University Physics — Year 1 · Bachelor Year 1

22The First Law of Thermodynamics

Pump up a bicycle tire and the barrel of the pump gets too hot to hold. A diesel engine has no spark plug: it squeezes its air twenty times smaller, and the air becomes hot enough to light the fuel by itself. A cylinder of compressed gas frosts over as it empties. In each case energy changes form — work becomes heat, heat becomes work, both become internal energy — and the first law of thermodynamics is the bookkeeping that makes the accounts balance. This chapter defines work and heat for a thermodynamic system, states the first law, introduces internal energy and enthalpy as the quantities it conserves, and applies it to the transformations of a perfect gas and to calorimetry.

A bicycle pump: the work of the hand on the trapped air raises its internal energy, and the barrel gets warm — the first law in the palm.

22.1 Transformations, work, and heat

Definition 22.1 (System, transformation)

A thermodynamic system is the matter inside a chosen boundary; everything else is the surroundings. A transformation takes it from an equilibrium state to another. It is quasi-static if the system passes through a succession of equilibrium states (slow enough for $P$ , $T$ to be defined throughout), and reversible if, in addition, reversing the external conditions reverses the path (no friction, no finite temperature or pressure gap with the surroundings). Names by what stays constant: isothermal ( $T$ ), isobaric ( $P$ ), isochoric ( $V$ ); adiabatic if no heat is exchanged; monobaric (monothermal) if the external pressure (temperature) is constant, whatever happens inside.

Proposition 22.2 (Work of pressure forces)

A system whose volume changes by $\dd V$ against an external pressure $P_{\mathrm{ext}}$ receives the work

\delta W = -P_{\mathrm{ext}}\,\dd V, \qquad W = -\int_{V_1}^{V_2}P_{\mathrm{ext}}\,\dd V .

For a quasi-static transformation $P_{\mathrm{ext}} = P$ (the system’s own pressure) and $W = -\int P\,\dd V$ is minus the area under the path in the Clapeyron diagram: positive when compressed, negative when expanding.

Proof. A piston of area $S$ pushed by the surroundings with the force $P_{\mathrm{ext}}S$ moves by $\dd x$ outward: the surroundings do the work $-P_{\mathrm{ext}}S\,\dd x = -P_{\mathrm{ext}}\,\dd V$ on the system. Any boundary is a collection of such pistons. Quasi-static: the piston is in equilibrium, $P_{\mathrm{ext}} = P$ up to a vanishing difference. ∎

Definition 22.3 (Heat)

Heat $Q$ is energy transferred to the system other than by macroscopic work — by molecular collisions at a wall (conduction), by the circulation of a fluid (convection), by radiation. A thermostat is a body so large that its temperature does not change whatever heat it exchanges. Sign convention throughout: $W$ and $Q$ are received by the system, positive when they enter.

22.2 The first law

Theorem 22.4 (First law of thermodynamics)

A closed system possesses a state function, its internal energy $U$ (extensive), such that for any transformation

\Delta U + \Delta E_{k} = W + Q ,

$E_k$ being the macroscopic kinetic energy (zero for a system at rest): the energy received as work and heat is stored as internal energy. $W$ and $Q$ each depend on the path; their sum does not.

Proof. Admitted at this level. ∎

Remark 22.5 (What it says)

It is the conservation of energy, with heat recognized as a transfer of energy (Joule’s paddle-wheel experiment: a measured work always raises the temperature of water as much as a measured heat). “ $U$ is a state function” means $\Delta U$ is fixed by the initial and final states — so a cycle has $\Delta U = 0$ and $W + Q = 0$ : an engine that delivers work must receive heat. The microscopic content of $U$ is that of Definition 20.10.

Definition 22.6 (Enthalpy; heat capacities)

The enthalpy $H = U + PV$ is a state function adapted to transformations at constant pressure. The heat capacities at constant volume and pressure are

C_V = \left(\frac{\partial U}{\partial T}\right)_V, \qquad C_P = \left(\frac{\partial H}{\partial T}\right)_P ,

in $\mathrm{J}/\mathrm{K}$ ; per mole $C_{V,m}$ , $C_{P,m}$ ; per kilogram the specific heats $c_V$ , $c_P$ .

Proposition 22.7 (Heat at constant volume and at constant pressure)

For a system at rest exchanging work only through pressure forces:

isochoric transformation: $W = 0$ and $Q_V = \Delta U$ ;
monobaric transformation between two states at the external pressure $P_0$ : $Q_P = \Delta H$ .

