Thermodynamics Notes for BSc 1st Year Physics (TU): Complete Study Guide with Derivations, Numericals & Exam Questions

Utsab Ojha July 14, 2022


Thermodynamics Notes for BSc 1st Year Physics (TU): Complete Study Guide with Derivations, Numericals & Exam Questions


Introduction

Thermodynamics is the branch of physics that studies the relationship between heat, work, temperature, and energy. It explains how energy moves and transforms in physical and chemical systems — from the burning of fuel in a car engine to the cooling effect of a refrigerator to the metabolism of living cells.

For BSc 1st Year Physics students, thermodynamics is one of the most conceptually important topics. It introduces ideas that reappear throughout physics, chemistry, engineering, and even biology: the impossibility of perfect efficiency, the tendency of systems to move toward disorder, and the mathematical relationships that govern energy exchange. It also generates a consistent set of exam questions — the Carnot cycle efficiency derivation, first law numericals, and entropy definitions appear in TU papers year after year.

This guide covers all four laws of thermodynamics, thermodynamic processes, heat engines, the Carnot cycle, entropy, Mayer's formula, key derivations, solved numerical problems, and important exam questions. By the end, you should be able to handle both theory questions and numerical problems on thermodynamics in university examinations.

1. Basic Concepts

System and Surroundings

A thermodynamic system is the part of the universe being studied. Everything outside the system is the surroundings. The boundary between them can be real (like the walls of a container) or imaginary.

Types of systems:

  • Open system: Exchanges both matter and energy with surroundings (e.g., boiling water in an open pot)
  • Closed system: Exchanges energy but not matter (e.g., gas in a sealed cylinder with a movable piston)
  • Isolated system: Exchanges neither matter nor energy (e.g., a perfectly insulated flask — idealized)

State Variables

State variables describe the condition of a system:

  • Temperature (T): Measure of average kinetic energy of particles (in Kelvin for thermodynamics)
  • Pressure (P): Force per unit area exerted by the system
  • Volume (V): Space occupied by the system
  • Internal Energy (U): Total kinetic and potential energy of all particles in the system

Thermodynamic Equilibrium

A system is in thermodynamic equilibrium when its macroscopic properties (T, P, V) are uniform throughout and not changing with time. All thermodynamic laws strictly apply at equilibrium.

Quasi-static Process

A quasi-static process is one that proceeds infinitely slowly through a continuous sequence of equilibrium states. Real processes are approximations; quasi-static processes are the idealized model for thermodynamic analysis.

2. Zeroth Law of Thermodynamics

Statement: If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

This law defines temperature. If object A is the same temperature as object C, and object B is the same temperature as object C, then A and B are the same temperature as each other. This seems obvious, but it is the foundation that makes thermometers meaningful.

Practical significance: The Zeroth Law allows us to use a thermometer as the "third body" — if the thermometer reads the same temperature for two objects, those objects are at the same temperature and will not exchange heat if placed in contact.

3. First Law of Thermodynamics

Statement: Energy cannot be created or destroyed. The total energy of a system plus its surroundings remains constant. Heat added to a system either increases its internal energy or does work, or both.

Mathematical Form:

Q = ΔU + W

Where:

  • Q = Heat added to the system (positive when heat flows in)
  • ΔU = Change in internal energy of the system
  • W = Work done by the system on the surroundings (positive when system expands)

Alternative form: ΔU = Q − W (change in internal energy = heat absorbed minus work done by system)

Derivation from energy conservation:

Consider a gas in a cylinder doing work by expanding against a piston. The energy supplied as heat (Q) must account for:

  1. The increase in internal energy of the gas (ΔU = U₂ − U₁)
  2. The work done by the gas in pushing the piston (W = P × Î”V for constant pressure)

Therefore: Q = ΔU + W, which is the First Law.

Sign conventions:

  • Q > 0: Heat flows into the system
  • Q < 0: Heat flows out of the system
  • W > 0: System does work on surroundings (expansion)
  • W < 0: Surroundings do work on system (compression)

4. Thermodynamic Processes

Isothermal Process (Constant Temperature)

Temperature remains constant throughout. For an ideal gas, internal energy depends only on temperature, so ΔU = 0.

First Law gives: Q = W (all heat absorbed goes into doing work)

Work done: W = nRT ln(V₂/V₁)

where n = moles, R = gas constant, T = temperature, V₁ and V₂ = initial and final volumes.

