Systems, surroundings and the bookkeeping of energy

A thermodynamic system is the part of the physical world we choose to study, focusing on its energy transfers and changes of state. The surroundings are everything outside that chosen system. Together, the system and surroundings form the thermodynamic “universe” for the problem. Choose the system first. A vague choice of system is where many sign errors in this topic start.

A closed system is a thermodynamic system that cannot exchange mass with its surroundings, although it can still exchange energy by heating or by doing work. The standard IB example is a fixed mass of gas under a piston: no gas particles enter or leave, but energy can cross the boundary.

The first law of thermodynamics states conservation of energy by linking the energy transferred into a closed system with the change in its internal energy and the work done by the system:

IB uses the Clausius sign convention. Positive $Q$ means energy is supplied to the system by heating. Positive $\Delta U$ means the system’s internal energy store increases. Positive $W$ means the system does work. So work done on a system is negative, while work done by a system is positive. Keep that convention throughout a calculation; if you mix conventions, you can get physically nonsense answers even when the algebra looks neat.

What the first law does and does not say

The first law says that work and heating are two ways to transfer energy across the system boundary. It does not say which processes can happen naturally. For example, conservation of energy alone would not rule out energy flowing spontaneously from a cold object to a hot object; the second law deals with that direction-of-change question later in this topic.

Joule’s paddle-wheel experiments mattered historically because they showed that mechanical work done on water can raise its temperature. In modern terms, doing work on a system can increase its internal energy, just as transferring energy by heating can.

Boundary work

When a gas pushes a movable piston, the system boundary shifts. The gas applies a force through a distance, so it does work. For a small change at constant pressure,

If the gas expands, $\Delta V$ is positive, so the gas does positive work on the surroundings. If the gas is compressed, $\Delta V$ is negative and the work done by the gas is negative; equivalently, work is being done on the gas.

The same result comes from looking at a piston of cross-sectional area $A$, where $A$ is the piston area ($m^2$), moving a distance $\Delta x$, where $\Delta x$ is the piston displacement (m). The force from the gas is $PA$ and the volume change is $\Delta V = A\Delta x$, so the work is force \times distance = $PA\Delta x = P\Delta V$.

$P$–$V$ diagrams and area

A $P$–$V$ diagram is a graph of pressure against volume used to represent a gas process. The work done by the gas equals the area under the process curve on a pressure–volume graph. That still holds when the pressure changes; in quantitative IB problems, you may have to estimate the area by counting squares or splitting the curve into thin vertical strips.

For a curved path, you can approximate the area by adding rectangular strip areas:

Thinner strips give a better estimate, but use practical judgement: in an exam, a sensible area estimate is usually what is being tested.

Watch out for false origins. If a P–V graph does not start from zero pressure and zero volume, the visible rectangle is not necessarily the whole area under the graph. Work is the full area down to the volume axis, not just the neat-looking part inside the plotted grid.

Internal energy of a monatomic ideal gas

Internal energy is the total microscopic kinetic and potential energy associated with the particles of a system. In the ideal gas model, there is no intermolecular potential energy, so internal energy depends only on temperature. For a monatomic ideal gas,

The equation is useful because it doesn’t depend on the path the gas takes between the initial temperature and the final temperature. Stir it, compress it, heat it slowly, heat it quickly: for a monatomic ideal gas, $\Delta T$ alone fixes the change in internal energy.

Combining with the first law

The first law shows how that temperature change happened. If energy is supplied to a gas while the gas expands, part of the energy increases its internal energy and part is transferred as work done on the surroundings. If a gas is compressed adiabatically, $Q = 0$ and the negative value of $W$ makes $\Delta U$ positive: work done on the gas raises its temperature.

Quantitative problems in this topic are restricted to monatomic ideal gases, including situations where pressure is not constant. When pressure varies, don’t force $W = P\Delta V$ with a single pressure value unless the process really is at constant pressure. Use the area under the $P$–$V$ curve for work, and use the temperature change equation for $\Delta U$.

Entropy and disorder

Entropy is a thermodynamic state quantity. It measures how many microscopic arrangements can produce the observed macroscopic state of a system. Its symbol is S, where S is entropy ($\mathrm{J\,K^{-1}}$).

The phrase “degree of disorder” can help, but only if used carefully. A gas spread through a whole container is more disordered than the same gas confined to one corner because the particles have far more possible positions. A crystal with a missing atom has more disorder than a perfect crystal because the vacancy could occupy many possible places.

Entropy is not a substance. It also isn’t “mess” in the everyday sense. It is a state property linked to probability: macroscopic states with many microscopic arrangements are overwhelmingly more likely than macroscopic states with very few microscopic arrangements. For that reason, thermodynamics can predict the future evolution of large systems statistically, even though individual particle motion is random.

Macroscopic definition

For a reversible energy transfer at constant temperature, calculate the entropy change with

Watch the sign. When energy is transferred into the system, $\Delta Q$ is positive, so the system’s entropy increases. When energy is transferred out of the system, $\Delta Q$ is negative, so the system’s entropy decreases. Use this equation when the temperature is constant or changes negligibly; calculus versions are not required for IB Physics.

