Physics · Secondary School · Thermodynamics

Thermodynamics

The engine that drives a car, the power station that supplies a city and the fridge that keeps the milk cold all exchange energy with a gas, sometimes as heat, sometimes as work. We pick up where the Heat and Temperature lesson left off, with ideal gases, to measure the work done by a gas and its internal energy, reach the first law, study the processes and cycles of heat engines and finish with the second law, which explains why no engine makes use of 100% of the heat and why the fridge warms up the kitchen.

1Work done by a gas
2Internal energy
3First law
4Processes
5Cycles and engines
6Second law
7Refrigerators
✓Challenges

STEP 1

When does a gas do work?

In the Heat and Temperature lesson we saw that the particles of a gas hit the walls of the container and produce pressure. If we swap one of these walls for a sliding piston, the gas pushes it as it expands and does work on it, like a force that moves a body.

With constant pressure, the force on the piston is \(F = p\,A\), and the product of the area and the displacement \(d\) is the change in volume. This leads to \(\tau = p\,\Delta V\), which on the \(p \times V\) graph is the area of the rectangle under the horizontal line.

\(\tau = p\,\Delta V\)\(\tau = \text{area under the curve } p \times V\)\(1\ \text{kPa} \cdot 1\ \text{L} = 1\ \text{J}\)The first formula holds at constant pressure. With the pressure in kPa and the volume in litres, the product comes out directly in joules, because \(10^3\ \text{Pa} \cdot 10^{-3}\ \text{m}^3 = 1\ \text{J}\). In the simulation, the cylinder holds about 0.24 mol of ideal gas, with \(n\,R = 2\ \text{J/K}\) and \(R = 8.3\ \text{J/(mol·K)}\). Changing the pressure with the piston at rest corresponds to changing the weight and, at the same time, heating or cooling the gas at constant volume.

When the pressure changes during the process, we split the path into tiny, almost horizontal pieces and add up the areas. The work then becomes the area under the curve on the \(p \times V\) graph, and so it depends on the path between two states, as well as on the start and the end.

The sign says who pushes whom. In an expansion, \(\Delta V > 0\) and the gas does work, \(\tau > 0\); in a compression, \(\Delta V < 0\), the gas receives work from outside and \(\tau < 0\). With the volume fixed, the work is zero, however large the pressure.

Let's discuss

Heat the gas at constant pressure until the volume reaches 8 L and check the \(\tau\) shown against \(p \cdot \Delta V\), reading the values off the graph.
Now cool the gas back to the initial volume. Does the accumulated work return to zero? Which area was subtracted?
Change the pressure with the piston at rest. What shape does this stretch have on the graph, and how much work does it add?
Restart and take the gas to the same final state by two paths, first heating and then raising the pressure, and then in the reverse order. Was the work the same?

Pressure (weight on the piston)

Pressure

Volume

Temperature

Accumulated work

ΔV since the start

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A gas expands at a constant pressure of \(2 \cdot 10^5\ \text{Pa}\), and its volume goes from \(0.010\ \text{m}^3\) to \(0.025\ \text{m}^3\). How much work does the gas do?

Show solution
With constant pressure, the work is the pressure times the change in volume, \(\tau = p\,\Delta V = 2 \cdot 10^5 \cdot 0.015\).
\(\tau = 3000\ \text{J}\)
basic
A gas is compressed at a constant pressure of \(1.0 \cdot 10^5\ \text{Pa}\), from 6 L to 2 L. What is the work done by the gas? Does it do work or receive work?

Show solution
We convert the change in volume to cubic metres, \(\Delta V = -4\ \text{L}\) \(= -4 \cdot 10^{-3}\ \text{m}^3\), and calculate \(\tau = p\,\Delta V\) \(= 1.0 \cdot 10^5 \cdot (-4 \cdot 10^{-3})\).
The negative sign indicates that the gas was compressed and received work from outside.
\(\tau = -400\ \text{J}\): the gas receives 400 J of work.
basic
A gas is heated inside a rigid gas cylinder, and its pressure doubles. Does it do work in this process? How does the process appear on the \(p \times V\) graph?

Show solution
The volume of the cylinder does not change, and with no change in volume the gas pushes no wall through any displacement.
On the \(p \times V\) graph, the process is a vertical segment, which has no area beneath it.
There is no work, \(\tau = 0\), however large the rise in pressure.
basic
We block the tip of a syringe with a finger and push the plunger in. Then we let go of the plunger, which springs back a little. In which of the two stages does the air in the syringe do work, and in which does it receive work?

Show solution
When we push the plunger in, the volume of the air decreases, \(\Delta V < 0\), and it is we who do work on the air.
When we let go, the air expands, \(\Delta V > 0\), and pushes the plunger back.
The air receives work in the compression and does work in the expansion.
intermediate
On the \(p \times V\) graph, a gas goes in a straight line from state A (2 L; 300 kPa) to state B (6 L; 100 kPa). How much work does it do? Remember that \(1\ \text{kPa} \cdot 1\ \text{L} = 1\ \text{J}\).

Show solution
The work is the area under the segment, a trapezium with parallel sides of 300 kPa and 100 kPa and a width of 4 L, \(\tau = \dfrac{(300 + 100) \cdot 4}{2}\).
Since the volume increased, the work is positive.
\(\tau = 800\ \text{J}\)
intermediate
A gas leaves state A (2 L; 100 kPa) and reaches state C (5 L; 300 kPa) by two paths. On path I, it is heated at constant volume up to 300 kPa and then expands at constant pressure. On path II, it first expands at 100 kPa and is then heated at constant volume. Calculate the work on each path.

Show solution
The constant-volume stretches involve no work. On path I, the expansion happens at 300 kPa, and \(\tau_I = 300 \cdot 3 = 900\ \text{J}\).
On path II, the expansion happens at 100 kPa, and \(\tau_{II} = 100 \cdot 3 = 300\ \text{J}\).
900 J along path I and 300 J along path II, which shows that the work depends on the path.
intermediate
In a cylinder, the piston has an area of \(0.02\ \text{m}^2\), and the gas pushes it at a constant pressure of \(1.5 \cdot 10^5\ \text{Pa}\), making it rise 10 cm. Calculate the work in two ways, from the force on the piston and from \(p\,\Delta V\).

Show solution
The force on the piston is \(F = p\,A\) \(= 1.5 \cdot 10^5 \cdot 0.02\) \(= 3000\ \text{N}\), and its work is \(F\,d = 3000 \cdot 0.10\).
From the volume, \(\Delta V = A\,d = 0.002\ \text{m}^3\), and \(p\,\Delta V = 1.5 \cdot 10^5 \cdot 0.002\).
Both calculations give \(\tau = 300\ \text{J}\).
intermediate
Two moles of an ideal gas are heated at constant pressure, from 300 K to 400 K. How much work does the gas do? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
From the ideal gas equation, \(p\,V = n\,R\,T\), and with the pressure fixed, \(p\,\Delta V = n\,R\,\Delta T\).
So \(\tau = n\,R\,\Delta T = 2 \cdot 8.3 \cdot 100\).
\(\tau = 1660\ \text{J}\)
challenge
We seal a syringe and push the plunger in slowly. The pressure of the air inside grows roughly linearly, from 100 kPa to 300 kPa, while the volume falls from 60 mL to 20 mL. Estimate the work done by the air in this compression.

