Thermodynamics begins with careful bookkeeping: draw a boundary around the part of the world you care about, describe its state, and track the energy that crosses the boundary as heat and as work. The result is the first law, $\Delta U=q+w$, and a way to calculate the work done when a gas expands or is compressed.
A pressure cooker on a gas stove is a small chemical plant. Heat flows in from the flame; steam builds up and pushes on the weight until it lifts and hisses; the food inside changes chemically. To make sense of what is going on we must first decide what we are talking about — just the water? the water and the food? the whole cooker? — and then keep an honest account of every joule of energy that goes in or comes out. That accounting is the subject of this section.
Energy can cross the boundary of a system in only two ways: as heat, driven by a temperature difference, and as work, for example when a gas pushes back a piston. Whatever enters and does not leave is stored as the system's internal energy. This simple statement, the first law of thermodynamics, is the law of conservation of energy written for chemists. In this section you will learn its language — systems, state functions, the sign convention — and how to calculate the work done when a gas expands, which turns out to depend very much on how the expansion is carried out.
Systems, Surroundings and State Functions
📘 Definition 5.1 — System, surroundings and boundary
The system is the part of the universe chosen for study. Everything else is the surroundings. The system and the surroundings together make up the universe. They are separated by a boundary, which may be real (the walls of a beaker, a piston) or imaginary (an imagined surface around a sample of air), and which may be fixed or movable.
$$\text{system}+\text{surroundings}=\text{universe}$$
In practice the surroundings are only the part of the rest of the universe that can exchange energy or matter with the system — the room around a beaker, not a distant galaxy.
📘 Definition 5.2 — Open, closed and isolated systems
An open system exchanges both energy and matter with its surroundings (reactants in an open beaker; a living cell).
A closed system exchanges energy but not matter (reactants in a sealed flask; gas in a cylinder with a piston).
An isolated system exchanges neither energy nor matter (an ideal sealed, insulated vacuum flask — real flasks only approximate this).
To describe a system we give the values of a few measurable quantities such as pressure $p$, volume $V$, temperature $T$ and amount $n$. Once enough of them are fixed, every other macroscopic property is fixed too: for a gas, $p$, $T$ and $n$ fix $V$. We do not need to know the position and speed of every molecule — only these few state variables.
📘 Definition 5.3 — State of a system and state functions
The state of a system is described by the values of its state variables ($p$, $V$, $T$, $n$, composition). A state function is a property whose value depends only on the present state of the system, not on how that state was reached. Pressure, volume, temperature, internal energy (and, later, enthalpy, entropy and Gibbs energy) are state functions. Consequently the change in a state function $X$ between two states depends only on those states,
$$\Delta X=X_{\text{final}}-X_{\text{initial}},$$
and for any cyclic process (one that returns to its starting state) $\Delta X=0$. Heat $q$ and work $w$ are not state functions: their values depend on the path. They are sometimes called path functions.
A mountain makes a good picture. Your altitude is a state function: going from base camp to the summit you gain the same height whichever route you take. The distance you walk, and the energy you burn, depend on the route — they are like $q$ and $w$.
📘 Definition 5.4 — Internal energy
The internal energy $U$ of a system is the total energy stored in it: the kinetic energy of its molecules (translation, rotation, vibration), the potential energy of the interactions between them, and the energy of the electrons and nuclei in chemical bonds. $U$ is a state function and an extensive property (Definition 5.7). Its absolute value cannot be measured, but its change $\Delta U=U_{f}-U_{i}$ can: it is the energy that crosses the boundary as heat and work.
✏️ Example 1 — Choosing and classifying systems
(a) Classify each system as open, closed or isolated: (i) tea in an open cup; (ii) a sealed bottle of soda water at room temperature; (iii) hot soup in a perfectly sealed, perfectly insulated vacuum flask; (iv) a human being. (b) Which of these are state functions: the temperature of a gas; the heat supplied to warm it; its volume; the work done on it; its internal energy? (c) A gas goes from state A to state B by two routes. Along route 1, $\Delta U=+120\ \mathrm{J}$. What is $\Delta U$ along route 2? What is $\Delta U$ for going A → B by route 1 and back B → A by route 2?
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(a) (i) Open: water evaporates (matter leaves) and heat is lost. (ii) Closed: no matter crosses the walls, but heat can flow through the glass. (iii) Isolated (an idealisation). (iv) Open: we breathe, eat, sweat and radiate heat.
(b) State functions: temperature, volume, internal energy. The heat supplied and the work done depend on how the change is made — they are path functions.
