B.4 Thermodynamics

First law for a closed system

Using the convention that W is work done by the system,

$$Q=\Delta U+W$$

So energy supplied as thermal transfer is split between internal-energy change and work done by the gas.

Rearrange for the unknown

$$\Delta U=Q-W$$

If heat is removed, Q is negative. If the gas expands and does work on its surroundings, W is positive. If work is done on the gas, W is negative in this convention.

Use the process information

For an adiabatic process Q=0, so $$\Delta U=-W$$. An expanding adiabatic gas does positive work and its internal energy decreases.

Sign boundary

Always state whether W means work done by the gas or work done on the gas before using a memorized first-law equation.

Work from pressure and volume

For a constant-pressure change, the work done by the gas is

$$W=P\Delta V$$

Expansion has $$\Delta V>0$$ and work done by the gas is positive in the first-law convention.

PV interpretation

On a pressure–volume diagram, work is the area under the process path. For constant pressure this is a rectangle; for changing pressure, use the area or integral specified by the model.

Units and direction

Use pressure in pascals and volume in cubic metres so $$P\Delta V$$ is in joules. Compression gives $$\Delta V<0$$, so work done by the gas is negative.

Common trap

Do not multiply pressure by the final volume alone. Work depends on the change in volume and on the process path.

Temperature controls internal energy

For an ideal monatomic gas,

$$\Delta U=\frac32Nk_B\Delta T=\frac32nR\Delta T$$

Only the temperature change and amount of gas are needed.

Use a PV route when useful

With $$PV=nRT$$, a change can sometimes be written as $$\Delta U=\frac32\Delta(PV)$$ for a fixed amount of monatomic ideal gas. Check which variable or path information the question supplies.

Sign

If temperature rises, $$\Delta U>0$$. If temperature falls, $$\Delta U<0$$. Internal energy is a state function, so its change depends only on the initial and final states.

Common trap

Do not add work or thermal transfer into $$\Delta U$$ directly. Use the first law to relate them, but calculate the state-function change from temperature or equivalent state data.

Entropy as multiplicity

Entropy measures how many microscopic arrangements, or microstates, are compatible with a system’s macroscopic state. More accessible microstates correspond to greater entropy and greater microscopic disorder.

Macroscopic reading

For a reversible transfer at temperature T, $$\Delta S=\frac{\Delta Q}{T}$$. Supplying thermal energy at constant temperature increases the number of accessible states and therefore increases entropy.

Microscopic reading

Boltzmann’s relation is $$S=k_B\ln\Omega$$, where Ω is the number of possible microstates. The logarithm means a multiplicative increase in Ω produces an additive entropy change.

Common trap

Entropy is not simply “how hot” a system is. Temperature appears in $$\Delta Q/T$$, but entropy also depends on how many microscopic arrangements are available.

Thermal entropy change

For a reversible transfer at constant absolute temperature,

$$\Delta S=\frac{\Delta Q}{T}$$

Use T in kelvin and energy in joules.

Statistical entropy

$$S=k_B\ln\Omega$$

If the number of microstates changes from Ω to Ω′, then $$\Delta S=k_B\ln(\Omega'/\Omega)$$.

Use the right temperature

For melting or vaporization, use the phase-change energy and the phase-change temperature in kelvin. For a microstate question, use logarithms rather than treating entropy as directly proportional to Ω.

Common trap

Entropy is not measured in joules alone; its SI unit is J K⁻¹. Do not use Celsius in $$\Delta Q/T$$.

Second law

The entropy of an isolated system never decreases:

$$\Delta S_{\mathrm{isolated}}\ge 0$$

Real spontaneous processes usually increase it.

Local decreases are allowed

A non-isolated subsystem can decrease in entropy, such as the contents of a refrigerator, but the surroundings must increase in entropy by at least as much. Judge the total isolated system.

Direction and engines

The second law gives the net direction of thermal transfer from hot to cold and prevents a heat engine from converting all input thermal energy into work. It leads to the Carnot efficiency limit.

Common trap

“Entropy never decreases” applies to an isolated system or the universe, not necessarily to every local subsystem.

Irreversibility

A real process is irreversible when it cannot be exactly undone so that both system and surroundings return to their original states. Real isolated processes are almost always irreversible.

Entropy signature

For a real isolated process, total entropy increases: $$\Delta S_{\mathrm{total}}>0$$. The process has a preferred direction even though energy is conserved.

Physical examples

Friction, free expansion, mixing and heat flow through a finite temperature difference are irreversible because they increase the number of accessible microstates or disperse energy.

