Thermodynamic Relations

Why it matters

Thermodynamic relations are crucial for engineers to analyze and design systems involving heat and work interactions, such as engines, refrigerators, and power plants. They provide the mathematical framework to relate different thermodynamic properties, enabling the prediction of system behavior under various conditions.

Key ideas

• Thermodynamic Properties: These include pressure (P), volume (V), temperature (T), and entropy (S), among others. Understanding how these properties interrelate is essential for system analysis.
• Maxwell's Relations: Derived from the second law of thermodynamics, these equations relate different partial derivatives of thermodynamic potentials, providing a way to calculate changes in entropy, volume, and other properties.
• Gibbs and Helmholtz Functions: These are thermodynamic potentials that help in understanding the energy changes in a system. Gibbs free energy is particularly useful for processes at constant temperature and pressure.
• Clapeyron Equation: This relates the pressure, volume, and temperature changes during phase transitions, such as from liquid to vapor.

The differential identities below apply to equilibrium states of a fixed-composition simple compressible system; additional work modes or composition changes add terms. Maxwell relations follow from equality of mixed second derivatives of smooth state potentials. For example (∂S/∂V)_T = (∂P/∂T)_V. These identities are not restricted to ideal gases.

Formulas

• dU = TdS - PdV
  • dU: Change in internal energy (Joules)
  • T: Temperature (Kelvin)
  • dS: Change in entropy (Joules per Kelvin)
  • P: Pressure (Pascals)
  • dV: Change in volume (cubic meters)
• dH = TdS + VdP
  • dH: Change in enthalpy (Joules)
  • V: Volume (cubic meters)
  • dP: Change in pressure (Pascals)
• dG = VdP - SdT
  • dG: Change in Gibbs free energy (Joules)
  • SdT: Entropy-temperature term (Joules)
• dA = -PdV - SdT
  • dA: Change in Helmholtz free energy (Joules)

Worked example

Problem: Calculate the change in Gibbs free energy when 2 m³ of an ideal gas at 300 K is compressed isothermally from 100 kPa to 200 kPa.

Given:

• Initial pressure, P1 = 100 kPa = 100,000 Pa
• Final pressure, P2 = 200 kPa = 200,000 Pa
• Volume, V = 2 m³
• Temperature, T = 300 K

1. At fixed temperature, dG = V dP, but V changes during compression.
2. For the fixed amount of ideal gas, V = nRT/P and nRT = P₁V₁ = 100000(2) = 200000 J.
3. Integrate: ΔG = integral from P₁ to P₂ of (nRT/P)dP = nRT ln(P₂/P₁).
4. ΔG = 200000 ln(2) = 138629 J = 138.63 kJ.

Answer: +138.63 kJ. Multiplying initial volume by a finite pressure change incorrectly treats V as constant. Final volume is 1 m³.

Common mistakes

• Confusing the signs in the thermodynamic equations, especially in Maxwell's relations.
• Incorrect unit conversions, particularly with pressure and volume.
• Misapplying the equations to non-ideal systems without considering necessary corrections.

For GATE ME

Questions often involve deriving or using Maxwell's relations, calculating changes in thermodynamic potentials, and applying the Clapeyron equation. Practice problems involving partial derivatives and understanding the physical significance of each term in the equations.

Quick check

1. What is the significance of Gibbs free energy in thermodynamics?
2. How does the Clapeyron equation relate to phase transitions?
3. What are Maxwell's relations used for?

Answers: 1. At constant temperature and pressure, the decrease in Gibbs free energy gives maximum reversible non-expansion work under the appropriate system constraints. 2. It relates pressure, volume, and temperature changes during phase transitions. 3. They relate different partial derivatives of thermodynamic potentials, aiding in property calculations.