For an ideal gas: PV = constant (Boyle's Law applies)

Example: Slow expansion of gas in a cylinder kept at constant temperature in a heat bath.

Adiabatic Process (No Heat Transfer)

No heat is exchanged with the surroundings: Q = 0.

First Law gives: ΔU = −W (work is done at the expense of internal energy)

When a gas expands adiabatically, it cools. When compressed adiabatically, it heats up.

Adiabatic relation: PVᵞ = constant, where γ = Cp/Cv (ratio of specific heats)

Work done: W = (P₁V₁ − P₂V₂)/(γ − 1) = nCv(T₁ − T₂)

Example: Rapid compression of a bicycle pump (so fast that no heat escapes), diesel engine compression stroke.

Isobaric Process (Constant Pressure)

Pressure remains constant. This is the most common everyday process — heating a gas in a container open to the atmosphere.

Work done: W = P(V₂ − V₁) = PΔV

Heat absorbed: Q = nCpΔT

Isochoric/Isovolumetric Process (Constant Volume)

Volume remains constant. No work is done (W = 0) since there is no displacement.

First Law gives: Q = ΔU (all heat goes into changing internal energy)

Heat absorbed: Q = nCvΔT

Example: Heating gas in a rigid, sealed container.

Comparison Table

Process Constant Q W ΔU
Isothermal Temperature Q = W W = nRT ln(V₂/V₁) 0
Adiabatic (no heat transfer) 0 −ΔU −W
Isobaric Pressure nCpΔT PΔV nCvΔT
Isochoric Volume ΔU 0 Q

5. Specific Heat Capacities and Mayer's Formula

Specific heat capacity is the heat required per unit mass per unit temperature rise.

For gases, two specific heats are defined:

  • Cv (at constant volume): Heat required to raise temperature by 1°C at constant volume
  • Cp (at constant pressure): Heat required to raise temperature by 1°C at constant pressure

Cp is always greater than Cv because at constant pressure, the gas expands and does extra work, so more heat is needed.

Mayer's Formula

Cp − Cv = R (for one mole of ideal gas)

Derivation:

At constant pressure: Q = nCpΔT, and ΔU = nCvΔT, and W = PΔV = nRΔT (from ideal gas law)

Applying First Law: nCpΔT = nCvΔT + nRΔT

Dividing by nΔT: Cp − Cv = R

This elegant result says the difference between the two specific heats equals the universal gas constant R = 8.314 J/mol·K.

The ratio γ = Cp/Cv (adiabatic index or heat capacity ratio):

  • Monatomic ideal gas (He, Ar): γ = 5/3 ≈ 1.67
  • Diatomic gas (O₂, N₂, air): γ = 7/5 = 1.40
  • Triatomic gas (CO₂): γ ≈ 1.30

6. Second Law of Thermodynamics

The First Law deals with energy conservation. The Second Law deals with the direction of energy transfer and the limits of efficiency.

Kelvin-Planck Statement: It is impossible to construct a heat engine that operates in a cycle and converts all the heat absorbed from a single heat reservoir entirely into work.

In other words: no heat engine can be 100% efficient. Some heat must always be rejected.

Clausius Statement: It is impossible for a process to transfer heat from a cooler body to a hotter body without any external work being done.

In other words: refrigerators require work input — heat doesn't spontaneously flow from cold to hot.

Both statements are equivalent — if one is violated, the other is too.

Why the Second Law matters: It explains why perpetual motion machines of the second kind are impossible, why all real engines are less than 100% efficient, and gives us the concept of entropy.

7. Heat Engine and Refrigerator

Heat Engine

A heat engine is a device that absorbs heat from a hot reservoir (Q₁), converts some of it to useful work (W), and rejects the rest as waste heat to a cold reservoir (Q₂).

Energy balance: Q₁ = W + Q₂

Efficiency: η = W/Q₁ = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁

Real efficiency is always less than 1 (less than 100%) because some heat is always rejected.

Refrigerator

A refrigerator does the reverse — it absorbs heat from a cold reservoir (Q₂) and, with input work (W), transfers heat to a hot reservoir (Q₁).

Energy balance: Q₁ = Q₂ + W

Coefficient of Performance (COP): COP = Q₂/W = Q₂/(Q₁ − Q₂)

A higher COP means a more efficient refrigerator.