Proof. Isochoric: $\dd V = 0$ . Monobaric: $W = -P_0(V_2 - V_1)$ , so $Q = \Delta U + P_0\Delta V = \Delta(U + PV)$ since $P_1 = P_2 = P_0$ . ∎

Proposition 22.8 (Perfect gas: $U$ , $H$ , Mayer’s relation)

For a perfect gas, $U$ and $H$ depend on $T$ alone (Joule’s laws): $\dd U = C_V\,\dd T$ , $\dd H = C_P\,\dd T$ , and

C_{P,m} - C_{V,m} = R, \qquad \gamma = \frac{C_P}{C_V}, \qquad C_{V,m} = \frac{R}{\gamma - 1}, \quad C_{P,m} = \frac{\gamma R}{\gamma - 1} ;

$\gamma = 5/3$ for monatomic gases, $7/5$ for diatomic ones at ordinary temperatures. For a condensed phase in the model of Proposition 20.13, $C_P \approx C_V = C$ and $\dd U \approx \dd H = C\,\dd T$ .

Proof. $H = U + nRT$ ; differentiate with respect to $T$ . $C_{V,m} = \tfrac32 R$ or $\tfrac52 R$ from Proposition 20.11. ∎

22.3 Transformations of a perfect gas

Proposition 22.9 (The four standard transformations)

For $n$ moles of perfect gas:

isochoric: $W = 0$ , $Q = \Delta U = nC_{V,m}\Delta T$ ;
isobaric (quasi-static): $W = -P\Delta V = -nR\Delta T$ , $Q = \Delta H = nC_{P,m}\Delta T$ ;
isothermal reversible: $\Delta U = 0$ , $W = -Q = -nRT\ln(V_2/V_1) = nRT\ln(P_2/P_1)$ ;
adiabatic reversible: $Q = 0$ , $W = \Delta U = nC_{V,m}(T_2 - T_1)$ , and along the path (Laplace’s law)

$PV^\gamma = \text{const}, \qquad TV^{\gamma - 1} = \text{const}, \qquad T^\gamma P^{1 - \gamma} = \text{const} .$

Proof. Isochoric and isobaric: the previous propositions. Isothermal: $W = -\int nRT\,\dd V/V$ . Adiabatic: $\dd U = \delta W$ , i.e. $nC_{V,m}\dd T = -P\,\dd V = -nRT\,\dd V/V$ , so $\dd T/T = -(\gamma - 1)\dd V/V$ (using $R/C_{V,m} = \gamma - 1$ ); integrate: $TV^{\gamma - 1}$ constant; the other forms by $PV = nRT$ . ∎

Left: in the Clapeyron diagram the work received in a quasi-static compression is the area under the path; starting from the same point, an adiabat is steeper than an isotherm (> 1): a gas compressed without heat loss heats up and climbs to a higher isotherm. Right: the work of the surroundings on a piston. — Left: in the Clapeyron diagram the work received in a quasi-static compression is the area under the path; starting from the same point, an adiabat is steeper than an isotherm ( $\gamma > 1$ ): a gas compressed without heat loss heats up and climbs to a higher isotherm. Right: the work of the surroundings on a piston.

Example 22.10 (Pump and diesel)

Air ( $\gamma = 1.4$ ) compressed adiabatically from $1\,\mathrm{bar}$ , $300\,\mathrm{K}$ to $7\,\mathrm{bar}$ (a bicycle pump): $T_2 = 300 \times 7^{0.286} = 523\,\mathrm{K}$ — $250{}^{\circ}\mathrm{C}$ for an instant, which the barrel feels. A diesel’s compression ratio of $20$ : $T_2 = 300 \times 20^{0.4} = 994\,\mathrm{K}$ , $P_2 = 20^{1.4} = 66\,\mathrm{bar}$ — above the self-ignition temperature of the fuel: the engine needs no spark (the weekend problem runs the whole cycle).

Remark 22.11 (Irreversible transformations)

When the transformation is not quasi-static, Laplace’s law and $W = -\int P\,\dd V$ do not apply; only $W = -\int P_{\mathrm{ext}}\,\dd V$ and the first law do. Two classic cases: a gas expanding into vacuum (Joule expansion) receives no work and, in adiabatic walls, no heat — $\Delta U = 0$ , and a perfect gas keeps its temperature; a gas suddenly compressed by an external pressure jumping to $P_2$ receives $W = -P_2\Delta V$ , more than the reversible work to the same pressure, and ends hotter (Exercise 22.11).

