Thermodynamic Processes and First Law

Definition: in thermodynamics, mechanical work is associated with volume changes of the system. When a gas expands, it pushes a piston and does positive work on the surroundings ($W>0$). When compressed, the surroundings do work on the gas ($W<0$).

Geometric interpretation: on a $P$-$V$ diagram, work is the area under the curve of the process: $W = \int P\,dV$. This is why work depends on the path (it is not a state function), unlike $U$ (internal energy) and $S$ (entropy), which are state functions — they depend only on the system's state (P, V, T), not on how it got there. Two different paths between the same states enclose different areas (different work), but give the same $\Delta U$.

• Isobaric ($P$ const): $W = P\Delta V = nR\Delta T$ — area of a rectangle.
• Isothermal ($T$ const): $W = nRT\ln(V_2/V_1)$ — area under a hyperbola.
• Adiabatic ($Q=0$): $W = -\Delta U = -nC_v\Delta T$ — work is done at the expense of internal energy.
• Isochoric ($V$ const): $W = 0$ — no mechanical work.

Sign convention (IUPAC): $W > 0$ if the system does work on the surroundings (expansion); $W < 0$ if the surroundings do work on the system (compression).

Real-world example: the work in an internal combustion engine is the area enclosed by the cycle on the $P$-$V$ diagram. The larger the area, the greater the work per cycle.

Statement: $\Delta U = Q - W$. The change in internal energy $\Delta U$ of a system equals the heat $Q$ absorbed from the surroundings minus the work $W$ done by the system on the surroundings. This is a formulation of the principle of energy conservation.

Key points:

• $U$ is a state function: it depends only on the thermodynamic state, not on the path taken. For an ideal gas, $U$ depends only on $T$: $\Delta U = nC_v\Delta T$.
• $Q$ and $W$ are not state functions: they depend on the path. This means that for the same initial and final states, $Q$ and $W$ can differ (but $Q - W$ will be the same, equal to $\Delta U$).
• Signs: $Q > 0$ if the system absorbs heat, $Q < 0$ if it releases heat. $W > 0$ if the system does work, $W < 0$ if work is done on it.

Important consequences:

• Thermodynamic cycle: $\Delta U = 0$ (final state = initial state) → $Q_{net} = W_{net}$. The net heat absorbed in a cycle is entirely converted into net work.
• Adiabatic process ($Q=0$): $W = -\Delta U$. Work is done at the expense of internal energy.
• Cyclic process: a heat engine cannot produce work without absorbing heat (otherwise it would violate energy conservation).

Real-world example: a pressure cooker — heat from the stove ($Q > 0$) increases the internal energy of the steam ($\Delta U > 0$), and the safety valve releases steam when $P$ exceeds the limit ($W$ negative for steam expanding into the environment).

A process is the passage of a gas from one equilibrium state to another. In the four fundamental processes one quantity is held constant ($T$, $P$ or $V$), or heat exchange is prevented (adiabatic). Fixing a variable has a precise purpose: it simplifies the first law $\Delta U = Q - W$ and makes the curve in the $P$-$V$ plane predictable. For each one we see what stays constant, why the formulas hold, and how heat, work and internal energy are distributed.

1. Isothermal ($T$ constant) — The gas exchanges heat with a thermostat that locks its temperature, so the process must be very slow, otherwise $T$ would change. From $PV=nRT$ with fixed $T$ follows $PV=\text{constant}$: in the $P$-$V$ plane the curve is an equilateral hyperbola. Since for an ideal gas $U$ depends only on $T$, if $T$ does not change then $\Delta U = 0$: the first law reduces to $Q = W$, so all the absorbed heat leaves as work, $Q = W = nRT\ln(V_2/V_1)$.

• Example: slow expansion of a gas immersed in a heat bath at fixed temperature.

2. Isobaric ($P$ constant) — The gas expands against an external pressure that stays constant (typical case: a free piston under atmospheric pressure). From $PV=nRT$ with fixed $P$ follows $V/T=\text{constant}$, i.e. volume and temperature are directly proportional. The work is the area of a rectangle, $W = P\Delta V = nR\Delta T$. Here heat does two things at once: it raises the internal energy and does expansion work. That is why $Q = nC_p\Delta T$ is larger than $\Delta U = nC_v\Delta T$.

• Example: heating a gas in a cylinder with a free piston (constant atmospheric pressure).

3. Isochoric ($V$ constant) — The gas is sealed in a rigid container: the volume cannot change, so there is no mechanical work, $W = 0$. From $PV=nRT$ with fixed $V$ follows $P/T=\text{constant}$. With $W=0$ the first law becomes $Q = \Delta U = nC_v\Delta T$: all the heat supplied ends up in the internal energy, and this is the most direct way to link heat and temperature.

• Example: heating a gas in a rigid closed container (pressure cooker without a valve).

4. Adiabatic ($Q = 0$) — There is no heat exchange with the surroundings, either because the system is insulated or because the process is too fast for heat to flow in time. With $Q=0$ the first law gives $\Delta U = -W$: the gas does work at the expense of its own internal energy, so on expanding it cools down and on compressing it heats up. The curve is steeper than the isotherm and follows $PV^\gamma=\text{constant}$, with $TV^{\gamma-1}=\text{constant}$ and $TP^{(1-\gamma)/\gamma}=\text{constant}$.

• Example: rapid compression in a Diesel engine (the gas heats up without exchanging heat).

Why do we need $\gamma$ and Mayer's relation?

• $\gamma = C_p/C_v$ is the adiabatic index and measures how steep the adiabat is compared with the isotherm: $\gamma_{mono}=5/3\approx1.67$, $\gamma_{di}=7/5=1.4$, $\gamma_{polyatomic} \approx 4/3$. The more degrees of freedom the molecule has, the lower $\gamma$ is.
• Mayer's relation: $C_p - C_v = R$. For the same $\Delta T$, more heat is needed at constant $P$ than at constant $V$, precisely because in the isobar part of the heat becomes expansion work (the isochor wastes none of it).

Sensible heat: $Q = mc\,\Delta T$. This raises or lowers the body's temperature without changing its state of aggregation. $c$ is the specific heat capacity (J/(kg·K)), characteristic of each material. Water has $c = 4186$ J/(kg·K), one of the highest values, making it an excellent coolant.

Latent heat: $Q = mL$. Energy needed to change the state of aggregation at constant temperature. The temperature does not change because the energy is used to break (fusion, vaporisation) or form (solidification, condensation) intermolecular bonds:

• Fusion/solidification: $L_f$ — ice melting: $L_f = 334$ kJ/kg.
• Vaporisation/condensation: $L_v$ — water vaporisation: $L_v = 2260$ kJ/kg (much larger than $L_f$, because gas molecules must completely overcome intermolecular forces).

The three mechanisms of heat transfer:

1. Conduction: transfer by direct contact — Fourier's law $Q = kA\Delta T t / d$, where $k$ is thermal conductivity. Example: a metal spoon in a hot cup of tea heats up at the far end.
2. Convection: heat transport through fluid motion. Hot air rises (less dense), cold air sinks. Example: convective currents in the atmosphere generating winds.
3. Radiation: energy transfer via electromagnetic waves (infrared). $P = \varepsilon\sigma A T^4$ (Stefan-Boltzmann law). Example: the Sun's heat reaches Earth through the vacuum by radiation.

Connection: specific heats of gases ($C_v$, $C_p$) are linked to molecular degrees of freedom (Theory §3), while latent heat is crucial for understanding phase transitions and phase diagrams.