CBSE Class 11 Chemistry Thermodynamics Chapter Notes

Introduction to Thermodynamics: Systems and Surroundings

Thermodynamics is a branch of science that deals with the relationship between heat and other forms of energy. It helps predict the feasibility and direction of chemical reactions and physical changes.

A system refers to the part of the universe under thermodynamic investigation. The surroundings constitute everything else in the universe outside the system. A boundary separates the system from its surroundings.

Systems can be classified based on matter and energy exchange:

• Open System: Exchanges both energy and matter with the surroundings (e.g., an open cup of coffee).
• Closed System: Exchanges energy but not matter with the surroundings (e.g., a sealed bottle of hot water).
• Isolated System: Exchanges neither energy nor matter with the surroundings (e.g., a perfectly insulated thermos flask).

State Functions are properties whose value depends only on the current state of the system, irrespective of how that state was reached. Examples include Pressure (P), Volume (V), Temperature (T), Internal Energy (U), Enthalpy (H), Entropy (S), and Gibbs Free Energy (G). Changes in state functions (ΔX) depend only on the initial and final states. Path Functions, on the other hand, depend on the path taken to reach a particular state. Heat (q) and Work (w) are classic examples of path functions.

Key Thermodynamic Definitions

Internal Energy (U)

The total energy contained within a thermodynamic system, including kinetic and potential energy of its constituent particles. It is a state function.

Enthalpy (H)

A thermodynamic property of a system, equal to the internal energy plus the product of pressure and volume (H = U + PV). It represents the heat content at constant pressure and is a state function.

Entropy (S)

A measure of the randomness or disorder of a system. The greater the disorder, the higher the entropy. It is a state function.

Gibbs Free Energy (G)

A thermodynamic potential that measures the 'useful' or process-initiating work obtainable from an isothermal, isobaric thermodynamic system. It determines the spontaneity of a process (G = H - TS).

Spontaneous Process

A process that occurs on its own without any external intervention. It proceeds in a definite direction towards equilibrium.

First Law of Thermodynamics

States that energy can neither be created nor destroyed, only converted from one form to another. Mathematically, ΔU = q + w.

Second Law of Thermodynamics

States that for a spontaneous process, the total entropy of the universe (system + surroundings) must increase (ΔS_total > 0).

Must Remember: Key Formulas and Principles

• First Law of Thermodynamics: ΔU = q + w (where ΔU = change in internal energy, q = heat, w = work).
• Work done (w) by gas expansion: w = -P_extΔV (for irreversible process) and w = -2.303 nRT log(V2/V1) (for isothermal reversible expansion).
• Enthalpy Change (ΔH): ΔH = ΔU + PΔV (at constant pressure). For reactions involving gases: ΔH = ΔU + Δn_gRT.
• Hess's Law of Constant Heat Summation: The total enthalpy change for a reaction is the same whether the reaction occurs in one step or in several steps.
• Entropy Change (ΔS): ΔS = q_rev/T. For a spontaneous process, ΔS_total = ΔS_system + ΔS_surroundings > 0.
• Gibbs Free Energy (ΔG): ΔG = ΔH - TΔS. This is the primary criterion for spontaneity at constant temperature and pressure.
• Spontaneity Criteria based on ΔG: ΔG < 0: Process is spontaneous. ΔG > 0: Process is non-spontaneous (reverse is spontaneous). * ΔG = 0: Process is at equilibrium.
• Relationship between ΔG° and Equilibrium Constant (K): ΔG° = -RTlnK, or K = e^(-ΔG°/RT).

Comparison of Thermodynamic Processes

Solved Example: Calculating Work Done

• {"title":"Example 1: Work done in Isothermal Reversible Expansion","bodyMarkdown":"Question: 1 mole of an ideal gas expands isothermally and reversibly from 10 L to 20 L at 298 K. Calculate the work done.\n\nSolution:\nGiven: n = 1 mole, V1 = 10 L, V2 = 20 L, T = 298 K.\nFor isothermal reversible expansion, w = -2.303 nRT log(V2/V1)\nSubstituting the values:\nw = -2.303 × 1 mol × 8.314 J K⁻¹ mol⁻¹ × 298 K × log(20 L / 10 L)\nw = -2.303 × 1 × 8.314 × 298 × log(2)\nw = -2.303 × 1 × 8.314 × 298 × 0.3010\nw ≈ -1718.4 J\n\nAnswer: The work done by the gas is -1718.4 J (negative sign indicates work done by the system)."}

Exam Tip: Sign Conventions are Crucial!

A common mistake in Thermodynamics is getting the signs wrong for heat (q) and work (w). Always remember:

• Heat (q):
• q > 0 (positive): Heat absorbed by the system from the surroundings (endothermic).
• q < 0 (negative): Heat released by the system to the surroundings (exothermic).

• Work (w):
• w > 0 (positive): Work done on the system by the surroundings (compression).
• w < 0 (negative): Work done by the system on the surroundings (expansion).

Consistently applying these conventions will help avoid errors in calculations involving the First Law of Thermodynamics and other energy changes. Pay close attention to keywords like 'absorbed', 'released', 'by the system', 'on the system'.

Practice Questions with Solutions

• Q: Distinguish between an extensive property and an intensive property with one example each. A: An extensive property depends on the amount of matter (e.g., mass, volume, internal energy). An intensive property is independent of the amount of matter (e.g., temperature, pressure, density).
• Q: Under what conditions is ΔH equal to ΔU? A: ΔH = ΔU under two main conditions: 1) When the process occurs at constant volume (ΔV=0), so PΔV=0. 2) When there is no change in the number of moles of gaseous reactants and products (Δn_g = 0).
• Q: What is the significance of a negative Gibbs Free Energy change (ΔG < 0) for a chemical reaction? A: A negative ΔG indicates that the reaction is spontaneous under the given conditions of temperature and pressure, meaning it will proceed in the forward direction to form products.
• Q: State the Third Law of Thermodynamics. A: The Third Law of Thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero (0 Kelvin) is zero. This provides a reference point for absolute entropy values.

Frequently Asked Questions