22.4 Calorimetry

Method 22.12 (Calorimetric balance)

Bodies put in thermal contact inside an insulated vessel at constant (atmospheric) pressure exchange heat only among themselves: the total $\Delta H = \sum Q_P = 0$ . For each body $\Delta H = mc\,\Delta T$ (no phase change) or $\pm mL$ for a change of state at its transition temperature, with $L$ the latent heat (enthalpy of fusion or vaporization per unit mass: water, $L_f = 334\,\mathrm{kJ}/\mathrm{kg}$ at $0{}^{\circ}\mathrm{C}$ , $L_v = 2260\,\mathrm{kJ}/\mathrm{kg}$ at $100{}^{\circ}\mathrm{C}$ ). Write the balance, solve for the final temperature, and check that each body’s assumed phase is consistent with it.

Example 22.13 (Ice in water)

$50\,\mathrm{g}$ of ice at $0{}^{\circ}\mathrm{C}$ into $200\,\mathrm{g}$ of water at $30{}^{\circ}\mathrm{C}$ : melting needs $16.7\,\mathrm{kJ}$ ; the water can give $25.1\,\mathrm{kJ}$ in cooling to $0{}^{\circ}\mathrm{C}$ , so all the ice melts and the balance $0.250 \times 4180 \times (T_f - 0) = 25.1 - 16.7$ kJ gives $T_f = 8{}^{\circ}\mathrm{C}$ . Had there been $100\,\mathrm{g}$ of ice, $33.4\,\mathrm{kJ}$ would be needed: only part melts and $T_f = 0{}^{\circ}\mathrm{C}$ — the check matters.

Heating one kilogram of water from ice at -20 C to steam: the temperature pauses during each change of state while the latent heat is supplied — 334\, kJ to melt, 2260\, kJ to boil, far more than the 418\, kJ that carries the liquid from 0\, to 100 C. — Heating one kilogram of water from ice at $-20{}^{\circ}\mathrm{C}$ to steam: the temperature pauses during each change of state while the latent heat is supplied — $334\,\mathrm{kJ}$ to melt, $2260\,\mathrm{kJ}$ to boil, far more than the $418\,\mathrm{kJ}$ that carries the liquid from $0\,$ to $100{}^{\circ}\mathrm{C}$ .

22.5 Exercises

Exercise 22.1 ★

One mole of air ( $C_{P,m} = 29.1\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ ) is heated from $300\,\mathrm{K}$ to $400\,\mathrm{K}$ at constant pressure. Heat received, work received, change of internal energy.

Solution

Solution of Exercise 22.1.

$Q = C_{P,m}\Delta T = 29.1 \times 100 = 2.91\,\mathrm{kJ}$ ; $W = -nR\Delta T = -0.83\,\mathrm{kJ}$ ; $\Delta U = Q + W = 2.08\,\mathrm{kJ} = C_{V,m}\Delta T$ .

Exercise 22.2 ★

One mole of gas at $300\,\mathrm{K}$ is compressed isothermally and reversibly from $2.0\,\mathrm{L}$ to $0.50\,\mathrm{L}$ . Work received, heat exchanged, $\Delta U$ .

Solution

Solution of Exercise 22.2.

$W = -nRT\ln(V_2/V_1) = -2494\ln(0.25) = +3.46\,\mathrm{kJ}$ ; $\Delta U = 0$ so $Q = -3.46\,\mathrm{kJ}$ (given to the bath).

Exercise 22.3 ★

Air at $300\,\mathrm{K}$ , $1\,\mathrm{bar}$ is compressed adiabatically and reversibly to a tenth of its volume. Final temperature and pressure.

Solution

Solution of Exercise 22.3.

$T_2 = T_1 \times 10^{\gamma - 1} = 300 \times 10^{0.4} = 754\,\mathrm{K}$ ; $P_2 = 10^{1.4} = 25\,\mathrm{bar}$ .

Exercise 22.4 ★

Give $C_{V,m}$ , $C_{P,m}$ and $\gamma$ for argon and for nitrogen. For liquid water ( $c = 4180\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ ), why is the distinction between $c_P$ and $c_V$ dropped?

Solution

Solution of Exercise 22.4.

