section 3 lets us connect heating, work and temperature change for an ideal monatomic gas:

ΔU = 3/2 Nk_BΔT = 3/2 nRΔT

Key idea. For an ideal monatomic gas, the change in internal energy, ΔU = 3/2 Nk_BΔT = 3/2 nRΔT, is related to the change in its temperature.

Check your understanding

13State the first law of thermodynamics, and explain what each of the four sign conventions (+Q, −Q, +W, −W) represents physically.

Q = ΔU + W. +Q means thermal energy is transferred into the gas; −Q means it is transferred out. +ΔU means the internal energy of the gas increases; −ΔU means it decreases. +W means the gas does work on the surroundings as it expands; −W means work is done on the gas as it is compressed.

14A gas is compressed from a volume of 0.060 m³ to 0.035 m³ at a constant pressure of 2.0 × 10⁵ Pa. At the same time, 3200 J of thermal energy is removed from the gas. Calculate (a) the work done on the gas, (b) the change in internal energy of the gas.

(a) W = PΔV = 2.0 × 10⁵ × (0.035 − 0.060). This is negative, showing that work is done on the gas as it is compressed.
(b) Q = ΔU + W, with Q = −3200 J and W as found in (a). ΔU = Q − W = (−3200) − [2.0 × 10⁵ × (0.035 − 0.060)]

150.60 mol of a monatomic ideal gas is heated at constant pressure. 940 J of thermal energy is transferred into the gas, and it does 310 J of work as it expands. Calculate the temperature change of the gas.

Q = ΔU + W, so ΔU = Q − W = 940 − 310.
ΔU = 3/2 nRΔT, so ΔT = ΔU / (1.5 × n × R) = (940 − 310) / (1.5 × 0.60 × 8.31)

16Explain why, for an ideal gas held at constant volume, all of the thermal energy transferred into the gas goes into increasing its internal energy.

At constant volume, ΔV = 0, so no work is done (W = PΔV = 0) — the gas cannot expand or be compressed. The first law then reduces to Q = ΔU, so every joule of thermal energy transferred in becomes an increase in internal energy (and so, for an ideal gas, an increase in temperature).

5. Four thermodynamic processes

Among all the possible ways a gas's state could change, it is useful to consider four extreme, special cases — each found by holding one of the four quantities in the first law fixed. In each case, the general equation Q = ΔU + W simplifies.

The prefix "iso-" comes from the Greek for "equal" — so isothermal, isovolumetric and isobaric each describe a process where one particular quantity stays exactly the same (constant) throughout.

ΔU = 0 — isothermal (constant temperature)
W = 0 — isovolumetric (constant volume)
Q = 0 — adiabatic (no thermal energy transfer)
ΔP = 0 — isobaric (constant pressure)

Fig. 5.1 The four thermodynamic processes, each starting from the same state A on a PV diagram. The dashed grey curves are neighbouring isotherms, shown for reference — the solid teal curve is the actual path followed in each case. Notice that the adiabatic curve (c) is steeper than an isotherm through the same points (compare to the dashed isotherm passing through A in that panel): because the temperature is also falling as an adiabatic expansion proceeds, the pressure drops faster than it would at constant temperature.

Isothermal (ΔU = 0)

The temperature — and so the internal energy — of the gas does not change. The first law reduces to Q = W: all the work done by the gas during an isothermal expansion is supplied by thermal energy transferred in; during a compression, the work done on the gas is all transferred away as thermal energy. For a process to genuinely approximate an isothermal change, it must happen slowly enough for the gas to stay in thermal equilibrium with its surroundings throughout. Isothermal changes obey Boyle's law: PV = constant.

Key idea. In an isothermal change: Q = 0 + W, or Q = W.

Isovolumetric (W = 0)

There is no change in volume, so no work is done by or on the gas. The first law reduces to Q = ΔU: any thermal energy transferred into the gas becomes an increase in internal energy (and temperature); any thermal energy transferred out becomes a decrease.

