How a Diesel Engine Lights Its Fuel Without a Spark

Compress air slowly in a cylinder kept in cool water, and its temperature hardly changes. Compress it suddenly, and it heats up sharply. The same start and end volumes, but completely different processes.

Thermodynamics names these processes, draws them on a P-V diagram, and calculates the work each one involves.

This part covers state variables, Mayer's relation, the main thermodynamic processes, and work in isothermal and adiabatic changes. Take $R=8.314$ J/mol K.

**State variables such as pressure, volume, temperature, internal energy and mass describe a system in equilibrium, and for an ideal gas they are linked by the equation of state $PV=\mu RT$.

- Intensive variables do not depend on the amount of substance: pressure, temperature, density
- Extensive variables** do: volume, mass, internal energy

Knowing any two of $P$, $V$ and $T$ for a given amount of ideal gas fixes the third.

Worked example 1 — finding temperature. $2$ mol of gas occupies $0.05$ m${}^3$ at $1\times 10^5$ Pa:

$$T=\frac{PV}{\mu R}=\frac{1\times 10^5\times 0.05}{2\times 8.314}\approx 301\text{ K}$$

Worked example 2 — halving the volume at that temperature. Pressure doubles to $2\times 10^5$ Pa.

An everyday example. The gauge on a sealed gas cylinder reads its pressure; with volume and temperature known, the equation of state tells how much gas remains.

The substance. State variables are defined only for equilibrium states — during a sudden explosion, a single "pressure" of the gas does not exist.

**Heating at constant pressure needs extra heat to do the work of expansion, $P\Delta V=\mu R\Delta T$, on top of the rise in internal energy, so the molar specific heats of an ideal gas satisfy $C_p-C_v=R$.

Derivation.** At constant volume, no work is done:

$$Q_v=\mu C_v\Delta T=\Delta U$$

At constant pressure, by the first law, and since $\Delta U$ of an ideal gas depends only on $\Delta T$:

$$\mu C_p\Delta T=\Delta U+P\Delta V=\mu C_v\Delta T+\mu R\Delta T\Rightarrow C_p-C_v=R$$

Values. Monatomic: $C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$, $\gamma =\frac{5}{3}$. Diatomic: $C_v=\frac{5}{2}R$, $C_p=\frac{7}{2}R$, $\gamma =1.4$.

Worked example. Heating $2$ mol of a diatomic gas by $50$ K:

$$Q_p=2\times \frac{7}{2}(8.314)\times 50\approx 2910\text{ J}{Q}_v=2\times \frac{5}{2}(8.314)\times 50\approx 2079\text{ J}$$

The difference, $\mu R\Delta T=2\times 8.314\times 50\approx 831$ J, is exactly the work of expansion.

An everyday example. Heating air in a sealed pressure cooker needs less heat for each degree than heating the same air in an open vessel, where it pushes outward as it warms.

The substance. Mayer's relation holds for molar specific heats of an ideal gas; per kilogram, the difference is $\frac{R}{M}$.

**An isothermal process keeps temperature constant ($PV$ constant), an adiabatic process allows no heat exchange ($P{V}^{\gamma }$ constant), an isobaric process keeps pressure constant, an isochoric process keeps volume constant, and a cyclic process returns the system to its starting state.

On a P-V diagram:

- Isothermal** — a hyperbola, $PV=$ constant
- Adiabatic — a steeper curve through the same point, since its slope is $\gamma$ times the isothermal slope
- Isobaric — a horizontal line
- Isochoric — a vertical line
- Cyclic — a closed loop whose enclosed area is the net work

**Worked example — halving the volume of air ($\gamma =1.4$) starting at $300$ K.

- Isothermally**: $P_2=2P_1$, $T$ stays $300$ K
- Adiabatically, using $P{V}^{\gamma }$ and $T{V}^{\gamma -1}$ constant:

$$P_2=P_1(2{)}^{1.4}\approx 2.64P_1{T}_2=300\times 2^{0.4}\approx 396\text{ K}$$

The adiabatic compression reaches **higher pressure and heats the gas by about $96$ K.

An everyday example. A quick stroke of a bicycle pump is nearly adiabatic, which is why the pump barrel warms.

The substance. Adiabatic does not mean constant temperature** — no heat flows, but work changes the internal energy.

**Integrating $W=\int PdV$ gives $W=\mu RT\ln \frac{V_2}{V_1}$ for a reversible isothermal process and $W=\frac{\mu R(T_1-T_2)}{\gamma -1}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$ for a reversible adiabatic process.

Isothermal derivation.** With $P=\frac{\mu RT}{V}$ and $T$ fixed:

$$W=\mu RT{\int }_{V_1}^{V_2}\frac{dV}{V}=\mu RT\ln \frac{V_2}{V_1}$$

Worked example 1 — isothermal. $1$ mol at $300$ K doubles its volume:

$$W=8.314\times 300\times \ln 2\approx 1.73\times 10^3\text{ J}$$

Since $\Delta U=0$, the gas absorbs exactly this much heat.

Worked example 2 — adiabatic. $2$ mol of a monatomic gas ($\gamma =\frac{5}{3}$) expands adiabatically and cools from $400$ K to $300$ K:

$$W=\frac{2\times 8.314\times 100}{\frac{2}{3}}\approx 2494\text{ J}$$

Check: $\Delta U=\frac{3}{2}\times 2\times 8.314\times (-100)\approx -2494$ J, and with $Q=0$, $W=-\Delta U$.

An everyday example. Gas rushing out of a punctured tyre feels cold, because it expands quickly, doing work at the expense of its internal energy.

The substance. For the same expansion, isothermal work is larger than adiabatic work, because pressure falls more slowly along an isotherm.