(c) $U$ is a state function, so $\Delta U=U_{B}-U_{A}=+120\ \mathrm{J}$ along route 2 as well. For the round trip the gas returns to A: $\Delta U=+120+(-120)=0$.
Check A cyclic process always has $\Delta U=0$ — but the heat and work over the cycle need not be zero individually (Exercise IV.1).
Work, Heat and the First Law
Take some water in a container with insulating walls, so that no heat can pass through. We can still change its state: turn a paddle wheel inside it, or pass an electric current through a heating coil immersed in it. James Joule did exactly this in the 1840s and found that the same amount of work always produced the same rise in temperature, whether the work was mechanical or electrical. Since nothing but work crossed the boundary, the work must equal the change of a property of the water — its internal energy.
📘 Definition 5.5 — Adiabatic process, work and heat
An adiabatic process is one in which no heat passes between the system and the surroundings (an adiabatic wall is a perfect insulator). For an adiabatic change, $\Delta U=w_{ad}$: the work done on the system equals its change in internal energy, whatever the kind of work.
Heat $q$ is energy transferred because of a temperature difference between the system and the surroundings, through walls that conduct heat (diathermic walls).
Work $w$ is energy transferred in any other way — by a force moving through a distance (a piston, a paddle) or by an electric current.
IUPAC sign convention: $q$ and $w$ are positive when energy is transferred to the system (heat absorbed by it, work done on it) and negative when energy is transferred from the system (heat given out, work done by it).
Heat and work are energy in transit. A system does not "contain" heat or work; it contains internal energy. Once energy has crossed the boundary, there is no way of telling whether it came in as heat or as work.
📐 Law 5.1 — First law of thermodynamics
The energy of an isolated system is constant: energy can be converted from one form to another but can be neither created nor destroyed. For a closed system that exchanges heat $q$ and work $w$ with its surroundings,
$$\Delta U=q+w$$
with $q$ and $w$ taken with the IUPAC signs. Although $q$ and $w$ separately depend on the path, their sum $q+w$ is the same for every path between the same two states, because $\Delta U$ is a change in a state function. For an isolated system $q=0$ and $w=0$, so $\Delta U=0$.
⚠️ Two sign conventions
Many older chemistry books, and most physics books (including Class 11 Physics), write the first law as $\Delta U=q-w$, where $w$ is the work done by the system. The physics is identical — only the meaning of the symbol $w$ changes. In this course $w$ always means work done on the system, and $\Delta U=q+w$. When you read a problem, decide first which convention it uses. Typical slips: writing $w=+150\ \mathrm{J}$ for a gas that does $150\ \mathrm{J}$ of work (it should be $-150\ \mathrm{J}$), and forgetting that heat released by a system is negative.
✏️ Example 2 — Using the first law
Find the missing quantity in each case and say whether the internal energy of the system rises or falls. (a) A system absorbs $250\ \mathrm{J}$ of heat and $120\ \mathrm{J}$ of work is done on it. (b) A gas absorbs $300\ \mathrm{J}$ of heat and does $450\ \mathrm{J}$ of work on the surroundings. (c) In an adiabatic process, $2.0\ \mathrm{kJ}$ of work is done on a gas. (d) A system gives out $80\ \mathrm{J}$ of heat and its internal energy falls by $30\ \mathrm{J}$; find $w$ and say who does work on whom.
(b) $q=+300\ \mathrm{J}$, work done by the gas so $w=-450\ \mathrm{J}$: $\Delta U=300-450=-150\ \mathrm{J}$; $U$ falls — the gas spends more energy than it receives.
(c) Adiabatic: $q=0$, so $\Delta U=w=+2.0\ \mathrm{kJ}$; $U$ (and the temperature) rises. This is why a bicycle pump warms up.
(d) $q=-80\ \mathrm{J}$, $\Delta U=-30\ \mathrm{J}$: $w=\Delta U-q=-30-(-80)=+50\ \mathrm{J}$. Positive $w$: the surroundings do $50\ \mathrm{J}$ of work on the system.
Check (b) in the physics convention: $\Delta U=q-w_{\text{by}}=300-450=-150\ \mathrm{J}$ — the same answer, as it must be.