Key idea. In an isovolumetric change: Q = ΔU + 0, or Q = ΔU.

Adiabatic (Q = 0)

No thermal energy is transferred between the gas and its surroundings at all. The first law reduces to 0 = ΔU + W, so ΔU = −W: in an adiabatic expansion the gas does work using only its own internal energy, so its temperature falls; in an adiabatic compression, all the work done on the gas becomes an increase in internal energy, so its temperature rises. For a process to approximate an adiabatic change, it must happen rapidly, in a well-insulated container, so there is no time for thermal energy to be transferred.

Key idea. In an adiabatic change: 0 = ΔU + W, so ΔU = −W for a compression, and −ΔU = W for an expansion. Adiabatic processes in monatomic ideal gases can be modelled by the equation PV^5/3 = constant.

Isobaric (ΔP = 0)

The pressure stays constant as the gas expands or is compressed — this happens whenever a gas is free to change volume while its temperature changes, keeping its pressure equal to that of its surroundings. The general first law equation applies with no simplification, and the work done is simply W = PΔV.

Key idea. In an isobaric change: Q = ΔU + W (and W = PΔV).

Process	Held constant	First law reduces to
Isothermal	temperature (ΔU = 0)	Q = W
Isovolumetric	volume (W = 0)	Q = ΔU
Adiabatic	— (Q = 0)	ΔU = −W
Isobaric	pressure (ΔP = 0)	Q = ΔU + W

Worked example 5.1

An ideal monatomic gas is compressed adiabatically, reducing its volume by a factor of 6.0. Determine the factor by which its pressure increases.

Answer:
PV^5/3 = constant, so P₁V₁^5/3 = P₂V₂^5/3
P₂/P₁ = (V₁/V₂)^5/3 = 6.0^5/3

Worked example 5.2

A gas is compressed isothermally, so that the work done on it is 640 J. State the thermal energy transferred, and whether it is transferred into or out of the gas.

Answer:
Isothermal: Q = W. Since work is done on the gas, W = −640 J.
Q = W = −640 J: 640 J of thermal energy is transferred out of the gas.

Live simulation: PV diagram explorer

Drive a gas along each process

Process

Drive the process: 0%

Choose a process and drag the slider.

Fig. 5.2 Choose a process and drag the slider to drive the gas from state A along that process. The readout shows the resulting P, V and the signs of Q, ΔU and W at each stage.

Check your understanding

17A gas is heated at constant volume. Explain what happens to its pressure and temperature, and state the values of Q, ΔU and W that apply (in terms of each other).

This is an isovolumetric process. W = 0, since there is no volume change. Q = ΔU: all the thermal energy transferred in increases the internal energy of the gas, so its temperature rises. Since PV = nRT and V is constant, P is proportional to T, so the pressure also rises.

18An ideal monatomic gas expands adiabatically. Using the first law, explain why its temperature must fall.

Adiabatic means Q = 0, so 0 = ΔU + W. As the gas expands it does positive work on its surroundings (+W), so ΔU = −W must be negative. Since ΔU = 3/2 nRΔT for an ideal gas, a negative ΔU means a negative ΔT — the temperature falls. Physically, the gas can only do work by drawing on its own internal energy, since no thermal energy is being supplied from outside.

19The volume of an ideal monatomic gas is increased in an adiabatic expansion from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³. If the initial pressure was 3.4 × 10⁵ Pa, calculate the final pressure.

P₁V₁^5/3 = P₂V₂^5/3
P₂ = P₁(V₁/V₂)^5/3 = 3.4 × 10⁵ × (2.0 × 10⁻³ / 5.0 × 10⁻³)^5/3

20Copy and complete: for each process below, state whether Q, ΔU and W are positive (+), negative (−), zero (0), or cannot be determined without more information (±). Isothermal expansion; adiabatic compression; isobaric expansion; isovolumetric pressure decrease.

