Physics 1

Solved thermodynamics exercises: processes, cycles, engines and entropy

Fourteen exercises with solutions and figures: ideal gas, calorimetry, isotherm, isobar, adiabat, Carnot, heat pump, entropy, rectangular cycle, Otto and Diesel cycles, lost work, free expansion, kinetic theory.

ielettronica 25 October 2026 10 min read 0 comments

Thermodynamics exercises of the kind set in Physics 1 and Engineering Thermodynamics courses, in three levels: basic, with the ideal gas, calorimetry and simple processes; intermediate, with adiabats, Carnot engines, heat pumps, entropy and a complete cycle; exam, with the Otto and Diesel cycles, entropy in irreversible processes and kinetic theory. The result is given straight away; the full solution, with the figure, opens with a click. The proofs and the table of processes are in the in-depth study linked at the top.

Data: $R = 8.314$ J/(mol·K); water $c = 4186$ J/(kg·K), latent heat of fusion of ice $L_f = 334$ kJ/kg; monatomic gas $c_V = \frac{3}{2}R$ , diatomic (air) $c_V = \frac{5}{2}R$ , $\gamma = 1.4$ .

Basic level

Exercise 1. The equation of state

2.0 mol of ideal gas at 300 K occupy 10 L. What is the pressure?

Result: $4.99\cdot 10^5$ Pa, about 4.9 atm.

Solution

$p = \dfrac{nRT}{V} = \dfrac{2.0\cdot 8.314\cdot 300}{0.010} = 4.99\cdot 10^5$ Pa. The volume must be in m³ (10 L = 0.010 m³): this is the most common mistake.

In atmospheres, $4.99\cdot 10^5/101\,325 = 4.9$ atm. Alternatively, with $R = 0.082\,06$ L·atm/(mol·K) and the volume in litres: $p = 2\cdot 0.082\,06\cdot 300/10 = 4.9$ atm.

The gas seen as molecules in motion — Pressure is the average effect of the molecules hitting the walls of the container.

Exercise 2. Mixing hot and cold water

200 g of water at 80 °C are mixed with 300 g at 20 °C, in an insulated container. Find the final temperature and the total entropy change.

Result: 44 °C; $\Delta S = +8.8$ J/K.

Solution

The heat given by the hot water equals that received by the cold water: $m_1c(T_1 - T_f) = m_2c(T_f - T_2)$ , so $T_f = \dfrac{m_1T_1 + m_2T_2}{m_1 + m_2} = \dfrac{0.2\cdot 80 + 0.3\cdot 20}{0.5} = 44$ °C.

Entropy, with temperatures in kelvin ( $T_f = 317.15$ K): hot water $0.2\cdot 4186\ln\frac{317.15}{353.15} = -90.0$ J/K; cold water $0.3\cdot 4186\ln\frac{317.15}{293.15} = +98.8$ J/K. Total $+8.8$ J/K: positive, as it must be in an irreversible process.

The temperatures of the two bodies approaching equilibrium — The hot body, in red, cools down and the cold one, in blue, warms up, until both reach the common final temperature.

Exercise 3. Isothermal expansion

1.0 mol of ideal gas expands reversibly at 300 K from 10 L to 20 L. Find the work, heat, change in internal energy and change in entropy.

Result: $W = Q = 1.73$ kJ; $\Delta U = 0$ ; $\Delta S = 5.76$ J/K.

Solution

In the isotherm $\Delta U = 0$ (the internal energy of an ideal gas depends only on $T$ ), so $Q = W$ .

$W = nRT\ln\dfrac{V_2}{V_1} = 8.314\cdot 300\cdot\ln 2 = 1729$ J.

$\Delta S = \dfrac{Q}{T} = nR\ln 2 = 5.76$ J/K. The gas took in heat and gave it all back as work: this does not violate Kelvin, because the process is not a cycle (the gas has changed, it occupies twice the volume).

Isothermal expansion in the pressure-volume plane — The gas goes from state 1 to state 2 along the hyperbola p V constant; the work is the red area under the curve.

Exercise 4. Heating at constant pressure

2.0 mol of monatomic gas are heated at constant pressure from 300 K to 400 K. Find the heat, the work and the change in internal energy.

Result: $Q = 4.16$ kJ; $W = 1.66$ kJ; $\Delta U = 2.49$ kJ.

Solution

$Q = nc_p\Delta T = 2\cdot\frac{5}{2}\cdot 8.314\cdot 100 = 4157$ J.

