Physics Fluids and Thermal - How to Solve: Thermodynamics Processes (Isothermal, Adiabatic, P-V Diagrams, Work Done) – IIT JEE Guide

How to Solve: Thermodynamics Processes (Isothermal, Adiabatic, P-V Diagrams, Work Done) – IIT JEE Guide

Introduction

Mastering thermodynamics processes unlocks 10-15% of IIT JEE’s thermodynamics questions—including high-weightage problems on work done, efficiency, and P-V diagrams. If you can solve these, you’re not just scoring marks; you’re beating the curve in one of the most predictable sections of the exam.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. First Law of Thermodynamics: ΔU = Q – W (where ΔU = change in internal energy, Q = heat added, W = work done by the system).
2. Ideal Gas Law: PV = nRT (P = pressure, V = volume, n = moles, R = gas constant, T = temperature).
3. Work Done in a P-V Diagram: Work = Area under the curve (for quasi-static processes).

KEY TERMS & FORMULAS

Key Terms

1. Isothermal Process: Temperature remains constant (ΔT = 0).
2. Adiabatic Process: No heat exchange (Q = 0).
3. Quasi-Static Process: Infinitely slow, reversible process (used in P-V diagrams).
4. Work Done (W): Energy transferred by a system via expansion/compression.
5. Internal Energy (U): For an ideal gas, U depends only on temperature (U = (3/2)nRT for monatomic gas).

Formulas

1. Work Done in an Isothermal Process

Formula: [ W = nRT \ln \left( \frac{V_f}{V_i} \right) ] - ( W ) = Work done by the gas (Joules) - ( n ) = Moles of gas - ( R ) = Universal gas constant (8.314 J/mol·K) - ( T ) = Temperature (K) - ( V_f ) = Final volume - ( V_i ) = Initial volume MEMORISE THIS (Not given on exam sheet).

2. Work Done in an Adiabatic Process

Formula: [ W = \frac{P_i V_i - P_f V_f}{\gamma - 1} ] - ( \gamma ) = Adiabatic index (Cp/Cv) - Monatomic gas: ( \gamma = 5/3 ) - Diatomic gas: ( \gamma = 7/5 ) MEMORISE THIS (Not given on exam sheet).

3. Adiabatic Relation (P-V Relation)

Formula: [ P V^\gamma = \text{constant} ] or [ T V^{\gamma - 1} = \text{constant} ] MEMORISE THIS (Not given on exam sheet).

4. First Law of Thermodynamics

Formula: [ \Delta U = Q - W ] - ( \Delta U ) = Change in internal energy - ( Q ) = Heat added to the system - ( W ) = Work done by the system GIVEN ON EXAM SHEET (but you must know how to apply it).

5. Work Done from P-V Diagram

Formula: [ W = \int_{V_i}^{V_f} P \, dV ] - For a straight line in P-V diagram: ( W = \text{Area under the curve} ) - For isobaric process (constant P): ( W = P \Delta V ) MEMORISE THIS (Not given on exam sheet).

STEP-BY-STEP METHOD

Step 1: Identify the Process

• Isothermal? → ΔT = 0 → Use ( W = nRT \ln(V_f/V_i) )
• Adiabatic? → Q = 0 → Use ( W = \frac{P_i V_i - P_f V_f}{\gamma - 1} ) or ( P V^\gamma = \text{constant} )
• Isobaric? → P = constant → Use ( W = P \Delta V )
• Isochoric? → V = constant → W = 0

Step 2: Check Given Data

• Are P, V, T given? → Use ideal gas law if needed.
• Is work or heat asked? → Apply First Law if needed.
• Is a P-V diagram given? → Calculate work as area under the curve.

Step 3: Apply the Correct Formula

• For isothermal: Plug into ( W = nRT \ln(V_f/V_i) ).
• For adiabatic: Use ( P V^\gamma = \text{constant} ) first, then ( W = \frac{P_i V_i - P_f V_f}{\gamma - 1} ).
• For P-V diagram: Break into geometric shapes (rectangles, triangles) to find area.

Step 4: Sign Convention

• Work done by the gas (expansion): +W
• Work done on the gas (compression): -W
• Heat added to the system: +Q
• Heat removed from the system: -Q

Step 5: Solve for Unknown

• If work is asked, compute W.
• If internal energy is asked, use ( \Delta U = Q - W ).
• If efficiency is asked, use ( \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} ).