Interactive 1 — Two Paths, One ΔU: a State-Function Explorer
One mole of a monatomic ideal gas goes from state A ($1.0\ \mathrm{bar}$, $10\ \mathrm{L}$) to a state B that you choose. Pick a path on the $p$–$V$ diagram: the shaded area under the path is the work done by the gas, so $w=-(\text{area})$ for an expansion. For this gas $U$ changes by $\tfrac32\Delta(pV)$, which depends only on A and B. Press ▶ Play to send the gas along each path in turn and compare $q$, $w$ and $\Delta U$.
2.0 bar25 L
ΔU (any path)—
w (this path)—
q = ΔU − w—
q + w—
Pressure–Volume Work
The work that matters most in chemistry is the work done when a gas expands against, or is compressed by, an external pressure. Picture a gas in a cylinder closed by a frictionless, weightless piston of area $A$, with an external pressure $p_{ex}$ acting on it. The external force is $F=p_{ex}A$. If the gas expands and the piston moves out by a distance $l$, the gas pushes the surroundings back and does work $F\,l=p_{ex}A\,l=p_{ex}\Delta V$. Energy leaves the system, so with the IUPAC sign $w=-p_{ex}\Delta V$. On compression $\Delta V\lt0$ and $w$ comes out positive, as it should.
The piston moves out from the dashed to the solid position. The swept volume $\Delta V=A\,l$ is shaded lightly; the work done by the gas against the external pressure is $p_{ex}\Delta V$.
If $p_{ex}$ changes during the change, we add up small pieces $-p_{ex}\,dV$: the magnitude of the work is the area under the graph of $p_{ex}$ against $V$. This is why the work depends on the path.
📘 Definition 5.6 — Reversible and irreversible processes
A reversible process is carried out so slowly, through a series of states each infinitesimally close to equilibrium, that it could be reversed at any moment by an infinitesimal change in the conditions. For a gas this means that the external pressure is always only infinitesimally different from the pressure of the gas: $p_{ex}=p\pm dp$. Any process that does not meet this condition — a sudden expansion, a quick compression, a gas rushing into a vacuum — is irreversible. Real processes are irreversible; the reversible process is an ideal limit, like a frictionless slope.
🧰 Tool from earlier work — the ideal gas equation
For an ideal gas $pV=nRT$, with $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}=0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$. At constant temperature $pV$ is constant, so $p_{i}V_{i}=p_{f}V_{f}$. Unit of work: $1\ \mathrm{L\,bar}=10^{-3}\ \mathrm{m^{3}}\times10^{5}\ \mathrm{Pa}=100\ \mathrm{J}$ (and $1\ \mathrm{L\,atm}=101.3\ \mathrm{J}$). The gas laws behind $pV=nRT$ are taught in the 📘 box of Section 1.2.
📐 Rule 5.2 — Work of expansion and compression
Against a constant external pressure (irreversible): $\;w=-p_{ex}(V_{f}-V_{i})=-p_{ex}\Delta V$.
Free expansion into a vacuum ($p_{ex}=0$): $\;w=0$, whatever the change in volume.
Reversible isothermal change of $n$ mol of an ideal gas at temperature $T$ (here $p_{ex}=p=nRT/V$ throughout):
$$w_{rev}=-nRT\ln\frac{V_{f}}{V_{i}}=-2.303\,nRT\log\frac{V_{f}}{V_{i}}=-2.303\,nRT\log\frac{p_{i}}{p_{f}}$$
The internal energy of an ideal gas depends only on its temperature, so in any isothermal change of an ideal gas $\Delta U=0$ and $q=-w$. For free isothermal expansion $q=w=\Delta U=0$.
For an expansion between the same two states, the reversible path gives the work of largest magnitude: at every step the gas pushes against the largest pressure it can overcome. In an irreversible expansion the gas pushes against a smaller pressure (in a free expansion, against nothing) and does less work. For a compression the reverse is true — the reversible path needs the least work. Interactive 2 shows how the stepped path approaches the smooth reversible curve as the number of steps grows.
✏️ Example 3 — Expansion against a constant pressure
(a) An ideal gas expands isothermally from $3.0\ \mathrm{L}$ to $8.0\ \mathrm{L}$ against a constant external pressure of $2.0\ \mathrm{bar}$. Find $w$, $q$ and $\Delta U$ in joules. (b) The same gas expands isothermally from $3.0\ \mathrm{L}$ to $8.0\ \mathrm{L}$ into a vacuum. Find $w$, $q$ and $\Delta U$. (c) A gas absorbs $500\ \mathrm{J}$ of heat and expands from $1.0\ \mathrm{L}$ to $2.5\ \mathrm{L}$ against a constant pressure of $2.0\ \mathrm{bar}$. Find $\Delta U$.