Show solution
On a \(p \times V\) graph, the path is a straight segment, and the area under it is a trapezium, \(\dfrac{(100 + 300) \cdot 40}{2}\) \(= 8000\ \text{kPa} \cdot \text{mL}\).
Since \(1\ \text{kPa} \cdot 1\ \text{mL}\) \(= 10^3\ \text{Pa} \cdot 10^{-6}\ \text{m}^3\) \(= 10^{-3}\ \text{J}\), the area is \(8000 \cdot 10^{-3}\) joules, with a negative sign because the volume decreased.
\(\tau \approx -8\ \text{J}\): the air receives about 8 J of work.
challenge
A vertical cylinder with air inside is closed by a 20 kg piston of area \(0.01\ \text{m}^2\), which slides without friction. Atmospheric pressure is \(1.0 \cdot 10^5\ \text{Pa}\), and \(g = 10\ \text{m/s}^2\). When heated, the air lifts the piston 5 cm slowly. What is the pressure of the air, how much work does it do and what part of this work goes into lifting the piston?

Show solution
In equilibrium, the air holds up the atmosphere and the weight of the piston, \(p = 1.0 \cdot 10^5 + \dfrac{20 \cdot 10}{0.01}\).
The volume grows by \(\Delta V = 0.01 \cdot 0.05 = 5 \cdot 10^{-4}\ \text{m}^3\), and \(\tau = p\,\Delta V\).
Lifting the piston costs \(m\,g\,h = 20 \cdot 10 \cdot 0.05\), and the rest of the work pushes back the air outside.
The pressure is \(1.2 \cdot 10^5\ \text{Pa}\) and the air does 60 J, of which 10 J lift the piston and 50 J push back the atmosphere.

STEP 2

What changes when a gas warms up?

The internal energy \(U\) of a gas is the sum of the energies of its particles. In a monatomic ideal gas, such as helium, neon or argon, we assume that the particles do not interact at a distance, and the internal energy comes down to the kinetic energy of translation.

Kinetic theory, which we sketched in the Heat and Temperature lesson, indicates that the average kinetic energy of each particle is proportional to the absolute temperature, \(\tfrac{3}{2}\,k\,T\). Adding over the \(n\) moles of particles, we arrive at \(U = \tfrac{3}{2}\,n\,R\,T\).

Something that may come as a surprise is what we do not find in this formula. Volume and pressure do not appear, and two containers with the same amount of gas at the same temperature have the same internal energy, even if one is much bigger than the other.

From this follow two consequences that we will use until the end of the lesson. In an isothermal process of an ideal gas, \(\Delta U = 0\). In a cycle, in which the gas returns to its starting state, the temperature returns to its initial value, and again \(\Delta U = 0\).

\(U = \tfrac{3}{2}\,n\,R\,T\)\(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\)\(\bar E_k = \tfrac{3}{2}\,k\,T\)\(R = 8.3\ \text{J/(mol·K)}\), \(k = 1.38 \cdot 10^{-23}\ \text{J/K}\) and \(T\) in kelvin. From here to the end of the lesson we assume a monatomic ideal gas. For air, made of two-atom molecules that also rotate, the factor 3/2 becomes close to 5/2, and the qualitative conclusions do not change. In the simulation, box A holds about 0.24 mol of gas in 4 L, and the number of dots is only representative.

Let's discuss

Leave the two boxes at the same temperature and change the volume of B. Does the internal energy of B change? And the pressure?
Heat box A only and compare the agitation of the particles and the average kinetic energy in the two boxes.
Double the amount of gas in B, at the same temperature as A. What happens to \(U\) and to the average kinetic energy of each particle?
Why, at the same temperature, does the pressure depend on the volume while the internal energy does not? Let us think about what changes when the same particles fill a bigger box.

Temperature of A

Temperature of B

Volume of B

Mean Ek in A

Mean Ek in B

U in A

U in B

Pressure in A

Pressure in B

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the internal energy of 2 mol of a monatomic ideal gas at 300 K? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
For a monatomic ideal gas, \(U = \tfrac{3}{2}\,n\,R\,T = 1.5 \cdot 2 \cdot 8.3 \cdot 300\).
\(U = 7470\ \text{J}\)
basic
One mole of helium, which we treat as a monatomic ideal gas, is heated from 300 K to 500 K. What is the change in its internal energy? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
The internal energy follows the temperature, \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\) \(= 1.5 \cdot 1 \cdot 8.3 \cdot 200\).
\(\Delta U = 2490\ \text{J}\)
basic
Two containers hold the same amount of the same ideal gas, at the same temperature. One of them has three times the volume of the other. Which of the two gases has more internal energy?

Show solution
For an ideal gas, \(U = \tfrac{3}{2}\,n\,R\,T\) depends only on the amount of gas and the temperature, and the volume does not enter the formula.
The pressures are different, but the particles have the same average kinetic energy in the two containers.
The internal energies are equal.
basic
An ideal gas expands slowly, and its temperature stays constant the whole time. What is the change in its internal energy?

Show solution
For an ideal gas, the internal energy depends only on the temperature, and if \(T\) does not change, \(U\) does not change either.
\(\Delta U = 0\)
intermediate
Half a mole of a monatomic ideal gas has an internal energy of 3735 J. What is its temperature? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
Making the temperature the subject of \(U = \tfrac{3}{2}\,n\,R\,T\), we get \(T = \dfrac{U}{1.5\,n\,R} = \dfrac{3735}{1.5 \cdot 0.5 \cdot 8.3}\).
\(T = 600\ \text{K}\)
intermediate
What is the average kinetic energy of a particle of an ideal gas at 300 K? And at 600 K? Use \(k = 1.38 \cdot 10^{-23}\ \text{J/K}\).

Show solution
The average translational kinetic energy is \(\bar E_k = \tfrac{3}{2}\,k\,T\). At 300 K, \(\bar E_k = 1.5 \cdot 1.38 \cdot 10^{-23} \cdot 300\), and at 600 K the value doubles, because it is proportional to \(T\).
About \(6.2 \cdot 10^{-21}\ \text{J}\) at 300 K and \(1.24 \cdot 10^{-20}\ \text{J}\) at 600 K.
intermediate
One balloon holds helium and another holds neon, both at 27 °C. In which of them do the particles have more average kinetic energy? In which do they move faster, on average?

Show solution
The average translational kinetic energy, \(\tfrac{3}{2}\,k\,T\), depends only on the temperature, and it is the same in the two balloons.
With the same kinetic energy, the particle of greater mass moves more slowly, and a neon atom has about five times the mass of a helium atom.
The average kinetic energy is the same in both, and the helium atoms move faster.
intermediate
A monatomic ideal gas goes from the state (100 kPa; 3 L) to the state (150 kPa; 2 L), by any path. What is the change in its internal energy?

Show solution
Since \(n\,R\,T = p\,V\), the internal energy can also be written as \(U = \tfrac{3}{2}\,p\,V\).
In both states, \(p\,V = 300\ \text{kPa} \cdot \text{L} = 300\ \text{J}\), and therefore the temperature is the same.
\(\Delta U = 0\), whatever the path taken.
challenge
Show that, for a monatomic ideal gas, \(U = \tfrac{3}{2}\,p\,V\). Use this relation to calculate the internal energy of a gas at \(2 \cdot 10^5\ \text{Pa}\) in \(0.03\ \text{m}^3\) and its change if the gas expands at constant pressure up to \(0.05\ \text{m}^3\).

Show solution
From the ideal gas equation, \(n\,R\,T = p\,V\), and substituting into \(U = \tfrac{3}{2}\,n\,R\,T\) we arrive at \(U = \tfrac{3}{2}\,p\,V\).
At the start, \(U = 1.5 \cdot 2 \cdot 10^5 \cdot 0.03\), and in the expansion, \(\Delta U = 1.5 \cdot 2 \cdot 10^5 \cdot 0.02\).
\(U = 9000\ \text{J}\) at the start and \(\Delta U = 6000\ \text{J}\) in the expansion.
challenge
In a rigid, insulated container, a partition separates 1 mol of argon at 300 K from 2 mol of argon at 450 K. We remove the partition and wait for equilibrium. Treating argon as a monatomic ideal gas, what is the final temperature?