Isothermal expansion: ΔU = 0, W = +, Q = +.
Adiabatic compression: Q = 0, W = −, ΔU = +.
Isobaric expansion: ΔP = 0, W = +, ΔU = +, Q = + (from Q = ΔU + W, both positive).
Isovolumetric pressure decrease: W = 0, ΔU = −, Q = − (temperature and pressure fall together at constant volume).

6. Heat engines and the Carnot cycle

A gas doing useful work by expanding — pushing a piston, say — cannot expand forever. To do continuous useful work, it must be compressed back to its starting state and expanded again, over and over: a thermodynamic cycle. A device that does this to convert a flow of thermal energy into useful mechanical work is a heat engine.

Fig. 6.1 Energy flow in a heat engine. A temperature difference (T_h − T_c) drives a flow of thermal energy: Q_h flows out of the hot reservoir into the engine, Q_c flows out of the engine into the cold reservoir, and the difference is transferred as useful mechanical work, W.

Key idea. Cyclic gas processes are used to run heat engines. Because the gas returns to its starting state at the end of each cycle, the net work done by the gas over one complete cycle is equal to the area enclosed by the loop on a PV diagram.

Efficiency of a heat engine

By conservation of energy, the thermal energy taken in from the hot reservoir either becomes useful work or is rejected to the cold reservoir: Q_h = W + Q_c. The efficiency of a heat engine compares the useful work produced with the energy that had to be supplied to produce it:

η = useful work / input energy = W / Q_h = (Q_h − Q_c)/Q_h = 1 − Q_c/Q_h

Key idea. A heat engine can respond to different cycles, and is characterized by its efficiency, η = useful work / input energy.

Worked example 6.1

A heat engine takes in 4.0 × 10⁴ J of thermal energy from its hot reservoir each cycle, and rejects 2.4 × 10⁴ J to its cold reservoir. Calculate (a) the useful work done per cycle, (b) the efficiency of the engine.

Answer:
(a) W = Q_h − Q_c = 4.0 × 10⁴ − 2.4 × 10⁴
(b) η = W/Q_h = (4.0 × 10⁴ − 2.4 × 10⁴) / 4.0 × 10⁴

The Carnot cycle

No real engine can convert 100% of the thermal energy it takes in into useful work — but how efficient could an engine possibly be, in principle, operating between two given reservoir temperatures? The idealized, reversible cycle that achieves the maximum theoretically possible efficiency is called the Carnot cycle, proposed by Sadi Carnot in 1824.

Key idea. The Carnot cycle is a four-stage process: an isothermal expansion at T_h (A→B), followed by an adiabatic expansion (B→C) that cools the gas to T_c; then an isothermal compression at T_c (C→D), followed by an adiabatic compression (D→A) that returns the gas to T_h, closing the cycle.

Fig. 6.2 The Carnot cycle plotted precisely on a PV diagram, for one mole of a monatomic ideal gas with T_h = 500 K and T_c = 250 K. A→B: isothermal expansion at T_h (teal). B→C: adiabatic expansion, cooling the gas to T_c (navy). C→D: isothermal compression at T_c (teal). D→A: adiabatic compression, returning the gas to T_h (navy), closing the cycle. Notice the adiabatic legs (B→C and D→A) are steeper and span a wider range of volume than the isothermal legs — because temperature is also falling as the gas expands adiabatically, pressure drops faster than it would at constant temperature. The area enclosed is the net useful work done per cycle.

Because it is idealized and fully reversible (every stage happens infinitely slowly, with no friction or other losses), the Carnot cycle achieves the maximum efficiency thermodynamically possible for any heat engine operating between temperatures T_h and T_c:

η_Carnot = 1 − T_c/T_h

Key idea. The Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs: η_Carnot = 1 − T_c/T_h. (Temperatures must be in kelvin.) We will see why this equation holds when we look at entropy in Section 7.

