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Thermodynamic potentials are special state functions derived from the fundamental laws of thermodynamics. They provide powerful tools to predict the behavior of systems under various constraints such as constant temperature, pressure, or volume. By using these potentials, engineers and scientists can determine equilibrium conditions, spontaneity of processes, and maximum work obtainable from a system without needing to track every microscopic detail.
Before diving into the different potentials, let's recall that a state function is a property that depends only on the current state of the system, not on the path taken to reach that state. Thermodynamic potentials are state functions that combine energy, entropy, pressure, volume, and temperature in specific ways to suit different practical scenarios.
Imagine you want to know if a chemical reaction will proceed spontaneously at room temperature and atmospheric pressure. Or, you want to calculate the maximum useful work from a heat engine operating between two temperatures. Thermodynamic potentials provide the framework to answer such questions efficiently.
Each potential has natural variables-the variables that are easiest to control or measure in a given process. Understanding these natural variables helps in selecting the right potential for a problem.
Internal energy is the total energy contained within a system due to the microscopic motion and interactions of its molecules. It includes kinetic energy of molecules, potential energy from molecular forces, chemical energy, and other microscopic forms.
Mathematically, internal energy is expressed as a function of entropy \(S\) and volume \(V\):
This means if you know the entropy and volume of a system, you can determine its internal energy. The differential form of internal energy, derived from the first law of thermodynamics, is:
This equation tells us that the internal energy increases if entropy increases (heat added) or decreases if volume increases (work done by the system).
Enthalpy is defined as the sum of internal energy and the product of pressure and volume:
Enthalpy is especially useful for processes occurring at constant pressure, such as many chemical reactions and heating/cooling in open systems like boilers and heat exchangers.
Its natural variables are entropy \(S\) and pressure \(P\), expressed as:
The differential form is:
The Helmholtz free energy is defined as:
Helmholtz free energy is most useful for systems held at constant temperature and volume, such as in many statistical mechanics problems and isothermal processes in rigid containers.
Its natural variables are temperature \(T\) and volume \(V\):
The differential form is:
The Gibbs free energy is defined as:
Gibbs free energy is the most widely used potential in chemical engineering and thermodynamics because many processes occur at constant temperature and pressure (e.g., atmospheric conditions).
Its natural variables are temperature \(T\) and pressure \(P\):
The differential form is:
Maxwell relations are a set of equations derived from the symmetry of second derivatives of thermodynamic potentials. They link different partial derivatives of thermodynamic properties, allowing us to express difficult-to-measure quantities in terms of more accessible ones.
These relations arise because the mixed second derivatives of state functions are equal, for example:
\[\frac{\partial^2 U}{\partial S \partial V} = \frac{\partial^2 U}{\partial V \partial S}\]Here is a summary table of Maxwell relations for each potential:
| Potential | Natural Variables | Maxwell Relation |
|---|---|---|
| Internal Energy \(U(S,V)\) | Entropy \(S\), Volume \(V\) | \(\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V\) |
| Enthalpy \(H(S,P)\) | Entropy \(S\), Pressure \(P\) | \(\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P\) |
| Helmholtz Free Energy \(A(T,V)\) | Temperature \(T\), Volume \(V\) | \(\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V\) |
| Gibbs Free Energy \(G(T,P)\) | Temperature \(T\), Pressure \(P\) | \(\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P\) |
These relations are invaluable for simplifying thermodynamic calculations and connecting measurable properties like pressure, volume, temperature, and entropy.
Step 1: Convert all units to SI. Pressure is 1 atm = 101325 Pa, temperature \(T = 373 \, K\), latent heat \(L = 2257 \times 10^3 \, J/kg\).
Step 2: Calculate entropy change during vaporization:
\[ \Delta S = \frac{L}{T} = \frac{2257 \times 10^3}{373} = 6053 \, J/(kg \cdot K) \]
Step 3: Calculate Gibbs free energy change:
At equilibrium phase change, \(\Delta G = 0\). To verify, use:
\[ \Delta G = \Delta H - T \Delta S = L - T \times \frac{L}{T} = 0 \]
This confirms the phase change occurs at equilibrium at 373 K and 1 atm.
Step 4: Calculate energy cost for vaporizing 1 kg of water:
Energy required = \(L = 2257 \, kJ = 0.627 \, kWh\) (since 1 kWh = 3600 kJ)
Cost = \(0.627 \times 5 = 3.14\) INR
Answer: The Gibbs free energy change is zero at boiling point, confirming equilibrium. The energy cost to vaporize 1 kg of water is approximately INR 3.14.
Step 1: For isothermal process, the maximum work done by the system equals the decrease in Helmholtz free energy:
\[ W_{\max} = -\Delta A \]
Step 2: For an ideal gas, Helmholtz free energy is:
\[ A = -nRT \ln V + \text{constant} \]
Step 3: Number of moles \(n\) can be found using ideal gas law at initial state assuming pressure \(P = 1 \, atm = 101325 \, Pa\):
\[ n = \frac{PV}{RT} = \frac{101325 \times 0.01}{8.314 \times 300} \approx 0.406 \, \text{mol} \]
Step 4: Calculate change in Helmholtz free energy:
\[ \Delta A = A_2 - A_1 = -nRT \ln \frac{V_2}{V_1} = -0.406 \times 8.314 \times 300 \times \ln \frac{0.02}{0.01} \]
\[ = -0.406 \times 8.314 \times 300 \times \ln 2 = -0.406 \times 8.314 \times 300 \times 0.693 = -702 \, J \]
Step 5: Maximum work output is:
\[ W_{\max} = -\Delta A = 702 \, J \]
Answer: The maximum work obtainable from the isothermal expansion is approximately 702 J.
Step 1: Recall the Maxwell relation from Gibbs free energy \(G(T,P)\):
\[ \left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P \]
This relation allows us to express the change in entropy with pressure at constant temperature in terms of the change in volume with temperature at constant pressure.
Step 2: This is useful because volume and temperature are often easier to measure experimentally.
Answer: \(\boxed{\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P}\)
Step 1: For constant pressure heating, enthalpy change is:
\[ \Delta H = m C_p \Delta T \]
Step 2: Substitute values:
\[ \Delta H = 1 \times 1.005 \times (400 - 300) = 1.005 \times 100 = 100.5 \, kJ \]
Answer: The enthalpy change is 100.5 kJ.
Step 1: For constant volume, internal energy change is:
\[ \Delta U = m C_v \Delta T \]
Step 2: Substitute values:
\[ \Delta U = 2 \times 0.718 \times (350 - 300) = 2 \times 0.718 \times 50 = 71.8 \, kJ \]
Answer: The internal energy increases by 71.8 kJ.
When to use: When deciding which thermodynamic potential applies to a given process or constraint.
When to use: When analyzing spontaneity and equilibrium in chemical and phase change problems.
When to use: When dealing with isothermal processes in closed, rigid containers.
When to use: When partial derivatives of thermodynamic properties are required but not directly available.
When to use: During calculations to avoid unit conversion errors.
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