Thermodynamic Process

Definition

A thermodynamic process is a procedure by which the initial state of a system changes to its final state. Equivalently, a process by which two or more of the state variables $p$, $V$, $T$ of a system are changed is called a thermodynamic process or a thermodynamic change.

Types of thermodynamic process (any two are enough when the board asks for two): quasi-static, isothermal, adiabatic, isochoric, isobaric, reversible, irreversible, cyclic.

• During the change, heat may be transferred into the system (positive heat) or released from it (negative heat), and work may be done by the system (positive work) or on the system (negative work).
• Processes in which the changes in the state variables occur infinitesimally slowly, so that the system is always in thermodynamic equilibrium, are called quasi-static processes.
• The series of intermediate states through which the system passes, plotted on a $p$–$V$ diagram, is called a path. The path tells you the way the system went through the change.

Work and Heat Depend on the Path

Consider a system taken from state A $(V_i, p_i)$ to state B $(V_f, p_f)$ by three different paths (Fig. 4.12).

• Path 1: both pressure and volume change together along a smooth curve; the pressure falls while the volume rises. The work done is positive and equals the area under curve 1.
• Path 2: the volume first increases to $V_f$ at constant pressure $p_i$, up to a point C; the pressure then falls to $p_f$ at constant volume. The work is the area under curve 2.
• Path 3: the pressure first falls from $p_i$ to $p_f$ at constant volume; the volume then increases to $V_f$ at the constant pressure $p_f$. The work is the area under curve 3.

The three areas are different, although the initial and final states are identical. Conclusion: the work done by a system depends not only on the initial and the final states but also on the intermediate states, that is, on the path along which the change takes place.

The same is true of heat (Fig. 4.13). Take a gas of volume $V_i$ to a final volume $V_f$ at constant temperature in two ways.

• Fig. 4.13 (a): the gas is heated slowly in a controlled manner and expands isothermally to $V_f$, pushing the piston. The system absorbs a finite amount of heat and does work: $Q > 0$, $W > 0$.
• Fig. 4.13 (b): the insulated cylinder is divided by a thin breakable partition into compartment X of volume $V_i$ holding the same gas and an empty compartment Y of volume $V_i'$, with $V_i + V_i' = V_f$. The partition is suddenly broken and the gas expands into the vacuum. No heat is exchanged because the cylinder is insulated, and no work is done because no piston or surface is pushed: $Q = 0$, $W = 0$. This is free expansion.

It is experimentally observed that an ideal gas undergoing free expansion has no change of temperature, so the final state is the same as in the first case. Same initial state, same final state, different paths, different heat. Conclusion: heat transferred to a system also depends on the path.

Reversible and Irreversible Processes

A reversible process is a change that can be retraced in the reverse direction; the path of a reversible process is the same in the forward and reverse directions (Fig. 4.14 (a)). Such changes are very slow, there is no loss of energy, and the system returns exactly to its initial state along the reverse path. Reversible processes are ideal; a real thermodynamic process always encounters some loss due to friction or other dissipative forces. Melting of ice, freezing of water, boiling of water and condensation of steam are treated as reversible.

An irreversible process is a change that cannot be retraced in the reverse direction; its forward and reverse paths differ, and the $p$–$V$ diagram shows a hysteresis loop (Fig. 4.14 (b)). There is a permanent loss of energy from the system due to friction or other dissipative forces, and the change of state depends on the path taken. Two objects at different temperatures reaching thermal equilibrium, puncturing an inflated balloon or tyre, rubbing palms together and burning a candle are irreversible.

Causes of irreversibility:

• Many processes, such as free expansion or an explosive chemical reaction, take the system to non-equilibrium states.
• Most processes involve friction, viscosity or other dissipative forces. An object sliding on a surface stops and loses its mechanical energy as heat. Dissipative forces are present everywhere and can at best be minimised, never fully eliminated.

All spontaneous natural processes are irreversible — heat always flows on its own from higher to lower temperature. An irreversible process therefore gives the preferred direction of a thermodynamic process and can be called a unidirectional process.

Assumptions Used From Here On

• Most of the processes discussed are reversible, that is, quasi-static: extremely slow, with the system undergoing infinitesimal change at every stage and remaining in thermodynamic equilibrium throughout — except the adiabatic processes.
• The system is an ideal gas enclosed in a cylinder with a movable, frictionless and massless piston. The walls are good thermal conductors for an isothermal process and thermally insulating for an adiabatic process.
• The ideal gas equation applies to the system.