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(a) work $w=-p_{ex}\Delta V=-(2.0\ \mathrm{bar})(8.0-3.0)\ \mathrm{L}=-10\ \mathrm{L\,bar}=-10\times100\ \mathrm{J}=-1.0\times10^{3}\ \mathrm{J}$.
(a) q, ΔU Isothermal, ideal gas: $\Delta U=0$, so $q=-w=+1.0\times10^{3}\ \mathrm{J}$. The gas absorbs $1.0\ \mathrm{kJ}$ of heat and spends all of it as work.
(b) $p_{ex}=0$: $w=0$. Isothermal: $\Delta U=0$, so $q=0$. The same change of state, but no heat and no work — $q$ and $w$ depend on the path, $\Delta U$ does not.
✏️ Example 4 — Reversible and irreversible isothermal expansion
$0.500\ \mathrm{mol}$ of an ideal gas at $298\ \mathrm{K}$ expands isothermally from $2.00\ \mathrm{L}$ to $20.0\ \mathrm{L}$. Calculate $w$, $q$ and $\Delta U$ (a) when the expansion is reversible; (b) when the gas expands in one step against a constant external pressure equal to its final pressure. Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
💡 Show solution▶
nRT $nRT=0.500\times8.314\times298=1238.8\ \mathrm{J}$ (keep the extra figure until the end).
(a) $w_{rev}=-2.303\,nRT\log\dfrac{20.0}{2.00}=-2.303\times1238.8\times1=-2853\ \mathrm{J}=-2.85\ \mathrm{kJ}$. $\Delta U=0$ (isothermal), so $q=+2.85\ \mathrm{kJ}$.
(b) Final pressure $p_{f}=nRT/V_{f}$, so $w=-p_{f}(V_{f}-V_{i})=-nRT\left(1-\dfrac{V_{i}}{V_{f}}\right)=-1238.8\times(1-0.100)=-1115\ \mathrm{J}=-1.11\ \mathrm{kJ}$; $\Delta U=0$, $q=+1.11\ \mathrm{kJ}$.
Compare Same initial and final states, same $\Delta U=0$, but the reversible path delivers $2.85/1.11\approx2.6$ times as much work.
Check In $\mathrm{L\,bar}$: $p_{f}=\dfrac{0.500\times0.0831\times298}{20.0}=0.619\ \mathrm{bar}$, $w=-0.619\times18.0=-11.1\ \mathrm{L\,bar}=-1.11\ \mathrm{kJ}$ ✓.
✏️ Example 5 — More steps, more work
The gas of Example 4 ($nRT=1238.8\ \mathrm{J}$) again expands isothermally from $2.00\ \mathrm{L}$ to $20.0\ \mathrm{L}$, but in two steps: first against a constant pressure equal to its pressure at $8.00\ \mathrm{L}$, until it reaches $8.00\ \mathrm{L}$; then against a constant pressure equal to its final pressure. (a) Find the total work. (b) Express the work for one step, two steps and the reversible path as multiples of $nRT$. (c) The gas is now compressed back from $20.0\ \mathrm{L}$ to $2.00\ \mathrm{L}$ in one step by a constant pressure equal to the final pressure. Find $w$ and compare it with the reversible compression.
Step 2 $p_{ex}=nRT/20.0$: $w_{2}=-\dfrac{nRT}{20.0}(20.0-8.00)=-0.600\,nRT$. Total $w=-1.35\,nRT=-1.35\times1238.8=-1672\ \mathrm{J}=-1.67\ \mathrm{kJ}$.
(b) One step: $-0.900\,nRT$; two steps: $-1.35\,nRT$; reversible: $-\ln10\;nRT=-2.30\,nRT$. Each extra step moves the stepped path closer to the isotherm and adds work.
(c) $p_{ex}=nRT/2.00$: $w=-\dfrac{nRT}{2.00}(2.00-20.0)=+9.00\,nRT=+11.1\ \mathrm{kJ}$. Reversible compression needs only $+2.30\,nRT=+2.85\ \mathrm{kJ}$: for compression the reversible path needs the least work.
Interactive 2 — Piston Lab: Reversible vs Irreversible Isothermal Work
$1.00\ \mathrm{mol}$ of an ideal gas at temperature $T$ expands (or is compressed) between $5.0\ \mathrm{L}$ and $25.0\ \mathrm{L}$. In each step the external pressure is dropped (or raised) suddenly to the pressure the gas will have at the end of that step. The red curve is the isotherm $p=nRT/V$; the shaded area under the stepped external pressure is $|w|$. Choose the number of steps — "∞" is the reversible path — and press ▶ Play to watch the piston move step by step.