Microscopic definition

A microstate is one particular microscopic arrangement of the particles that produces the same observed macroscopic state. For example, if ten particles are divided between the left and right halves of a box, “five on the left” is a macrostate; the particular choice of which five particles are on the left is a microstate.

Entropy can also be written as

In the simple models used here, all microstates are assumed to be equally probable. That makes counting useful: a macrostate with more microstates is more likely to be observed.

A coin or counter model shows the idea neatly. If each counter can be in either of two boxes, the total number of arrangements doubles for every extra counter. The evenly split macrostate has many more microstates than the “all counters in one box” macrostate, so it is much more likely. For a mole of gas particles, the difference in probabilities is so extreme that the unlikely macroscopic states are never seen in practice.

Linking the two views

The macroscopic equation $\Delta S = \Delta Q / T$ and the microscopic equation $S = k_B \ln \Omega$ describe the same physical idea in two ways. When a gas expands into a larger volume, $\Omega$ increases because the particles have more possible positions. For an isothermal expansion, energy supplied to the gas is converted into work while the temperature remains fixed, so the same entropy increase can be described either by the energy transfer or by the increased number of microstates.

Three equivalent ways to say the second law

An isolated system is a thermodynamic system that cannot exchange either mass or energy with its surroundings. With an isolated system, there’s no “outside” energy transfer available to make an impossible process happen.

The second law of thermodynamics restricts the direction of real thermodynamic processes. Energy conservation is not the whole story: the law also tells us which energy transfers can happen spontaneously.

In the Clausius form: energy cannot be transferred spontaneously from a colder body to a hotter body. To do that, work has to be done on a device or system. That is why a fridge needs electrical work to cool its interior.

In the Kelvin form: no heat engine operating in a cycle can take energy from a single hot reservoir and turn all of it into work. Some energy must be rejected to a colder reservoir. That’s the deep reason no cyclic engine is 100% efficient.

In entropy form: the entropy of an isolated system cannot decrease. For a reversible process in an isolated system, the entropy change is zero. For an irreversible process in an isolated system, entropy increases.

Consequences for the universe

If we treat the universe as the largest isolated system, the second law says that its total entropy tends to increase. Temperature differences act as the “fuel” for useful processes: engines, weather systems, stars radiating to space. In the very long term, if all temperature differences were smoothed out and entropy reached a maximum, no heat engine-like process could extract useful work. This idea is often called the heat death hypothesis. It is a physical extrapolation from the second law, not an experiment we can actually run on the universe.

Why real processes do not simply run backwards

A reversible process is an idealised process that can be reversed by an infinitesimal change, with no net change left in the system and surroundings. To be reversible, a process must be quasi-static: the system is effectively in thermodynamic equilibrium at every stage.

An irreversible process is a real process that cannot be undone without causing a net change somewhere else. Friction, turbulence, unrestrained expansion, mixing and heat transfer across a finite temperature difference are all irreversible features.

Most real isolated systems behave irreversibly, because energy spreads out among microscopic degrees of freedom. A piston moving with friction warms its surroundings; a gas expanding freely into a vacuum fills the available volume; two gases mix. Energy is still conserved in each case. The problem is that running the process backwards would need a fantastically unlikely microscopic rearrangement.

For an isolated system, irreversible change means entropy increases. “Almost always” matters here: the laws of mechanics do not forbid all particles from accidentally moving into a more ordered arrangement, but with macroscopic numbers of particles the probability is so tiny that it is physically irrelevant.

Local order is allowed

A non-isolated system is a thermodynamic system that can exchange energy, and possibly mass, with its surroundings. Its own entropy can decrease, as long as the surroundings gain at least the same amount of entropy.

Freezing water gives a neat example. The water becomes more ordered, so its entropy decreases. At the same time, it releases energy to colder surroundings, which gain more entropy than the water loses. For the water plus surroundings, the total entropy change is positive.

Living organisms are a useful check on the phrase “entropy always increases”. A plant builds ordered structures, but it isn’t an isolated system. Energy from the Sun, along with energy transfers to the environment, makes the total entropy change of the larger Sun–Earth–surroundings system positive. The second law forbids a total entropy decrease for an isolated system; it does not forbid local organisation.

Four standard gas processes

A gas process is a change from one thermodynamic state of a gas to another. On a $P$–$V$ diagram, each process has its own shape, and the shape shows what happens to the work done.

An isovolumetric process is a gas process in which the volume stays constant. Since $\Delta V = 0$, no boundary work is done. The first law becomes $Q = \Delta U$. For an ideal gas, $\frac{P}{T} = \text{constant}$ during a constant-volume process.

An isobaric process is a gas process in which the pressure stays constant. Its $P$–$V$ graph is horizontal, and the work done is simply $W = P\Delta V$. For an ideal gas,

An isothermal process is a gas process in which absolute temperature stays constant. For an ideal gas, the internal energy does not change, so $\Delta U = 0$ and the first law gives $Q = W$. The ideal gas equation is $PV = nRT$, so for fixed $n$ and constant $T$, $PV = \text{constant}$. On a $P$–$V$ graph, an isothermal expansion follows a downward-curving isotherm.