Isothermal Process

Definition: a process in which change in pressure and volume takes place at a constant temperature is called an isothermal process or isothermal change.

It is possible only when the system is in good thermal contact with its environment and the transfer of heat is extremely slow, so that thermal equilibrium is maintained throughout. Melting of ice, and indeed any change of phase, is isothermal. Its $p$–$V$ curve (Fig. 4.15) is an isotherm, a rectangular hyperbola.

Equation of state (Boyle's law):

$$pV = \text{constant}, \qquad p_iV_i = p_fV_f$$

Derivation of the work done (the board sets this):

• Consider the isothermal expansion of an ideal gas. Let its initial volume be $V_i$ and the final volume be $V_f$.
• The work done in an infinitesimally small isothermal expansion is $dW = p\,dV$.
• The total work done in bringing about the expansion from $V_i$ to $V_f$ is

$$W = \int_{V_i}^{V_f} p\,dV \qquad \ldots (1)$$

• But for an ideal gas $pV = nRT$, so $p = nRT/V$. Using this in equation (1),

$$W = nRT\int_{V_i}^{V_f} \dfrac{dV}{V}$$

$$\therefore\ W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right) \qquad \ldots (2)$$

Since the internal energy of an ideal gas depends only on its temperature, and $\Delta T = 0$, the internal energy remains constant: $\Delta U = 0$. The first law $Q = \Delta U + W$ then gives

$$Q = W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right)$$

So the heat transferred to the gas is completely converted into work. When the gas absorbs heat it does positive work and expands; when it is compressed it does negative work and releases heat. Using $p_iV_i = p_fV_f$, the same work may also be written $W = nRT\ln(p_i/p_f)$.

Remember for an isothermal process: $pV =$ constant; $\Delta T = 0$; $\Delta U = 0$; $Q = W$; the change is very slow and the system stays in thermal equilibrium with its environment throughout. Neither $Q$ nor $W$ is zero. Note also that the total amount of heat of the system does not remain constant in an isothermal change.

Isobaric Process

A constant pressure process, $\Delta p = 0$. Boiling water at atmospheric pressure is an example. Its $p$–$V$ curve (Fig. 4.16) is an isobar, a horizontal straight line parallel to the volume axis. (The curves on a meteorological map are isobars in the same sense — lines joining places of equal pressure.)

$$W = p\,(V_f - V_i) = p\,\Delta V = nR\,(T_f - T_i)$$

$$\Delta U = n\,C_V\,\Delta T = n\,C_V\,(T_f - T_i)$$

From the first law $Q = \Delta U + W$:

$$Q = nC_V(T_f - T_i) + nR(T_f - T_i) = n\,(C_V + R)(T_f - T_i)$$

$$\therefore\ Q = n\,C_p\,(T_f - T_i), \qquad C_p = C_V + R$$

The temperature of the system changes, so its internal energy changes too. The heat exchanged is used partly to raise the temperature and partly to do work.

Remember for an isobaric process: $\Delta p = 0$; $\Delta T \ne 0$; $Q = \Delta U + W$; $W = p\,\Delta V$. None of $\Delta U$, $Q$, $W$ is zero.

Isochoric Process

A constant volume process, $\Delta V = 0$. Its $p$–$V$ curve (Fig. 4.17) is a vertical straight line parallel to the pressure axis. Heating a gas in a constant-volume container, or diffusion of a gas in a closed chamber, are examples.

$$W = p\,\Delta V = 0, \qquad \Delta U = n\,C_V\,\Delta T, \qquad Q = \Delta U = n\,C_V\,\Delta T$$

A system does no work on its environment during an isochoric change, so all the energy added as heat remains in the system and increases its internal energy.

Remember for an isochoric process: $\Delta V = 0$; $W = 0$; $Q = \Delta U$; $\Delta T \ne 0$.

Adiabatic Process

Definition: an adiabatic process is a process during which there is no transfer of heat from or to the system. For an adiabatic change $Q = 0$.