$W = p\Delta V = nR\Delta T = 2\cdot 8.314\cdot 100 = 1663$ J.

$\Delta U = nc_V\Delta T = 2\cdot\frac{3}{2}\cdot 8.314\cdot 100 = 2494$ J. Check with the first law: $Q - W = 4157 - 1663 = 2494$ J. 40 % of the heat goes into work and 60 % into internal energy, ratio $R/c_p = 2/5$ .

Isobaric heating — The gas goes from isotherm T1 to isotherm T2 along a horizontal segment; the work is the area of the rectangle under the segment.

Intermediate level

Exercise 5. Adiabatic compression of air

Air at 1.0 atm and 300 K is compressed adiabatically and reversibly to one tenth of its volume. Find the final temperature and pressure and the work done on the gas per mole.

Result: $T_2 = 754$ K; $p_2 = 25.1$ atm; 9.4 kJ/mol.

Solution

$TV^{\gamma - 1}$ constant: $T_2 = 300\cdot 10^{0.4} = 754$ K (481 °C). $pV^\gamma$ constant: $p_2 = 1.0\cdot 10^{1.4} = 25.1$ atm.

Work on the gas: it all goes into internal energy, $W = c_V\Delta T = \frac{5}{2}\cdot 8.314\cdot 454 = 9.4$ kJ/mol.

Comparison: an isotherm with the same volume ratio would have brought the pressure only to 10 atm. This is the principle of the Diesel engine: the compressed air becomes hot enough to ignite the fuel without a spark.

Adiabat and isotherm compared — Starting from the same state 1, the adiabat in red rises more steeply than the dashed isotherm, because the gas heats up during compression.

Exercise 6. A Carnot engine

A Carnot engine works between 500 K and 300 K and absorbs 1000 J per cycle. Find the efficiency, the work, the heat given out and the entropy change of the universe in one cycle.

Result: 40 %; 400 J; 600 J; 0.

Solution

$\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 0.40$ . $W = \eta Q_1 = 400$ J; $Q_2 = Q_1 - W = 600$ J.

Entropy: the hot reservoir loses $1000/500 = 2$ J/K, the cold one gains $600/300 = 2$ J/K, the gas returns to its initial state. Total zero: this is the hallmark of a reversible cycle. A real engine between the same reservoirs would give out more than 600 J, and the entropy of the universe would increase.

Carnot cycle in the temperature-entropy plane — The work is the area of the rectangle, the heat given out the area under the lower isotherm.

Exercise 7. The heat pump

A heat pump must deliver 10 kW to a house at 20 °C taking heat from the outside air at 0 °C. What is the minimum electrical power needed? And the real one, with a coefficient of performance of 3.5?

Result: minimum 0.68 kW (ideal COP 14.7); real 2.86 kW.

Solution

For a heat pump the coefficient of performance is the heat delivered divided by the work: $COP = Q_1/W$ . The maximum, for a reversible machine, is $\dfrac{T_1}{T_1 - T_2} = \dfrac{293.15}{20} = 14.7$ .

Minimum power: $10/14.7 = 0.68$ kW. With a real COP of 3.5: $10/3.5 = 2.86$ kW, against 10 kW for an electric heater. The heat pump does not “create” energy: it moves heat from the cold outside to the warm inside, and the work only serves to make it go against its natural direction.

Diagram of a heat pump — The machine P takes the heat Q2 from the cold reservoir, receives the work W and gives the hot reservoir the heat Q1, the sum of the two.

Exercise 8. The entropy of melting ice

100 g of ice at 0 °C slowly melt in a room at 25 °C. Compute the entropy change of the ice, of the room and of the universe (neglect the warming of the water after melting).

Result: ice +122.3 J/K; room −112.0 J/K; universe +10.3 J/K.

Solution

Heat absorbed: $Q = mL_f = 0.1\cdot 334\,000 = 33.4$ kJ, at constant temperature.

Ice: $\Delta S = Q/T = 33\,400/273.15 = +122.3$ J/K. Room (which stays practically at 298.15 K): $-33\,400/298.15 = -112.0$ J/K.

Universe: $+10.3$ J/K. The same heat weighs more in entropy at the lower temperature: this is why heat flows spontaneously from hot to cold.

Heat flows from the room to the ice — The room at 298 K gives the heat Q to the ice at 273 K: the entropy of the room decreases, that of the ice increases more.

