Module 1 · NEET Physics

Thermodynamics

Thermal properties, laws of thermodynamics, heat transfer, kinetic theory.
Thermal Physics · Laws of TD · Heat Transfer · Kinetic Theory
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1. Thermal Properties of Matter

This section covers the fundamental concepts of temperature, thermal expansion, calorimetry, and phase changes. These topics form the foundation for understanding how heat energy interacts with matter and are essential for solving NEET numerical problems in thermodynamics. The study of thermal properties of matter bridges the gap between macroscopic observations (like a metal rod expanding when heated) and the microscopic behaviour of atoms and molecules.

Importance for NEET & UPSC: Questions from this section appear regularly in both NEET (3–5 questions) and UPSC GS Prelims (1–2 questions). Common question types include: calculating expansion gaps in structures, finding final temperature of mixtures, determining heat required for phase changes, and analyzing thermal stress in composite materials. A strong grasp of these fundamentals is essential before moving to the laws of thermodynamics.

Section 1: Key Topics at a Glance
TopicKey ConceptEssential Formula
Temperature & HeatTemperature measures average KE; Heat is energy in transitT(K) = T(°C) + 273
Thermal ExpansionSubstances expand when heatedΔL = αL0ΔT
CalorimetryHeat lost = Heat gainedQ = mcΔT
Latent HeatPhase changes at constant temperatureQ = mL

Temperature & Heat

Temperature is a measure of the average kinetic energy of the molecules in a substance. Heat is the energy transferred between systems due to a temperature difference. The SI unit of heat is the joule (J), while temperature is measured in kelvin (K), degree Celsius (°C), or degree Fahrenheit (°F). The relationship between temperature and molecular kinetic energy is fundamental: at a given temperature, all gases have the same average molecular kinetic energy, (3/2)kT per molecule for monatomic gases.

Heat vs Temperature: It is crucial to distinguish between these two. Heat is energy in transit (measured in joules), while temperature is a measure of thermal intensity (measured in degrees or kelvin). A body can have a high temperature but low heat content (e.g., a spark from a flint has high temperature but negligible heat capacity), and vice versa (e.g., a large lake at 30°C contains enormous heat despite its moderate temperature).

Thermal equilibrium occurs when two systems in thermal contact cease to exchange heat, reaching the same temperature. The concept of temperature is fundamental to the Zeroth Law of Thermodynamics. When a thermometer is placed in contact with a body, heat flows between them until thermal equilibrium is reached, allowing the thermometer to display the body's temperature.

Temperature conversion formulas:

ScaleConversionRelation
Celsius to KelvinK = °C + 273.15T(K) = T(°C) + 273
Celsius to Fahrenheit°F = (°C × 9/5) + 32C/100 = (F−32)/180
Fahrenheit to KelvinK = (°F + 459.67) × 5/9Rankine: °R = °F + 459.67
NEET Tip
For NEET, always convert Celsius to Kelvin by adding 273 (not 273.15) unless precision is specified. Most problems use: T(K) = T(°C) + 273.

Thermal Expansion

Most substances expand when heated. For solids, we consider three types of thermal expansion:

Linear, Areal, Volume Expansion

Linear expansion: ΔL = α L0 ΔT where α is the coefficient of linear expansion (K−1). The expanded length is L = L0(1 + αΔT).

Areal (superficial) expansion: ΔA = β A0 ΔT where β = 2α. The expanded area is A = A0(1 + βΔT) = A0(1 + 2αΔT).

Volume expansion: ΔV = γ V0 ΔT where γ = 3α. The expanded volume is V = V0(1 + γΔT) = V0(1 + 3αΔT).

Thermal stress: If a rod is prevented from expanding or contracting when its temperature changes, it experiences thermal stress. The stress developed is: Stress = YαΔT, where Y is Young's modulus. The corresponding force is F = YAαΔT. This is why railway tracks and long bridges require expansion joints — without them, the thermal stress could cause buckling or structural failure.

Applications of thermal expansion: Bimetallic strips (two metals with different α bonded together) bend when heated and are used in thermostats and thermometers. The radius of curvature of a bimetallic strip is: r = d / [(α2 − α1)ΔT], where d is the total thickness of the two strips. The different expansion rates cause the strip to curve towards the metal with the lower expansion coefficient. This principle is used in fire alarms, temperature controllers, and automobile indicators. Other engineering applications include: fitting metal tyres onto wooden wheels (the metal tyre is heated, expands, placed over the wheel, and contracts on cooling to form a tight fit), and using expansion joints in bridges, pipelines, and railway tracks to accommodate thermal expansion without structural damage.

Anomalous expansion of water revisited: Water has a unique density maximum at 4°C. When water cools from 4°C to 0°C, it expands (its density decreases). This means ice at 0°C is less dense than water at 4°C, which is why ice floats on water. Bodies of water freeze from the top down because the 4°C water is densest and sinks to the bottom, while the 0°C water stays at the surface and freezes. This insulating layer of ice protects aquatic life in cold climates — the bottom of a lake remains at 4°C even when the surface is frozen solid. Without this anomalous property, lakes would freeze from the bottom up, destroying aquatic ecosystems. This is a frequently tested concept in both NEET and UPSC GS.

