Thermodynamic Processes

A thermodynamic process is a transition of a thermodynamic system from an initial equilibrium state to a final equilibrium state. During a process, the state variables of the system change due to interactions with the surroundings (heat transfer, work done). The path of a process is the series of intermediate states through which the system passes.

Different types of thermodynamic processes are distinguished based on which state variables are kept constant or how the energy transfer occurs.

Quasi-Static Process

A quasi-static process (or quasi-equilibrium process) is an idealised thermodynamic process that proceeds infinitely slowly, such that the system is always infinitesimally close to a state of thermodynamic equilibrium at every step along the path.

In a truly quasi-static process, the system passes through a sequence of equilibrium states. This means that at any point during the process, the system's properties (pressure, temperature, density, etc.) are uniform throughout the system, and there are no unbalanced forces or temperature gradients within the system.

A real process can approximate a quasi-static process if it occurs slowly enough. For example, slowly compressing a gas in a cylinder with a piston is approximately quasi-static, whereas rapidly compressing it would lead to pressure and temperature gradients within the gas, making it non-quasi-static.

Significance:

Well-defined Path: Only quasi-static processes can be represented as continuous paths on thermodynamic diagrams (like P-V diagrams) because only equilibrium states can be plotted.
Maximum Work: Quasi-static processes are often reversible processes (processes that can be reversed by an infinitesimal change in conditions, returning the system and surroundings to their original states). Reversible processes are ideal processes that yield the maximum possible work output (or require minimum work input) compared to any irreversible process between the same two end states.
Calculation of Work: For a quasi-static process where a gas changes volume, the work done by the gas is $W = \int P \, dV$. Since the process is quasi-static, the pressure $P$ at each step is well-defined and corresponds to the equilibrium pressure of the gas. This integral can be calculated if the relationship between P and V along the path is known.

Most thermodynamic processes analyzed theoretically are assumed to be quasi-static to simplify calculations and represent them on diagrams.

Isothermal Process ($ T = \text{constant} $)

An isothermal process is a thermodynamic process that occurs at a constant temperature. During an isothermal process, the system is in thermal contact with a large reservoir (heat bath) at a constant temperature, allowing heat to be exchanged freely between the system and the reservoir to maintain the system's temperature constant despite changes in other state variables.

For an ideal gas, the internal energy $U$ depends only on temperature ($U = \frac{f}{2}nRT$). Therefore, in an isothermal process involving an ideal gas, the change in internal energy is zero: $\Delta U = 0$.

Applying the First Law of Thermodynamics ($\Delta U = Q - W$) to an isothermal process for an ideal gas:

$ 0 = Q - W $

$ Q = W $

This means that in an isothermal process, any heat absorbed by the ideal gas is completely converted into work done by the gas, and vice versa.

Work Done in an Isothermal Process (Ideal Gas)

Consider an isothermal expansion of $n$ moles of an ideal gas from initial volume $V_1$ to final volume $V_2$ at constant temperature $T$. The work done by the gas is $W = \int_{V_1}^{V_2} P \, dV$. From the ideal gas law, $P = \frac{nRT}{V}$. Since $T$ is constant, $nRT$ is constant.

$ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV $

$ W = nRT [\ln V]_{V_1}^{V_2} = nRT (\ln V_2 - \ln V_1) $

$ W = nRT \ln\left(\frac{V_2}{V_1}\right) $

Using $PV = nRT$, we can also write $nRT = P_1 V_1 = P_2 V_2$. Also, for isothermal process, $P_1 V_1 = P_2 V_2$, so $V_2/V_1 = P_1/P_2$.

$ W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) = P_2 V_2 \ln\left(\frac{P_1}{P_2}\right) $

For expansion ($V_2 > V_1$), $W > 0$ and $Q > 0$ (heat is absorbed). For compression ($V_2 < V_1$), $W < 0$ and $Q < 0$ (work is done on the gas, heat is released).

P-V diagram for an isothermal process (ideal gas).

(Image Placeholder: A P-V diagram. Y-axis is Pressure (P), X-axis is Volume (V). Show a curve (hyperbola, PV=constant) starting from (P1, V1) and going to (P2, V2) where V2>V1 and P2

Adiabatic Process ($ Q = 0 $)

An adiabatic process is a thermodynamic process that occurs without any heat transfer into or out of the system. The system is thermally insulated from its surroundings.

Applying the First Law of Thermodynamics ($\Delta U = Q - W$) to an adiabatic process:

$ \Delta U = 0 - W $

$ \Delta U = -W $

This means that in an adiabatic process, any work done by the system comes at the expense of its internal energy ($\Delta U$ is negative, temperature decreases), and any work done on the system increases its internal energy ($\Delta U$ is positive, temperature increases).

Adiabatic Process for an Ideal Gas

For an ideal gas, the relationship between pressure and volume during a quasi-static adiabatic process is given by:

$ P V^\gamma = \text{constant} $

where $\gamma = C_p/C_v$ is the ratio of specific heats (adiabatic index), which is greater than 1 for all gases. This equation describes the path of an adiabatic process on a P-V diagram.

The work done in a quasi-static adiabatic process from $(P_1, V_1)$ to $(P_2, V_2)$ for an ideal gas is $W = \int_{V_1}^{V_2} P \, dV$. Using $P V^\gamma = K$ (constant), $P = K V^{-\gamma}$.