8. Carnot Cycle and Carnot Efficiency

The Carnot cycle is an ideal reversible thermodynamic cycle that achieves the maximum possible efficiency for a heat engine operating between two temperatures. It was conceived by Sadi Carnot in 1824.

Four Steps of the Carnot Cycle

              P
              │    Isothermal Expansion (A→B)
              │    at Temperature T₁ (absorbs Q₁)
              A ╮
              │   ╮ Adiabatic Expansion (B→C)
              │     ╮
              │       B
              │         ╮
              │     D     ╮ C
              │     │       │
              │     ╰───────╯
              │    Isothermal Compression (C→D)
              │    at Temperature T₂ (rejects Q₂)
              │    Adiabatic Compression (D→A)
              └─────────────────────────────── V

  1. A → B (Isothermal expansion at T₁): Gas absorbs heat Q₁ from hot reservoir at temperature T₁ and expands isothermally. Work done by gas.

  2. B → C (Adiabatic expansion): Gas expands adiabatically. Temperature drops from T₁ to T₂. No heat exchange.

  3. C → D (Isothermal compression at T₂): Gas is compressed isothermally at temperature T₂. Heat Q₂ is rejected to cold reservoir.

  4. D → A (Adiabatic compression): Gas is compressed adiabatically. Temperature rises from T₂ back to T₁. No heat exchange.

Derivation of Carnot Efficiency

For a Carnot engine operating between temperatures T₁ (hot) and T₂ (cold):

For the isothermal processes in an ideal gas:

  • Q₁ = nRT₁ ln(V_B/V_A) (heat absorbed at T₁)
  • Q₂ = nRT₂ ln(V_C/V_D) (heat rejected at T₂)

For the adiabatic processes: T₁V_Bᵞ⁻¹ = T₂V_Cᵞ⁻¹ and T₁V_Aᵞ⁻¹ = T₂V_Dᵞ⁻¹

Dividing these: V_B/V_A = V_C/V_D

Therefore: Q₁/Q₂ = T₁/T₂

Carnot Efficiency:

η_Carnot = 1 − Q₂/Q₁ = 1 − T₂/T₁ = (T₁ − T₂)/T₁

Key conclusions:

  • Carnot efficiency depends only on the temperatures T₁ and T₂, not on the working substance
  • Efficiency increases as T₁ increases or T₂ decreases
  • η < 1 always (since T₂ > 0 K in practice)
  • No real engine can exceed Carnot efficiency between the same two temperatures

9. Entropy

Entropy (S) is a thermodynamic state function that measures the degree of disorder or randomness in a system.

Definition: For a reversible process at temperature T, the change in entropy is:

ΔS = Q_reversible / T

Units: J/K (Joules per Kelvin)

Second Law in terms of entropy:

  • For a reversible process: ΔS_universe = 0
  • For an irreversible (real) process: ΔS_universe > 0
  • Total entropy of the universe never decreases: ΔS_universe ≥ 0

Physical meaning: Entropy always increases in natural (spontaneous) processes. A hot-cold system always evolves toward equal temperatures, not away from it. Gas always expands to fill a container, never spontaneously compresses itself. Entropy is "time's arrow" in physics — it tells us which direction time runs.

Entropy change for an ideal gas:

ΔS = nCv ln(T₂/T₁) + nR ln(V₂/V₁)

10. Third Law of Thermodynamics

Statement: The entropy of a perfect crystal at absolute zero temperature (0 K) is zero.

Practical implication: Absolute zero (0 K = −273.15°C) cannot be reached by any finite number of steps. You can approach it arbitrarily closely but never attain it. This is why liquid helium cryogenics labs get to within a few millikelvin of absolute zero but never reach it.

11. Real-Life Applications of Thermodynamics

Steam Power Plants: Coal, gas, or nuclear heat boils water to steam that drives turbines. The Carnot efficiency equation sets the theoretical maximum efficiency. Higher steam temperatures mean higher efficiency.

Internal Combustion Engines: Car engines follow a thermodynamic cycle (Otto cycle for petrol, Diesel cycle for diesel engines). Fuel combustion releases heat, which expands gas to push pistons.

Refrigerators and Air Conditioners: Operate as heat pumps — using work input to move heat from cold to hot, the opposite of natural direction. The compressor does work; refrigerant absorbs heat inside and releases it outside.