Materialα (×10−6 K−1)β (×10−6 K−1)γ (×10−6 K−1)
Aluminium234669
Brass193857
Copper173451
Iron / Steel11–1222–2433–36
Glass (Pyrex)3.36.69.9
Invar (Ni-Fe alloy)1.22.43.6
Example 1 — Linear Expansion
A steel rail of length 10 m is laid at 15°C. If the maximum temperature reaches 45°C, what gap should be left between rails? (αsteel = 1.2 × 10−5 K−1)
Solution: ΔL = α L0 ΔT = (1.2 × 10−5)(10)(45 − 15) = 1.2 × 10−5 × 10 × 30 = 3.6 × 10−3 m = 3.6 mm. So a gap of about 3.6 mm should be left between successive rails.
Quick Memory Aid
For isotropic solids, the expansion coefficients are related as: β = 2α and γ = 3α. This holds because expansion is proportional to length in each dimension, so area (∝ L²) gives factor 2 and volume (∝ L³) gives factor 3.
Example 1b — Areal Expansion
A copper sheet has an area of 2 m² at 20°C. Find its area at 120°C. (αCu = 1.7 × 10−5 K−1)
Solution: β = 2α = 3.4 × 10−5 K−1. ΔA = β A0 ΔT = (3.4 × 10−5)(2)(100) = 6.8 × 10−3 m² = 68 cm². Final area A = 2 + 0.0068 = 2.0068 m².
Anomalous Expansion of Water
Water has a unique property: its density is maximum at 4°C. Between 0°C and 4°C, water contracts on heating (negative expansion coefficient). Above 4°C, it expands normally. This is why ice floats and lakes freeze from the top down — a frequently tested concept in NEET.

Calorimetry

Calorimetry is the measurement of heat transfer. The principle of calorimetry states: heat lost by hotter bodies equals heat gained by colder bodies, assuming no heat exchange with the surroundings.

Specific heat capacity (c): Q = m c ΔT, where Q is heat energy, m is mass, and ΔT is the temperature change. Units: J kg−1 K−1.

Molar specific heat (C): Q = n C ΔT, where n is the number of moles. For gases, Cv (constant volume) and Cp (constant pressure) differ.

Water equivalent: The mass of water that has the same thermal capacity as a given body = m c / cwater.

SubstanceSpecific Heat (J kg−1 K−1)cal g−1 °C−1
Water41861.00
Ice21000.50
Aluminium9000.215
Copper3900.093
Iron4500.107
Glass8400.20
Example 2 — Calorimetry
200 g of water at 80°C is mixed with 100 g of water at 30°C. Find the final temperature of the mixture.
Solution: Heat lost by hot water = Heat gained by cold water. m1c(T1 − Tf) = m2c(Tf − T2). 200(80 − T) = 100(T − 30) ⇒ 16000 − 200T = 100T − 3000 ⇒ 300T = 19000 ⇒ T = 63.33°C.
NEET Shortcut — Mixture Temperature
When mixing two masses of the same substance, the final temperature is: Tf = (m1T1 + m2T2) / (m1 + m2), which is a weighted average of the temperatures. For the example above: T = (200×80 + 100×30)/300 = 19000/300 = 63.33°C.

Specific Heat of Solids & Dulong-Petit Law

The Dulong-Petit law states that the molar specific heat of most solid elements at room temperature is approximately 3R ≈ 25 J mol−1 K−1. This is because each atom in a solid has 6 degrees of freedom (3 kinetic + 3 potential), giving Cv = 3R. Exceptions include light elements like carbon (diamond) and beryllium, which have lower specific heats at room temperature.

SolidMolar Mass (g/mol)Specific Heat (J/kg·K)Molar Specific Heat (J/mol·K)
Aluminium2790024.3
Copper63.539024.8
Iron5645025.2
Lead20712826.5
Diamond (C)125106.1 (exception)
Dulong-Petit for NEET
For NEET, remember that the product of specific heat (in J/kg·K) and molar mass (in g/mol) is approximately 25 J/mol·K for most solid elements. This lets you quickly estimate the specific heat if you know the molar mass, or vice versa. For example, specific heat of silver (M = 108 g/mol) ≈ 25/108 × 1000 ≈ 231 J/kg·K.

Change of State & Latent Heat

When a substance changes from one state (solid, liquid, gas) to another, heat is absorbed or released without a change in temperature. This heat is called latent heat.

Latent heat of fusion (Lf): Heat required to convert 1 kg of solid to liquid at its melting point. For ice: Lf = 3.36 × 105 J/kg = 80 cal/g.

Latent heat of vaporization (Lv): Heat required to convert 1 kg of liquid to vapour at its boiling point. For water: Lv = 2.26 × 106 J/kg = 540 cal/g.

Heat required for phase change: Q = m L.

SubstanceMelting Point (°C)Lf (×105 J/kg)Boiling Point (°C)Lv (×106 J/kg)
Water (ice)03.361002.26
Ethanol−1141.04780.85
Aluminium6603.97246711.4
Copper10832.0625624.79
Example 3 — Latent Heat
How much heat is required to convert 20 g of ice at 0°C to steam at 100°C? (cwater = 4.2 J/g°C, Lf = 336 J/g, Lv = 2260 J/g)
Solution: Total heat = Q1 (melting ice) + Q2 (heating water) + Q3 (vaporizing water). Q1 = mLf = 20 × 336 = 6720 J. Q2 = mcΔT = 20 × 4.2 × 100 = 8400 J. Q3 = mLv = 20 × 2260 = 45200 J. Total = 6720 + 8400 + 45200 = 60,320 J = 60.32 kJ.
Heating Curve Concept
The heating curve of a substance shows plateaus at phase change temperatures (melting, boiling) where temperature remains constant despite heat input. The slope of the rising portions is determined by specific heat (Q = mcΔT), while the plateau length is determined by latent heat (Q = mL). This is a popular diagram-based NEET question.
Example 3b — Ice & Water Mixture
200 g of water at 50°C is mixed with 50 g of ice at 0°C. Find the final temperature and composition. (Lf = 336 J/g, cwater = 4.2 J/g°C)
Solution: Check if all ice melts: Heat required to melt all ice = mLf = 50 × 336 = 16800 J. Heat available from water cooling to 0°C = mcΔT = 200 × 4.2 × 50 = 42000 J. Since 42000 > 16800, all ice melts and the resulting water warms up. Let T be final temperature. Heat lost by warm water = 200 × 4.2 × (50 − T). Heat gained by ice = 50 × 336 (melting) + 50 × 4.2 × (T − 0). Equating: 200 × 4.2 × (50 − T) = 16800 + 50 × 4.2 × T ⇒ 840(50−T) = 16800 + 210T ⇒ 42000 − 840T = 16800 + 210T ⇒ 25200 = 1050T ⇒ T = 24°C. Final: 250 g water at 24°C.
Heating Curve of Water (0°C to 100°C) Heat Added T 0°C (ice) Melting Water heating 100°C Vaporization
Heating curve of water showing plateaus at 0°C (melting) and 100°C (vaporization). The rising sections represent temperature increase of a single phase; flat sections represent phase change at constant temperature.