Step 6: Verify Units

• Work (W) → Joules (J)
• Pressure (P) → Pascals (Pa) or atm (convert if needed)
• Volume (V) → m³ or liters (1 m³ = 1000 L)
• Temperature (T) → Kelvin (K) (always convert °C to K)

WORKED EXAMPLES

Example 1 – Basic (Isothermal Expansion)

Problem: 1 mole of an ideal gas expands isothermally at 300 K from 10 L to 20 L. Calculate the work done by the gas.

Solution:
1. Identify the process: Isothermal (ΔT = 0).
2. Given: - n = 1 mol - T = 300 K - V_i = 10 L = 0.01 m³ - V_f = 20 L = 0.02 m³
3. Formula: ( W = nRT \ln(V_f/V_i) )
4. Plug in values: [ W = (1)(8.314)(300) \ln(0.02/0.01) ] [ W = 2494.2 \ln(2) ] [ W = 2494.2 \times 0.693 ] [ W = 1728.5 \, \text{J} ]
5. Sign: Expansion → Work done by gas → +1728.5 J

What we did and why: - Recognized isothermal process → used ( W = nRT \ln(V_f/V_i) ). - Converted liters to m³ (1 L = 0.001 m³) for SI units. - Applied natural log (ln) correctly.

Example 2 – Medium (Adiabatic Compression)

Problem: A diatomic gas (γ = 1.4) is compressed adiabatically from 2 atm, 5 L to 1 L. Calculate the final pressure.

Solution:
1. Identify the process: Adiabatic (Q = 0).
2. Given: - P_i = 2 atm - V_i = 5 L - V_f = 1 L - γ = 1.4
3. Formula: ( P_i V_i^\gamma = P_f V_f^\gamma )
4. Plug in values: [ 2 \times (5)^{1.4} = P_f \times (1)^{1.4} ] [ P_f = 2 \times 5^{1.4} ] [ 5^{1.4} = 5^{1} \times 5^{0.4} \approx 5 \times 1.903 = 9.515 ] [ P_f = 2 \times 9.515 = 19.03 \, \text{atm} ]

What we did and why: - Used adiabatic relation ( P V^\gamma = \text{constant} ). - Calculated ( 5^{1.4} ) by breaking into ( 5^1 \times 5^{0.4} ). - Final pressure increases because volume decreases (compression).

Example 3 – Exam-Style (P-V Diagram + Work Done)

Problem: An ideal gas undergoes the process shown in the P-V diagram below. Calculate the net work done by the gas. - A → B: Isobaric expansion (P = 2 atm, V from 1 L to 3 L) - B → C: Isochoric cooling (V = 3 L, P drops to 1 atm) - C → A: Linear process (straight line in P-V diagram)

Solution:
1. Break into segments: - A → B: Isobaric (P = constant) [ W_{AB} = P \Delta V = 2 \times (3 - 1) = 4 \, \text{atm·L} ] Convert to Joules: ( 1 \, \text{atm·L} = 101.325 \, \text{J} ) [ W_{AB} = 4 \times 101.325 = 405.3 \, \text{J} ] - B → C: Isochoric (V = constant) → ( W_{BC} = 0 ) - C → A: Linear process → Work = Area under the line - Trapezoid area: ( \frac{1}{2} (P_A + P_C)(V_A - V_C) ) [ W_{CA} = \frac{1}{2} (2 + 1)(1 - 3) = \frac{1}{2} \times 3 \times (-2) = -3 \, \text{atm·L} ] [ W_{CA} = -3 \times 101.325 = -303.975 \, \text{J} ]
2. Net work: [ W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 405.3 + 0 - 303.975 = 101.325 \, \text{J} ]

What we did and why: - Split the process into segments and calculated work for each. - For linear process, used trapezoid area formula. - Converted atm·L to Joules for final answer.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

Listen up—this is your last-minute lifeline.

1. Isothermal? → ΔT = 0 → Use ( W = nRT \ln(V_f/V_i) ).
2. Adiabatic? → Q = 0 → Use ( PV^\gamma = \text{constant} ) and ( W = \frac{P_i V_i - P_f V_f}{\gamma - 1} ).
3. P-V diagram? → Work = Area under the curve. Break into shapes if needed.
4. First Law? → ΔU = Q – W. Remember: Work done by the gas is positive.
5. Units? → Always convert to Joules, Pascals, and m³.

Pro tip: If the problem mentions "insulated," it’s adiabatic. If it says "slow process," it’s quasi-static (use P-V diagrams).

You’ve got this. Now go solve those problems like a JEE topper. ?