Show solution
The container is rigid and insulated, so no work or heat is exchanged with the outside, and the total internal energy is conserved.
With \(U = \tfrac{3}{2}\,n\,R\,T\), conservation gives \(1 \cdot 300 + 2 \cdot 450 = 3\,T_f\), where the factor \(\tfrac{3}{2}\,R\) cancels out.
\(T_f = 400\ \text{K}\)

STEP 3

Where does the heat a gas receives go?

When we heat a gas in a cylinder with a piston, the energy that comes in as heat can follow two paths. Part of it stays in the gas, raising the internal energy and the temperature, and part leaves as work, if the gas pushes the piston.

We call the conservation of energy written for this balance the first law of thermodynamics, \(Q = \tau + \Delta U\). The heat \(Q\) that the gas receives equals the work \(\tau\) that it does plus the change \(\Delta U\) in its internal energy.

We use the signs in the table below, which deserve attention, because a good share of the mistakes in exercises tends to come from them. The heat that the gas gives out enters with a negative sign, and so does the work it receives in a compression.

With the right signs, the law allows cases that look strange at first sight. A gas can receive heat and cool down, if it does more work than the heat it received, and it can warm up without receiving any heat at all, if it is compressed.

\(Q = \tau + \Delta U\)\(\Delta U = Q - \tau\)\(1\ \text{cal} \approx 4.2\ \text{J}\)When the heat is given in calories, we convert to joules with 1 cal ≈ 4.2 J. In the simulation, the gas is the same as in step 1, with \(n\,R = 2\ \text{J/K}\), and it starts at 300 K. Since \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T = 3\,\Delta T\), in joules, the change in temperature comes straight out of the balance.

Sign convention
Quantity	Positive	Negative
\(Q\)	the gas receives heat	the gas gives out heat
\(\tau\)	the gas expands and does work	the gas is compressed and receives work
\(\Delta U\)	the temperature rises	the temperature falls

Let's discuss

Choose heating at constant pressure and see how the 500 J are shared out. What fraction becomes work?
Switch to heating at constant volume. Why is \(\Delta U = Q\) now?
In the adiabatic case, \(Q = 0\). Where does the increase in internal energy come from?
Adjust the controls until the gas receives heat and still cools down. What condition must the work meet?

Heat Q

Work τ

Heat Q

Work τ

Change ΔU

Change in temperature

Final temperature

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A gas receives 500 J of heat and does 200 J of work. What is the change in its internal energy?

Show solution
By the first law, \(\Delta U = Q - \tau = 500 - 200\).
\(\Delta U = 300\ \text{J}\)
basic
A gas is compressed and receives 300 J of work, while it gives out 100 J of heat to the surroundings. What is the change in its internal energy? Does the temperature go up or down?

Show solution
With the signs from the table, \(Q = -100\ \text{J}\) and \(\tau = -300\ \text{J}\), and \(\Delta U = Q - \tau = -100 - (-300)\).
\(\Delta U = +200\ \text{J}\), and the temperature goes up.
basic
A gas gives out 50 J of heat at constant volume. What is the change in its internal energy, and what happens to the temperature?

Show solution
At constant volume, \(\tau = 0\), and the first law becomes \(\Delta U = Q = -50\ \text{J}\).
\(\Delta U = -50\ \text{J}\), and the temperature decreases.
basic
A gas receives 100 cal of heat and does 120 J of work. What is the change in its internal energy, in joules? Use \(1\ \text{cal} \approx 4.2\ \text{J}\).

Show solution
In joules, the heat received is \(Q = 100 \cdot 4.2 = 420\ \text{J}\), and \(\Delta U = 420 - 120\).
\(\Delta U = 300\ \text{J}\)
intermediate
One mole of a monatomic ideal gas is heated at constant pressure, and the temperature rises by 100 K. Calculate the work, the change in internal energy and the heat received. What fraction of the heat becomes work? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
At constant pressure, \(\tau = n\,R\,\Delta T = 8.3 \cdot 100\), and \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\) \(= 1.5 \cdot 8.3 \cdot 100\).
By the first law, \(Q = \tau + \Delta U = \tfrac{5}{2}\,n\,R\,\Delta T\), and the fraction is \(\dfrac{\tau}{Q} = \dfrac{n\,R\,\Delta T}{\tfrac{5}{2}\,n\,R\,\Delta T}\) \(= \dfrac{2}{5}\).
\(\tau = 830\ \text{J}\), \(\Delta U = 1245\ \text{J}\) and \(Q = 2075\ \text{J}\), with 40% of the heat becoming work.
intermediate
A gas receives 1000 J of heat at a constant pressure of \(2 \cdot 10^5\ \text{Pa}\), and its internal energy increases by 600 J. By how much did the volume increase, in litres?

Show solution
The work comes from the first law, \(\tau = Q - \Delta U\) \(= 400\ \text{J}\).
With constant pressure, \(\Delta V = \dfrac{\tau}{p} = \dfrac{400}{2 \cdot 10^5}\) \(= 2 \cdot 10^{-3}\ \text{m}^3\).
The volume increased by 2 L.
intermediate
Is it possible for a gas to receive heat and still cool down? And to warm up without receiving any heat at all? Justify with the first law.

Show solution
By the first law, \(\Delta U = Q - \tau\). If the gas receives heat, \(Q > 0\), but does more work than this heat, \(\Delta U\) is negative and the gas cools down.
If there is no heat, \(Q = 0\), and the gas is compressed, \(\tau < 0\), then \(\Delta U = -\tau > 0\) and the gas warms up.
Both cases are possible, because the temperature follows \(\Delta U\), which depends on the heat and the work together.
intermediate
A syringe with 0.05 mol of argon, which we treat as a monatomic ideal gas, is compressed so quickly that there is almost no heat exchange. The gas receives 50 J of work. By how much does the temperature rise? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
With no heat exchange, \(Q = 0\) and \(\Delta U = -\tau = +50\ \text{J}\).
With \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\), we have \(\Delta T = \dfrac{50}{1.5 \cdot 0.05 \cdot 8.3}\).
The temperature rises by about 80 K.
challenge
We supply 498 J of heat to 0.4 mol of a monatomic ideal gas, first at constant volume and then, starting from the same state, at constant pressure. In which case does the temperature rise more, and by how much does it rise in each? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
At constant volume, all the heat becomes internal energy, \(\Delta T = \dfrac{Q}{1.5\,n\,R} = \dfrac{498}{1.5 \cdot 0.4 \cdot 8.3}\).
At constant pressure, \(Q = \tfrac{5}{2}\,n\,R\,\Delta T\), because part of the heat leaves as work, and \(\Delta T = \dfrac{498}{2.5 \cdot 0.4 \cdot 8.3}\).
It rises by 100 K at constant volume and by 60 K at constant pressure.
challenge
A monatomic ideal gas receives 500 cal of heat at a constant pressure of \(1.4 \cdot 10^5\ \text{Pa}\). How much work does it do, by how much does its internal energy increase and by how much does the volume grow? Use \(1\ \text{cal} \approx 4.2\ \text{J}\).

Show solution
In joules, \(Q = 500 \cdot 4.2 = 2100\ \text{J}\). In an isobaric process of a monatomic gas, \(\tau = n\,R\,\Delta T\) and \(Q = \tfrac{5}{2}\,n\,R\,\Delta T\), so that \(\tau = \tfrac{2}{5}\,Q = \tfrac{2}{5} \cdot 2100\).
The internal energy increases by \(\Delta U = Q - \tau\), and the volume grows by \(\Delta V = \dfrac{\tau}{p}\), with \(\tau\) in joules and \(p\) in pascals.
\(\tau = 840\ \text{J}\), \(\Delta U = 1260\ \text{J}\) and the volume increases by 6 L.