300 K
w (this path)—
w (reversible)—
q = −w—
this path / reversible—
Extensive and Intensive Properties
📘 Definition 5.7 — Extensive and intensive properties
An extensive property depends on the amount of matter in the system: mass, volume, amount $n$, internal energy, heat capacity (and enthalpy, entropy, Gibbs energy). An intensive property does not: temperature, pressure, density, concentration, molar volume, specific and molar heat capacity, boiling point. If a system is divided into two equal halves, each half has half the value of every extensive property but the same value of every intensive property.
The ratio of two extensive properties is intensive: density $=m/V$, molar volume $V_{m}=V/n$, molar heat capacity $C_{m}=C/n$, molar internal energy $U_{m}=U/n$. This is why tables of data list molar or specific quantities, and why thermochemical values are quoted per mole of reaction.
✏️ Example 6 — Extensive or intensive?
(a) Classify: mass, volume, temperature, density, pressure, internal energy, molar mass, molar volume, heat capacity, molar heat capacity, concentration in $\mathrm{mol\,L^{-1}}$. (b) Two identical flasks each contain $1.0\ \mathrm{mol}$ of nitrogen at $300\ \mathrm{K}$ and $1.0\ \mathrm{bar}$, with internal energy $U$ each (measured from the same reference). They are connected. State the temperature, pressure, total amount and total internal energy of the combined system.
(a) intensive temperature, density, pressure, molar mass, molar volume, molar heat capacity, concentration (each is a property of the material or a ratio of two extensive properties).
1 Classify as open, closed or isolated: (a) coffee in a sealed, well-insulated vacuum flask; (b) a growing plant; (c) water boiling in an uncovered pan; (d) gas in a sealed steel cylinder.
2 Which of these depend only on the initial and final states of a system: $q$, $w$, $q+w$, $\Delta U$, $\Delta T$?
3 Classify as extensive or intensive: mass, density, molar mass, volume, temperature, internal energy, molar volume, concentration.
4 Show with one example that the ratio of two extensive properties is an intensive property.
5 Explain why the temperature of a gas is a state function but the heat supplied to it is not.
6 For the system "a reaction mixture in a beaker", name the boundary and say whether it is real or imaginary, fixed or movable.
7 List four kinds of energy that contribute to the internal energy of a sample of a gas.
8 Two samples of the same gas, at the same temperature and pressure, have volumes $2\ \mathrm{L}$ and $6\ \mathrm{L}$. Compare their temperature, pressure, density and internal energy.
4 Density $=m/V$: doubling the sample doubles both $m$ and $V$, so the ratio is unchanged (any similar example — molar volume $V/n$, molar heat capacity $C/n$ — is fine).
5 The temperature depends only on the present state. The heat supplied depends on the path: the same rise in temperature can be produced by heating, by doing work (stirring, compression) or by a mixture of both.
6 The walls and base of the beaker and an imaginary surface across its open top; the walls are real, the top is imaginary; the boundary is fixed (the open top lets matter escape, so the system is open).
7 Translational, rotational and vibrational kinetic energy of the molecules; potential energy of intermolecular interactions; electronic (bond) energy; nuclear energy (any four).
8 Same temperature, pressure and density (intensive). The $6\ \mathrm{L}$ sample contains three times the amount of gas and has three times the internal energy (extensive).
II
The first law and the sign convention. Use $\Delta U=q+w$ (IUPAC).
2 A gas absorbs $3.5\ \mathrm{kJ}$ of heat and does $1.2\ \mathrm{kJ}$ of work on its surroundings. Find $\Delta U$.
3 A gas gives out $600\ \mathrm{J}$ of heat while $250\ \mathrm{J}$ of work is done on it. Find $\Delta U$.
4 A gas does $1.50\ \mathrm{kJ}$ of work in an adiabatic expansion. Find $\Delta U$. Does its temperature rise or fall?
5 $\Delta U=+50\ \mathrm{J}$ and the system does $30\ \mathrm{J}$ of work. Find $q$.
6 A system loses $200\ \mathrm{J}$ of heat and its internal energy falls by $500\ \mathrm{J}$. Find $w$ and say whether work is done on or by the system.
