'Thermodynamics'

Introduction

Thermodynamics deals with the relationships between heat, work, and internal energy of systems.

Thermodynamic System

A specified portion of the universe under study.

• Open system: Exchanges both matter and energy with surroundings.
• Closed system: Exchanges only energy, not matter.
• Isolated system: Exchanges neither matter nor energy.

Zeroth Law of Thermodynamics

If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

This law establishes the concept of temperature. It allows the use of thermometers for temperature measurement.

Internal Energy (U)

The sum of all molecular kinetic and potential energies of a system.

• Internal energy is a state function (depends only on state, not path).
• Delta U = U_f - U_i depends only on initial and final states.

First Law of Thermodynamics

Delta Q = Delta U + Delta W

Where:

• Delta Q = heat supplied to the system
• Delta U = change in internal energy
• Delta W = work done BY the system

Equivalent statement: Energy cannot be created or destroyed; it can only be transformed from one form to another.

Sign Convention

• Heat added to system: Q > 0
• Heat removed from system: Q < 0
• Work done by system: W > 0
• Work done on system: W < 0
• Internal energy increases: Delta U > 0

Application to Various Processes

Isothermal Process (constant temperature): Delta U = 0, so Q = W. Work done: W = nRT ln(V_2/V_1) = nRT ln(P_1/P_2).

Adiabatic Process (no heat exchange): Q = 0, so Delta U = -W. PV^gamma = constant, where gamma = C_p/C_v. TV^(gamma-1) = constant.

Isobaric Process (constant pressure): W = P Delta V = nR Delta T. Q = n C_p Delta T.

Isochoric Process (constant volume): W = 0, so Q = Delta U = n C_v Delta T.

Second Law of Thermodynamics

Kelvin-Planck Statement

It is impossible to construct a heat engine that converts all heat from a source into work without rejecting some heat to a sink.

Clausius Statement

Heat cannot flow from a colder body to a hotter body without external work being done.

Heat Engines

A device that converts heat into mechanical work.

Carnot Engine

The most efficient reversible heat engine operating between two temperatures.

Carnot Cycle: Isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.

Efficiency: eta = 1 - T_2/T_1 = 1 - Q_2/Q_1 Where T_1 = source temperature, T_2 = sink temperature.

Key points:

• Efficiency is 100% only when T_2 = 0 K (absolute zero), which is impossible.
• Carnot engine is the most efficient engine operating between two temperatures.
• No real engine can have efficiency greater than Carnot efficiency.

Refrigerator

Reversed heat engine: work is done to transfer heat from cold to hot body.

Coefficient of Performance (COP): beta = Q_2/W = Q_2/(Q_1 - Q_2) = T_2/(T_1 - T_2)

Entropy

A measure of disorder or randomness of a system.

Delta S = int dQ_rev/T

Second Law in Terms of Entropy

The entropy of an isolated system always increases (or remains constant for reversible processes). Delta S >= 0

Entropy Changes

• Reversible process: Delta S_(system) + Delta S_(surroundings) = 0
• Irreversible process: Delta S_(system) + Delta S_(surroundings) > 0

Worked Examples

Example 1: In an isothermal process, 2 moles of an ideal gas expand from 1 L to 5 L at 300 K. Find work done. (R = 8.314 J/mol K). Solution: W = nRT ln(V_2/V_1) = 2*8.314*300*ln(5) = 4988.4*1.609 = 8025 J.

Example 2: A Carnot engine operates between 500 K and 300 K. Find its efficiency. Solution: eta = 1 - 300/500 = 1 - 0.6 = 0.4 = 40%.

Example 3: In an adiabatic process, the temperature of an ideal gas increases from 300 K to 400 K. If gamma = 1.4, find the ratio of volumes. Solution: T_1 V_1^(gamma-1) = T_2 V_2^(gamma-1) => V_2/V_1 = (T_1/T_2)^(1/(gamma-1)) = (300/400)^(1/0.4) = (0.75)^2.5 = 0.488.

Common Mistakes

1. Q = U + W (not W - U): Internal energy increase uses some of the heat, rest goes to work.
2. Sign convention for W: Work done BY system is positive, work done ON system is negative.
3. gamma = C_p/C_v: Always greater than 1. For monatomic gas = 5/3, diatomic = 7/5.
4. Carnot efficiency is ideal: No real engine can achieve this efficiency; it is the theoretical maximum.

ISC Exam Focus

• Theory (70%): First and second laws, thermodynamic processes, Carnot cycle, entropy.
• Application (30%): Numerical problems on work done, efficiency, entropy change.
• ISC frequently asks: "Derive expression for work done in isothermal/adiabatic process."
• Carnot efficiency and PV diagrams are commonly tested.

Self-Test Questions

Q1: State the first law of thermodynamics. Answer: Delta Q = Delta U + Delta W. Energy is conserved in thermodynamic processes.

Q2: In which process is work done maximum: isothermal or adiabatic expansion? Answer: Isothermal expansion does more work than adiabatic for the same volume change (since no internal energy loss).

Q3: A Carnot engine has 40% efficiency and sink temperature 300 K. Find source temperature. Answer: 0.4 = 1 - 300/T_1 => 300/T_1 = 0.6 => T_1 = 500 K.

Q4: Define entropy. What is the change in entropy for a reversible process? Answer: Delta S = int dQ_rev/T. For a reversible process, Delta S_total = 0.

Q5: For isochoric process, what is the work done? Answer: W = 0 because volume is constant. Q = Delta U = n C_v Delta T.

Q6: Differentiate between isothermal and adiabatic processes. Answer: Isothermal: constant T, Delta U = 0, Q = W, PV = constant. Adiabatic: no heat exchange, Delta U = -W, PV^gamma = constant.