Human Metabolism: The human body is a heat engine — food provides chemical energy, the body does work (movement, organ function) and releases waste heat. The efficiency of the body in converting food energy to mechanical work is approximately 20–25%.

Weather and Climate: Atmospheric thermodynamics governs convection, cloud formation, hurricane energy, and global climate. The ocean-atmosphere system is a vast heat engine.

Rocket Propulsion: Combustion of fuel produces hot gases that expand through a nozzle. Thermodynamic efficiency principles determine the specific impulse and fuel consumption.

Industrial Boilers and Heat Exchangers: Industrial processes use thermodynamic principles to maximize energy recovery and minimize waste heat.

12. Solved Numerical Problems

Problem 1 — First Law Application

A gas absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

Q = 500 J, W = 200 J

ΔU = Q − W = 500 − 200 = 300 J

The internal energy increases by 300 J.

Problem 2 — Isothermal Work Done

One mole of ideal gas at 300 K expands isothermally from 2 L to 6 L. Find the work done. (R = 8.314 J/mol·K)

W = nRT ln(V₂/V₁) = 1 × 8.314 × 300 × ln(6/2) = 2494.2 × ln(3) = 2494.2 × 1.0986 = 2740 J

Since Q = W for isothermal and ΔU = 0: the gas also absorbed 2740 J of heat.

Problem 3 — Carnot Efficiency

A Carnot engine operates between 600 K and 300 K. Find: (a) efficiency, (b) work done if it absorbs 1000 J of heat.

(a) η = 1 − T₂/T₁ = 1 − 300/600 = 1 − 0.5 = 50%

(b) W = η × Q₁ = 0.5 × 1000 = 500 J

Heat rejected to cold reservoir: Q₂ = Q₁ − W = 1000 − 500 = 500 J

Problem 4 — Entropy Change

5 J of heat is reversibly added to a system at constant temperature 500 K. Find the entropy change.

ΔS = Q/T = 5/500 = 0.01 J/K

Problem 5 — Mayer's Formula

The specific heat at constant volume for a gas is Cv = 20.8 J/mol·K. Find Cp. (R = 8.314 J/mol·K)

Using Mayer's Formula: Cp = Cv + R = 20.8 + 8.314 = 29.114 J/mol·K

Also: γ = Cp/Cv = 29.114/20.8 = 1.40 (diatomic gas, as expected)

13. Important Derivations

These derivations appear frequently in TU examinations and should be practiced until they can be written from memory:

1. Derivation of First Law of Thermodynamics — from energy conservation (Q = ΔU + W)

2. Work Done in Isothermal Process — W = nRT ln(V₂/V₁), derived from W = ∫P dV with PV = nRT

3. Work Done in Adiabatic Process — W = (P₁V₁ − P₂V₂)/(γ − 1) = nCv(T₁ − T₂)

4. Carnot Efficiency — η = 1 − T₂/T₁, derived from Q₁/Q₂ = T₁/T₂ using isothermal and adiabatic relations

5. Mayer's Formula: Cp − Cv = R — derived from First Law applied to constant pressure heating of ideal gas

6. Entropy Change — ΔS = Q_rev/T, and the principle that ΔS_universe ≥ 0

14. Important Exam Questions

Long Answer (8–10 Marks)

  1. State and prove the First Law of Thermodynamics. Apply it to isothermal, adiabatic, isobaric, and isochoric processes.
  2. Explain the Carnot cycle in detail with a PV diagram. Derive the expression for Carnot efficiency.
  3. State the Second Law of Thermodynamics in both Kelvin-Planck and Clausius forms. Prove their equivalence.
  4. Define entropy. Explain its physical significance and state its relation to the Second Law.
  5. Derive Mayer's Formula (Cp − Cv = R). What is the significance of γ = Cp/Cv?
  6. Explain the working of a heat engine. Derive its efficiency. Why can no real engine achieve 100% efficiency?

Short Answer (3–5 Marks)

  1. State the Zeroth Law. Why is it called "zeroth"?
  2. Differentiate between isothermal and adiabatic processes.
  3. Define coefficient of performance (COP) of a refrigerator.
  4. State the Third Law of Thermodynamics and its significance.
  5. Differentiate between Cv and Cp. Which is greater and why?
  6. What is a quasi-static process?