7 Why is $\Delta U=0$ for any process in an isolated system?
8 A gas absorbs $400\ \mathrm{J}$ of heat and does $150\ \mathrm{J}$ of work. Calculate $\Delta U$ using $\Delta U=q+w$ and using the physics form $\Delta U=q-w$, stating what $w$ means in each.
6 $q=-200$, $\Delta U=-500$: $w=-500+200=-300\ \mathrm{J}$; the system does $300\ \mathrm{J}$ of work on the surroundings.
7 No energy crosses its boundary: $q=0$ and $w=0$, so $\Delta U=0$ (energy may change form inside, but the total is constant).
8 IUPAC: $w$ = work done on the gas $=-150\ \mathrm{J}$; $\Delta U=400+(-150)=+250\ \mathrm{J}$. Physics: $w$ = work done by the gas $=+150\ \mathrm{J}$; $\Delta U=400-150=+250\ \mathrm{J}$. Same result.
III
Pressure–volume work. Take $1\ \mathrm{L\,bar}=100\ \mathrm{J}$, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, ideal gases; give answers to 3 significant figures unless the data allow fewer.
1 A gas expands from $1.5\ \mathrm{L}$ to $4.0\ \mathrm{L}$ against a constant pressure of $1.2\ \mathrm{bar}$. Find $w$.
2 A gas is compressed from $10.0\ \mathrm{L}$ to $4.0\ \mathrm{L}$ by a constant external pressure of $3.0\ \mathrm{bar}$. Find $w$.
3 A gas expands isothermally from $2\ \mathrm{L}$ to $6\ \mathrm{L}$ into an evacuated vessel. Find $w$, $q$ and $\Delta U$.
4 $1.00\ \mathrm{mol}$ of gas at $300\ \mathrm{K}$ expands reversibly and isothermally from $5.0\ \mathrm{L}$ to $50.0\ \mathrm{L}$. Find $w$ and $q$.
5 $2.00\ \mathrm{mol}$ of gas at $350\ \mathrm{K}$ is compressed reversibly and isothermally from $30.0\ \mathrm{L}$ to $10.0\ \mathrm{L}$. Find $w$.
6 $3.00\ \mathrm{mol}$ of gas at $400\ \mathrm{K}$ expands reversibly and isothermally from $10.0\ \mathrm{bar}$ to $2.00\ \mathrm{bar}$. Find $w$.
7 The gas of part 4 makes the same change in one step against a constant pressure equal to its final pressure. Find $w$ and compare it with part 4.
8Applied. A car airbag inflates from nearly zero volume to $60\ \mathrm{L}$ against an atmospheric pressure of $1.0\ \mathrm{bar}$. How much work does the gas do on the atmosphere?
7 $w=-nRT(1-V_{i}/V_{f})=-2494.2\times0.900=-2245\ \mathrm{J}=-2.24\ \mathrm{kJ}$, about $39\%$ of the reversible $5.74\ \mathrm{kJ}$.
8 $w=-p_{ex}\Delta V=-1.0\times60=-60\ \mathrm{L\,bar}=-6.0\ \mathrm{kJ}$; the gas does $6.0\ \mathrm{kJ}$ of work pushing back the atmosphere.
IV
Thinking with the first law (harder).
1 Show that for any cyclic process $q=-w$. Does this mean $q=0$?
2 Explain why the work done by a gas in a reversible isothermal expansion is greater than in any irreversible isothermal expansion between the same two states.
3 $1.00\ \mathrm{mol}$ of gas at $300\ \mathrm{K}$ expands isothermally from $2.00\ \mathrm{L}$ to $8.00\ \mathrm{L}$: (a) in one step against its final pressure; (b) in two steps, stopping at $4.00\ \mathrm{L}$, each step against the pressure at the end of the step; (c) reversibly. Find the work done by the gas in each case.
4 A gas goes from state A to state B along path 1 with $q=+80\ \mathrm{J}$ and $w=-30\ \mathrm{J}$. Along path 2, $q=+20\ \mathrm{J}$. Find $w$ along path 2.
5 In one cycle of an engine the working gas absorbs $500\ \mathrm{J}$ of heat from a hot source and gives $300\ \mathrm{J}$ to a cold sink. Find $w$ for the cycle and the net work done by the gas.
6 Show that $1\ \mathrm{L\,bar}=100\ \mathrm{J}$ and $1\ \mathrm{L\,atm}=101.3\ \mathrm{J}$ ($1\ \mathrm{atm}=1.013\times10^{5}\ \mathrm{Pa}$).
