Thermodynamics

Thermodynamics deals with heat, work, and the relationship between them. It’s fundamental to understanding engines, refrigerators, and natural processes.

Overview

graph TD
    A[Thermodynamics] --> B[Zeroth Law]
    A --> C[First Law]
    A --> D[Second Law]
    B --> B1[Thermal Equilibrium]
    C --> C1[Energy Conservation]
    C --> C2[Processes]
    D --> D1[Heat Engines]
    D --> D2[Entropy]

Basic Concepts

System and Surroundings

• System: Part of universe under study
• Surroundings: Everything outside the system
• Boundary: Separates system from surroundings

Types of Systems

Type	Energy Exchange	Matter Exchange
Open	Yes	Yes
Closed	Yes	No
Isolated	No	No

State Variables

• Intensive: Independent of amount (T, P, density)
• Extensive: Depends on amount (V, U, S, m)

Zeroth Law

If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.

This law defines temperature as a measurable quantity.

First Law of Thermodynamics

$$\boxed{Q = \Delta U + W}$$

or equivalently:

$$\boxed{dU = dQ - dW}$$

where:

• $Q$ = Heat added to system
• $\Delta U$ = Change in internal energy
• $W$ = Work done by system

Sign Convention

Quantity	Positive	Negative
Q	Heat absorbed	Heat released
W	Work done by system	Work done on system
ΔU	Internal energy increases	Internal energy decreases

Internal Energy

For an ideal gas:

$$U = \frac{f}{2}nRT$$

where $f$ = degrees of freedom

Gas Type	f	$C_V$	$C_P$	$\gamma$
Monatomic	3	$\frac{3}{2}R$	$\frac{5}{2}R$	1.67
Diatomic	5	$\frac{5}{2}R$	$\frac{7}{2}R$	1.4
Polyatomic	6+	$\frac{6}{2}R+$	$\frac{8}{2}R+$	1.33

Mayer’s Relation

$$\boxed{C_P - C_V = R}$$ $$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$

Thermodynamic Processes

Isobaric Process (Constant Pressure)

$$W = P\Delta V = nR\Delta T$$ $$Q = nC_P\Delta T$$ $$\Delta U = nC_V\Delta T$$

Isochoric Process (Constant Volume)

$$W = 0$$ $$Q = \Delta U = nC_V\Delta T$$

Isothermal Process (Constant Temperature)

$$\Delta U = 0$$ $$Q = W = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{P_1}{P_2}$$

Adiabatic Process (No Heat Exchange)

$$Q = 0$$ $$\Delta U = -W$$

Equations:

$$PV^\gamma = \text{constant}$$ $$TV^{\gamma-1} = \text{constant}$$ $$P^{1-\gamma}T^\gamma = \text{constant}$$

Work done:

$$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$

Process Summary

graph TD
    A[Thermodynamic Processes] --> B["Isobaric: P constant
W = PΔV"]
    A --> C["Isochoric: V constant
W = 0"]
    A --> D["Isothermal: T constant
ΔU = 0"]
    A --> E["Adiabatic: Q = 0
PVᵞ = constant"]

Process	Condition	W	Q	ΔU
Isobaric	P = const	$P\Delta V$	$nC_P\Delta T$	$nC_V\Delta T$
Isochoric	V = const	0	$nC_V\Delta T$	$nC_V\Delta T$
Isothermal	T = const	$nRT\ln\frac{V_2}{V_1}$	W	0
Adiabatic	Q = 0	$\frac{nR\Delta T}{1-\gamma}$	0	$-W$

Second Law of Thermodynamics

Kelvin-Planck Statement

It is impossible to construct an engine operating in a cycle that converts all heat into work.

Clausius Statement

It is impossible to transfer heat from cold body to hot body without external work.

Heat Engines

Efficiency

$$\boxed{\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}}$$

where:

• $Q_H$ = Heat absorbed from hot reservoir
• $Q_C$ = Heat rejected to cold reservoir
• $W$ = Work done by engine

Carnot Engine

Maximum efficiency heat engine operating between two temperatures.

$$\boxed{\eta_{Carnot} = 1 - \frac{T_C}{T_H}}$$

Carnot Cycle:

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression

Refrigerator

A heat engine running in reverse.

Coefficient of Performance (COP)

$$\boxed{COP = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}}$$

For Carnot refrigerator:

$$COP_{Carnot} = \frac{T_C}{T_H - T_C}$$

Entropy

Definition

$$dS = \frac{dQ_{rev}}{T}$$

Change in Entropy

$$\Delta S = \int \frac{dQ_{rev}}{T}$$

For isothermal process:

$$\Delta S = nR\ln\frac{V_2}{V_1}$$

Second Law in Terms of Entropy

For isolated system:

$$\Delta S_{universe} \geq 0$$

• Reversible process: $\Delta S = 0$
• Irreversible process: $\Delta S > 0$

Practice Problems

1. One mole of ideal gas at 300 K expands isothermally to double its volume. Find work done and heat absorbed.

2. An engine operating between 500 K and 300 K has efficiency 30%. Is it a reversible engine?

3. A gas is compressed adiabatically to half its volume. If initial temperature is 27°C, find final temperature. (γ = 1.4)

4. Calculate entropy change when 1 kg of ice at 0°C melts. (L = 334 kJ/kg)

Further Reading

• Kinetic Theory of Gases
• Properties of Solids and Liquids