2. Kinetic Theory of Gases

Section 2: Key Topics at a Glance
TopicKey ConceptEssential Formula
Gas LawsBehaviour of ideal gasesPV = nRT
Kinetic TheoryMolecular basis of pressure & temperatureP = (1/3)ρ<v²>
Molecular SpeedsDistribution of molecular velocitiesvrms = √(3RT/M)
Degrees of FreedomEnergy storage modesCv = (f/2)R

Gas Laws

Ideal gases obey the following laws under specific conditions. These laws form the foundation of the ideal gas equation and are essential for solving NEET problems on gas behaviour. The term "ideal gas" refers to a hypothetical gas whose molecules occupy negligible volume, have no intermolecular forces, and undergo perfectly elastic collisions. Real gases approach ideal behaviour at low pressures and high temperatures.

Historical context: Boyle's law (1662) was the first quantitative gas law, discovered by Robert Boyle using a J-shaped tube. Charles' law (1787) was discovered by Jacques Charles during his pioneering hot-air balloon flights. Gay-Lussac's law (1802) was published by Joseph Louis Gay-Lussac, who also established the law of combining volumes for chemical reactions. Avogadro's hypothesis (1811) provided the crucial link between gas volumes and the number of molecules.

LawRelationConstantStatement
Boyle's LawP ∝ 1/VPV = constantAt constant T, P × V = constant
Charles' LawV ∝ TV/T = constantAt constant P, V ∝ T (absolute)
Gay-Lussac's LawP ∝ TP/T = constantAt constant V, P ∝ T (absolute)
Avogadro's LawV ∝ nV/n = constantAt constant P,T, equal V contains equal n
Example 3c — Boyle's Law
A gas occupies 500 mL at 1 atm pressure. If the pressure is increased to 2.5 atm at constant temperature, what will be the new volume?
Solution: Boyle's law: P1V1 = P2V2. V2 = P1V1/P2 = (1 × 500) / 2.5 = 200 mL. The volume decreases as pressure increases, as expected.
Example 3d — Charles' Law
A gas occupies 2 L at 27°C. To what temperature must it be heated at constant pressure to occupy 3 L?
Solution: Charles' law: V1/T1 = V2/T2. T2 = V2T1/V1 = (3 × 300) / 2 = 450 K = 177°C. Remember to convert to Kelvin first!
Combined Gas Law
The three gas laws can be combined into: PV/T = constant. This is the combined gas law: (P1V1)/T1 = (P2V2)/T2. Memorise this single equation — it covers all three laws. If any one variable is constant, cancel it out to get the relevant individual law.

Real Gases & Van der Waals Equation

Real gases deviate from ideal behaviour at high pressures and low temperatures. The Van der Waals equation corrects for two factors ignored in the ideal gas model:

(P + a n²/V²)(V − nb) = nRT

where 'a' corrects for intermolecular attraction (reduces pressure), and 'b' corrects for the finite volume of molecules (reduces available volume).

Compressibility factor (Z): Z = PV/nRT. For an ideal gas, Z = 1. For real gases: Z < 1 at low pressures (attraction dominates), Z > 1 at high pressures (volume dominates). At Boyle temperature, Z ≈ 1 over a range of pressures.

Gasa (L² atm/mol²)b (L/mol)Boyle Temperature (K)
He0.0340.023724
H20.2440.0266110
O21.360.0318405
CO23.590.0427710
Van der Waals — NEET Relevance
For NEET, the most common question on van der Waals equation is identifying the correct units of 'a' and 'b' or comparing Z values for different gases. Remember: 'a' has units of pressure×volume² (atm L² mol−2), 'b' has units of volume (L mol−1). Gases with stronger intermolecular forces (higher 'a') have lower Boyle temperatures and show greater deviation at room temperature.

Ideal Gas Equation

The ideal gas equation combines all gas laws into one:

PV = nRT

where P = pressure (Pa), V = volume (m³), n = number of moles, R = universal gas constant = 8.314 J mol−1 K−1, T = absolute temperature (K).

In terms of the number of molecules N: PV = NkT, where k = R/NA = 1.38 × 10−23 J/K is the Boltzmann constant. The ideal gas equation is accurate for real gases at low pressures and high temperatures (when intermolecular forces and molecular volume become negligible).

STP conditions: At standard temperature and pressure (0°C = 273 K, 1 atm = 1.013 × 105 Pa), 1 mole of an ideal gas occupies 22.4 L (2.24 × 10−2 m³). This is a useful reference point for NEET problems.

Example 4 — Ideal Gas Equation
A gas occupies 5 L at 27°C and 2 atm pressure. How many moles of gas are present? (R = 0.0821 L atm mol−1 K−1)
Solution: PV = nRT. P = 2 atm, V = 5 L, T = 27 + 273 = 300 K. n = PV/RT = (2 × 5) / (0.0821 × 300) = 10 / 24.63 = 0.406 moles.
Example 4b — Density from Ideal Gas
Find the density of nitrogen gas (M = 28 g/mol) at 27°C and 2 atm pressure. (R = 0.0821 L atm mol−1 K−1)
Solution: From PV = (m/M)RT, we get density ρ = m/V = PM/RT. ρ = (2 × 28) / (0.0821 × 300) = 56 / 24.63 = 2.27 g/L. Note that gas density increases with pressure and decreases with temperature.
Value of R to Remember
R = 8.314 J mol−1 K−1 (SI units). Also remember: R = 0.0821 L atm mol−1 K−1 (for P in atm, V in L). Use the one that matches your units! In caloric units: R = 2 cal mol−1 K−1.