7 $1.00\ \mathrm{mol}$ of an ideal gas at $300\ \mathrm{K}$ and $2.00\ \mathrm{L}$ expands reversibly and isothermally until it has done $1.00\ \mathrm{kJ}$ of work. Find its final volume.
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1 The system returns to its initial state, so $\Delta U=0=q+w$ and $q=-w$. No: $q$ and $w$ can each be non-zero (an engine takes in net heat and gives out the same amount of net work).
2 At every stage of the reversible path the gas pushes against the largest pressure it can still overcome ($p_{ex}=p-dp$), so the area under $p_{ex}$ against $V$ is the whole area under the isotherm. Any irreversible path has $p_{ex}\lt p$ somewhere, and its area lies below the isotherm.
3 $nRT=2494.2\ \mathrm{J}$. (a) $-nRT(1-\tfrac{2}{8})=-0.75\,nRT=-1.87\ \mathrm{kJ}$; (b) $-nRT[(1-\tfrac24)+(1-\tfrac48)]=-1.00\,nRT=-2.49\ \mathrm{kJ}$; (c) $-nRT\ln4=-1.386\,nRT=-3.46\ \mathrm{kJ}$. The gas does $1.87$, $2.49$ and $3.46\ \mathrm{kJ}$ of work.
4 $\Delta U=80-30=+50\ \mathrm{J}$ for every path; along path 2, $w=50-20=+30\ \mathrm{J}$ (work is done on the gas).
5 Net $q=500-300=+200\ \mathrm{J}$; cycle: $\Delta U=0$, so $w=-200\ \mathrm{J}$: the gas does $200\ \mathrm{J}$ of net work per cycle.
E1.NEET Which of the following is not a state function? (A) internal energy (B) volume (C) work (D) temperature
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Idea A state function depends only on the state; work depends on the path (compare free and reversible expansion between the same states).
Answer(C).
E2.JEE MainNumerical answer. $2.00\ \mathrm{mol}$ of an ideal gas expands reversibly and isothermally at $300\ \mathrm{K}$ from $1.00\ \mathrm{L}$ to $10.0\ \mathrm{L}$. The magnitude of the work done by the gas, in kJ, to the nearest integer is ____ ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$).
💡 Show solution▶
Formula $|w|=2.303\,nRT\log\tfrac{V_{f}}{V_{i}}=2.303\times2.00\times8.314\times300\times\log10$.
E3.Board · 3 marks (a) State the first law of thermodynamics and write it in mathematical form for a closed system. (b) A system absorbs $650\ \mathrm{J}$ of heat and does $250\ \mathrm{J}$ of work on its surroundings. Find $\Delta U$. (c) Heat and work are not state functions, yet $q+w$ is the same for every path between two states. Explain.
💡 Show solution▶
(a) 1 mark Energy can be neither created nor destroyed; the energy of an isolated system is constant. $\Delta U=q+w$ ($q$ heat absorbed by, $w$ work done on the system).
(b) 1 mark $q=+650\ \mathrm{J}$, $w=-250\ \mathrm{J}$; $\Delta U=+400\ \mathrm{J}$.
(c) 1 mark $q+w=\Delta U$, the change in a state function, which depends only on the initial and final states; how the total is split between heat and work depends on the path.
E4.JEE Main $1.00\ \mathrm{mol}$ of an ideal gas at $300\ \mathrm{K}$ is compressed isothermally from $10.0\ \mathrm{L}$ to $5.00\ \mathrm{L}$ in a single step by a constant external pressure equal to the final pressure of the gas. The work done on the gas is (A) $-2.49\ \mathrm{kJ}$ (B) $+2.49\ \mathrm{kJ}$ (C) $+1.73\ \mathrm{kJ}$ (D) zero
Work $w=-p_{ex}(V_{f}-V_{i})=-\dfrac{nRT}{5.00}(5.00-10.0)=+nRT=+1.00\times8.314\times300=+2494\ \mathrm{J}$.
Answer(B) $+2.49\ \mathrm{kJ}$. (C) is the reversible value $nRT\ln2=1.73\ \mathrm{kJ}$ — the least work that could achieve this compression.
E5.JEE AdvancedOne or more options correct. An ideal gas expands isothermally from $V_{1}$ to $V_{2}$. (A) If it expands into a vacuum, $q=w=\Delta U=0$. (B) The work done by the gas is larger for the reversible expansion than for an expansion against a constant pressure equal to the final pressure. (C) In the irreversible expansion $\Delta U\lt0$ because the gas does work. (D) For both the reversible and the irreversible expansion, $q=-w$.
