Physics · Ch 11 — Thermodynamics

Thermodynamic Processes

11.8

Thermodynamic Processes

The Meaning of a Thermodynamic Process

A thermodynamic process is any change that a thermodynamic system undergoes — a change from one equilibrium state (initial state) to another equilibrium state (final state). The path the system takes through its state variables (pressure $P$ , volume $V$ , temperature $T$ , internal energy $U$ ) defines the process. The key is that the system is not in equilibrium during the actual change; we only know its state at the start and at the end, and sometimes at carefully chosen intermediate points.

The textbook introduces four fundamental types of processes, each defined by which state variable is held constant. These are the building blocks for solving problems.

Table 11.2 Some special thermodynamic processes

Process	Feature
Isothermal	Temperature constant
Isobaric	Pressure constant
Isochoric	Volume constant
Adiabatic	No heat flow between the system and the surroundings ( $\Delta Q = 0$ )

1. Isothermal Process

Definition: A process in which the temperature of the system remains constant throughout the change ( $\Delta T = 0$ ).

For an ideal gas, internal energy depends only on temperature. Therefore, in an isothermal process:

\Delta U = 0

From the first law of thermodynamics ( $\Delta Q = \Delta U + \Delta W$ ), this gives:

\Delta Q = \Delta W

Important

In an isothermal process for an ideal gas, the heat supplied to the system is entirely converted into work done by the system (or, if work is done on the system, an equal amount of heat is released).

Work done in an isothermal process (ideal gas):

We start from the definition of work:

\Delta W = \int_{V_i}^{V_f} P \, dV

For an ideal gas, $P = \frac{nRT}{V}$ . Since $T$ is constant, $nRT$ is a constant:

\Delta W = \int_{V_i}^{V_f} \frac{nRT}{V} \, dV = nRT \int_{V_i}^{V_f} \frac{dV}{V}

\Delta W = nRT \, \ln \left( \frac{V_f}{V_i} \right)

Since $P_i V_i = P_f V_f$ (Boyle's law), we can also write:

\frac{V_f}{V_i} = \frac{P_i}{P_f}

Thus:

\Delta W = nRT \, \ln \left( \frac{P_i}{P_f} \right)

\Delta W_{\text{isothermal}} = nRT \, \ln \left( \frac{V_f}{V_i} \right) = nRT \, \ln \left( \frac{P_i}{P_f} \right)

Graphical representation: On a $P$ – $V$ diagram, an isothermal process for an ideal gas follows a rectangular hyperbola ( $PV = \text{constant}$ ). This curve is called an isotherm. The work done is the area under the curve.

Watch out

A common mistake is to think isothermal means "slow". While a slow process allows heat exchange to keep temperature constant, the defining feature is $\Delta T = 0$ , not the speed. A process can be isothermal only if the system is in thermal contact with a large reservoir.

2. Adiabatic Process

Definition: A process in which no heat is exchanged between the system and its surroundings ( $\Delta Q = 0$ ).

From the first law:

0 = \Delta U + \Delta W \quad \Rightarrow \quad \Delta W = -\Delta U

Important

In an adiabatic process, any work done by the system comes at the cost of its internal energy (the gas cools down). Any work done on the system increases its internal energy (the gas heats up).

Adiabatic relation for an ideal gas:

For an ideal gas, $dU = n C_V dT$ . The first law in differential form is:

dQ = dU + P dV = 0

n C_V dT + P dV = 0

Using the ideal gas law $P = \frac{nRT}{V}$ :

n C_V dT + \frac{nRT}{V} dV = 0

Divide through by $nT$ :

C_V \frac{dT}{T} + R \frac{dV}{V} = 0

Recall that $C_P - C_V = R$ for an ideal gas. Define the ratio of specific heats $\gamma = \frac{C_P}{C_V}$ . Then $R = C_P - C_V = C_V (\gamma - 1)$ . Substituting:

C_V \frac{dT}{T} + C_V (\gamma - 1) \frac{dV}{V} = 0

\frac{dT}{T} + (\gamma - 1) \frac{dV}{V} = 0

Integrating:

\ln T + (\gamma - 1) \ln V = \text{constant}

\ln (T V^{\gamma - 1}) = \text{constant}

T V^{\gamma - 1} = \text{constant}

Using the ideal gas law $T = \frac{PV}{nR}$ , we get:

\frac{PV}{nR} V^{\gamma - 1} = \text{constant} \quad \Rightarrow \quad P V^{\gamma} = \text{constant}

Similarly, eliminating $V$ gives:

T^{\gamma} P^{1 - \gamma} = \text{constant}

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

Work done in an adiabatic process (ideal gas):

Since $\Delta Q = 0$ , $\Delta W = -\Delta U = -n C_V \Delta T = n C_V (T_i - T_f)$ .

We can also derive it from $P V^{\gamma} = \text{constant} = K$ . Then $P = \frac{K}{V^{\gamma}}$ .

\Delta W = \int_{V_i}^{V_f} P \, dV = \int_{V_i}^{V_f} \frac{K}{V^{\gamma}} \, dV = K \left[ \frac{V^{1 - \gamma}}{1 - \gamma} \right]_{V_i}^{V_f}

\Delta W = \frac{K}{1 - \gamma} \left( V_f^{1 - \gamma} - V_i^{1 - \gamma} \right)

Since $K = P_i V_i^{\gamma} = P_f V_f^{\gamma}$ , we can write:

\Delta W = \frac{1}{1 - \gamma} \left( P_f V_f^{\gamma} V_f^{1 - \gamma} - P_i V_i^{\gamma} V_i^{1 - \gamma} \right) = \frac{1}{1 - \gamma} \left( P_f V_f - P_i V_i \right)

\Delta W_{\text{adiabatic}} = \frac{P_f V_f - P_i V_i}{1 - \gamma} = \frac{nR (T_f - T_i)}{1 - \gamma} = n C_V (T_i - T_f)

Graphical representation: On a $P$ – $V$ diagram, an adiabatic curve ( $PV^{\gamma} = \text{constant}$ ) is steeper than an isotherm ( $PV = \text{constant}$ ) because $\gamma > 1$ .

Note

| Feature | Isothermal | Adiabatic |

| :--- | :--- | :--- |

| Condition | $\Delta T = 0$ | $\Delta Q = 0$ |

| $P$ – $V$ relation | $PV = \text{const}$ | $PV^{\gamma} = \text{const}$ |

| Slope ( $\frac{dP}{dV}$ ) | $-\frac{P}{V}$ | $-\gamma \frac{P}{V}$ (steeper) |

| $\Delta U$ | $0$ | $n C_V \Delta T$ |

| $\Delta Q$ | $\Delta W$ | $0$ |

3. Isochoric Process

Definition: A process in which the volume of the system remains constant ( $\Delta V = 0$ ).

Since volume does not change, no work is done:

\Delta W = 0

From the first law:

\Delta Q = \Delta U

Important

In an isochoric process, all heat added to the system goes entirely into increasing its internal energy (and thus its temperature). No work is done.

For an ideal gas, $\Delta U = n C_V \Delta T$ , so:

\Delta Q = n C_V \Delta T

Graphical representation: On a $P$ – $V$ diagram, an isochoric process is a vertical line (constant volume). The area under the curve is zero, confirming zero work.

4. Isobaric Process

Definition: A process in which the pressure of the system remains constant ( $\Delta P = 0$ ).

Work done is straightforward:

\Delta W = \int_{V_i}^{V_f} P \, dV = P \int_{V_i}^{V_f} dV = P (V_f - V_i) = P \Delta V

From the first law:

$\Delta Q = \Delta U + P \Delta V$ …

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