Kinetic Theory & Molecular Speeds

The kinetic theory of gases makes these assumptions: gas molecules are point particles in constant random motion; collisions are perfectly elastic; no intermolecular forces except during collisions; the volume of molecules is negligible compared to gas volume; the duration of collisions is negligible; and the gas obeys Newton's laws of motion.

Pressure from kinetic theory derivation: Consider N molecules in a cubical box of side L. A molecule with velocity component vx collides with a wall, and the change in momentum per collision is 2mvx. The time between collisions with the same wall is 2L/vx, giving force F = Δp/Δt = (2mvx) / (2L/vx) = mvx²/L. Summing over all molecules and averaging: P = F/A = (m/L³) Σvx² = (Nm/V) <vx²>. Since <v²> = <vx²> + <vy²> + <vz²> = 3<vx²>, we get: P = (1/3)(Nm/V)<v²> = (1/3)ρ<v²>.

Root mean square speed: vrms = √(3RT/M) = √(3kT/m), where M = molar mass (kg/mol), m = mass per molecule. The RMS speed is proportional to the square root of temperature and inversely proportional to the square root of molar mass.

Average speed: vavg = √(8RT/πM) = 0.921 × vrms

Most probable speed: vmp = √(2RT/M) = 0.816 × vrms

Relation between speeds: vrms : vavg : vmp = √3 : √(8/π) : √2 ≈ 1.732 : 1.596 : 1.414. For NEET, remember that vrms is the largest and vmp is the smallest of the three speeds.

Mean free path (λ): The average distance a molecule travels between collisions. λ = kT / (√2 π d² P), where d is the molecular diameter. At constant T, λ ∝ 1/P. At constant P, λ ∝ T. The mean free path is independent of the number of different gas species in a mixture (at a given P and T).

Maxwell-Boltzmann distribution: The distribution of molecular speeds in a gas follows the Maxwell-Boltzmann distribution function f(v) = 4πN(m/2πkT)3/2 v² e−mv²/(2kT). The area under the curve gives the total number of molecules. As temperature increases, the curve becomes broader and flatter, with the peak shifting to higher speeds.

Example 5 — RMS Speed
Calculate the RMS speed of oxygen molecules (M = 32 g/mol) at 27°C. (R = 8.314 J mol−1 K−1)
Solution: vrms = √(3RT/M). T = 27 + 273 = 300 K. M = 32 g/mol = 0.032 kg/mol. vrms = √(3 × 8.314 × 300 / 0.032) = √(7482.6 / 0.032) = √(233,831) ≈ 484 m/s.
Speed Formula Shortcut
For NEET, remember: vrms ∝ √T/M. If temperature doubles, vrms increases by √2. If gas is changed from O2 (32) to He (4), vrms increases by √(32/4) = √8 = 2.83 times.

Degrees of Freedom, Specific Heat & Equipartition Theorem

The degrees of freedom (f) of a gas molecule is the number of independent coordinates required to specify its position and configuration fully. According to the equipartition theorem, energy is equally distributed among all degrees of freedom, with each degree contributing (1/2)kT of energy per molecule:

  • Monatomic (He, Ne, Ar): f = 3 (only 3 translational). No rotational or vibrational modes because a single atom has negligible moment of inertia.
  • Diatomic (O2, N2, H2): f = 5 at moderate T (3 translational + 2 rotational). At high temperatures (>500 K), 2 vibrational modes (kinetic + potential) are also excited, giving f = 7. At low temperatures (<100 K), rotational modes may freeze out, giving f = 3.
  • Polyatomic linear (CO2): f = 5 (3 translational + 2 rotational) — similar to diatomic because linear molecules have only 2 rotational axes.
  • Polyatomic non-linear (NH3, H2O): f = 6 (3 translational + 3 rotational). With vibration, additional modes contribute.

Temperature dependence of specific heats: For diatomic gases like H2, Cv = (3/2)R at very low T (only translation), (5/2)R at moderate T (translation + rotation), and (7/2)R at high T (translation + rotation + vibration). This variation is observed experimentally and confirms the quantum nature of energy storage — rotational and vibrational modes are "frozen out" at low temperatures.

Equipartition of energy: Each degree of freedom contributes (1/2)kT of energy per molecule or (1/2)RT per mole.

Internal energy: U = (f/2) nRT

Molar specific heats: Cv = (f/2)R, Cp = Cv + R = (f/2 + 1)R, γ = Cp/Cv

TypefCvCpγ = Cp/CvExamples
Monatomic33R/25R/25/3 = 1.67He, Ne, Ar
Diatomic (rigid)55R/27R/27/5 = 1.40N2, O2, H2
Polyatomic (non-linear)63R4R4/3 = 1.33CO2, NH3
γ Values to Memorise
γ = 5/3 for monatomic gases (He, Ne, Ar). γ = 7/5 for diatomic gases (N2, O2, H2). γ = 4/3 for polyatomic gases (CO2). These are frequently tested in NEET.
Example 5b — Specific Heat Ratio
The molar specific heat at constant pressure for a gas is 7R/2. Find the degrees of freedom and the value of γ.
Solution: Cp = 7R/2. Since Cp = (f/2 + 1)R, we have (f/2 + 1)R = 7R/2 ⇒ f/2 + 1 = 7/2 ⇒ f/2 = 5/2 ⇒ f = 5. This is a diatomic gas. Cv = 5R/2. γ = Cp/Cv = (7R/2)/(5R/2) = 7/5 = 1.40.
Maxwell-Boltzmann Speed Distribution Speed f(v) T3 > T2 > T1 vmp vavg vrms
Maxwell-Boltzmann speed distribution for a gas at three different temperatures. As T increases, the curve flattens and shifts right. Note: vmp < vavg < vrms.

3. Thermodynamics

Section 3: Key Topics at a Glance
TopicKey ConceptEssential Formula
Zeroth LawThermal equilibrium & temperature concept
First LawConservation of energyΔU = Q − W
Thermodynamic ProcessesFour types with different constantsPVγ = const (adiabatic)
Second LawEntropy & direction of processesΔS ≥ 0
Heat EnginesConverting heat to workη = 1 − Tc/Th

Zeroth Law of Thermodynamics

The Zeroth Law states: 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 as a fundamental property. It allows us to use thermometers — a thermometer (system C) placed in contact with a body (A) reaches thermal equilibrium, and the thermometer reading gives the temperature.