Argon: $C_{V,m} = \tfrac32 R = 12.5$ , $C_{P,m} = \tfrac52 R = 20.8$ J/(mol K), $\gamma = 1.67$ ; nitrogen: $20.8$ , $29.1$ , $\gamma = 1.40$ . For water the expansion on heating is tiny, so the work $P\,\dd V$ at constant pressure is negligible against the heat: $c_P - c_V \approx 0$ .

Exercise 22.5 ★★

$200\,\mathrm{g}$ of water at $80{}^{\circ}\mathrm{C}$ are mixed with $300\,\mathrm{g}$ at $20{}^{\circ}\mathrm{C}$ in an insulated cup. Final temperature; what changes if the cup itself has a heat capacity of $80\,\mathrm{J}/\mathrm{K}$ and starts at $20{}^{\circ}\mathrm{C}$ ?

Solution

Solution of Exercise 22.5.

$T_f = (200 \times 80 + 300 \times 20)/500 = 44{}^{\circ}\mathrm{C}$ . With the cup: $(200 \times 4.18 \times 80 + (300 \times 4.18 + 80) \times 20)/(500 \times 4.18 + 80) = 43{}^{\circ}\mathrm{C}$ .

Exercise 22.6 ★★

$50\,\mathrm{g}$ of ice at $-10{}^{\circ}\mathrm{C}$ ( $c_{\text{ice}} = 2100\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ ) are dropped into $200\,\mathrm{g}$ of water at $30{}^{\circ}\mathrm{C}$ . Final state and temperature. Same question with $150\,\mathrm{g}$ of ice.

Solution

Solution of Exercise 22.6.

Ice: warm to $0$ : $0.050 \times 2100 \times 10 = 1.05\,\mathrm{kJ}$ ; melt: $16.7\,\mathrm{kJ}$ ; total $17.8\,\mathrm{kJ}$ . Water can give $25.1\,\mathrm{kJ}$ down to $0$ : all melts; then $0.250 \times 4180\,T_f = 25.1 - 17.8$ kJ: $T_f = 7{}^{\circ}\mathrm{C}$ . With $150\,\mathrm{g}$ : needs $53.3\,\mathrm{kJ}$ $>$ $25.1\,\mathrm{kJ}$ : only $(25.1 - 3.15)/334 = 66\,\mathrm{g}$ of ice melt, $T_f = 0{}^{\circ}\mathrm{C}$ , $84\,\mathrm{g}$ of ice remain.

Exercise 22.7 ★★

A gas follows a rectangular cycle in the Clapeyron diagram: $(V_1, P_1) \to (V_2, P_1) \to (V_2, P_2) \to (V_1, P_2) \to (V_1, P_1)$ with $V_2 > V_1$ , $P_2 > P_1$ . Work received over the cycle, as a function of the area; sign, and meaning of the sign.

Solution

Solution of Exercise 22.7.

$W = -P_1(V_2 - V_1) + 0 - P_2(V_1 - V_2) + 0 = (P_2 - P_1)(V_2 - V_1) > 0$ : the cycle is traversed counterclockwise (expansion at low pressure, compression at high), so the gas receives net work, equal to the enclosed area; clockwise it would deliver it — an engine.

Exercise 22.8 ★★

A bicycle pump compresses air from $1\,\mathrm{bar}$ , $300\,\mathrm{K}$ to $7\,\mathrm{bar}$ quickly enough to be adiabatic. Temperature reached; why the pump warms up; and why the tire’s pressure drops a little after the pumping stops.

Solution

Solution of Exercise 22.8.

$T_2 = 300 \times 7^{0.286} = 523\,\mathrm{K}$ . The hot compressed air heats the barrel by conduction at each stroke. In the tire the air cools to ambient at fixed volume: $P \propto T$ , the pressure falls by the ratio of temperatures — top it up.

Exercise 22.9 ★★

A perfect gas expands into an evacuated, insulated vessel (Joule expansion). Work, heat, $\Delta U$ , $\Delta T$ . Is the transformation reversible? What would a real gas do?

Solution

Solution of Exercise 22.9.

$W = 0$ (no external pressure to push against), $Q = 0$ : $\Delta U = 0$ , hence $\Delta T = 0$ for a perfect gas. Irreversible: the gas never flows back by itself. A real gas, whose molecules attract, cools slightly (part of the kinetic energy becomes potential energy as they separate).

