CBSE Class 11 Physics Chapter Notes: Thermodynamics

Essential Thermodynamic Definitions

Thermodynamic System

A collection of matter or a region of space chosen for study, separated from its surroundings by a boundary.

Surroundings

Everything external to the thermodynamic system with which it can exchange energy or matter.

Boundary

The real or imaginary surface that separates the system from its surroundings. Can be fixed or movable, permeable or impermeable.

State Variables

Macroscopic properties (like pressure P, volume V, temperature T, internal energy U) that describe the equilibrium state of a thermodynamic system.

State Function

A property whose value depends only on the state of the system, not on the path taken to reach that state (e.g., U, P, V, T). Heat (Q) and Work (W) are NOT state functions.

Thermodynamic Process

The path or sequence of states through which a system changes from one equilibrium state to another.

Internal Energy (U)

The total energy contained within a thermodynamic system, including kinetic and potential energies of its constituent particles. For an ideal gas, it depends only on temperature.

Heat (Q)

Energy transferred between a system and its surroundings due to a temperature difference.

Work (W)

Energy transferred between a system and its surroundings by means other than temperature difference, typically due to a change in volume (P-V work).

Thermodynamic Systems and Equilibrium

In thermodynamics, understanding the system, surroundings, and boundary is crucial. A system can be classified into three types:

1. Open System: Exchanges both matter and energy with its surroundings. Example: Boiling water in an open container.
2. Closed System: Exchanges energy but not matter with its surroundings. Example: Boiling water in a sealed container.
3. Isolated System: Exchanges neither matter nor energy with its surroundings. Example: An ideal thermos flask.

The state of a thermodynamic system is described by its state variables such as pressure (P), volume (V), and temperature (T). These variables are macroscopic properties that can be measured. When these variables do not change over time, and there are no macroscopic flows of matter or energy within the system, the system is said to be in thermodynamic equilibrium. This implies thermal, mechanical, and chemical equilibrium. Any change in these state variables leads to a thermodynamic process.

Internal Energy (U) is a key state function. For an ideal gas, internal energy depends only on its temperature, meaning ΔU = 0 for an isothermal process in an ideal gas. Heat (Q) and Work (W) are path functions; their values depend on the specific process connecting the initial and final states, not just the states themselves. This distinction is vital for applying the First Law of Thermodynamics correctly. The sign conventions for Q and W are critical: Heat absorbed by the system (Q > 0), Heat given out by the system (Q < 0). Work done by the system (W > 0) (expansion), Work done on the system (W < 0) (compression).

The Laws of Thermodynamics

Key Thermodynamic Processes

Worked Example: Work Done by an Ideal Gas

• {"title":"Problem","bodyMarkdown":"An ideal gas expands isothermally from an initial volume of 2 L to 4 L at a constant temperature of 300 K. If the initial pressure is 10 atm, calculate the work done by the gas during expansion. (Given R = 8.314 J/mol·K and 1 L·atm = 101.3 J)"}
• {"title":"Solution","bodyMarkdown":"1. Identify the process: Isothermal expansion (T = constant).\n2. Formula for work done: For an isothermal process, W = nRT ln(Vf/Vi).\n3. Find number of moles (n): Using PV = nRT, n = PV/(RT). P = 10 atm, V = 2 L, T = 300 K. We need R in L·atm/mol·K. Use R ≈ 0.0821 L·atm/mol·K for consistency with P in atm and V in L, or convert everything to SI units. Let's use SI units for final calculation with given R in J/mol.K. P = 10 101325 Pa = 1013250 Pa. V = 2 10^-3 m^3. n = (1013250 Pa 2 10^-3 m^3) / (8.314 J/mol·K 300 K) ≈ 0.811 mol.\n4. Calculate W: W = (0.811 mol) (8.314 J/mol·K) (300 K) ln(4 L / 2 L)\n W = 0.811 8.314 300 ln(2)\n W ≈ 0.811 8.314 300 0.693 ≈ 1402 J\n \n Alternatively, using P-V relationship for isothermal process: W = P_i V_i ln(V_f/V_i) = (10 atm 2 L) ln(2) = 20 L·atm ln(2) = 20 0.693 L·atm = 13.86 L·atm. Converting to Joules: 13.86 L·atm * 101.3 J/L·atm ≈ 1404 J. (Slight difference due to R value and rounding)."}

Key Points to Remember for Exams

• Sign Convention: Always be careful with the sign convention for heat (Q) and work (W) in the First Law (ΔU = Q - W). Q is positive when heat is added to the system; W is positive when work is done by the system.
• Internal Energy of Ideal Gas: For an ideal gas, internal energy (U) is a function of temperature only (U ∝ T). Hence, ΔU = 0 for an isothermal process.
• Specific Heat Capacities: Cv (at constant volume) and Cp (at constant pressure) are related by Mayer's Formula: Cp - Cv = R (for one mole of ideal gas).
• Ratio of Specific Heats (γ): γ = Cp/Cv. It's crucial for adiabatic processes (PV^γ = constant). γ > 1 for all gases (monatomic: 5/3, diatomic: 7/5).
• Work Done on P-V Diagram: The area under the P-V curve represents the work done. For a cyclic process, the area enclosed by the loop is the net work done.
• Carnot Engine Efficiency: Maximum possible efficiency for a heat engine operating between two temperatures T1 (source) and T2 (sink) is η = 1 - (T2/T1). Temperatures MUST be in Kelvin.
• Refrigerator Coefficient of Performance (COP): β = Q2/W = Q2/(Q1-Q2) = T2/(T1-T2). Higher COP means more efficient cooling.

Exam Trap: Sign Conventions & Units

A common mistake in Thermodynamics problems is incorrect application of sign conventions for Heat (Q) and Work (W). Remember: Heat absorbed by the system is positive (+Q), heat rejected is negative (-Q). **Work done by the system (expansion) is positive (+W), work done on the system (compression) is negative (-W). Always ensure consistency in units; convert everything to SI units** (Joules for energy, Pascals for pressure, cubic meters for volume, Kelvin for temperature) before numerical calculations to avoid errors, especially when using the ideal gas constant (R) in J/mol·K.

Quick Check: Thermodynamics Concepts

• Q: What is the main difference between an isothermal and an adiabatic process in terms of heat exchange? A: In an isothermal process, the temperature remains constant, and heat exchange occurs to maintain this. In an adiabatic process, there is no heat exchange with the surroundings (Q=0).
• Q: For an ideal gas, what is the change in internal energy during a cyclic process? A: For an ideal gas undergoing a cyclic process, the system returns to its initial state, so the net change in internal energy (ΔU) is zero.
• Q: According to the First Law of Thermodynamics, if 100 J of heat is added to a system and the system does 30 J of work, what is the change in internal energy? A: Using ΔU = Q - W, ΔU = 100 J - 30 J = 70 J. The internal energy increases by 70 J.
• Q: Why can't the efficiency of a heat engine ever be 100% according to the Second Law of Thermodynamics? A: According to the Kelvin-Planck statement of the Second Law, it is impossible for a heat engine to convert all the heat absorbed from a single reservoir completely into work. Some heat must always be rejected to a colder sink, meaning efficiency can never reach 100%.

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