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1.1. Course Introduction

See also: Course-wide Conventions & Notation

Overview and Learning Objectives

Thermodynamics and statistical mechanics connect microscopic models of matter to macroscopic observables such as temperature, pressure, and free energy. This section introduces course conventions, the language of systems and surroundings, and the role of state variables in equilibrium descriptions. It also reviews common forms of energy, energy transfer, and the unit conventions used throughout the course.

This course develops the core concepts of thermodynamics and statistical mechanics and applies them to chemical systems. It builds on the thermodynamic principles introduced in Chem 106 and Chem 112, and on the quantum mechanics developed in Chem 105, Chem 111, and Chem 401. By connecting molecular-level behavior to macroscopic thermodynamic observations, you will see how theory underpins real-world chemical processes.


Learning objectives:

Core Ideas and Derivations

Why Study Thermodynamics and Statistical Mechanics?

Bridging the Microscopic and Macroscopic Worlds

Typical chemical systems contain on the order of Avogadro’s number of particles (i.e., NA=6.022×1023N_{\mathrm{A}} = 6.022 \times 10^{23}). Thermodynamics abstracts this complexity into a framework for predicting, among other things,

Modern Applications

Thermodynamics and statistical mechanics are central to diverse fields, including industrial chemistry, materials science, and biochemistry. The examples below illustrate a few representative applications:

Industrial Chemistry

Predicting reaction spontaneity and modeling large-scale chemical processes.

Hydrogen production via steam–methane reforming.

Figure 1:Hydrogen production via steam–methane reforming.[1]

Materials Chemistry

Designing advanced materials for optical, electronic, or mechanical applications.

Electricity generation using solar panels.

Figure 2:Electricity generation using solar panels.[2]

Biochemistry

Investigating protein folding and enzyme activity in biological systems.

Predicting protein structures with AlphaFold.

Figure 3:Predicting protein structures with AlphaFold.[3]


Key Definitions

Thermodynamic Systems

Source
import matplotlib.pyplot as plt
import matplotlib.patches as mpatches
from myst_nb import glue

# Helper function to plot a system
def plot_system(ax, title, annotations, boundary_color='b'):
    box = mpatches.FancyBboxPatch((0, 0), 1, 1, boxstyle='roundtooth', ec=boundary_color, fc='w')
    ax.add_patch(box)
    ax.set_title(title, fontsize=14)
    ax.text(0.5, 0.5, 'System', ha='center', va='center', fontsize=12)
    ax.text(0.5, -0.65, 'Surroundings', ha='center', va='center', fontsize=12)
    ax.text(0.5, 1.3, 'Boundary', ha='center', va='bottom', fontsize=12, color=boundary_color)
    for annotation in annotations:
        if "arrowprops" in annotation:  # Arrow annotations
            ax.annotate('', **annotation)
        else:  # Text annotations
            ax.text(**annotation)
    ax.set_xlim(-1, 2)
    ax.set_ylim(-1, 2)
    ax.set_aspect('equal')
    ax.axis('off')

# Define annotations for each system
annotations = [
    [],  # Isolated system (no arrows)
    [  # Closed system (energy arrow)
        dict(xy=(-0.6, 0.15), xytext=(0.15, 0.15), arrowprops=dict(arrowstyle='<->', color='r')),
        dict(x=-1, y=0.3, s='Energy', ha='left', va='bottom', fontsize=12, color='r'),
    ],
    [  # Open system (energy + matter arrows)
        dict(xy=(-0.6, 0.15), xytext=(0.15, 0.15), arrowprops=dict(arrowstyle='<->', color='r')),
        dict(xy=(0.85, 0.15), xytext=(1.6, 0.15), arrowprops=dict(arrowstyle='<->', color='m')),
        dict(x=-1, y=0.3, s='Energy', ha='left', va='bottom', fontsize=12, color='r'),
        dict(x=2, y=0.3, s='Matter', ha='right', va='bottom', fontsize=12, color='m'),
    ],
]

titles = ["Isolated system", "Closed system", "Open system"]

fig, axes = plt.subplots(1, 3, figsize=(12, 4))
for i, ax in enumerate(axes):
    plot_system(ax, titles[i], annotations[i])

plt.show()
plt.close(fig)
<Figure size 1200x400 with 3 Axes>

Types of thermodynamic systems. (a) Isolated—no exchange of energy or matter; (b) Closed—exchanges energy but not matter; (c) Open—exchanges both energy and matter.

System
The portion of the universe chosen for study, separated from its surroundings by a boundary.
Surroundings
Everything external to the system that can exchange energy or matter with it.
Boundary
The interface separating a system from its surroundings.
Isolated system
Exchanges neither energy nor matter with its surroundings.
Closed system
Exchanges energy but not matter with its surroundings.
Open system
Exchanges both energy and matter with its surroundings.