Exercise 22.10 ★★★

Boiling $1.0\,\mathrm{kg}$ of water at $100{}^{\circ}\mathrm{C}$ and $1\,\mathrm{atm}$ (steam volume $1.67\,\mathrm{m}^{3}$ ). Heat supplied, work exchanged with the atmosphere, $\Delta U$ . Where does most of the latent heat go?

Solution

Solution of Exercise 22.10.

$Q = L_v = 2.26\,\mathrm{MJ}$ ; $W = -P(V_g - V_l) = -1.013\times10^5 \times 1.67 = -0.17\,\mathrm{MJ}$ ; $\Delta U = 2.09\,\mathrm{MJ}$ : $93\%$ of the heat goes into separating the molecules (internal energy), $7\%$ into pushing back the atmosphere.

Exercise 22.11 ★★★

One mole of diatomic gas at $300\,\mathrm{K}$ , $1\,\mathrm{bar}$ , in an insulated cylinder, is compressed by suddenly setting the external pressure to $5\,\mathrm{bar}$ and waiting for equilibrium. Final temperature and volume (first law with $W = -P_{\mathrm{ext}}\Delta V$ ); compare with the reversible adiabatic compression to $5\,\mathrm{bar}$ .

Solution

Solution of Exercise 22.11.

$W = -P_2(V_2 - V_1)$ , $\Delta U = nC_{V,m}(T_2 - T_1)$ , $V_2 = nRT_2/P_2$ , $V_1 = nRT_1/P_1$ : $\tfrac52(T_2 - 300) = -T_2 + 300 \times 5$ , $3.5T_2 = 2250$ , $T_2 = 643\,\mathrm{K}$ , $V_2 = 10.7\,\mathrm{L}$ . Reversible: $T_2 = 300 \times 5^{0.286} = 475\,\mathrm{K}$ , $V_2 = 7.9\,\mathrm{L}$ . The sudden compression does more work on the gas (the full $5\,\mathrm{bar}$ acts from the start) and ends hotter and less compressed.

Exercise 22.12 ★★★

Sound is a fast compression, hence adiabatic: its speed is $c = \sqrt{\gamma P/\rho}$ (the isothermal value $\sqrt{P/\rho}$ was Newton’s mistake). Compute both for air at $20{}^{\circ}\mathrm{C}$ and compare with the measured $343\,\mathrm{m}/\mathrm{s}$ . Show that $c = \sqrt{\gamma RT/M}$ and compare with the rms speed of Chapter 20.

Solution

Solution of Exercise 22.12.

$\rho = PM/RT = 1.20\,\mathrm{kg}/\mathrm{m}^{3}$ : $\sqrt{\gamma P/\rho} = \sqrt{1.4 \times 1.013\times10^5/1.20} = 343\,\mathrm{m}/\mathrm{s}$ ; $\sqrt{P/\rho} = 290\,\mathrm{m}/\mathrm{s}$ , $15\%$ low. With $P/\rho = RT/M$ : $c = \sqrt{\gamma RT/M}$ ; against $u = \sqrt{3RT/M}$ = $502\,\mathrm{m}/\mathrm{s}$ : $c/u = \sqrt{\gamma/3} = 0.68$ .

22.6 Problem: From the bicycle pump to the diesel engine

Problem 22.1

Weekend problem — a hand pump that burns the fingers, a cylinder that lights its own fuel, and a steel ball that bounces on a cushion of air: the first law in three compressions

Air: perfect diatomic gas, $\gamma = 1.40$ , $C_{V,m} = \tfrac52 R$ , $C_{P,m} = \tfrac72 R$ , $M = 29\,\mathrm{g}/\mathrm{mol}$ ; $R = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ . Initial state everywhere: $P_1 = 1.00\,\mathrm{bar}$ , $T_1 = 300\,\mathrm{K}$ .

Part I — The bicycle pump. The pump’s cylinder has a volume $V_1 = 212\,\mathrm{cm}^{3}$ ; the air in it is compressed reversibly and adiabatically until it reaches the tire’s pressure, $7.0\,\mathrm{bar}$ .

Amount of air in the cylinder.
Volume at the end of the compression.
Temperature at the end of the compression.
Work received by the air during one stroke.
The tire ( $2.0\,\mathrm{L}$ ) must be brought from $1\,\mathrm{bar}$ to $7\,\mathrm{bar}$ at $300\,\mathrm{K}$ : how many strokes, and how much work in all (take the work per stroke as constant)?
The hot air then cools in the tire to $300\,\mathrm{K}$ : what happens to the tire’s pressure, and what does the cyclist do?
Compare the adiabatic work per stroke with the work of an isothermal compression to the same $7\,\mathrm{bar}$ . Why is the isothermal work larger although the gas ends cooler?