STEP 4

Four ways of changing a gas

Some processes come up so often that we give them names of their own. In an isobaric process, the pressure stays constant, the \(p \times V\) graph is a horizontal line and \(\tau = p\,\Delta V\). In an isochoric, or isovolumetric, process the volume does not change, \(\tau = 0\) and the heat exchanged equals \(\Delta U\).

In an isothermal process, the temperature stays constant, \(\Delta U = 0\) and \(Q = \tau\), so that the gas gives back as work the heat it receives. The curve is the hyperbola \(p\,V = \text{constant}\) that we saw in Heat and Temperature.

isobaric: \(\tau = p\,\Delta V\)isochoric: \(\tau = 0,\ Q = \Delta U\)isothermal: \(\Delta U = 0,\ Q = \tau\)adiabatic: \(Q = 0,\ \tau = -\Delta U\)In the simulation, all four start from state A, with 100 kPa, 6 L and 300 K, and the gas has \(n\,R = 2\ \text{J/K}\). In the adiabatic process of a monatomic ideal gas, \(p\,V^{5/3} = \text{constant}\) holds, a relation that only the simulation uses.

In an adiabatic process, we assume that the gas exchanges no heat, \(Q = 0\), and \(\tau = -\Delta U\). As it expands, it pays for the work with its own internal energy and cools down, and that is why the adiabatic curve falls more steeply than the isotherm on the graph.

Fast processes tend to be almost adiabatic, because the heat does not have time to cross the walls. This may help to explain why a bicycle pump warms up when we pump up a tyre quickly and why the spray from an aerosol can comes out cold, since the gas expands as it leaves the can (the evaporation of the liquid also contributes).

Let's discuss

Run the four processes taking the gas to 9 L, or to 450 K in the isochoric one. In which of them does the gas do the most work, and in which does the temperature fall?
In the isothermal process, compare \(Q\) and \(\tau\). Where did the energy that came in as heat go?
Compress the gas adiabatically down to 3 L, as in a bicycle pump, and follow the temperature.
In the isochoric process, why does the work stay at zero even though the pressure rises?

Final volume

Final temperature (isochoric)

Pressure

Volume

Temperature

Heat Q

Work τ

Change ΔU

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A gas receives 300 J of heat in an isochoric process. What is the work, and by how much does the internal energy increase?

Show solution
In an isochoric process the volume does not change, and \(\tau = 0\). By the first law, \(\Delta U = Q - 0\).
\(\tau = 0\) and \(\Delta U = 300\ \text{J}\)
basic
In an isothermal expansion, an ideal gas does 500 J of work. How much heat does it receive?

Show solution
In an isothermal process of an ideal gas, \(\Delta U = 0\), and the first law becomes \(Q = \tau\).
\(Q = 500\ \text{J}\)
basic
A gas is compressed adiabatically and receives 200 J of work. By how much does its internal energy change, and what happens to the temperature?

Show solution
In an adiabatic process, \(Q = 0\), and \(\Delta U = -\tau = -(-200)\).
\(\Delta U = +200\ \text{J}\), and the gas warms up.
basic
Anyone who pumps up a bicycle tyre quickly notices that the body of the pump warms up. Why does this happen?

Show solution
The compression is fast, and the air has almost no time to exchange heat with the walls, as in an adiabatic process.
With \(Q \approx 0\), the work we do on the air becomes internal energy, \(\Delta U = -\tau > 0\).
The compressed air warms up, and part of this heat later passes to the body of the pump.
intermediate
When we press the valve of a spray deodorant, the jet comes out cold. Explain this cooling with the first law.

Show solution
As it leaves the can, the gas expands very quickly and hardly exchanges any heat with the surroundings at that instant, which comes close to an adiabatic expansion.
With \(Q \approx 0\), the work of expansion comes out of the internal energy, \(\Delta U = -\tau < 0\), and the temperature falls. The evaporation of the liquid that comes out with it also helps to cool the jet.
The gas expands with almost no heat received, pays for the work with its own internal energy and cools down.
intermediate
Two moles of a monatomic ideal gas are heated at constant volume from 300 K to 350 K. Calculate the work, the change in internal energy and the heat received. Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
In an isochoric process, \(\tau = 0\), and \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\) \(= 1.5 \cdot 2 \cdot 8.3 \cdot 50\).
By the first law, \(Q = \Delta U\).
\(\tau = 0\), \(\Delta U = 1245\ \text{J}\) and \(Q = 1245\ \text{J}\)
intermediate
A monatomic ideal gas expands at a constant pressure of 100 kPa, from 2 L to 5 L. Calculate the work, the change in internal energy and the heat received. Remember that \(1\ \text{kPa} \cdot 1\ \text{L} = 1\ \text{J}\).

Show solution
The work is \(\tau = p\,\Delta V = 100 \cdot 3\). Since \(U = \tfrac{3}{2}\,p\,V\), at constant pressure \(\Delta U = 1.5 \cdot 100 \cdot 3\).
By the first law, \(Q = \tau + \Delta U\).
\(\tau = 300\ \text{J}\), \(\Delta U = 450\ \text{J}\) and \(Q = 750\ \text{J}\)
intermediate
On the \(p \times V\) graph, an isotherm and an adiabatic curve start from the same state, and the gas expands to the same final volume. Which curve ends at the lower pressure? Which expansion produces more work?

Show solution
On the isotherm, the temperature stays constant, and the pressure falls only because the volume increases.
On the adiabatic curve, the gas pays for the work with its internal energy and cools down, and the pressure falls for both reasons, the larger volume and the lower temperature.
The adiabatic curve ends at the lower pressure, and since it lies below the isotherm, the area under it is smaller, and the isothermal expansion produces more work.
challenge
One mole of a monatomic ideal gas expands adiabatically, and its temperature falls from 400 K to 300 K. How much work does the gas do? Use \(R = 8.3\ \text{J/(mol·K)}\).

Show solution
The internal energy changes by \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\) \(= 1.5 \cdot 1 \cdot 8.3 \cdot (-100)\) \(= -1245\ \text{J}\).
In an adiabatic process, \(Q = 0\) and \(\tau = -\Delta U\).
\(\tau = 1245\ \text{J}\), taken from the internal energy of the gas.
challenge
A gas is compressed to half its volume, always starting from the same state, once isothermally and once adiabatically. In which of the compressions do we need to do more work on the gas? Use the \(p \times V\) graph to justify your answer.

Show solution
In the adiabatic compression, the work we do becomes internal energy, the temperature rises and the pressure grows more than in the isothermal one. For a monatomic gas it becomes about 3.2 times larger, against 2 times in the isothermal case.
On the graph, the adiabatic curve runs above the isotherm during the compression, and the area under it, which measures the work received, is larger.
The adiabatic compression requires more work, because the gas warms up and resists more.

STEP 5

How does an engine turn heat into work?

A heat engine repeats a cycle. The gas receives heat \(Q_h\) from a hot reservoir, does work, gives out heat \(Q_c\) to a cold reservoir and returns to its starting state, ready to begin again.

Since the final state is exactly the same as the initial one, \(\Delta U = 0\) over the cycle, and the first law says that the net work is the difference between the heat received and the heat given out. On the \(p \times V\) graph, this work is the area enclosed by the cycle, positive when the path runs clockwise, the direction of an engine.

\(\Delta U_{\text{cycle}} = 0\)\(\tau = Q_h - Q_c\)\(\eta = \dfrac{\tau}{Q_h} = 1 - \dfrac{Q_c}{Q_h}\)\(Q_h\) and \(Q_c\) are positive values, the heat received and the heat given out over the cycle. In the simulation, the gas is the same as in the previous steps, and the heat on each side of the rectangle comes from the first law, with \(\tau = 0\) on the vertical sides. The red sides receive heat, and the blue ones give it out.