First Law & Internal Energy

The First Law of Thermodynamics is the law of conservation of energy for thermodynamic systems:

ΔU = Q − W

where ΔU = change in internal energy, Q = heat added to the system, W = work done BY the system.

Sign convention (important for NEET):

  • Q > 0: Heat flows INTO the system
  • Q < 0: Heat flows OUT of the system
  • W > 0: Work done BY the system (expansion)
  • W < 0: Work done ON the system (compression)
  • ΔU > 0: Internal energy increases

Internal energy (U) is a state function — it depends only on the state of the system, not on the path taken. Heat (Q) and work (W) are path functions.

Example 6 — First Law
A gas is compressed by doing 500 J of work on it, and it absorbs 300 J of heat. Find the change in internal energy.
Solution: ΔU = Q − W. Work is done ON the system, so W = −500 J (negative by convention). Heat absorbed Q = +300 J. ΔU = 300 − (−500) = 300 + 500 = 800 J. The internal energy increases by 800 J.
Example 6b — Cyclic Process
A gas undergoes a cyclic process ABCA. In process AB (isobaric expansion), 400 J of heat is absorbed and 200 J of work is done. In BC (isochoric), 150 J of heat is released. In CA (adiabatic compression), find the work done.
Solution: For a cycle, ΔU = 0. So Qnet = Wnet. QAB = +400 J, QBC = −150 J, QCA = 0 (adiabatic). Qnet = 400 − 150 + 0 = 250 J. WAB = +200 J, WBC = 0 (isochoric). So Wnet = 200 + WCA = 250 ⇒ WCA = 50 J. Since it is compression, work is done ON the system, so WCA = −50 J (using sign convention, W = −50 J means 50 J work done on system).
Cyclic Process Rule
For any cyclic process (system returns to initial state), ΔU = 0, therefore Qnet = Wnet. This is a powerful NEET shortcut — if you know the net heat exchange, you directly know the net work output, and vice versa.
Sign Convention Trick
Remember: ΔU = Q − W. If work is done BY the system (expansion), W is positive and internal energy decreases (system loses energy doing work). If work is done ON the system (compression), W is negative and internal energy increases. Always ask: "Who is doing work on whom?"

Thermodynamic Processes

Isobaric, Isochoric, Isothermal, Adiabatic

ProcessConstantWΔUQKey Relation
IsobaricPPΔVnCvΔTnCpΔTV/T = constant
IsochoricV0QnCvΔTP/T = constant
IsothermalTnRT ln(V2/V1)0WPV = constant
AdiabaticQ = 0−ΔU−W0PVγ = constant

Additional adiabatic relations: TVγ−1 = constant, P1−γ Tγ = constant.

Work in adiabatic process: W = (P1V1 − P2V2)/(γ − 1) = nR(T1 − T2)/(γ − 1). For expansion, W > 0 (work done by system) and T decreases. For compression, W < 0 (work done on system) and T increases.

Isobaric process details: In an isobaric process, pressure remains constant. The work done is W = PΔV = nRΔT. For an ideal gas, the heat added is Q = nCpΔT, and the change in internal energy is ΔU = nCvΔT ≠ 0. The fraction of heat used for work is W/Q = R/Cp = 1/γ * (γ−1) = (γ−1)/γ. For monatomic gas (Cp=5R/2), only 40% of the heat goes into work; the remaining 60% increases internal energy (temperature).

Isochoric process details: In an isochoric process, volume is constant, so no work is done (W = 0). All the heat added goes into increasing internal energy: Q = ΔU = nCvΔT. The pressure increases linearly with temperature: P/T = constant.

Example 6c — Isobaric Process
An ideal gas at 2 atm pressure occupies 3 L at 27°C. It is heated at constant pressure until its volume doubles. Find (a) the final temperature, (b) the work done, and (c) the heat supplied. (Cp = 7R/2, R = 0.0821 L atm/mol·K)
Solution: (a) For isobaric process: V1/T1 = V2/T2 ⇒ 3/300 = 6/T2 ⇒ T2 = 600 K. (b) W = PΔV = 2 × (6−3) = 6 L atm = 6 × 101.3 = 607.8 J. (c) n = PV/RT = (2×3)/(0.0821×300) = 6/24.63 = 0.244 mol. Q = nCpΔT = 0.244 × (7/2 × 8.314) × 300 = 0.244 × 29.1 × 300 = 2130 J. Check: ΔU = Q − W = 2130 − 608 = 1522 J. Also ΔU = nCvΔT = 0.244 × (5R/2) × 300 = 0.244 × 20.785 × 300 = 1522 J. ✓
Example 7 — Isothermal Process
An ideal gas at 27°C expands isothermally from 2 L to 8 L. Calculate the work done if the initial pressure is 5 atm. (Given: 1 L atm = 101.3 J)
Solution: For isothermal process, W = nRT ln(V2/V1). First find n: n = PV/RT = (5 × 2)/(0.0821 × 300) = 10/24.63 = 0.406 mol. W = 0.406 × 8.314 × 300 × ln(8/2) = 0.406 × 8.314 × 300 × 1.386 = 1403 J. (Alternatively, W = P1V1 ln(V2/V1) = 10 L atm × 1.386 × 101.3 = 1404 J.)
Example 8 — Adiabatic Process
A gas (γ = 5/3) at 300 K is compressed adiabatically to 1/8 of its initial volume. Find the final temperature.
Solution: TVγ−1 = constant. T2 = T1(V1/V2)γ−1 = 300 × 82/3 = 300 × (23)2/3 = 300 × 22 = 300 × 4 = 1200 K = 927°C.
Process Identification
To identify the process in NEET problems, look at what is held constant: ΔP = 0 ⇒ isobaric; ΔV = 0 ⇒ isochoric; ΔT = 0 ⇒ isothermal; Q = 0 ⇒ adiabatic. Also: isothermal curves on PV diagram are rectangular hyperbolas; adiabatic curves are steeper.

