Thermo ICE 06FEB

Internal Energy and Enthalpy

  • Definition of Internal Energy: Internal energy (U) is the total energy contained within a system due to both its molecular motion and interactions between molecules. It can be expressed as a function of temperature for an ideal gas.
    • Internal energy changes are determined by changes in temperature.
  • Enthalpy (H) is similarly defined as a function of temperature.
    • For an ideal gas, variations in enthalpy can be calculated from initial and final temperatures.

Important Relationships

  • Key relationship discussed:
    C<em>P=C</em>V+RC<em>P = C</em>V + R
    where:

    • $C_P$ = heat capacity at constant pressure
    • $C_V$ = heat capacity at constant volume
    • $R$ = universal gas constant

  • The relationship for isothermal processes:
    Q=−WQ = -W

    • The heat added to a system (Q) is equal to the negative of the work done by the system (-W).
    • This holds true for isothermal processes (constant temperature).

  • For a reversible process at constant pressure, replace $dW$ with $pdV$:

    • Using the ideal gas law:
    • P dV = rac{RT}{V}

Isothermal and Isochoric Processes

  • In isothermal processes, the internal energy change ($ riangle U$) remains zero because temperature is constant.
  • In isochoric processes (constant volume):
    • Work done is always zero: W=0W = 0
    • Therefore, heat added is equal to the change in internal energy:
      Q=riangleUQ = riangle U
  • For an ideal gas, knowing the initial and final temperatures is sufficient to determine internal energy and enthalpy changes.

Adiabatic Processes

  • Characteristics: Adiabatic processes mean that there is no heat exchange with the environment ($ riangle Q = 0$).
  • Equation for adiabatic processes includes the adiabatic expansion coefficient:
    extwhere<br/>eqC<em>P/C</em>Vext{where} <br /> eq C<em>P/C</em>V
  • Key relationships established for ideal gas state based on $P$, $V$, and $T$

Example Problem - Ideal Gas Pathways

  • Discussed a problem involving an ideal gas transitioning between states with paths including:
    • Isochoric compression (Path A)
    • Adiabatic compression followed by isochoric cooling (Path B)

Path A: Isochoric Compression

  • Initial state ($P1$, $V1$) changes to final state ($P2$, $V2$).
  • Work done is zero; the change in internal energy ($ riangle U$) equals Q.

Path B: Adiabatic Compression followed by Isochoric Cooling

  • An intermediate state B is established before reaching the final state.
    • Temperature changes (affecting internal energy and enthalpy).
  • Work ($W$) needs to be found analytically using different approaches (i.e., first law of thermodynamics).

First Law of Thermodynamics

  • Fundamental equation:
    riangleU=Q+Wriangle U = Q + W
  • Applicable to both adiabatic and isochoric processes. If no heat exchange occurs, the work leads directly to a change in internal energy.

Summary of Steps for Calculation

  • For each pathway, calculate
    • $Q$, $W$, $ riangle U$, and $ riangle H$.
  • Keep in mind that for ideal gas scenarios, internal energy and enthalpy are state functions and dependent on temperature alone.

Practical Implications and Applications

  • Understanding the relationships between work, heat, and changes in state allows for predicting how gases will behave under various thermodynamic processes.
  • Different pathways lead to varying energies, but state functions indicate that internal energy and enthalpy changes only depend on initial and final states, not the path taken.

Review and Practice

  • Revisiting derivations and pathways during office hours may reinforce concepts. Be ready to solve similar numerical problems using ideal gas equations and thermodynamic relationships.

Homework Assignments

  • Complete assigned homework focusing on the questions' thermodynamic principles and calculations.
  • Prepare for the upcoming exam, ensuring mastery over fundamental laws and concepts of thermodynamics in ideal gas states.