We measure the performance of an engine by its efficiency \(\eta = \tau/Q_h\), which tells us what fraction of the heat received became work. In a car engine, which roughly follows the Otto cycle, the mixture of air and fuel is compressed, explodes, pushes the piston in the expansion and leaves hot through the exhaust, carrying \(Q_c\) away, and the efficiency typically lies between 25% and 30%.

In a thermal power station, burning coal, gas or sugar-cane bagasse boils water, the steam turns a turbine and is then condensed with water from a river or from cooling towers. This water receives \(Q_c\), and it generally carries away more energy than leaves through the wires.

Let's discuss

Drag a corner of the cycle and follow the area. Which change increases the work per cycle the most?
In the initial cycle, check that \(Q_h - Q_c\) equals the area of the rectangle.
Try to push the efficiency above 30%. What did you have to do with the pressures and the volumes?
Reverse the direction of the cycle. What happens to the sign of the work and to the roles of the heat received and the heat given out?

Work (area)

Heat received Q_h

Heat given out Q_c

Efficiency

ΔU over the cycle

Temperatures

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Over a cycle, a gas receives 1200 J of heat in total and gives out 800 J. What is the change in internal energy over the cycle, and how much work does it do?

Show solution
The gas returns to its initial state, and over the cycle \(\Delta U = 0\).
By the first law, \(\tau = Q_{\text{net}} = 1200 - 800\).
\(\Delta U = 0\) and \(\tau = 400\ \text{J}\)
basic
An engine receives 1000 J of heat per cycle and rejects 700 J. What is the work per cycle and what is the efficiency?

Show solution
The work is the difference between the heats, \(\tau = 1000 - 700\), and the efficiency is \(\eta = \dfrac{\tau}{Q_h}\).
\(\tau = 300\ \text{J}\) and \(\eta = 30\%\)
basic
On the \(p \times V\) graph, a cycle is a rectangle with volumes between 2 L and 6 L and pressures between 100 kPa and 300 kPa, traversed clockwise. What is the work per cycle?

Show solution
The work of the cycle is the area enclosed by the rectangle, \(\tau = (300 - 100) \cdot (6 - 2)\), in kPa times litres, which gives joules. Clockwise, it is positive.
\(\tau = 800\ \text{J}\)
basic
Does a cycle traversed clockwise on the \(p \times V\) graph correspond to an engine or to a refrigerator? And anticlockwise?

Show solution
Clockwise, the expansion happens at higher pressures than the compression, and the gas does more work than it receives, \(\tau > 0\).
Anticlockwise, the opposite occurs, and the gas receives net work.
Clockwise is an engine; anticlockwise is a refrigerator, which needs to receive work.
intermediate
An engine has an efficiency of 25% and does 500 J of work per cycle. How much heat does it receive and how much does it reject per cycle?

Show solution
From the efficiency, \(Q_h = \dfrac{\tau}{\eta} = \dfrac{500}{0.25}\), and the heat rejected is \(Q_c = Q_h - \tau\).
It receives 2000 J and rejects 1500 J per cycle.
intermediate
A car engine, with an efficiency of 25%, delivers 30 kW to the wheels and to the parts it turns. How much heat must the burning of the fuel release per second, and how much goes out through the exhaust and the radiator?

Show solution
Per second, \(Q_h = \dfrac{P_{\text{useful}}}{\eta} = \dfrac{30}{0.25} = 120\ \text{kW}\), and the remainder is \(120 - 30\).
The burning releases 120 kW, and 90 kW are rejected as heat.
intermediate
A thermal power station has an efficiency of 40% and supplies 500 MW to the grid. How much heat does the burning release per second, and how much is handed over to the cooling water and the air?

Show solution
The thermal power is \(\dfrac{500}{0.40} = 1250\ \text{MW}\), and the rejected power is \(1250 - 500\).
The burning releases 1250 MW, and 750 MW go to the surroundings, more than the power station delivers to the grid.
intermediate
The petrol engine roughly follows the Otto cycle, in four strokes: intake, compression, combustion with expansion, and exhaust. In which stroke does the gas do the work that moves the car, and in which does it receive work? Where does \(Q_c\) go?

Show solution
In the compression stroke, the piston rises and compresses the mixture of air and fuel, which receives work and warms up, almost adiabatically.
Just after the spark, the burning raises the temperature and pressure a great deal, and in the expansion the hot gases push the piston and do more work than was received.
The gas does work in the expansion and receives work in the compression; \(Q_c\) leaves with the hot exhaust gases and through the cooling system.
challenge
A monatomic ideal gas goes clockwise round the rectangular cycle A (2 L; 100 kPa), B (2 L; 300 kPa), C (6 L; 300 kPa), D (6 L; 100 kPa). Calculate the heat exchanged on each stretch, the heat received \(Q_h\), the work and the efficiency.

Show solution
On the constant-volume stretches, \(Q = \Delta U = \tfrac{3}{2}\,V\,\Delta p\), and on the constant-pressure stretches, \(Q = \tfrac{5}{2}\,p\,\Delta V\).
On AB, \(Q = 1.5 \cdot 2 \cdot 200 = 600\ \text{J}\); on BC, \(Q = 2.5 \cdot 300 \cdot 4 = 3000\ \text{J}\); on CD, \(Q = -1800\ \text{J}\); and on DA, \(Q = -1000\ \text{J}\).
The positive heats add up to \(Q_h = 600 + 3000\), the negative ones to \(Q_c = 2800\ \text{J}\), and the work is the difference, equal to the area of the rectangle, \(200 \cdot 4\). The efficiency is \(\eta = \tau/Q_h\).
\(Q_h = 3600\ \text{J}\), \(\tau = 800\ \text{J}\) and \(\eta \approx 22\%\)
challenge
A car drives for 1 hour with the engine delivering, on average, 15 kW. The engine's efficiency is 25%, and each litre of petrol releases about \(3.2 \cdot 10^7\ \text{J}\) when it burns. How many litres of petrol does the car use in this hour?

Show solution
The useful energy is \(15\,000 \cdot 3600 = 5.4 \cdot 10^7\ \text{J}\), and the heat the burning must release is \(\dfrac{5.4 \cdot 10^7}{0.25} = 2.16 \cdot 10^8\ \text{J}\).
Dividing by the energy of one litre, \(\dfrac{2.16 \cdot 10^8}{3.2 \cdot 10^7}\).
About 6.8 L of petrol.

STEP 6

Why does no engine make use of all the heat?

We can state the second law of thermodynamics in equivalent ways. One of them says that no machine operating in cycles turns into work all the heat it receives, and part of it must go to a colder reservoir. Another says that heat does not pass spontaneously from a cold body to a hot one.

In 1824, Sadi Carnot showed that no machine operating between two temperatures beats an ideal machine, with no friction and no abrupt heat exchanges, that runs through what we now call the Carnot cycle. Its efficiency depends only on the temperatures of the reservoirs, in kelvin, \(\eta_C = 1 - T_c/T_h\).

\(\eta_C = 1 - \dfrac{T_c}{T_h}\)\(\eta \le \eta_C\)The temperatures go in in kelvin, \(T = \theta + 273\). In the simulation, the temperatures assigned to each machine and the real efficiencies are typical, approximate values, which vary a lot from one model to another. In engines, \(T_h\) is the temperature that the combustion gases reach for an instant.

Real engines and power stations fall below this limit, because of friction, heat losses and processes that are too fast. The formula also suggests the way to improve a machine, which is to move the temperatures of the hot and cold reservoirs further apart.

Behind the second law lies an idea of irreversibility. When we open the partition of a box with gas on one side only, the particles spread out and do not gather together again on their own, although no law of motion forbids it. Entropy measures the number of possible arrangements of the particles, which we usually associate with disorder, and in an isolated system it tends to increase.