Enthalpy (H)

Enthalpy is defined as H = U + PV. It is a state function that represents the total heat content of a system. For a constant-pressure process (the most common scenario in chemistry and many physics problems):

ΔH = ΔU + PΔV = Qp

For isobaric processes, the heat exchanged equals the change in enthalpy: Qp = nCpΔT = ΔH. For isochoric processes, Qv = ΔU = nCvΔT. The relationship between ΔH and ΔU for ideal gases is: ΔH = ΔU + ΔngRT, where Δng is the change in the number of moles of gas.

Enthalpy vs Internal Energy
In NEET, remember: ΔH = ΔU + ΔngRT. For reactions where the number of gas moles does not change (Δng = 0), ΔH = ΔU. For reactions producing gas (Δng > 0), ΔH > ΔU. For reactions consuming gas (Δng < 0), ΔH < ΔU.

Thermodynamic Potentials (Advanced)

Thermodynamic potentials are state functions that describe the energy content of a system under different constraints. The four fundamental thermodynamic potentials are:

PotentialDefinitionNatural VariablesInfinitesimal Change
Internal Energy (U)Total energy of systemS, VdU = TdS − PdV
Enthalpy (H)H = U + PVS, PdH = TdS + VdP
Helmholtz Free Energy (F)F = U − TST, VdF = −SdT − PdV
Gibbs Free Energy (G)G = H − TST, PdG = −SdT + VdP

Gibbs free energy (G) is particularly important for determining the spontaneity of a process at constant temperature and pressure: if ΔG < 0, the process is spontaneous; if ΔG = 0, the system is at equilibrium; if ΔG > 0, the process is non-spontaneous. The change in Gibbs free energy is related to enthalpy and entropy changes by: ΔG = ΔH − TΔS.

Maxwell's relations are a set of four equations derived from the equality of mixed partial derivatives of the thermodynamic potentials. They provide useful relationships between seemingly unrelated thermodynamic quantities. For NEET, the key takeaway is that state functions have exact differentials, which means the order of partial differentiation does not matter.

Second Law of Thermodynamics & Entropy

Kelvin-Planck statement: It is impossible to construct a heat engine that converts heat completely into work without any other effect. In other words, no heat engine can have 100% efficiency because some heat must always be rejected to a cold reservoir.

Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body without external work being done. This is why refrigerators require work input — they pump heat against the natural direction of flow.

Equivalence of statements: The Kelvin-Planck and Clausius statements are equivalent. If one were false, the other would also be false. A violation of the Kelvin-Planck statement (a 100% efficient engine) could be used to drive a refrigerator that transfers heat from cold to hot without any net work input, violating the Clausius statement.

Entropy (S): A measure of disorder or randomness of a system. The change in entropy for a reversible process is: ΔS = ∫ dQrev/T. For an isothermal process at temperature T, ΔS = Q/T. For a reversible process, ΔSuniverse = 0. For an irreversible (spontaneous) process, ΔSuniverse > 0. The Second Law can be stated as: the entropy of an isolated system never decreases. In statistical mechanics, entropy is related to the number of microstates (W) by Boltzmann's formula: S = k ln W.

Example 8c — Entropy Change
Calculate the entropy change when 100 g of ice melts at 0°C. (Lf = 336 J/g)
Solution: Melting is a reversible isothermal process at T = 273 K. Q = mLf = 100 × 336 = 33600 J. ΔS = Q/T = 33600/273 = 123.1 J/K. The entropy of the system increases because the liquid state is more disordered (higher entropy) than the solid state.
Entropy & Spontaneity
Natural processes always increase the total entropy of the universe. For example, when a hot object cools in a cold room, heat flows from hot to cold, and the total entropy increases. The reverse process (heat flowing from cold to hot) would decrease entropy and is forbidden by the Second Law. Entropy is often called "time's arrow" because it gives a direction to thermodynamic processes.

Heat Engines & Carnot Cycle

A heat engine is a device that converts heat into work. It operates in a cyclic process, absorbing heat Q1 from a hot reservoir, doing work W, and rejecting heat Q2 to a cold reservoir.

Efficiency: η = W/Q1 = 1 − Q2/Q1

Carnot engine: The most efficient heat engine operating between two temperatures. It consists of four reversible processes:

  • Isothermal expansion (T1) — absorbs heat Q1
  • Adiabatic expansion (T1 → T2)
  • Isothermal compression (T2) — rejects heat Q2
  • Adiabatic compression (T2 → T1)

Carnot efficiency: ηmax = 1 − T2/T1 (both temperatures in Kelvin). No engine operating between two temperatures can be more efficient than a Carnot engine.

Refrigerator / Heat pump: Coefficient of Performance (COP) = Q2/W = Q2/(Q1 − Q2). For a Carnot refrigerator: COP = T2/(T1 − T2).