$ W = \int_{V_1}^{V_2} K V^{-\gamma} \, dV = K \left[ \frac{V^{-\gamma+1}}{-\gamma+1} \right]_{V_1}^{V_2} = \frac{K}{1-\gamma} [V_2^{1-\gamma} - V_1^{1-\gamma}] $

Since $K = P_1 V_1^\gamma = P_2 V_2^\gamma$, we can write:

$ W = \frac{P_2 V_2^\gamma V_2^{1-\gamma} - P_1 V_1^\gamma V_1^{1-\gamma}}{1-\gamma} = \frac{P_2 V_2 - P_1 V_1}{1-\gamma} $

$ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} $

Using the ideal gas law, $PV = nRT$, we can also express work in terms of temperature: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.

Also, $\Delta U = n C_V (T_2 - T_1) = -n C_V (T_1 - T_2)$. Since $\gamma = C_p/C_v = 1 + R/C_v$, $\gamma - 1 = R/C_v$, so $C_v = R/(\gamma - 1)$.

$ \Delta U = n \frac{R}{\gamma - 1} (T_2 - T_1) = \frac{nR(T_2 - T_1)}{\gamma - 1} = -W $

This confirms $\Delta U = -W$ for an adiabatic process.

On a P-V diagram, adiabatic curves are steeper than isothermal curves passing through the same point (because $\gamma > 1$).

(Image Placeholder: A P-V diagram. Show a higher temperature isotherm (PV=const). Show a lower temperature isotherm (PV=const). Show an adiabatic curve (PV^gamma=const) starting on the higher isotherm and ending on the lower isotherm, steeper than the isotherms.)

Adiabatic processes occur when processes are very rapid (not enough time for significant heat transfer) or when the system is very well insulated.

Isochoric Process ($ V = \text{constant}, W=0 $)

An isochoric process (or isometric or isovolumetric process) is a thermodynamic process that occurs at a constant volume. In such a process, the system's volume does not change, so no work is done by or on the system due to volume change.

For an isochoric process, $dV = 0$, so the work done $W = \int P \, dV = 0$.

Applying the First Law of Thermodynamics ($\Delta U = Q - W$) to an isochoric process:

$ \Delta U = Q - 0 $

$ \Delta U = Q $

This means that in an isochoric process, the heat added to the system is entirely used to increase its internal energy (and thus its temperature, for an ideal gas), and heat removed from the system equals the decrease in its internal energy (and temperature).

On a P-V diagram, an isochoric process is represented by a vertical line.

(Image Placeholder: A P-V diagram. Show a vertical line at a constant volume V. The pressure changes from P1 to P2. No area under the curve, so W=0.)

Examples include heating a gas in a sealed, rigid container.

Isobaric Process ($ P = \text{constant}, W=P\Delta V $)

An isobaric process is a thermodynamic process that occurs at a constant pressure. Changes in volume and temperature can occur, but the pressure is maintained constant, usually by allowing the system boundary (e.g., a piston) to move freely against a constant external force (like atmospheric pressure and the weight of the piston).

For an isobaric process, $dP = 0$. The work done by the system as it expands from volume $V_1$ to $V_2$ at constant pressure $P$ is:

$ W = \int_{V_1}^{V_2} P \, dV = P \int_{V_1}^{V_2} dV = P (V_2 - V_1) $

$ W = P\Delta V $

Applying the First Law of Thermodynamics ($\Delta U = Q - W$) to an isobaric process:

$ \Delta U = Q - P\Delta V $

$ Q = \Delta U + P\Delta V $

This shows that the heat added to the system in an isobaric process is used partly to increase its internal energy and partly to do work of expansion. For an ideal gas, $P\Delta V = nR\Delta T$. Also $\Delta U = nC_V\Delta T$. So $Q = nC_V\Delta T + nR\Delta T = n(C_V+R)\Delta T = nC_P\Delta T$. This is consistent with the definition of $C_P$.

On a P-V diagram, an isobaric process is represented by a horizontal line.

(Image Placeholder: A P-V diagram. Show a horizontal line at a constant pressure P. The volume changes from V1 to V2 where V2>V1. The area under the line (rectangle) represents the work done.)

Examples include heating water in an open container (pressure is atmospheric pressure), or a gas expanding in a cylinder with a freely moving piston under constant external pressure.

Cyclic Process ($ \Delta U = 0 $)

A cyclic process is a thermodynamic process (or a sequence of processes) in which the system returns to its initial equilibrium state. The system completes a cycle of changes in its state variables.

Since the internal energy $U$ is a state function, its value depends only on the current state. In a cyclic process, the final state is the same as the initial state, so the change in internal energy over one complete cycle is always zero:

$ \Delta U_{cycle} = U_{final} - U_{initial} = 0 $

Applying the First Law of Thermodynamics ($\Delta U = Q - W$) to a complete cycle:

$ 0 = Q_{net} - W_{net} $

$ Q_{net} = W_{net} $

where $Q_{net}$ is the net heat transferred into the system during the entire cycle (sum of heat absorbed and released in different parts of the cycle) and $W_{net}$ is the net work done by the system during the entire cycle (sum of work done by and on the system in different parts of the cycle).

This means that in a cyclic process, the net heat absorbed by the system is equal to the net work done by the system. This principle is fundamental to the operation of heat engines, where a working substance undergoes a cycle, absorbs heat from a high-temperature reservoir, rejects heat to a low-temperature reservoir, and performs net work.

On a P-V diagram, a cyclic process is represented by a closed loop. The net work done during the cycle is equal to the area enclosed by the loop on the P-V diagram (clockwise loop indicates net work done by the system, counterclockwise loop indicates net work done on the system).

(Image Placeholder: A P-V diagram showing a closed loop, e.g., starting at A, going to B, then C, then D, and back to A. Arrows indicate the direction of the process. The area enclosed by the loop is shaded, representing the net work done.)