Worth knowing No real engine reaches η_Carnot. Real engines are never perfectly reversible — friction, turbulence and rapid (non-quasi-static) changes all make real cycles less efficient than the Carnot limit for the same T_h and T_c. The Carnot efficiency is a ceiling, not a target that can be reached in practice.

Worked example 6.2

Calculate the maximum theoretical efficiency of a heat engine operating between a hot reservoir at 500 °C and a cold reservoir at 50 °C.

Answer:
η_Carnot = 1 − T_c/T_h = 1 − (50 + 273)/(500 + 273)

Live simulation: Carnot efficiency calculator

Vary the reservoir temperatures

Hot reservoir, T_h: 500 K

Cold reservoir, T_c: 250 K

η = 50%

Fig. 6.3 Drag the sliders to see how the reservoir temperatures set the maximum possible efficiency of a heat engine. Notice how much a lower T_c helps compared with the same increase in T_h — can you see why, from the equation?

Check your understanding

21Describe the four stages of a Carnot cycle, in order, and state what happens to the gas's temperature during each.

A→B: isothermal expansion at T_h — temperature constant. B→C: adiabatic expansion — temperature falls from T_h to T_c. C→D: isothermal compression at T_c — temperature constant. D→A: adiabatic compression — temperature rises from T_c back to T_h, closing the cycle.

22A heat engine takes in 8.5 × 10⁵ J of thermal energy per cycle and has an efficiency of 0.35. Calculate the useful work done per cycle and the thermal energy rejected to the cold reservoir per cycle.

η = W/Q_h, so W = ηQ_h = 0.35 × 8.5 × 10⁵.
Q_c = Q_h − W = 8.5 × 10⁵ − (0.35 × 8.5 × 10⁵)

23A power station's steam turbine operates with an inlet steam temperature of 560 °C and rejects heat to a condenser at 35 °C. Calculate the maximum theoretical (Carnot) efficiency of the plant.

η_Carnot = 1 − T_c/T_h = 1 − (35 + 273)/(560 + 273)

24Using η_Carnot = 1 − T_c/T_h, explain whether it is more effective to increase efficiency by raising T_h by 50 K, or by lowering T_c by 50 K, starting from T_h = 600 K, T_c = 300 K.

Starting efficiency: 1 − 300/600 = 0.500. Raising T_h to 650 K: 1 − 300/650 ≈ 0.538. Lowering T_c to 250 K: 1 − 250/600 ≈ 0.583. Lowering T_c gives the bigger improvement, because T_c/T_h is more sensitive to changes in the smaller of the two temperatures — this is why power stations put so much engineering effort into cooling their condensers as effectively as possible.

7. Entropy and the second law

Section 6 stated, without proof, that no heat engine can be more efficient than a Carnot cycle operating between the same two temperatures, and that even a Carnot engine can never be 100% efficient. Why not? The answer lies in a quantity called entropy, which turns out to govern which processes can happen at all, and in which direction.

Irreversibility

Watch a video of coffee cooling, an egg breaking, or ice melting in a warm room, played backwards — you would recognize immediately that something is wrong. These processes are irreversible: they happen spontaneously in one direction only. Nothing in the first law of thermodynamics (conservation of energy) forbids a cup of coffee spontaneously drawing thermal energy back out of the room and getting hotter — energy would still be conserved — yet it never happens.

Key idea. Almost all real, macroscopic processes are irreversible. Entropy, S, is a thermodynamic quantity that relates to the degree of disorder of the particles in a system, and it is entropy — not energy — that explains why processes run in only one direction.

Entropy and disorder

Entropy is best understood not as a measure of "messiness," but as a measure of how many different microscopic arrangements — microstates — of a system's particles all look identical from the outside. The short video below explains this idea clearly, using a simple example of a small number of particles distributed between two halves of a container.

What is entropy? — TED-Ed video thumbnail

Watch "What is entropy?" — a short video explaining entropy in terms of microstates and disorder, the same idea introduced above.