Let's discuss

Choose the coal-fired power station and compare the real efficiency with the Carnot efficiency at the same temperatures. Do the same with the other machines.
With the cold reservoir at 300 K, what hot-reservoir temperature gives a Carnot efficiency of 50%? And of 75%?
Bring the two temperatures closer together. What happens to the work that comes out of each 1000 J of heat?
Open the partition and wait a good while. How many times did you see all the particles back on the left-hand side?

Hot reservoir T_h

Cold reservoir T_c

Carnot efficiency

Typical real efficiency

Maximum work per 1000 J

Minimum heat to the cold reservoir

Particles on the left

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the efficiency of a Carnot engine operating between 600 K and 300 K?

Show solution
The Carnot efficiency is \(\eta_C = 1 - \dfrac{T_c}{T_h} = 1 - \dfrac{300}{600}\).
\(\eta_C = 50\%\)
basic
A Carnot engine operates between 327 °C and 27 °C. A student calculates the efficiency as \(1 - 27/327\) and gets about 92%. Where is the mistake, and what is the right value?

Show solution
The Carnot formula calls for absolute temperatures, in kelvin. Adding 273, the reservoirs are at 600 K and 300 K, and \(\eta_C = 1 - \dfrac{300}{600}\).
The mistake was using degrees Celsius; the correct efficiency is 50%.
basic
An inventor claims to have built an engine that, operating in cycles, turns into work all the heat it receives from a single reservoir. What does the second law say about this?

Show solution
The second law states that no machine operating in cycles turns the heat received from a reservoir entirely into work. Part of the heat must be given out to a colder reservoir.
The engine described is impossible, because it would violate the second law.
basic
We open a bottle of perfume in a corner of the room, and after a while the smell spreads around. Why do we not see the perfume go back into the bottle on its own?

Show solution
The molecules move at random, and there are many more ways for them to be spread around the room than gathered in the bottle.
Going back into the bottle does not violate the conservation of energy, but it is so improbable that, in practice, it is never observed.
The spreading is irreversible; the entropy increases, and the reverse process does not happen spontaneously.
intermediate
An inventor announces an engine that works between 500 K and 300 K with an efficiency of 60%. Would you invest in it?

Show solution
The maximum possible between these temperatures is the Carnot efficiency, \(\eta_C = 1 - \dfrac{300}{500} = 40\%\).
No real machine can beat this value.
No, because 60% goes beyond the Carnot limit of 40%.
intermediate
A Carnot engine has an efficiency of 30% and its cold reservoir is at 280 K. What is the temperature of the hot reservoir?

Show solution
Making \(T_h\) the subject of \(\eta_C = 1 - T_c/T_h\), we get \(T_h = \dfrac{T_c}{1 - \eta_C} = \dfrac{280}{0.7}\).
\(T_h = 400\ \text{K}\)
intermediate
A Carnot engine operates between 800 K and 300 K and receives 2000 J of heat per cycle. How much work does it do and how much heat does it reject?

Show solution
The efficiency is \(\eta_C = 1 - \dfrac{300}{800} = 0.625\), and the work is \(\tau = 0.625 \cdot 2000\). The rest goes to the cold reservoir, \(Q_c = 2000 - \tau\).
\(\tau = 1250\ \text{J}\) and \(Q_c = 750\ \text{J}\)
intermediate
In a coal-fired power station, the steam reaches the turbine at about 830 K, and the cooling water is at 310 K. The power station has a real efficiency of 35%. Compare it with the Carnot efficiency and say what fraction of the maximum possible it achieves.

Show solution
The Carnot limit is \(\eta_C = 1 - \dfrac{310}{830} \approx 0.63\), and the ratio between the real and the maximum is \(\dfrac{0.35}{0.63}\).
The maximum would be about 63%, and the power station reaches some 56% of this limit.
challenge
A Carnot engine works between 500 K and 300 K. To raise the efficiency to 50%, we can heat the hot reservoir or cool the cold reservoir. Calculate the new temperature in each case and discuss which seems more feasible in practice.

Show solution
At present, \(\eta_C = 1 - \dfrac{300}{500} = 40\%\). Keeping \(T_c = 300\ \text{K}\), we need \(T_h = \dfrac{300}{0.5}\). Keeping \(T_h = 500\ \text{K}\), we need \(T_c = 0.5 \cdot 500\).
The cold reservoir is usually the surroundings, a river or the air, and cooling it below that would require a refrigerator, which uses up work.
\(T_h = 600\ \text{K}\) or \(T_c = 250\ \text{K}\); in practice, power stations tend to raise the temperature of the hot reservoir, limited by the strength of the materials.
challenge
A box with a partition holds \(N\) particles that move at random. Once the partition is opened, each particle has a 1/2 chance of being on the left-hand side at any given instant. What is the chance of finding all of them on the left-hand side, for \(N = 4\), \(N = 10\) and \(N = 40\)? What does this suggest for a real gas, with about \(10^{22}\) particles?

Show solution
The particles move independently, and the chances multiply, \(\left(\tfrac{1}{2}\right)^N\). We just raise \(\tfrac{1}{2}\) to the powers 4, 10 and 40, and in the last case \(2^{40} \approx 1.1 \cdot 10^{12}\).
For \(10^{22}\) particles, the number is so small that the event would never be seen in the age of the universe.
1/16, 1/1024 and about 1 in \(10^{12}\); in a real gas, the spontaneous return is so improbable that we treat the spreading as irreversible.

STEP 7

The fridge is an engine run backwards

If we go round a cycle anticlockwise, the gas receives work and the heat flows the other way, from a cold reservoir to a hot one. This is what the fridge, the air conditioner and the heat pump do, and the second law explains why this uses up work, since heat would not pass from cold to hot on its own.

In the fridge, a refrigerant fluid circulates in pipes, evaporates inside the appliance taking the heat \(Q_c\) from the food, is compressed by the compressor, which receives the work \(\tau\) from the mains, and hands the heat \(Q_h\) to the kitchen through the hot grille at the back of the appliance. We measure its performance by the coefficient of performance \(e = Q_c/\tau\), which is generally greater than 1.

By the conservation of energy, \(Q_h = Q_c + \tau\), and the kitchen always receives more heat than is taken from the food. That is why leaving the door open does not cool the kitchen. The cold air that comes out is warmed again at the hot grille, and on balance the kitchen gains the energy that the compressor uses.

Air conditioning does the same with the room and dumps the heat outside. The heat pump swaps the roles in winter, takes heat from the cold air outside and delivers it indoors, and with 1 J from the mains we can put several joules of heat into the house.

\(e = \dfrac{Q_c}{\tau}\)\(Q_h = Q_c + \tau\)\(e_C = \dfrac{T_c}{T_h - T_c}\)\(e_C\) is the greatest possible coefficient of performance between two temperatures, in kelvin. In the simulation, the fridge works between an inside at 5 °C and a kitchen at 25 °C, and the powers of each appliance are typical, approximate values. The arrows have a width proportional to the energy per second.

Let's discuss

With the fridge selected and the coefficient of performance at 3, check on the screen that \(Q_h = Q_c + \tau\).
Open the fridge door and read the kitchen's energy balance. Does it cool down?
Reduce the coefficient of performance to 1.5. How much does the compressor now use to take out the same heat?
Switch to the heat pump and compare the heat delivered to the house with the electrical energy used.

Coefficient of performance e

Heat taken out per second

Heat taken out Q_c

Work τ

Heat delivered Q_h

Coefficient of performance e

Carnot limit

Electrical energy in 1 h

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A fridge takes 300 J out of its inside using 100 J of work from the compressor. What is its coefficient of performance, and how much heat does it deliver to the kitchen?

