reversible irreversible and adiabatic processes

A process is called reversible if it can be run backwards so that both the system and the surroundings return to their initial states; this requires the process to be quasi-static (a continuous sequence of equilibrium states) and free of dissipation — friction, viscosity, heat transfer across a finite temperature difference.

All real processes are irreversible: free expansion of a gas into vacuum, heat exchange at $\Delta T \neq 0$, friction in a moving piston.

An adiabatic process is one with no heat exchange with the surroundings, $Q = 0$.

By the first law of thermodynamics $Q = \Delta U + A$, where $A$ is the work done by the gas, so for $Q = 0$ we get $A = -\Delta U$: expanding adiabatically, a gas does work at the expense of its internal energy and cools, while compression heats it.

For an ideal gas with constant heat capacity a quasi-static adiabatic process obeys the Poisson equation $pV^{\gamma} = \text{const}$ with $\gamma = C_p/C_V > 1$, so on a $pV$ diagram the adiabat is steeper than the isotherm.

A reversible adiabatic process is isentropic ($\Delta S = 0$), whereas an irreversible adiabatic one with the same $Q = 0$ gives $\Delta S > 0$ and ends in a different final state.

An ideal gas with $\gamma = 1.4$ is compressed adiabatically from volume $V_1$ to $V_1/8$. The initial temperature is $T_1 = 300$ K. Find the final temperature $T_2$.

Two moles of an ideal gas ($\gamma = 1.4$, $R = 8.31$ J/(mol·K)) expand adiabatically, and the temperature drops from $T_1 = 400$ K to $T_2 = 300$ K. Find the work done by the gas and the change of internal energy. Check the result with two formulas.

A monatomic ideal gas ($\gamma = 5/3$) occupies volume $V_1$ at temperature $T_1$ and pressure $p_1$. It is brought to volume $2V_1$ adiabatically in two ways: (a) reversibly, by slowly pulling the piston; (b) irreversibly, by free expansion into vacuum. Find the final $T$ and $p$ in each case, the entropy changes, and explain the difference.

Self-study with an interactive ideal-gas simulation (for example PhET "Gas Properties" or an equivalent). At home do not use pressurized vessels, do not heat sealed containers, and do not work with open flames — perform all experiments inside the simulation.

1. Take a cylinder with insulated walls and a movable piston. Move the piston slowly in small steps and record $p$, $V$, $T$ in a table at each step.
2. Plot the points on a $pV$ diagram and check that $pV^{\gamma} \approx \text{const}$ with $\gamma = 1.4$; compare with the isotherm through the initial point.
3. Compress the same gas quickly to the same final volume. Since the walls are insulated, both compressions are adiabatic ($Q = 0$): the slow one is quasi-static and reversible, while the fast one is irreversible. Compare the final temperatures: the fast compression gives a hotter gas, because dissipation (friction in the piston, eddies) adds to the work done by external forces, so the internal energy increases more than in the reversible compression. Write down the observation and explain it through $A = -\Delta U$.
4. Let the gas expand into the evacuated part of the simulation (free expansion) and verify that $T$ does not change while the entropy indicator grows. State a conclusion about the link between irreversibility and the growth of $S$.

To show on a single $pV$ diagram that the adiabat is steeper than the isotherm and that reversible and irreversible adiabatic expansions to the same volume end in different final states.

Sketch of a $pV$ diagram: through point $A(V_1,\,p_1)$ a dashed isotherm $pV = \text{const}$ and a solid adiabat $pV^{\gamma} = \text{const}$ with $\gamma = 5/3$ are drawn. From $A$ two arrows go to the right toward volume $2V_1$: the solid one is the reversible adiabat to point $B$ with $p_B = 0.315\,p_1$, the dashed one is the irreversible free expansion to point $C$ with $p_C = 0.500\,p_1$. Labels beside them: $T_B = 0.630\,T_A < T_A$, $T_C = T_A$; $\Delta S_{AB} = 0$, $\Delta S_{AC} = \nu R \ln 2 > 0$.

A cyclist quickly inflates a tire. In $5\ \mathrm{s}$ he pushes the pump piston and compresses the air from a volume of $100\ \mathrm{cm^3}$ to $20\ \mathrm{cm^3}$. The initial air temperature is $20^\circ\mathrm{C}$, the pressure is $1.0\cdot10^5\ \mathrm{Pa}$; after the rapid compression the air in the pump warms to about $60^\circ\mathrm{C}$, and the metal cylinder becomes warm.

The rapid compression occurs in a time during which heat exchange with the air is negligible, so the process is treated as adiabatic: $Q=0$. The first law of thermodynamics for a gas becomes $\Delta U=Q-W_{\mathrm{by}}=-W_{\mathrm{by}}.$ During compression the gas does negative work ($W_{\mathrm{by}}<0$), so $\Delta U>0$: internal energy and temperature increase.

A car of mass $1200\ \mathrm{kg}$ brakes from $20\ \mathrm{m/s}$ to rest. The kinetic energy $\frac{1}{2}mv^2=\frac{1}{2}\cdot1200\cdot20^2=2.4\cdot10^5\ \mathrm{J}$ turns into internal energy of the brake discs and air; the discs heat up by tens of degrees.

This is a typical irreversible process: mechanical energy is dissipated by friction, and the same energy cannot spontaneously accelerate the car again. The first law of thermodynamics still holds: $Q$ and $W$ account for the transferred energy, but the reverse process would require external work; entropy increases.

An air mass rises along a mountain slope by $1000\ \mathrm{m}$. The pressure falls from about $100\ \mathrm{kPa}$ to $90\ \mathrm{kPa}$, and the air expands. The dry adiabatic cooling rate is about $10^\circ\mathrm{C}$ per $1\ \mathrm{km}$, so the temperature drops from $20^\circ\mathrm{C}$ to $10^\circ\mathrm{C}$; with further rise, water vapor condenses and a cloud forms.

The rise is fast compared with heat exchange, so it is approximately treated as adiabatic: $Q=0$. Expansion means $W_{\mathrm{by}}>0$, and the first law $\Delta U=Q-W_{\mathrm{by}}=-W_{\mathrm{by}}$ gives $\Delta U<0$ — the air cools. A real cloud with turbulence and mixing is an irreversible process, but the adiabatic model describes the cooling well.