Part II — The diesel compression. A cylinder of $V_1 = 0.50\,\mathrm{L}$ compresses its air reversibly and adiabatically by a factor $20$ .

Amount of air, and temperature $T_2$ after compression.
Pressure $P_2$ .
The fuel self-ignites above about $500\,\mathrm{K}$ : explain why a diesel needs no spark plug, and why a petrol engine (ratio $\approx 10$ , see Exercise 22.3) must not self-ignite.
Work received by the air during the compression.
At $3000\,\mathrm{rpm}$ one cylinder compresses $25$ times per second: power absorbed by the compression (it is returned later in the cycle).
Real compressions are fast (irreversibility) and lose heat to the walls: in which direction does each effect move $T_2$ ?

Part III — Combustion and expansion. At the top of the stroke $20\,\mathrm{mg}$ of fuel (heating value $43\,\mathrm{MJ}/\mathrm{kg}$ ) burn while the piston starts down, at constant pressure $P_2$ ; then the gas expands reversibly and adiabatically back to $V_1$ . Treat the gas as air throughout.

Heat released by the fuel.
Temperature $T_3$ and volume $V_3$ at the end of the isobaric combustion.
Work exchanged during the isobaric phase (sign!).
Temperature $T_4$ at the end of the adiabatic expansion, and the work exchanged during it.
Net work delivered by the gas over the four strokes, and the ratio to the heat released (the ideal efficiency).
Compare that efficiency with $1 - T_1/T_3$ , the bound that Chapter 24 will establish for any engine working between these extreme temperatures.
Heat rejected with the exhaust, by the first law over the cycle; check the balance.

Part IV — Measuring $\gamma$ with a bouncing ball. A steel ball of mass $m = 16.5\,\mathrm{g}$ fits a vertical glass tube of cross-section $S = 2.0\,\mathrm{cm}^{2}$ closed at the bottom by a flask of volume $V = 10\,\mathrm{L}$ of air at $P = 1.0\,\mathrm{bar}$ ; displaced by $x$ from its equilibrium position, it oscillates.

Why can the compressions and expansions of the air be taken as adiabatic? Write the relation between a small displacement $x$ and the pressure change $\dd P$ (Laplace’s law, linearized).
Deduce the restoring force on the ball and show that its motion is harmonic, with $\omega_0^2 = \gamma PS^2/mV$ .
Compute the period, and the value of $\gamma$ one would deduce from a measured period of $1.10\,\mathrm{s}$ .
Why does friction between ball and tube, and leakage of air, bias the result, and in which direction?
Summarize the three compressions: which quantity the first law conserved in each, and what the adiabatic exponent $\gamma$ governed.

Solution

Solution of Problem 22.1.

1. $n = P_1V_1/RT_1 = 10^5 \times 2.12\times10^{-4}/2494 = 8.5 \times 10^{-3}\,\mathrm{mol}$ .

2. $PV^\gamma$ constant: $V_2 = V_1(P_1/P_2)^{1/\gamma} = 212 \times 7^{-0.714} = 53\,\mathrm{cm}^{3}$ .

3. $T_2 = T_1(P_2/P_1)^{(\gamma - 1)/\gamma} = 300 \times 7^{0.286} = 523\,\mathrm{K}$ .

4. $W = nC_{V,m}(T_2 - T_1) = 8.5\times10^{-3} \times 20.8 \times 223 = 39\,\mathrm{J}$ .

5. Air to add: $\Delta n = \Delta P\,V/RT = 6\times10^5 \times 2\times10^{-3}/ 2494 = 0.48\,\mathrm{mol}$ : $0.48/8.5\times10^{-3} = 57$ strokes, about $2.2\,\mathrm{kJ}$ — a minute of honest effort.

6. At constant volume $P \propto T$ : the pressure falls as the air cools (by up to $300/523$ for the last stroke’s air); the cyclist adds a few strokes after a pause.

7. $W_{\text{iso}} = nRT\ln(P_2/P_1) = 8.5\times10^{-3} \times 2494 \times 1.95 = 41\,\mathrm{J} > 39\,\mathrm{J}$ : the isothermal path compresses the gas further (to $30\,\mathrm{cm}^{3}$ instead of $53\,\mathrm{cm}^{3}$ ) to reach $7\,\mathrm{bar}$ , because it stays cool — more volume swept, more work.