Common trap

Irreversible does not mean energy disappears. The first law still holds; irreversibility describes entropy and the impossibility of complete reversal.

Choose the system boundary

A non-isolated subsystem can have $$\Delta S_{\mathrm{local}}<0$$. The second law is satisfied if the surroundings increase by at least as much, so that

$$\Delta S_{\mathrm{total}}=\Delta S_{\mathrm{local}}+\Delta S_{\mathrm{surr}}\ge0$$

Examples

A refrigerator reduces the entropy of its contents but releases more entropy to the room. A growing organism becomes more ordered locally while chemical processes increase entropy in the environment.

Explain, do not just label

When a local entropy decrease is proposed, identify the energy or matter transfer across the boundary and explain where the compensating entropy increase occurs.

Common trap

“Entropy can never decrease” is incomplete. The correct statement applies to an isolated system or to the total system plus surroundings.

Process map

Connect to the first law

Use $$Q=\Delta U+W$$ after identifying the fixed quantity. For an isothermal ideal-gas process, $$\Delta U=0$$, so Q=W. For an adiabatic process, Q=0, so $$\Delta U=-W$$.

Read the PV path

An isovolumetric process is vertical on a PV diagram; an isobaric process is horizontal. Isothermal and adiabatic curves both change P and V, but the adiabatic curve is steeper than the isothermal curve for an ideal gas.

Common trap

“Adiabatic” means no thermal energy transfer, not constant temperature. Temperature usually changes when an ideal gas expands adiabatically.

Adiabatic monatomic gas

For an adiabatic process of a monatomic ideal gas,

$$PV^{5/3}=\text{constant}$$

so $$P_1V_1^{5/3}=P_2V_2^{5/3}$$.

Solve a state change

Write the two-state equation first, keep volumes in the same units, and rearrange for the unknown pressure or volume. The exponent 5/3 belongs to a monatomic ideal gas in this syllabus.

Interpret the expansion

During adiabatic expansion, the gas does work without receiving thermal energy, so its internal energy and temperature fall. The pressure drops more steeply with volume than along an isothermal path.

Common trap

Do not use $$PV=\text{constant}$$ for an adiabatic change; that is the isothermal relation. Do not use 5/3 for a non-monatomic gas unless the model specifies it.

A cyclic engine

A heat engine uses a working substance that undergoes a sequence of thermodynamic processes and returns to its initial state. Over one complete cycle, $$\Delta U=0$$.

Net work from the PV loop

The net work done by the gas during a cycle is the signed area enclosed by the path on a P–V diagram. A clockwise cycle gives positive net work by the gas; a counterclockwise cycle gives net work on the gas.

Energy bookkeeping

The engine absorbs energy from a hot reservoir, converts part of it to useful work, and rejects the remainder to a colder reservoir. The first law applies to every stage and to the full cycle.

Common trap

Returning to the initial state means zero net change in internal energy, not zero work or zero thermal energy transfer during the cycle.

Efficiency

For a heat engine,

$$\eta=\frac{W_{\mathrm{useful}}}{Q_H}=1-\frac{Q_C}{Q_H}$$

where $$Q_H$$ is energy absorbed from the hot reservoir and $$Q_C$$ is energy rejected to the cold reservoir.

Use a complete cycle

Over one cycle, $$\Delta U=0$$, so net work is related to net thermal transfer. On a PV diagram, useful net work is the signed enclosed area; input energy is the heat taken from the hot reservoir.

Sanity checks

Efficiency is dimensionless and lies between 0 and 1 for a physical engine. Use the same energy units for numerator and denominator and allow method marks by showing the substituted values.

Common trap

Do not divide useful work by rejected heat. The denominator is the energy input from the hot reservoir.

Carnot limit

For an ideal reversible engine operating between hot and cold reservoirs,

$$\eta_{\mathrm C}=1-\frac{T_C}{T_H}$$

Both reservoir temperatures must be in kelvin.

What the limit means

No real engine operating between the same reservoir temperatures can exceed $$\eta_{\mathrm C}$$. Real friction, finite temperature differences and other irreversibilities make actual efficiency lower.

Read temperature changes

At fixed $$T_H$$, lowering $$T_C$$ increases the Carnot efficiency. At fixed $$T_C$$, increasing $$T_H$$ also increases the limit.

Common trap

Carnot efficiency is a maximum, not the efficiency every engine achieves. Never use Celsius in the temperature ratio.