B.4 Thermodynamics

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.

Worked example from the mapped local textbook

A gas receives $Q=+120\,\mathrm{J}$ and does $W=+80\,\mathrm{J}$ of work. With work done by the gas defined as positive,

$$\Delta U=Q-W=120-80=+40\,\mathrm{J}$$

The positive result means the gas's internal energy increases; the remaining input energy left the system as work.

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.

Worked example from the mapped local textbook

A gas expands at constant pressure $1.0\times10^5\,\mathrm{Pa}$ from $0.020\,\mathrm{m^3}$ to $0.050\,\mathrm{m^3}$.

$$W=P\Delta V=(1.0\times10^5)(0.050-0.020)=3.0\times10^3\,\mathrm{J}$$

The work is positive because the gas expands and does work on its surroundings.

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.

Worked example from the mapped local textbook

An ideal monatomic gas contains $3.4\times10^{24}$ particles and warms from $297\,\mathrm{K}$ to $348\,\mathrm{K}$, so $\Delta T=51\,\mathrm{K}$.

$$\Delta U=\frac32Nk_B\Delta T=\frac32(3.4\times10^{24})(1.38\times10^{-23})(51)=3.6\times10^3\,\mathrm{J}$$

The positive value follows from the temperature rise; the path used to reach the final state does not change this $\Delta U$.

Why disorder usually wins

Imagine three freely moving particles and divide their container into left and right halves. Only two arrangements place all three particles together on one side; six arrangements distribute particles across both sides. The spread-out macrostate is therefore more likely because more microscopic arrangements produce it.

With enormous numbers of particles, the imbalance becomes overwhelming: systems tend toward macrostates compatible with the greatest number of microstates.

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.

Thermal example from local Question Bank row 32740

A gas reversibly receives $540\,\mathrm{J}$ at constant temperature $620\,\mathrm{K}$.

$$\Delta S=\frac{\Delta Q}{T}=\frac{540}{620}=0.87\,\mathrm{J\,K^{-1}}$$

The sign is positive because energy enters the gas.

Statistical example from the mapped local textbook

For $\Omega=1.0\times10^{22}$ accessible microstates,

$$S=k_B\ln\Omega=(1.38\times10^{-23})\ln(1.0\times10^{22})=7.0\times10^{-22}\,\mathrm{J\,K^{-1}}$$

The logarithm, not $\Omega$ itself, determines how multiplicity contributes to entropy.

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.

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.

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.

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.

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.

Worked example from the mapped local textbook

A monatomic ideal gas is compressed adiabatically to one eighth of its initial volume, so $V_1/V_2=8$.

$$\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^{5/3}=8^{5/3}=32$$

The pressure increases by a factor of 32. An isothermal compression by the same volume factor would increase pressure only by a factor of 8.

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.

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.

Worked example from local Question Bank row 28983

An engine produces $300\,\mathrm{J}$ of work and rejects $900\,\mathrm{J}$ per cycle. Energy conservation gives $Q_H=W+Q_C=1200\,\mathrm{J}$, so

$$\eta=\frac{W}{Q_H}=\frac{300}{1200}=0.25=25\%$$

The rejected energy is part of the input bookkeeping, not the denominator by itself.

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.

Worked example from local Question Bank row 31493

A plant operates between reservoirs at $T_H=885\,\mathrm{K}$ and $T_C=622\,\mathrm{K}$.

$$\eta_C=1-\frac{T_C}{T_H}=1-\frac{622}{885}=0.297\approx29.7\%$$

This is the theoretical maximum. A real plant between the same temperatures must have lower efficiency because its processes are irreversible.