An adiabatic process is a gas process in which no energy is transferred by heating across the system boundary, so $Q = 0$. In practice, you get close to this with good insulation, or with a process so rapid that there is no time for significant thermal energy transfer.

Slow, well-conducted changes give the best approximation to isothermal behaviour because the gas has to remain at the same temperature while energy crosses the boundary. Rapid, insulated changes give the best approximation to adiabatic behaviour because energy transfer by heating must be negligible.

The adiabatic model

For a monatomic ideal gas in an adiabatic process,

The exponent $5/3$ applies specifically to a monatomic ideal gas. In this topic, IB quantitative questions use monatomic ideal gases, so use this adiabatic model unless the question says otherwise.

In adiabatic expansion, the gas does positive work on the surroundings. Since $Q = 0$, the first law gives $\Delta U = -W$, so the internal energy and temperature decrease. In adiabatic compression, $W$ is negative because work is done on the gas; $\Delta U$ is positive, and the temperature rises.

On a P–V diagram, an adiabatic curve is steeper than an isothermal curve passing through the same state. For expansion, pressure drops faster in an adiabatic process because the gas cools as it expands. For compression, pressure rises faster because work done on the gas heats it.

Cycles and net work

A cyclic process is a sequence of thermodynamic changes that brings the system back to its initial state. Since internal energy is a state property, a gas that completes a full cycle has total $\Delta U = 0$ over the cycle.

On a P–V diagram, the net work done by the gas in one cycle equals the area enclosed by the loop. If the loop is clockwise, the gas does positive net work; if it is anticlockwise, net work is done on the gas.

A heat engine is a cyclic device that takes energy from a high-temperature reservoir, produces useful work, and rejects some energy to a lower-temperature reservoir. For a heat engine, $Q_h$ is the energy transferred from the hot reservoir to the engine per cycle (J), $Q_c$ is the energy rejected by the engine to the cold reservoir per cycle (J), $T_h$ is the absolute temperature of the hot reservoir (K), and $T_c$ is the absolute temperature of the cold reservoir (K).

After one complete cycle, the engine has returned to its starting internal energy. The useful work output is therefore the energy taken in minus the energy rejected:

Steam engines, internal combustion engines and gas-turbine systems all use cyclic ideas, even though real engineering cycles are messier than neat textbook curves. The central idea stays the same: return the working substance to its initial state so the device can keep operating.

Efficiency as useful output divided by input

Efficiency is a dimensionless ratio: useful energy output divided by total energy input. For a heat engine,

For a cyclic heat engine, the expression is usually written as

Engine cycles do not all have the same efficiency. They follow different paths on a P–V diagram, do different amounts of work, and reject different amounts of energy. Engineers often compare cycles using the area inside the P–V loop, but they also look at how much input energy is needed to produce that loop.

Why efficiency matters beyond gas diagrams

Efficiency matters in motors and generators because any energy transfer that is not useful ends up as heating, sound, turbulence, frictional losses or some other wasted output. An electric motor is not usually treated as a heat engine, but its useful mechanical output still has to compete with unwanted heating. In a power station, a generator is often driven by a turbine. That turbine may sit inside a thermal cycle, so thermodynamic efficiency limits how much electrical energy can be obtained from the fuel or nuclear energy source.

Steam power shows how physics can change society. As people understood enough thermodynamics to improve engines, steam engines began driving factories, pumps, locomotives and ships. This was more than a better machine; it helped move many societies from local agricultural production toward industrial production, urbanisation and large-scale transport. The same physics also brought costs: pollution, fossil-fuel dependence and new social pressures.

The reversible benchmark

A Carnot cycle is an ideal reversible heat-engine cycle made from two isothermal processes and two adiabatic processes. Engineers don’t copy it as a practical engine design. Instead, it acts as a benchmark: it gives the maximum possible efficiency for any engine working between two reservoir temperatures.

For a Carnot engine,

Use kelvin for the temperatures in this equation. A higher $T_h$ or a lower $T_c$ raises the theoretical maximum efficiency. That is why power-station engineers aim for high-temperature steam at the turbine inlet, then reject waste energy at as low a temperature as practical.

Why there is an upper limit

The Kelvin form of the second law says that a cyclic engine cannot convert all input energy into work. Put in reservoir terms, $Q_c$ cannot be zero for a real engine operating between finite temperatures. The Carnot equation says the same thing another way: $\eta_\text{Carnot}$ would reach $1$ only if $T_c$ were $0\ \text{K}$, which cannot be achieved physically as an operating cold reservoir.

Real engines fall below the Carnot limit because their processes involve friction, turbulence, finite-rate heat transfer and other irreversible effects. There is a practical trade-off as well. The reversible isothermal stages of a Carnot cycle would need to be carried out infinitely slowly, so the cycle gives an ideal efficiency limit, not a useful high-power machine.