Show solution
The coefficient of performance is \(e = \dfrac{Q_c}{\tau} = \dfrac{300}{100}\), and by the conservation of energy, \(Q_h = Q_c + \tau\).
\(e = 3\) and \(Q_h = 400\ \text{J}\)
basic
Over a certain interval, a fridge takes 600 J from the food and delivers 800 J to the kitchen. How much work did the compressor do, and what is the coefficient of performance?

Show solution
From the conservation of energy, \(\tau = Q_h - Q_c = 800 - 600\), and \(e = \dfrac{Q_c}{\tau}\).
\(\tau = 200\ \text{J}\) and \(e = 3\)
basic
On a hot day, someone leaves the fridge door open to cool down the closed kitchen. Does the kitchen cool down?

Show solution
With the door open, the inside and the kitchen form a single space. The heat taken from one side comes back through the hot grille on the other, together with the work of the compressor.
On balance, the kitchen receives the electrical energy that the compressor uses.
The kitchen does not cool down; it warms up a little.
basic
Behind or underneath the fridge there is a grille of pipes that gets hot when the appliance is running. What is it for?

Show solution
The refrigerant fluid, compressed by the compressor, arrives hot at this grille, the condenser, and hands heat over to the air in the kitchen.
This is where \(Q_h = Q_c + \tau\) leaves, the heat taken from the food added to the energy from the compressor.
The grille delivers to the kitchen the heat that the fridge takes from inside plus the work of the compressor.
intermediate
An air conditioner with a coefficient of performance of 3 draws 1000 W from the mains. How much heat does it take out of the room per second, and how much does it dump outside?

Show solution
Per second, \(Q_c = e\,\tau = 3 \cdot 1000 = 3000\ \text{J}\), and \(Q_h = Q_c + \tau\).
It takes 3000 W out of the room and dumps 4000 W outside.
intermediate
A fridge with a coefficient of performance of 2.5 takes 150 W out of its inside while the compressor is on. What power does the compressor use? If it runs 8 h a day, how much electrical energy does the fridge use per day, in kWh?

Show solution
The power of the compressor is \(\tau = \dfrac{Q_c}{e} = \dfrac{150}{2.5}\).
Over 8 h a day, \(E = \tau \cdot 8\ \text{h}\), in watt-hours, and we divide by 1000 to get kWh.
The compressor uses 60 W, and the fridge uses 0.48 kWh per day.
intermediate
A heat pump has a cooling coefficient of performance \(e = 3\) and draws 1 kW. How much heat does it deliver indoors? Compare it with a resistance heater that draws the same 1 kW.

Show solution
The pump takes \(Q_c = e\,\tau = 3\ \text{kW}\) from the air outside and delivers indoors \(Q_h = Q_c + \tau\).
In the resistance heater, all the electrical energy becomes heat, and it delivers 1 kW.
The pump delivers 4 kW, four times the heat of the heater, for the same consumption.
intermediate
The freezer of a fridge is at −5 °C, and the kitchen at 25 °C. What is the greatest possible coefficient of performance of a refrigerating machine between these temperatures?

Show solution
In kelvin, \(T_c = 268\ \text{K}\) and \(T_h = 298\ \text{K}\), and the Carnot limit is \(e_C = \dfrac{T_c}{T_h - T_c} = \dfrac{268}{30}\).
\(e_C \approx 8.9\); real fridges fall well below this.
challenge
In a closed, insulated kitchen, a fridge stands with its door open for 2 hours, and the compressor draws 150 W the whole time. How much energy does the kitchen gain over this period, in joules and in calories? Use \(1\ \text{cal} \approx 4.2\ \text{J}\).

Show solution
With the door open, the heat that the fridge takes from one side goes back to the other, and the kitchen's energy balance is the work of the compressor, \(E = 150 \cdot 7200\) joules.
In calories, we divide by 4.2.
The kitchen gains \(1.08 \cdot 10^6\ \text{J}\), about \(2.6 \cdot 10^5\ \text{cal}\), and warms up.
challenge
A freezer with a coefficient of performance of 2 freezes 1 kg of water that is already at 0 °C. The latent heat of fusion of water is 80 cal/g. How much work does the compressor do, and how much heat goes to the kitchen? Use \(1\ \text{cal} \approx 4.2\ \text{J}\).

Show solution
The heat that has to be taken out is \(Q_c = m\,L\) \(= 1000 \cdot 80\) \(= 80\,000\ \text{cal}\), or \(336\,000\ \text{J}\).
The work is \(\tau = \dfrac{Q_c}{e} = \dfrac{336\,000}{2}\), and the kitchen receives \(Q_h = Q_c + \tau\).
\(\tau = 1.68 \cdot 10^5\ \text{J}\) and \(Q_h = 5.04 \cdot 10^5\ \text{J}\)

WRAP-UP

Challenges

Work in a cycle

On the \(p \times V\) graph, a gas goes clockwise round the triangle with corners (1 L; 100 kPa), (4 L; 100 kPa) and (1 L; 400 kPa). How much work does it do per cycle?

Show solution

The work of the cycle is the area of the triangle, with a base of 3 L and a height of 300 kPa, \(\dfrac{3 \cdot 300}{2}\) in kPa times litres.
Since the direction is clockwise, the work is positive and comes to 450 J.

Internal energy

A balloon with 0.5 mol of helium rises and cools from 300 K to 270 K. By how much does the internal energy of the gas change? Treat helium as a monatomic ideal gas, with \(R = 8.3\ \text{J/(mol·K)}\).

Show solution

The internal energy follows the temperature, \(\Delta U = \tfrac{3}{2}\,n\,R\,\Delta T\) \(= 1.5 \cdot 0.5 \cdot 8.3 \cdot (-30)\).
This gives about \(-187\ \text{J}\), a loss of internal energy.

First law

A monatomic ideal gas receives 1000 J of heat at constant pressure. How much of this heat becomes work, and how much stays as internal energy?

Show solution

In an isobaric process, \(\tau = n\,R\,\Delta T\) and \(Q = \tfrac{5}{2}\,n\,R\,\Delta T\), and the work is \(\tfrac{2}{5}\) of the heat.
That makes 400 J of work, and the remaining 600 J stay as internal energy.

Adiabatic

When pumping up a tyre quickly, we do 60 J of work on the air in one stroke of the pump, with no time for heat exchange. By how much does the internal energy of the air change?

Show solution

With no heat exchange, \(Q = 0\), and the air receives work, \(\tau = -60\ \text{J}\).
By the first law, \(\Delta U = -\tau\), and the internal energy increases by 60 J, which warms the air.

Efficiency

An engine receives 5 kJ of heat per cycle, has an efficiency of 30% and completes 20 cycles per second. What is its useful power, and how much heat does it reject per second?

Show solution

Per cycle, \(\tau = 0.30 \cdot 5000 = 1500\ \text{J}\), and at 20 cycles per second the power is \(1500 \cdot 20\) watts, or 30 kW.
The heat rejected per cycle is \(5000 - 1500\), and per second it comes to 70 kW.

Carnot

A power station receives steam at 300 °C and rejects heat at 30 °C, with a real efficiency of 33%. What fraction of the Carnot efficiency does it achieve?

Show solution

In kelvin, \(T_h = 573\ \text{K}\) and \(T_c = 303\ \text{K}\), and \(\eta_C = 1 - \dfrac{303}{573} \approx 0.47\).
The ratio is \(\dfrac{0.33}{0.47}\), and the power station achieves about 70% of the limit.

Air conditioning

An air conditioner with a coefficient of performance of 3 takes 2.1 kW out of a room. What electrical power does it use, and how much heat does it dump outside per second?

Show solution

The electrical power is \(\tau = \dfrac{Q_c}{e} = \dfrac{2.1}{3}\), or 0.7 kW.
Outside, the outdoor unit delivers \(Q_h = Q_c + \tau\), which comes to 2.8 kW.