Carnot Cycle — PV Diagram V P A B C D T1 (Hot) T2 (Cold)
Carnot cycle on a PV diagram: AB (isothermal expansion), BC (adiabatic expansion), CD (isothermal compression), DA (adiabatic compression).
Example 9 — Carnot Efficiency
A Carnot engine operates between 127°C and 27°C. Find its efficiency and the work output per cycle if it absorbs 1000 J of heat.
Solution: T1 = 127 + 273 = 400 K, T2 = 27 + 273 = 300 K. η = 1 − T2/T1 = 1 − 300/400 = 0.25 = 25%. W = η × Q1 = 0.25 × 1000 = 250 J. Heat rejected Q2 = Q1 − W = 1000 − 250 = 750 J.
Efficiency Shortcut
Carnot efficiency η = 1 − Tc/Th. Remember: convert °C to K first! Many students forget and get the wrong answer. Efficiency is ALWAYS less than 1 (or 100%) for any real engine.
Important Thermodynamic Processes on PV Diagram V P Isobaric (P = const) Isochoric (V = const) Isothermal (T = const) Adiabatic (Q = 0)
PV diagram showing all four thermodynamic processes: isobaric (horizontal line), isochoric (vertical line), isothermal (hyperbolic curve), and adiabatic (steeper hyperbolic curve). Each process has a distinct graphical representation.
Example 9b — Refrigerator COP
A Carnot refrigerator extracts 400 J of heat from a cold reservoir at 250 K while doing 200 J of work. Find the coefficient of performance and the heat rejected to the hot reservoir.
Solution: COP = Q2/W = 400/200 = 2. From first law for the cycle: W = Q1 − Q2 ⇒ 200 = Q1 − 400 ⇒ Q1 = 600 J. Alternatively, using Carnot COP: COP = T2/(T1 − T2) ⇒ 2 = 250/(T1 − 250) ⇒ T1 = 250 + 125 = 375 K.
Heat Engine vs Refrigerator
Heat engine: absorbs heat from hot reservoir, does work, rejects rest to cold reservoir (η = W/Qh). Refrigerator: absorbs work, extracts heat from cold reservoir, rejects sum to hot reservoir (COP = Qc/W). They are thermodynamic opposites!
DevicePurposeEfficiency / COPRelation
Heat EngineConvert heat to workη = 1 − Qc/Qhη ≤ 1 − Tc/Th
RefrigeratorExtract heat from cold bodyCOP = Qc/WCOP ≤ Tc/(Th − Tc)
Heat PumpDeliver heat to hot bodyCOP = Qh/WCOP ≤ Th/(Th − Tc)
Heat Engine & Refrigerator — Schematic Heat Engine Qh → W + Qc Qh Qc W Refrigerator W + Qc → Qh W Qh Qc
Left: Heat engine absorbs Qh from hot reservoir, produces work W, rejects Qc to cold reservoir. Right: Refrigerator absorbs work W, extracts Qc from cold reservoir, rejects Qh = Qc + W to hot reservoir.
Example 8b — Adiabatic Relations
A diatomic gas (γ = 1.4) initially at 27°C and 1 atm is compressed adiabatically to half its volume. Find the final pressure and temperature.
Solution: PVγ = constant ⇒ P2 = P1(V1/V2)γ = 1 × 21.4 = 1 × 2.64 = 2.64 atm. TVγ−1 = constant ⇒ T2 = T1(V1/V2)γ−1 = 300 × 20.4 = 300 × 1.32 = 396 K = 123°C.
PV Diagram — Isothermal vs Adiabatic V P Isothermal (PV = const) Adiabatic (PVγ = const) Adiabatic is steeper because γ > 1
Comparison of isothermal and adiabatic curves on a PV diagram. For the same initial point, the adiabatic curve is steeper than the isothermal curve since γ > 1. During expansion, adiabatic curve falls faster because temperature also decreases.

4. Heat Transfer

Section 4: Key Topics at a Glance
TopicKey ConceptEssential Formula
ConductionHeat transfer through solidsQ/t = kA(ΔT)/d
ConvectionHeat transfer through fluid motionQ/t = hAΔT
RadiationHeat transfer via EM wavesP = εσAT4
Wien's LawPeak wavelength vs temperatureλmT = b
Newton's CoolingRate of cooling ∝ ΔT−dT/dt = k(T−T0)

Conduction

Conduction is the transfer of heat through a material without bulk motion of the material. It occurs due to the transfer of kinetic energy between adjacent molecules (in solids) or via free electrons (in metals). Good electrical conductors (like copper and silver) are also good thermal conductors because the same free electrons carry both charge and thermal energy.

Fourier's law of heat conduction states that the rate of heat flow through a material is proportional to the temperature gradient and the cross-sectional area:

Q/t = k A (T1 − T2) / d

where k = thermal conductivity (W m−1 K−1), A = cross-sectional area (m²), d = thickness (m), T1 − T2 = temperature difference (K).

Thermal resistance: Rth = d/(kA). The unit of thermal resistance is K/W. For series combination of slabs: Req = R1 + R2 + ... (same heat current through each, temperature drops add). For parallel combination: 1/Req = 1/R1 + 1/R2 + ... (same temperature difference across each, heat currents add). The temperature at the junction of two slabs in series with temperatures T1 and T2 at the outer ends is: Tj = (k1T1/d1 + k2T2/d2) / (k1/d1 + k2/d2). This formula is derived from the condition that the heat current through both slabs is equal.

Heat conduction through a composite wall: For a wall made of n layers in series, the total thermal resistance is R = Σ(di/kiAi). If the areas are equal, the heat current is: Q/t = (Th − Tc) / Σ(di/kiA). The temperature at any interface can be found by calculating the temperature drop across each layer: ΔTi = (Q/t) × (di/kiA).

Materialk (W m−1 K−1)Materialk (W m−1 K−1)
Silver428Water0.6
Copper401Glass (Pyrex)1.0
Aluminium237Wood0.12
Iron80Air (still)0.024

Convection

Winds & Monsoons — Large-Scale Convection: The differential heating of the Earth's surface creates global convection currents that drive weather patterns. During summer, the Indian subcontinent heats more than the surrounding Indian Ocean, causing air to rise (low pressure) and drawing in moist ocean air from the south-west — this is the mechanism of the Indian summer monsoon. Land and sea breezes are smaller-scale examples: during the day, land heats faster than the sea, so air rises over land and cooler sea air moves in (sea breeze). At night, the land cools faster, and the pattern reverses (land breeze). These concepts are important for both NEET physics (convection heat transfer) and UPSC GS geography.

Convection is heat transfer by the bulk movement of fluids (liquids and gases). It involves two mechanisms simultaneously: heat conduction within the fluid and the macroscopic motion of the fluid itself. Convection can be classified into two types:

Natural (free) convection occurs due to density changes caused by temperature gradients. When a fluid is heated, it expands, becomes less dense, and rises. Cooler, denser fluid sinks to replace it, creating a convection current. Examples include: the circulation of air in a room (hot air rises to the ceiling), sea breezes (land heats faster than sea during the day, causing air to rise over land and cooler air from the sea to move in), and the movement of magma in the Earth's mantle (driving plate tectonics).

