B4-1. Thermodynamic Systems and The First LawHL Only

1. Core Classifications of Thermodynamic Systems

In thermal physics, a system represents the specific region of space or quantity of matter under investigation, separated from the rest of the universe (the surroundings) by a real or imaginary boundary. Systems are strictly classified by their capacity to exchange energy and matter:

  • Open Systems: Can freely exchange both matter and energy across their boundaries with the surroundings (e.g., an open beaker of boiling water).
  • Closed Systems: Can exchange energy in the forms of heat and work, but the mass of the working substance remains strictly trapped inside (e.g., gas enclosed in a sealed piston-cylinder assembly).
  • Isolated Systems: Completely incapable of exchanging either matter or energy with their environment (e.g., an ideal, perfectly sealed vacuum flask, or the universe taken as a whole).

2. Microscopic Definition of Internal Energy

The total internal energy ($U$) of a substance is the macroscopic sum of the microscopic kinetic energies ($E_K$) and microscopic potential energies ($E_P$) of all its constituent particles. For a monatomic ideal gas, we assume that intermolecular forces are entirely non-existent. Consequently, the particles possess zero potential energy ($E_P = 0$), meaning the internal energy is purely a function of temperature.

Internal Energy Equations:

$$U = \dfrac{3}{2}nRT = \dfrac{3}{2}Nk_BT$$

When a system undergoes a change in temperature ($\Delta T$), its change in internal energy ($\Delta U$) is directly proportional:

$$\Delta U = \dfrac{3}{2}nR\Delta T = \dfrac{3}{2}Nk_B\Delta T$$
  • $\Delta U$: Change in internal energy in Joules (J)
  • $n$: Amount of substance in moles (mol)
  • $R$: Universal gas constant ($8.31$ J K⁻¹ mol⁻¹)
  • $N$: Total number of gas particles
  • $k_B$: Boltzmann constant ($1.38 \times 10^{-23}$ J K⁻¹)
  • $\Delta T$: Change in absolute temperature in Kelvin (K)

3. Boundary Work Done by a Gas

When a gas expands or contracts within a closed boundary (such as a piston moving through a cylinder of cross-sectional area $A$), it performs mechanical work ($W$) on its surroundings via pressure forces.

Work Formula at Constant Pressure:

$$W = p\Delta V$$
  • $W$: Work done by or on the gas in Joules (J)
  • $p$: Constant pressure of the gas in Pascals (Pa)
  • $\Delta V$: Change in volume ($V_{\text{final}} - V_{\text{initial}}$) in cubic metres (m³)

The Strict Sign Convention:

  • Expansion ($\Delta V > 0$): The gas performs positive work by the system on the surroundings ($W > 0$). Energy leaves the gas system as mechanical work.
  • Compression ($\Delta V < 0$): The surroundings perform work on the system ($W < 0$). Mechanical energy is injected into the gas system.

4. The First Law of Thermodynamics

The First Law is a statement of the fundamental principle of the conservation of energy applied directly to thermal and mechanical systems. It dictates that any energy delivered to a system must balance across internal kinetic changes and mechanical outputs.

The First Law Equation:

$$Q = \Delta U + W$$
  • $Q$: Net thermal energy transferred into the system by heating in Joules (J). If heat is removed from the system, $Q < 0$.
  • $\Delta U$: Net change in internal energy in Joules (J).
  • $W$: Work done by the system in Joules (J).

Example 1: Advanced First Law Application

Problem: A closed cylinder contains $0.40$ mol of a monatomic ideal gas. The gas is compressed at a constant external pressure of $1.50 \times 10^5$ Pa, causing its volume to decrease from $8.0 \times 10^{-3}$ m³ to $3.5 \times 10^{-3}$ m³. Simultaneously, the gas is cooled, losing $950$ J of thermal energy to the surroundings. Calculate the final change in internal energy ($\Delta U$) and determine if the final temperature of the gas increases or decreases.

Solution:

  • Step 1: Calculate the work done ($W$).
    $$\Delta V = V_f - V_i = (3.5 \times 10^{-3}) - (8.0 \times 10^{-3}) = -4.5 \times 10^{-3} \text{ m³}$$ $$W = p\Delta V = (1.50 \times 10^5 \text{ Pa}) \times (-4.5 \times 10^{-3} \text{ m³}) = \mathbf{-675 \text{ J}}$$
  • Step 2: Account for the sign of thermal energy ($Q$).
    Since thermal energy is lost to the surroundings, $Q$ must be negative:
    $$Q = \mathbf{-950 \text{ J}}$$
  • Step 3: Rearrange and apply the First Law to solve for $\Delta U$.
    $$Q = \Delta U + W \quad \Rightarrow \quad \Delta U = Q - W$$ $$\Delta U = (-950) - (-675) = -950 + 675 = \mathbf{-275 \text{ J}}$$
  • Step 4: Analyze the temperature trend.
    Since $\Delta U = \dfrac{3}{2}nR\Delta T$ and our calculated $\Delta U < 0$, the absolute temperature of the gas must decrease.