Think, no calculation

Could a ship move by taking heat from the ocean water and turning this heat into work, without burning fuel?

Show solution

The first law does not forbid it, because the energy of the ocean is enormous. The second law, however, requires a colder reservoir to send part of the heat to, and the ship has nothing nearby much colder than the ocean itself.
Without this temperature difference, the maximum Carnot efficiency is practically zero, and the ship would not move.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with a base text, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on thermodynamics, which may well be the best practice once these are done.

Work done by a gas · Step 1
In a school laboratory, Otávio followed the gas in a cylinder with a piston and recorded the graph below. The gas goes from A to B at constant pressure, while it is heated, and from B to C with the piston locked, while it cools. Let us remember that \(1\ \text{kPa} \cdot 1\ \text{L} = 1\ \text{J}\).

How much work did the gas do between states A and C?
1. 200 J
2. 400 J
3. 600 J
4. 900 J
5. \(6 \cdot 10^5\ \text{J}\)
Show solution

Answer: C.
From A to B, the pressure stays at 300 kPa and the volume grows by 2 L, and the work is the area of the rectangle under the stretch, \(\tau_{AB} = 300 \cdot 2\), in joules.
From B to C the volume does not change, and the vertical stretch has no area, \(\tau_{BC} = 0\). The total is 600 J.
Option A uses the final pressure, of 100 kPa, and B uses the average of the pressures. Option D multiplies the pressure at A by the volume at B, which is not work, and E multiplies pascals by litres without converting the litres to cubic metres.
Adiabatic process · Step 4
When using a spray deodorant, Júlia notices that the jet comes out very cold, although the can is at room temperature. Let us treat the gas that leaves the can as an ideal gas and set aside, in this question, the evaporation of the liquid that comes out with it.

The cooling of the jet can be explained because the gas
1. receives heat from the surroundings as it leaves, which reduces its internal energy.
2. is compressed as it passes through the valve, and compression cools a gas.
3. loses pressure, and the temperature of an ideal gas depends only on the pressure.
4. expands so quickly that it hardly exchanges any heat, and the work of the expansion comes out of its internal energy.
5. expands at constant temperature and turns into work the heat it receives.
Show solution

Answer: D.
The expansion is very fast, and the gas has almost no time to exchange heat, as in an adiabatic process. With \(Q \approx 0\), the first law gives \(\Delta U = -\tau\), and since the gas does work as it expands, \(\tau > 0\), the internal energy and the temperature decrease.
Option A reverses the effect of the heat received, which would increase the internal energy, and B swaps expansion for compression, which would warm the gas. Option C forgets that the temperature depends on the product \(p\,V\), and E describes an isothermal process, in which the temperature would not change.

Efficiency · Step 5

A news report compares four thermal power stations. For each one, the table shows the thermal power released by burning the fuel and the electrical power that the station delivers to the grid.

Powers of the power stations
Station	Burning (MW)	Electrical (MW)
I	1000	350
II	800	320
III	1200	360
IV	600	180

Which power station has the highest efficiency, and how much heat does it reject to the surroundings each second?

Station I, which rejects 650 MW.
Station II, which rejects 480 MW.
Station II, which rejects 320 MW.
Station III, which rejects 840 MW.
Station IV, which rejects 420 MW.

Show solution

Answer: B.
The efficiency is the ratio of the useful power to the thermal power, \(\eta = \dfrac{\tau}{Q_h}\). Stations I, II, III and IV have 35%, 40%, 30% and 30%, and the highest is II.
The heat rejected per second is the difference, \(Q_c = Q_h - \tau = 800 - 320\), or 480 MW.
Option D picks the station that delivers the most electrical power, which is not the one with the highest efficiency, and C confuses the heat rejected with the electrical power. Options A and E do the right calculation for stations of lower efficiency.

Second law · Step 6
A company announces an experimental engine that makes use of the hot gases from a factory chimney. The engine would take heat from these gases, at 400 K, reject heat to the air, at 300 K, and have an efficiency of 30%. An engineer is suspicious of the announcement and decides to check it against the second law.

Based on the second law of thermodynamics, the announcement
1. is impossible, because the maximum efficiency between these temperatures is 25%.
2. is plausible, because 30% is less than 100%.
3. is plausible, because the maximum efficiency between these temperatures is 75%.
4. is impossible, because no heat engine can reject heat to the air.
5. is plausible, provided the engine runs without friction.
Show solution

Answer: A.
The highest possible efficiency between two reservoirs is the Carnot efficiency, \(\eta_C = 1 - \dfrac{T_c}{T_h} = 1 - \dfrac{300}{400}\), or 25%, and the 30% announced goes beyond this limit.
Option C uses the ratio \(T_c/T_h\) in place of \(1 - T_c/T_h\), and B forgets that the limit is not 100%. Option D contradicts the second law itself, which requires heat to be rejected to a cold reservoir, and E ignores that not even a frictionless machine beats the Carnot efficiency.
Air conditioning · Step 7
To cool his bedroom, Fernando installed an air conditioner that takes 2400 J of heat out of the room each second. The manual gives a cooling coefficient of performance of 3.0, which is the ratio of the heat taken out of the room to the electrical energy used.

What electrical power does the appliance use, and how much heat does the outdoor unit deliver outside each second?
1. 7200 W and 9600 J
2. 800 W and 1600 J
3. 2400 W and 3200 J
4. 800 W and 2400 J
5. 800 W and 3200 J
Show solution

Answer: E.
From the definition of the coefficient of performance, \(\tau = \dfrac{Q_c}{e} = \dfrac{2400}{3.0}\), or 800 J per second, a power of 800 W.
By the conservation of energy, the outdoor unit delivers \(Q_h = Q_c + \tau = 2400 + 800\), or 3200 J per second.
Option A multiplies by the coefficient of performance instead of dividing, and B subtracts the work from the heat taken out. Option C confuses the electrical power with the heat taken out, and D forgets that the heat delivered outside includes the energy from the compressor.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2009, Day 1, blue booklet, question 20. The diagram of a fossil-fuel thermal power station shows the boiler, the turbine and the condenser, and the question asks which action would improve the efficiency of the station without reducing its output.
ENEM 2009, Day 1, blue booklet, question 39. Starting from the cycle of compression and expansion of the gas in a fridge, we choose the correct statement about the heat exchanges between the inside and the outside of the appliance.
ENEM 2012, Day 1, blue booklet, question 83. A text on the efficiency of combustion engines asks for the factor that limits this pursuit, and the answer goes through the second law of thermodynamics.
ENEM 2015, Day 1, blue booklet, question 63. A person closes the fridge and has trouble opening it again straight away, and the question asks for the explanation, which involves the air cooling at constant volume inside the appliance.
ENEM 2020, Day 2, blue booklet, question 105 (printed paper). Manuals advise against installing the fridge near sources of heat, and we need to explain why this increases energy consumption.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Work at constant pressure\(\tau = p\,\Delta V\)

Work in generalarea under the \(p \times V\) curve

Internal energy\(U = \tfrac{3}{2}\,n\,R\,T\)

First law\(Q = \tau + \Delta U\)

Isochoric\(\tau = 0\), \(Q = \Delta U\)

Isothermal\(\Delta U = 0\), \(Q = \tau\)

Adiabatic\(Q = 0\), \(\tau = -\Delta U\)

Cycle\(\Delta U = 0\), \(\tau = Q_h - Q_c\)

Efficiency\(\eta = \tau/Q_h = 1 - Q_c/Q_h\)

Carnot\(\eta_C = 1 - T_c/T_h\)

Refrigerator\(e = Q_c/\tau\), \(Q_h = Q_c + \tau\)

Constants\(R = 8.3\ \text{J/(mol·K)}\), \(1\ \text{cal} \approx 4.2\ \text{J}\)