8. $n = 10^5 \times 5\times10^{-4}/2494 = 0.020\,\mathrm{mol}$ ; $T_2 = 300 \times 20^{0.4} = 994\,\mathrm{K}$ .

9. $P_2 = 20^{1.4} = 66\,\mathrm{bar}$ .

10. $994\,\mathrm{K}$ is well above the fuel’s self-ignition point: the injected fuel lights on contact. A petrol engine compresses its air–fuel mixture only to about $750\,\mathrm{K}$ so that it does not ignite before the spark; too high a ratio gives “knock”.

11. $W = nC_{V,m}(T_2 - T_1) = 0.020 \times 20.8 \times 694 = 289\,\mathrm{J}$ .

12. $25 \times 289 = 7.2\,\mathrm{kW}$ per cylinder, stored in the hot gas and given back in the expansion.

13. Irreversibility (the gas does not follow the quasi-static path; extra work is dissipated) raises $T_2$ ; heat lost to the walls lowers it. In a real engine the second usually wins: $T_2$ is a little below the ideal value.

14. $Q = 20\times10^{-6} \times 43\times10^6 = 860\,\mathrm{J}$ .

15. Isobaric: $Q = nC_{P,m}(T_3 - T_2)$ : $T_3 - T_2 = 860/(0.020 \times 29.1) = 1480\,\mathrm{K}$ , $T_3 = 2470\,\mathrm{K}$ ; $V_3 = V_2T_3/T_2 = 25 \times 2.48 = 62\,\mathrm{cm}^{3}$ .

16. $W = -P_2(V_3 - V_2) = -66\times10^5 \times 37\times10^{-6} = -244\,\mathrm{J}$ : work delivered by the gas.

17. $T_4 = T_3(V_3/V_1)^{\gamma - 1} = 2470 \times (0.124)^{0.4} = 1070\,\mathrm{K}$ ; $W = nC_{V,m}(T_4 - T_3) = 0.020 \times 20.8 \times (-1400) = -582\,\mathrm{J}$ .

18. Net work delivered $= 244 + 582 - 289 = 537\,\mathrm{J}$ ; ratio $537/860 = 62\%$ (the ideal Diesel efficiency at these settings; real engines reach about $40\%$ ).

19. $1 - 300/2470 = 88\%$ : the ideal Diesel cycle stays well below the Carnot bound, because its heat is received over a range of temperatures rather than at the highest one.

20. Over a cycle $\Delta U = 0$ : $Q_{\text{out}} = -(860 - 537) = -323\,\mathrm{J}$ , i.e. $323\,\mathrm{J}$ leave with the exhaust; check: $nC_{V,m}(T_4 - T_1) = 0.020 \times 20.8 \times 770 = 320\,\mathrm{J}$ .

21. The oscillation is fast (a second) compared with heat conduction through $10\,\mathrm{L}$ of air: adiabatic. Linearizing $PV^\gamma$ : $\dd P/P = -\gamma\,\dd V/V = -\gamma Sx/V$ .

22. Force on the ball $S\,\dd P = -\gamma PS^2x/V$ : $m\ddot x = -(\gamma PS^2/V)x$ , harmonic, $\omega_0^2 = \gamma PS^2/mV$ .

23. $\omega_0^2 = 1.4 \times 10^5 \times 4\times10^{-8}/(0.0165 \times 0.010) = 33.9\,\mathrm{s}^{-2}$ : $T = 2\pi/5.83 = 1.08\,\mathrm{s}$ . From $T = 1.10\,\mathrm{s}$ : $\gamma = 4\pi^2mV/(T^2PS^2) = 4\pi^2 \times 0.0165 \times 0.010/(1.21 \times 10^5 \times 4\times10^{-8}) = 1.35$ .

24. Friction damps the motion and, if it is dry, shifts the equilibrium; leakage past the ball lets air escape during a compression, lowering the restoring force: both lengthen the period and bias $\gamma$ low — as the $1.35$ above.

25. Pump and diesel: $Q = 0$ , so $\Delta U = W$ — work became internal energy and temperature, by Laplace’s law with exponent $\gamma$ . Ball: a fast compression is adiabatic, and $\gamma$ sets the stiffness of an air spring.

Terms defined in this chapter

See all 393 terms in the glossary