JEE Physics: Thermodynamics Complete Guide

Thermodynamics is a fundamental branch of physics focused on heat, work, energy, and their transformations. For JEE aspirants, mastering thermodynamics concepts, laws, formulas, and problem-solving strategies is crucial to scoring high marks. This comprehensive guide covers the first and second laws of thermodynamics, thermodynamic processes, entropy, engines, and related formulas with detailed explanations and examples.

1. Basics and Terminology

System: The part of the universe under study (e.g., gas in a cylinder).
Surroundings: Everything outside the system.
Boundary: Real or imaginary surface separating system and surroundings.
State Variables: Pressure \(P\), volume \(V\), temperature \(T\), internal energy \(U\), etc.
Process: Change from one state to another.
Cycle: A sequence of processes returning the system to its initial state.

2. Zeroth Law of Thermodynamics

If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law defines temperature.

3. First Law of Thermodynamics

It is the law of conservation of energy applied to thermodynamics:

\( \displaystyle \Delta U = Q - W \)

Where \( \Delta U \) = change in internal energy,
\( Q \) = heat added to system (positive if added),
\( W \) = work done by system on surroundings (positive if done by system).

3.1 Internal Energy (\(U\))

Internal energy is the total microscopic kinetic and potential energy of molecules. For an ideal gas, \( U \) depends only on temperature.

3.2 Work Done by Gas

Work done during volume change:

\( \displaystyle W = \int_{V_i}^{V_f} P \, dV \)

3.3 Sign Conventions

Heat absorbed by the system: \( Q > 0 \)
Heat released by the system: \( Q < 0 \)
Work done by the system: \( W > 0 \)
Work done on the system: \( W < 0 \)

4. Thermodynamic Processes

Important idealized processes are:

4.1 Isothermal Process (\( T = \text{constant} \))

Temperature remains constant, so \( \Delta U = 0 \).

\( \displaystyle W = nRT \ln \frac{V_f}{V_i} \quad ; \quad Q = W \)

4.2 Adiabatic Process (\( Q = 0 \))

No heat exchange with surroundings.

\( \displaystyle P V^\gamma = \text{constant}, \quad TV^{\gamma-1} = \text{constant}, \quad T^{\gamma} P^{1-\gamma} = \text{constant} \)

Work done:

\( \displaystyle W = \frac{P_f V_f - P_i V_i}{\gamma - 1} \)

4.3 Isobaric Process (\( P = \text{constant} \))

\( \displaystyle W = P (V_f - V_i) \)

4.4 Isochoric Process (\( V = \text{constant} \))

No work done: \( W = 0 \).

5. Specific Heat Capacities

For ideal gases:

At constant volume \( C_V = \left(\frac{\partial Q}{\partial T}\right)_V \)
At constant pressure \( C_P = \left(\frac{\partial Q}{\partial T}\right)_P \)

Relation:

\( \displaystyle C_P - C_V = R \quad ; \quad \gamma = \frac{C_P}{C_V} \)

6. Second Law of Thermodynamics

The second law introduces the concept of entropy and directionality of natural processes:

Heat cannot spontaneously flow from cold to hot body.
It is impossible to convert all heat into work without other effects (no 100% efficient heat engine).
Entropy of an isolated system never decreases.

6.1 Clausius Statement

Heat cannot flow spontaneously from cold to hot.

6.2 Kelvin-Planck Statement

Impossible to construct a heat engine operating in a cycle with 100% efficiency.

7. Entropy (\( S \))

Entropy measures the disorder or randomness of a system.

\( \displaystyle dS = \frac{dQ_{\text{rev}}}{T} \)

For a reversible process:

\( \displaystyle \Delta S = \int \frac{dQ_{\text{rev}}}{T} \)

7.1 Entropy Change for Ideal Gas

\( \displaystyle \Delta S = n C_V \ln \frac{T_f}{T_i} + n R \ln \frac{V_f}{V_i} \)

8. Heat Engines and Refrigerators

8.1 Heat Engine

Converts heat energy into mechanical work.

Efficiency:

\( \displaystyle \eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H} \)

Where \( Q_H \) = heat absorbed from hot reservoir, \( Q_C \) = heat rejected to cold reservoir.

8.2 Carnot Engine

Theoretical engine with maximum efficiency operating between two reservoirs.

\( \displaystyle \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \)

8.3 Refrigerator and Heat Pump

Refrigerator transfers heat from cold to hot reservoir using work input.

Coefficient of Performance (COP):

\( \displaystyle COP_{\text{refrigerator}} = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C} \quad ; \quad COP_{\text{heat pump}} = \frac{Q_H}{W} = \frac{T_H}{T_H - T_C} \)

9. Thermodynamic Potentials

9.1 Internal Energy \( U \)

Energy contained within the system.

9.2 Enthalpy \( H \)

\( \displaystyle H = U + PV \)

9.3 Helmholtz Free Energy \( F \)

\( \displaystyle F = U - TS \)

9.4 Gibbs Free Energy \( G \)

\( \displaystyle G = H - TS = U + PV - TS \)

10. Important Formulas Summary

Concept	Formula	Remarks
First Law	\( \Delta U = Q - W \)	Energy conservation
Work done (Isothermal)	\( W = nRT \ln \frac{V_f}{V_i} \)	Constant temperature
Work done (Adiabatic)	\( W = \frac{P_f V_f - P_i V_i}{\gamma - 1} \)	No heat exchange
Entropy change (ideal gas)	\( \Delta S = n C_V \ln \frac{T_f}{T_i} + n R \ln \frac{V_f}{V_i} \)	Disorder measure
Carnot efficiency	\( \eta = 1 - \frac{T_C}{T_H} \)	Max theoretical efficiency
COP Refrigerator	\( \frac{T_C}{T_H - T_C} \)	Performance measure

11. Sample Problems and Solutions

Problem 1: Work Done in Isothermal Expansion

One mole of an ideal gas expands isothermally at 300 K from 10 L to 20 L. Calculate the work done.

Solution:

\( W = nRT \ln \frac{V_f}{V_i} = 1 \times 8.314 \times 300 \times \ln \frac{20}{10} \)

\( W = 2494.2 \times 0.693 = 1728.5 \, \text{J} \)

Problem 2: Efficiency of Carnot Engine

Calculate the efficiency of a Carnot engine working between 500 K and 300 K.

Solution:

\( \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 0.4 = 40\% \)

12. Tips for JEE Thermodynamics

Understand all thermodynamic processes and their formulas thoroughly.
Practice work and heat calculations in different processes.
Memorize relations between specific heats and their physical significance.
Focus on entropy and its calculation in various scenarios.
Get familiar with Carnot engines and refrigerators and their efficiencies.
Solve as many previous year problems and mock tests as possible.

13. Conclusion

Thermodynamics is a key topic in JEE physics requiring conceptual clarity and practice. This guide compiles all essential laws, processes, formulas, and problem-solving strategies to boost your confidence and performance. Stay consistent with your studies, revise regularly, and apply these concepts to excel in the exam.

Good luck with your preparation and keep practicing!