15. Past Question Trend Analysis

Appears almost every year:

  • Carnot cycle with PV diagram and efficiency derivation
  • First Law of Thermodynamics — statement, mathematical form, and application
  • Differentiate between isothermal and adiabatic processes
  • Mayer's Formula derivation
  • Numerical using First Law (Q = ΔU + W)

Frequently appears:

  • Second Law — Kelvin-Planck and Clausius statements
  • Entropy definition and change calculation
  • Heat engine efficiency
  • Isothermal work done (numerical)
  • Cp and Cv difference and γ ratio

Occasionally appears:

  • Zeroth Law explanation
  • Third Law and absolute zero
  • Carnot engine numerical
  • COP of refrigerator

Formula Quick-Reference

Concept Formula
First Law Q = ΔU + W
Isothermal work W = nRT ln(V₂/V₁)
Adiabatic relation PVᵞ = constant
Adiabatic work W = nCv(T₁ − T₂) = (P₁V₁ − P₂V₂)/(γ − 1)
Mayer's Formula Cp − Cv = R
Adiabatic index γ = Cp/Cv
Heat engine efficiency η = W/Q₁ = 1 − Q₂/Q₁
Carnot efficiency η = 1 − T₂/T₁ = (T₁ − T₂)/T₁
COP of refrigerator COP = Q₂/W = Q₂/(Q₁ − Q₂) = T₂/(T₁ − T₂)
Entropy change ΔS = Q_rev/T
Ideal gas law PV = nRT

Frequently Asked Questions

What is thermodynamics? Thermodynamics is the branch of physics that studies the relationships between heat, work, temperature, and energy, and how they govern the behaviour of macroscopic systems.

What are the four laws of thermodynamics? The Zeroth Law (defines temperature), First Law (energy conservation), Second Law (limits on efficiency and direction of processes), and Third Law (entropy at absolute zero is zero).

What is entropy? Entropy is a state function measuring the disorder or randomness of a system. In any spontaneous natural process, the total entropy of the universe increases — it never decreases.

What is the Carnot cycle? An ideal reversible thermodynamic cycle operating between two fixed temperatures, representing the maximum possible efficiency for a heat engine. Real engines are always less efficient than the Carnot engine operating between the same temperatures.

Which thermodynamic process has no heat transfer? Adiabatic process (Q = 0). All temperature change comes from work done on or by the gas.

Which process has constant temperature, pressure, and volume respectively? Isothermal (constant T), isobaric (constant P), isochoric/isovolumetric (constant V).

Why is Cp always greater than Cv? At constant pressure, when heat is added, the gas both increases its internal energy AND expands, doing work against the atmosphere. This extra work requires extra heat. At constant volume, no expansion occurs, so all heat goes directly into internal energy. Hence Cp = Cv + R.

Can absolute zero ever be reached? No — the Third Law of Thermodynamics states it cannot be attained by any finite number of steps. Modern experiments reach temperatures within fractions of a nanokelvin above absolute zero, but never zero itself.

Is the Carnot efficiency the maximum possible efficiency? Yes — the Carnot theorem states that no heat engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between the same two temperatures. This is a consequence of the Second Law.

What is the difference between a heat engine and a refrigerator? A heat engine absorbs heat from a hot reservoir and converts part of it to work, rejecting waste heat to a cold reservoir. A refrigerator runs the reverse process — using work input to transfer heat from a cold reservoir to a hot reservoir. Both obey the same thermodynamic laws.

Conclusion

Thermodynamics is one of the most intellectually rich topics in first-year physics because it connects everyday experience — hot and cold, engines and refrigerators, ice melting and fuel burning — to deep mathematical principles about energy and disorder. The four laws of thermodynamics form a logical hierarchy: the Zeroth defines temperature, the First establishes energy conservation, the Second introduces entropy and limits efficiency, and the Third anchors everything to absolute zero.

For TU examinations, focus especially on the First Law applications to different thermodynamic processes, the Carnot cycle efficiency derivation, Mayer's Formula, and entropy. These topics consistently generate both theory questions and numericals across past papers.

For related subjects: Conservation of Linear Momentum, Newton's Laws of Motion — Complete Guide, Collision in Physics, and Kinematics: Complete Notes for BSc 1st Year Physics.

In this page you can find BSC Physics Handwritten notes for TU 1st year students.You can download pdf file below.

Also read Collision notes pdf


Thank you for viewing this page. If you like this post, share this article with your friends.
Tags:

Post a Comment

0 Comments