Watch video ↗

"What is entropy?" — lesson by Jeff Phillips, TED-Ed.

Key idea. The greater the number of possible microstates of a system, the greater its disorder, and the greater its entropy:

S = k_B ln Ω

where Ω is the number of possible microstates of the system.

Worked example 7.1

Calculate the entropy of a system that has 1 × 10²² possible microstates.

Answer:
S = k_B ln Ω = (1.38 × 10⁻²³) × ln(1 × 10²²)

The second law of thermodynamics

Key idea. The second law of thermodynamics refers to the change in entropy of an isolated system, and sets constraints on possible physical processes and on the overall evolution of the system: in every process, the total entropy of any isolated system, or the Universe as a whole, always increases.

Because a random system is overwhelmingly more likely to become more disordered than less, and because everything that happens is ultimately driven by the random behaviour of enormous numbers of particles, the total entropy of an isolated system essentially never decreases. This single statistical fact can be expressed in several equivalent ways:

Version	Statement
General	The total entropy of an isolated system (or the Universe as a whole) always increases.
Clausius	Thermal energy cannot spontaneously transfer from a region of lower temperature to a region of higher temperature.
Kelvin–Planck	When extracting energy from a heat reservoir, it is impossible to convert all of it into work.

Worth knowing

The second law has consequences far beyond engines and gases. Every living organism maintains a highly ordered, low-entropy internal structure — but the second law still applies to the organism and its surroundings together, so this local decrease in entropy must be paid for by an even greater increase in the entropy of its environment, for example as waste heat and waste products.

"Pollution is not, as we are so often told, a product of moral turpitude. It is an inevitable consequence of life at work. The second law of thermodynamics clearly states that the low entropy and intricate, dynamic organization of a living system can only function through the excretion of low-grade products and low-grade energy to the environment."

— Lovelock, James. Gaia: A New Look at Life on Earth. Oxford University Press, 2009.

Key idea. Processes in real isolated systems are almost always irreversible, and consequently the entropy of a real isolated system always increases. The entropy of a non-isolated system (such as the coffee, or the contents of a refrigerator) can decrease locally — but only if it is compensated by an equal or greater increase in the entropy of the surroundings.

Calculating entropy change

For real systems, tracking every microstate is impossible — but changes in entropy can be found much more directly from macroscopic quantities: thermal energy and temperature.

ΔS = ΔQ / T

Key idea. Entropy can be determined in terms of macroscopic quantities such as thermal energy and temperature, as given by ΔS = ΔQ/T, and also in terms of the properties of individual particles of the system, as given by S = k_B ln Ω.

Worked example 7.2

Calculate the entropy change when 5000 J of thermal energy flows out of a cup of coffee at 70 °C into the surrounding room at 25 °C. (Assume both temperatures stay constant.)

Answer:
Entropy decrease of the coffee: ΔS = ΔQ/T = −5000 / (273 + 70)
Entropy increase of the room: ΔS = ΔQ/T = +5000 / (273 + 25)
Because the room is cooler than the coffee, the increase in the room's entropy is larger than the decrease in the coffee's entropy — so the total (coffee + room) entropy increases, consistent with the second law.

Worth knowing Why real engines fall short of the Carnot limit. Every real process — friction, turbulence, rapid (non-quasi-static) expansion — is irreversible, and every irreversible process increases the total entropy of the engine and its surroundings. That extra entropy corresponds to potential work that has been irretrievably "wasted" as low-grade thermal energy, rather than converted to useful work. A Carnot cycle is the theoretical limit precisely because it is perfectly reversible — every stage happens infinitely slowly, generating no extra entropy at all. This is also why heat energy can never be converted to work with 100% efficiency: doing so would require extracting entropy from a heat reservoir with nowhere for that entropy to go, violating the second law.

Check your understanding

25State the second law of thermodynamics, and explain what is meant by an "isolated system" in this context.