Exercise 9. A rectangular cycle

1 mol of monatomic gas performs the cycle A→B→C→D→A of the figure, with A at $(p_0, V_0)$ , B at $(2p_0, V_0)$ , C at $(2p_0, 2V_0)$ , D at $(p_0, 2V_0)$ . Express in terms of $p_0V_0$ the work, the heat absorbed and the efficiency, and compare with a Carnot engine between the extreme temperatures.

Rectangular cycle in the pressure-volume plane — The cycle is made of two isochores and two isobars; the dashed curves are the isotherms T0, 2 T0 and 4 T0 through the vertices.

Result: $W = p_0V_0$ ; $Q_{abs} = 6.5\,p_0V_0$ ; $\eta = 15.4$ %, against 75 % for Carnot.

Solution

Work. It is the area of the rectangle: $(2p_0 - p_0)(2V_0 - V_0) = p_0V_0$ .

Temperatures. With $pV = RT$ : $T_A = T_0 = p_0V_0/R$ , $T_B = T_D = 2T_0$ , $T_C = 4T_0$ .

Heats. A→B isochoric: $Q = c_V(T_B - T_A) = \frac{3}{2}R\,T_0 = 1.5\,p_0V_0$ (absorbed). B→C isobaric: $Q = c_p(T_C - T_B) = \frac{5}{2}R\cdot 2T_0 = 5\,p_0V_0$ (absorbed). C→D and D→A give out heat. Absorbed in total: $6.5\,p_0V_0$ .

Efficiency. $\eta = \dfrac{p_0V_0}{6.5\,p_0V_0} = 15.4$ %. Carnot between $4T_0$ and $T_0$ : $1 - \frac{1}{4} = 75$ %. The rectangular cycle exchanges heat at temperatures far from the extreme ones, and this makes it inefficient.

Typical mistake: dividing the work by the net heat (which equals the work, and would give 100 %) instead of by the heat absorbed.

Exam level

Exercise 10. The Otto cycle

An ideal petrol engine has a compression ratio of 9. Air enters at 1 bar and 300 K, and after combustion reaches 1800 K. Find the temperatures at the four vertices, the heat absorbed and given out and the work per mole, and the efficiency.

Result: $T_2 = 722$ K, $T_4 = 747$ K; $Q_{in} = 22.4$ kJ/mol, $Q_{out} = 9.3$ kJ/mol, $W = 13.1$ kJ/mol; $\eta = 58.5$ %.

Solution

Adiabatic compression 1→2: $T_2 = T_1r^{\gamma - 1} = 300\cdot 9^{0.4} = 722$ K.

Isochoric combustion 2→3: $T_3 = 1800$ K (given). $Q_{in} = c_V(T_3 - T_2) = 20.79\cdot 1078 = 22.4$ kJ/mol.

Adiabatic expansion 3→4: $T_4 = T_3/r^{\gamma - 1} = 1800/2.408 = 747$ K.

Isochoric exhaust 4→1: $Q_{out} = c_V(T_4 - T_1) = 20.79\cdot 447 = 9.3$ kJ/mol.

$W = 22.4 - 9.3 = 13.1$ kJ/mol, $\eta = 13.1/22.4 = 58.5$ %, equal to $1 - 9^{-0.4}$ (in-depth study). Carnot between 1800 K and 300 K would give 83 %: the Otto cycle absorbs heat at temperatures between 722 and 1800 K, not all at the maximum.

Exercise 11. The Diesel cycle

An ideal Diesel engine has a compression ratio of 18 and a cut-off ratio of 2. Compute the efficiency and compare it with that of an Otto cycle with the same compression ratio.

Result: 63.2 %; the Otto would give 68.5 %.

Solution

From the formula of the in-depth study:

$\eta = 1 - \frac{1}{18^{0.4}}\cdot\frac{2^{1.4} - 1}{1.4\cdot(2 - 1)} = 1 - \frac{1}{3.178}\cdot\frac{1.639}{1.4} = 1 - 0.315\cdot 1.171 = 0.632$

Otto with $r = 18$ : $1 - 18^{-0.4} = 68.5$ %. For the same compression the Otto is more efficient, but a petrol engine cannot reach 18 without detonation: in practice it works around 10 (60 %), and the Diesel wins.

Diesel cycle — Heat enters at constant pressure along the horizontal stretch 2 3; the rest of the cycle is as in the Otto.