(B) True: $nRT\ln\tfrac{V_{2}}{V_{1}}\gt nRT\left(1-\tfrac{V_{1}}{V_{2}}\right)$ for $V_{2}\gt V_{1}$ (the reversible area is the whole area under the isotherm).
(C) False: the temperature is constant, so $\Delta U=0$; the work is paid for by heat absorbed.
(D) True: $\Delta U=0$ in every isothermal change of an ideal gas.
Answer(A), (B), (D).
E6.Board · Case study $2.00\ \mathrm{mol}$ of nitrogen is held at $300\ \mathrm{K}$ in a steel cylinder fitted with a frictionless piston, immersed in a large water bath. (i) Is the gas an open, closed or isolated system? Give a reason. (1 mark) (ii) The gas expands slowly and reversibly from $10.0\ \mathrm{L}$ to $20.0\ \mathrm{L}$, its temperature staying at $300\ \mathrm{K}$. Calculate $w$ (treat nitrogen as ideal; $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$). (2 marks) (iii) Find $q$ and state where this heat comes from. (1 mark)
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(i) Closed: no gas can enter or leave, but heat passes through the steel walls and work is done through the piston.
(ii) $w=-2.303\,nRT\log\tfrac{V_{f}}{V_{i}}=-2.303\times2.00\times8.314\times300\times\log2=-2.303\times4988.4\times0.3010=-3.46\ \mathrm{kJ}$.
(iii) Isothermal ideal gas: $\Delta U=0$, so $q=-w=+3.46\ \mathrm{kJ}$, absorbed from the water bath (the surroundings).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Wind and solar farms often produce electricity when nobody needs it. One way to store it is to compress air into an underground salt cavern and let it expand through a turbine later. How much of the work put in can, at best, be recovered — and why does it matter how the air is compressed?
Why it matters. Compressed-air energy storage plants have run for decades at Huntorf in Germany and McIntosh in the United States, and new ones are being built in China and elsewhere as grids add renewable power. Their efficiency is decided by exactly the ideas of this section: reversible versus irreversible work, and what happens to the heat released on compression.
What to do.
Treat air as an ideal gas at $300\ \mathrm{K}$. Calculate the reversible isothermal work needed to compress $1.00\ \mathrm{mol}$ of air from $1.0\ \mathrm{bar}$ to $70\ \mathrm{bar}$ (a typical cavern pressure).
Calculate the work if the same compression is done in one step by a constant external pressure of $70\ \mathrm{bar}$. Then try two steps (to $10\ \mathrm{bar}$, then to $70\ \mathrm{bar}$). What pattern do you see? Use Interactive 2 to check the idea qualitatively.
In an isothermal compression $\Delta U=0$, so all the work you put in leaves as heat. If that heat is simply lost to the surroundings, can you get the work back when the gas expands? Explain using the first law.
Find out (from an engineering or energy-agency summary) the round-trip efficiency of a conventional plant and of an "adiabatic" plant that stores the heat of compression. Which idea from this section explains the difference?
Tools: Rule 5.2, Interactive 2, a calculator, and an energy-agency or engineering summary of compressed-air storage.
Hand in: Your three work calculations in a small table, and a paragraph (about 150 words) explaining why staged compression with heat storage is used in modern plants.
Going further. Real compressors cannot be reversible. Suggest two practical features (more stages, cooling between stages…) that bring a real compressor closer to the reversible limit.
Instructor note: $nRT=2494\ \mathrm{J}$. Reversible: $w=2.303\times2494\times\log70=+10.6\ \mathrm{kJ}$. One step at $70\ \mathrm{bar}$: $w=-p_{ex}\Delta V=nRT(70/1-1)=69\,nRT=+172\ \mathrm{kJ}$. Two steps: $nRT[(10-1)+(7-1)]=15\,nRT=+37.4\ \mathrm{kJ}$. More steps → less work, approaching the reversible value. Plants that dump the heat of compression recover only about 40–55% of the input energy; designs that store the heat aim at about 70%.
🧪 Self-check quiz
0/5 correct
Q1. Which of these is (ideally) an isolated system?
Q2. A gas absorbs $200\ \mathrm{J}$ of heat and does $80\ \mathrm{J}$ of work on the surroundings. $\Delta U$ is
Q3. An ideal gas expands isothermally into a vacuum. Which is correct?
Q4. Which property is intensive?
Q5. An ideal gas expands isothermally between the same two volumes in four ways. The gas does the most work when it expands