State of a System

Particle
A microscopic entity such as an atom, molecule, or ion.
Microscopic state (classical)
Positions and momenta of all particles in the system.
Microscopic state (quantum)
The wavefunction describing the system’s particles.
Equilibrium
A condition in which macroscopic properties remain constant over time.
Thermodynamic equilibrium
Simultaneous mechanical, thermal, and chemical equilibrium.
Thermodynamic state
A set of macroscopic variables defining a system in equilibrium.
State variable
A property that defines a system’s state.
State function
A property depending only on the system’s state, not on the path taken.
Equation of state
A mathematical relationship among state variables.
Path function
A property depending on the process or path taken between states.
Process
A transformation changing a system from one state to another.
Extensive property
A property proportional to system size (e.g., volume, entropy).
Intensive property
A property independent of system size (e.g., temperature, pressure).

Basic Forms of Energy and Energy Transfer

Energy

Kinetic energy
Energy due to motion (e.g., a moving particle).
Potential energy
Energy due to position or configuration (e.g., a stretched spring).
Internal energy
The total microscopic kinetic and potential energy of a system, averaged over its microstates.

Energy Transfer

Work
Energy transferred when a force acts over a distance (e.g., lifting a mass).
Heat
Energy transferred because of a temperature difference (e.g., conduction from hot to cold).

Important Units

SI Units[4]

Table 1:Base SI Units

Quantity

Unit

Symbol

Time

second

s

Length

meter

m

Mass

kilogram

kg

Temperature

kelvin

K

Table 2:Derived SI Units

Quantity

Unit

Symbol

Conversion

Frequency

hertz

Hz

1 Hz=1 s−11 \,\text{Hz} = 1 \,\text{s}^{-1}

Force

newton

N

1 N=1 kg m s−21 \,\text{N} = 1 \,\text{kg m s}^{-2}

Pressure

pascal

Pa

1 Pa=1 N m−21 \,\text{Pa} = 1 \,\text{N m}^{-2}

Energy

joule

J

1 J=1 N m1 \,\text{J} = 1 \,\text{N m}

Non-SI Units[5][6][7]

Table 3:Non-SI Units

Quantity

Unit

Symbol

Conversion

Pressure

bar

bar

1 bar=1×105 Pa1 \,\text{bar} = 1 \times 10^5 \,\text{Pa}

Pressure

atmosphere

atm

1 atm≈1.01325 bar1 \,\text{atm} \approx 1.01325 \,\text{bar}

Pressure

torr

torr

1 torr=1760 atm1 \,\text{torr} = \frac{1}{760}\,\text{atm}

Pressure

millimeters of mercury

mmHg

1 mmHg=1 torr1 \,\text{mmHg} = 1 \,\text{torr}

Energy

electronvolt

eV

1 eV=1.602×10−19 J1 \,\text{eV} = 1.602 \times 10^{-19} \,\text{J}

Energy

calorie

cal

1 cal=4.184 J1 \,\text{cal} = 4.184 \,\text{J}

Worked Example

Unit conversions you will use constantly

A gas sample has pressure P=1.50 atmP=1.50\ \mathrm{atm} at T=25.0∘CT=25.0^{\circ}\mathrm{C}. Convert PP to bar and Pa, and convert TT to kelvin.

Assumptions. Use 1 atm=1.01325×105 Pa1\ \mathrm{atm}=1.01325\times 10^{5}\ \mathrm{Pa} and 1 bar=105 Pa1\ \mathrm{bar}=10^{5}\ \mathrm{Pa}.

  1. Temperature to kelvin

    T(K)=T(∘C)+273.15⇒T=25.0+273.15=298.15 K.T(\mathrm{K}) = T(^{\circ}\mathrm{C}) + 273.15 \quad\Rightarrow\quad T = 25.0 + 273.15 = 298.15\ \mathrm{K}.
  2. Pressure to pascals

    P=(1.50 atm)(1.01325×105 Paatm)=1.52×105 Pa.P = (1.50\ \mathrm{atm})\left(1.01325\times 10^{5}\ \frac{\mathrm{Pa}}{\mathrm{atm}}\right) = 1.52\times 10^{5}\ \mathrm{Pa}.
  3. Pressure to bar

    P=1.52×105 Pa105 Pa/bar=1.52 bar.P = \frac{1.52\times 10^{5}\ \mathrm{Pa}}{10^{5}\ \mathrm{Pa/bar}} = 1.52\ \mathrm{bar}.

Result. T=298.15 KT=298.15\ \mathrm{K}, P=1.52×105 Pa=1.52 barP=1.52\times10^{5}\ \mathrm{Pa}=1.52\ \mathrm{bar}.

Concept Checks

  1. Why does thermodynamics emphasize state functions rather than path-dependent quantities?

  2. Give one real-world example each of an isolated, closed, and open system. What crosses the boundary in each case?

  3. Which variables would you choose as “independent” to describe a gas in a rigid, sealed container? In a piston open to the atmosphere?

  4. Why is dimensional analysis a useful error-checking tool in thermodynamic derivations?

Key Takeaways

Footnotes