The second law states that the total entropy of an isolated system, or of the Universe as a whole, always increases in every process. An isolated system is one that cannot exchange either energy or matter with its surroundings — so nothing outside it can supply or remove entropy, and the total entropy inside can only increase (or, for a perfectly reversible process, stay the same).

26Determine the increase in entropy of 500 g of ice when it melts at 0 °C. (Specific latent heat of fusion of water, L_f = 3.34 × 10⁵ J kg⁻¹.)

ΔQ = mL_f = 0.500 × 3.34 × 10⁵
ΔS = ΔQ/T = (0.500 × 3.34 × 10⁵) / (273 + 0)

27Calculate the entropy of a system that has 4 × 10³⁰ possible microstates.

S = k_B ln Ω = (1.38 × 10⁻²³) × ln(4 × 10³⁰)

28Using the idea of entropy, explain why no real heat engine can achieve the Carnot efficiency for its operating temperatures.

The Carnot cycle is perfectly reversible, generating no extra entropy anywhere in the process. Real engines always involve irreversible effects — friction, turbulence, rapid changes — each of which increases the total entropy of the engine and its surroundings. That increase in entropy corresponds to energy that becomes unavailable for doing useful work, so a real engine converts a smaller fraction of Q_h into work than the Carnot limit allows, for the same T_h and T_c.

Glossary

Pressure, P: Force acting perpendicular to a surface, per unit area: P = F/A. SI unit: pascal, Pa.
Amount of substance, n: A measure of the number of particles in a sample, in moles: n = N/N_A.
Avogadro constant, N_A: The number of particles in one mole of a substance: N_A = 6.02 × 10²³ mol⁻¹.
Ideal gas: A theoretical model of a gas whose particles have negligible volume, exert no forces on each other except during perfectly elastic collisions, and obey PV = nRT exactly.
Universal (molar) gas constant, R: The constant in PV = nRT; R = 8.31 J K⁻¹ mol⁻¹.
Boltzmann constant, k_B: The "per particle" version of the gas constant, k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹.
Internal energy, U: The total random kinetic and potential energy of all the particles in a system. For an ideal monatomic gas, U = 3/2 Nk_BT = 3/2 nRT.
First law of thermodynamics: Q = ΔU + W: the thermal energy transferred into a system either increases its internal energy or is used as it does work on its surroundings, or both.
Work done, W: The mechanical energy transferred as a system's volume changes; at constant pressure, W = PΔV. Positive when done by the gas (expansion); negative when done on the gas (compression).
Isothermal process: A change at constant temperature (ΔU = 0), so Q = W.
Isovolumetric process: A change at constant volume (W = 0), so Q = ΔU.
Adiabatic process: A change with no thermal energy transfer (Q = 0), so ΔU = −W. For a monatomic ideal gas, PV^5/3 = constant.
Isobaric process: A change at constant pressure (ΔP = 0).
Heat engine: A device that uses a flow of thermal energy from a hot reservoir to a cold reservoir to do useful mechanical work.
Efficiency, η: The fraction of the energy supplied to a heat engine that is converted to useful work: η = W/Q_h.
Carnot cycle: An idealized, reversible four-stage thermodynamic cycle (isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression) that achieves the maximum theoretically possible efficiency between two given temperatures: η_Carnot = 1 − T_c/T_h.
Reversible process: A process that could be reversed so that the system and all of its surroundings return exactly to their original states, with no change in total entropy. An idealization — real macroscopic processes are never perfectly reversible.
Irreversible process: A process that cannot be undone without leaving some other change in the surroundings; entropy always increases during an irreversible process.
Entropy, S: A thermodynamic quantity that relates to the degree of disorder of the particles in a system. Can be found from macroscopic quantities, ΔS = ΔQ/T, or from the number of microstates, S = k_B ln Ω.
Microstate: One possible detailed arrangement of the particles of a system consistent with its overall (macroscopic) state.
Second law of thermodynamics: The total entropy of an isolated system, or of the Universe as a whole, always increases.