Exercise 12. Two blocks in contact and the lost work

Two 1.0 kg copper blocks ( $c = 385$ J/(kg·K)), at 400 K and 300 K, are put in contact and insulated. Find the final temperature and the entropy change. What is the maximum work that could have been obtained by running a reversible engine between the two blocks until equilibrium?

Result: 350 K; $\Delta S = +7.9$ J/K; maximum work 2.76 kJ.

Solution

Direct contact. Equal masses and specific heats: $T_f = 350$ K. $\Delta S = mc\left(\ln\frac{350}{400} + \ln\frac{350}{300}\right) = 385\cdot 0.0206 = 7.9$ J/K.

Reversible engine. Now the total entropy does not change: $mc\ln\frac{T_f'}{400} + mc\ln\frac{T_f'}{300} = 0$ , so $T_f'^2 = 400\cdot 300$ and $T_f' = \sqrt{120\,000} = 346.4$ K, the geometric mean.

The work is the heat given by the hot block minus that received by the cold one: $W = mc(400 - 346.4) - mc(346.4 - 300) = mc(700 - 692.8) = 385\cdot 7.2 = 2.76$ kJ.

In direct contact this work is lost: the final temperature is higher (350 against 346.4 K) because the energy that could have become work has remained as heat.

Two bodies reaching equilibrium — The temperatures of the two blocks converge towards the common final value.

Exercise 13. Free expansion

1.0 mol of ideal gas at 300 K, in an insulated container, expands into a vacuum doubling its volume. Find $Q$ , $W$ , $\Delta U$ , $\Delta T$ and $\Delta S$ . How much work could have been obtained with a reversible isothermal expansion?

Result: $Q = W = \Delta U = \Delta T = 0$ ; $\Delta S = 5.76$ J/K; 1.73 kJ.

Solution

The gas pushes nothing (the vacuum has no pressure): $W = 0$ . The container is insulated: $Q = 0$ . So $\Delta U = 0$ and, for an ideal gas, $\Delta T = 0$ .

Entropy is a state function: it is computed along any reversible path between the same states, for example the isotherm: $\Delta S = nR\ln 2 = 5.76$ J/K, although no heat has flowed.

Lost work: a reversible isotherm would have given $nRT\ln 2 = T\Delta S = 1.73$ kJ. In general the energy “wasted” by an irreversible process at temperature $T$ is $T\Delta S_{universe}$ .

Free expansion — Once the wall is opened, the gas fills the whole container with no work and no heat.

Exercise 14. The speed of molecules

Compute the root-mean-square speed of nitrogen molecules ( $M = 28$ g/mol) and of hydrogen molecules ( $M = 2$ g/mol) at 300 K. Why has the Earth’s atmosphere lost almost all its free hydrogen, although the escape velocity is 11.2 km/s?

Result: nitrogen 517 m/s; hydrogen 1930 m/s.

Solution

$v_{rms} = \sqrt{\dfrac{3RT}{M}}$ : nitrogen $\sqrt{3\cdot 8.314\cdot 300/0.028} = 517$ m/s; hydrogen $\sqrt{3\cdot 8.314\cdot 300/0.002} = 1930$ m/s. At the same temperature the mean kinetic energies are equal, so the speeds go as $1/\sqrt M$ .

Both are far below the escape velocity, but speeds are distributed (Maxwell distribution): a small fraction of molecules is much faster than average. In the upper atmosphere, where the temperature reaches 1000 K and more, the tail of the hydrogen distribution exceeds the escape velocity often enough to empty it over geological times; for nitrogen the fraction is negligible.

Maxwell speed distribution at 300 K — The hydrogen curve, in red, is much wider and shifted towards high speeds than the nitrogen one, in blue. The dots mark the root-mean-square speeds.

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Basic level

Exercise 1. The equation of state

Exercise 2. Mixing hot and cold water

Exercise 3. Isothermal expansion

Exercise 4. Heating at constant pressure

Intermediate level

Exercise 5. Adiabatic compression of air

Exercise 6. A Carnot engine

Exercise 7. The heat pump

Exercise 8. The entropy of melting ice

Exercise 9. A rectangular cycle

Exam level

Exercise 10. The Otto cycle

Exercise 11. The Diesel cycle

Exercise 12. Two blocks in contact and the lost work

Exercise 13. Free expansion

Exercise 14. The speed of molecules

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