Heat transfer is prevented either by perfectly insulating the system from its environment, or by carrying out the change so rapidly that there is no time for any exchange of heat. Puncturing an inflated balloon or tyre is a familiar example. From the first law with $Q = 0$,

$$\Delta U = -W$$

• Adiabatic expansion: $W$ is positive (work done by the system), $\Delta U$ is negative, the internal energy and hence the temperature fall — the gas cools.
• Adiabatic compression: $W$ is negative (work done on the system), $\Delta U$ is positive, the internal energy and hence the temperature rise — the gas warms.

Equation of state:

$$pV^{\gamma} = \text{constant} = C, \qquad \gamma = \dfrac{C_p}{C_V}$$

$\gamma$ is the adiabatic ratio. For moderate temperature changes, $\gamma = 5/3$ for monatomic gases, $7/5$ for diatomic gases and $4/3$ for triatomic gases.

Derivation of the work done:

$$W = \int_{V_i}^{V_f} p\,dV = \int_{V_i}^{V_f} \dfrac{C}{V^{\gamma}}\,dV = C\left[\dfrac{V^{1-\gamma}}{1-\gamma}\right]_{V_i}^{V_f}$$

$$\therefore\ W = \dfrac{C}{1-\gamma}\left(V_f^{1-\gamma} - V_i^{1-\gamma}\right)$$

Since $pV^{\gamma} = C$, that is $p_iV_i^{\gamma} = p_fV_f^{\gamma} = C$, we may write $C\,V_f^{1-\gamma} = p_fV_f$ and $C\,V_i^{1-\gamma} = p_iV_i$, giving

$$W = \dfrac{p_fV_f - p_iV_i}{1-\gamma} = \dfrac{p_iV_i - p_fV_f}{\gamma - 1} = \dfrac{nR\,(T_i - T_f)}{\gamma - 1}$$

When the gas expands, $W > 0$ and $T_i > T_f$, so the gas cools down; when the gas is compressed, $W < 0$ and $T_i < T_f$, so the gas warms up.

Two further forms, obtained from $pV^{\gamma} =$ constant together with $pV = nRT$:

$$TV^{\gamma-1} = \text{constant}, \qquad p^{1-\gamma}T^{\gamma} = \text{constant}$$

$$p_f = p_i\left(\dfrac{V_i}{V_f}\right)^{\gamma}, \qquad T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}, \qquad \left(\dfrac{p_2}{p_1}\right)^{\gamma-1} = \left(\dfrac{T_2}{T_1}\right)^{\gamma}$$

Remember for an adiabatic process: $pV^{\gamma} =$ constant; $Q = 0$; $\Delta U = -W$; $\Delta T \ne 0$; adiabatic expansion cools and adiabatic compression heats the system; most of the time an adiabatic change is a sudden change, during which the system has no time to exchange heat with its environment.

Two standard oral questions. The adiabatic $p$–$V$ curve is steeper than the isothermal curve through the same point because for the adiabat $pV^{\gamma} =$ constant with $\gamma > 1$, so pressure falls faster with volume than for the isotherm $pV =$ constant. Clouds form at high altitude because rising moist air expands adiabatically against the lower pressure, cools, and the water vapour condenses.

Cyclic Process

A thermodynamic process that returns a system to its initial state is a cyclic process (Fig. 4.19); the initial and final states are the same. Hence

$$\Delta U = 0 \qquad \text{and} \qquad Q = W$$

The work done in one complete cycle is the area enclosed by the closed loop, being the difference between the area under the outward curve and the area under the return curve. The total work done in a cyclic process is positive if the process goes around the cycle in a clockwise direction, and negative if it goes around in a counterclockwise direction. If the total work in a cycle is $-1000$ J, then $\Delta U = 0$ and $Q = W = -1000$ J, that is, 1000 J of heat is rejected by the system. Reversing the direction of the cycle makes the work positive and the system does work. The working of all heat engines is a cyclic process.

Free Expansion

Free expansion is an adiabatic expansion with no exchange of heat and no work done either on or by the system:

$$Q = 0, \qquad W = 0, \qquad \therefore\ \Delta U = 0$$

Examples: a balloon ruptured suddenly, or a tyre suddenly punctured — the air inside rushes out rapidly but no piston or surface is displaced. Free expansion differs from the other processes because it is an uncontrolled, instantaneous change during which the system is not in thermodynamic equilibrium. A free expansion cannot be plotted on a $p$–$V$ diagram; only its initial and final states can be plotted.

Standard Results