Forced convection uses external means like fans, pumps, or blowers to enhance fluid motion and heat transfer. Examples include: cooling fans in computers, car radiators using coolant pumps, and air conditioning systems using blowers. Forced convection is generally much more efficient than natural convection because the fluid velocity is higher.

The rate of convective heat transfer is given by Newton's law of cooling for convection: Q/t = h A ΔT, where h is the convective heat transfer coefficient (W m−2 K−1), A is the surface area, and ΔT is the temperature difference between the surface and the fluid. The value of h depends on the fluid properties, flow velocity, and geometry, and is typically higher for forced convection than for natural convection.

Convection — NEET Relevance
For NEET, remember: convection requires a fluid medium (cannot occur in vacuum). Natural convection is driven by gravity (buoyancy). Hot air rises — this is why room heaters are placed near the floor and air conditioners near the ceiling. In the absence of gravity (space), natural convection does not occur, and only conduction and radiation are relevant.

Radiation

Radiation is the transfer of heat via electromagnetic waves (infrared radiation). It requires no medium and can occur through vacuum. All bodies above absolute zero emit thermal radiation. The rate of emission depends on temperature, surface area, and nature of the surface.

Emissive power (E): The energy radiated per unit area per unit time by a body at a given temperature. For a black body, Eb = σT4 (Stefan-Boltzmann law). For a real body, E = εσT4, where ε is the emissivity (0 < ε < 1).

Absorptive power (a): The fraction of incident radiation absorbed by a body. For a black body, a = 1 (perfect absorber). For a real body, 0 < a < 1. Kirchhoff's law states that at thermal equilibrium, the emissivity and absorptive power are equal: ε = a. This means good absorbers are also good emitters, and good reflectors are poor emitters.

Key properties of thermal radiation:

  • It travels at the speed of light (3 × 108 m/s)
  • It follows the inverse square law (intensity ∝ 1/r²)
  • It can be reflected, refracted, and absorbed like light
  • The wavelength distribution depends on temperature (Wien's law)
  • A black body is an ideal absorber (absorbs all incident radiation) and also an ideal emitter

Stefan-Boltzmann Law

The power radiated by a black body is proportional to the fourth power of its absolute temperature. This was discovered experimentally by Stefan and derived theoretically by Boltzmann:

P = ε σ A T4

where ε = emissivity (0 to 1, ε = 1 for a perfect black body), σ = Stefan-Boltzmann constant = 5.67 × 10−8 W m−2 K−4, A = surface area, T = absolute temperature (K).

Net power radiated (if surroundings at T0): Pnet = ε σ A (T4 − T04). If T > T0, the body loses net energy (net cooling). If T < T0, the body gains net energy (net warming). At thermal equilibrium (T = T0), Pnet = 0 (detailed balance). This principle explains why a hot cup of coffee cools down — it radiates more energy than it absorbs until room temperature is reached. Conversely, a cold drink placed in a warm room warms up as it absorbs more radiation than it emits.

Kirchhoff's law of radiation: At a given temperature, the ratio of emissive power to absorptive power is constant for all bodies. For a black body (absorptive power = 1), emissive power is maximum. Therefore, good absorbers are also good emitters at any given temperature.

Wien's Displacement Law

The wavelength at which the radiation intensity is maximum is inversely proportional to temperature:

λm T = b

where b = Wien's constant = 2.898 × 10−3 m K. As temperature increases, the peak wavelength shifts to shorter values (e.g., a hot star appears blue, a cooler star appears red). This law is used in:

  • Estimating stellar surface temperatures from their colour
  • Designing infrared thermometers
  • Understanding the cosmic microwave background radiation (T ≈ 2.7 K, λm ≈ 1 mm)

Newton's Law of Cooling

The rate of cooling of a body is proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small (ΔT < 30°C):

−dT/dt = k (T − T0)

where k is a constant that depends on the surface area, nature of the surface, and the medium. The negative sign indicates that temperature decreases with time when T > T0.

For small temperature differences (ΔT < 30°C), Newton's law in its approximate form is: (T1 − T2)/t = K [(T1 + T2)/2 − T0], where T1 and T2 are temperatures at the start and end of the time interval, and K = 4k is approximately constant. The exact solution of the differential equation gives: ln[(T − T0)/(Ti − T0)] = −kt, where Ti is the initial temperature.

Verification of Newton's Law: If a body cools from T1 to T2 in time t, and from T2 to T3 in the same time t, then using the exact logarithmic form: ln[(T1 − T0)/(T2 − T0)] = ln[(T2 − T0)/(T3 − T0)] = kt. This implies (T1 − T0)/(T2 − T0) = (T2 − T0)/(T3 − T0), i.e., the temperature differences decay geometrically. In the approximate linear form, we use the arithmetic mean temperature instead.

Limitations of Newton's Law: It is accurate only when the temperature difference is small (<30°C). For large temperature differences, the Stefan-Boltzmann law must be used, which gives a more complex cooling curve.

The Greenhouse Effect — An Application of Radiation: The Earth's atmosphere is transparent to visible light from the Sun (short wavelength, ≈500 nm at 5800 K). The Earth's surface absorbs this radiation and re-emits it as infrared radiation (long wavelength, ≈10 μm at 288 K). Greenhouse gases (CO2, H2O, CH4) absorb this infrared radiation and re-emit it in all directions, including back toward the Earth's surface. This trapping of heat keeps the Earth's average temperature at about 15°C instead of −18°C (which it would be without an atmosphere). This natural greenhouse effect is essential for life, but human activities (burning fossil fuels) are increasing greenhouse gas concentrations, leading to enhanced global warming — a critical topic for both NEET and UPSC GS.

Solar Energy & Solar Constant: The solar constant (S ≈ 1366 W/m²) is the amount of solar radiation received per unit area at the top of the Earth's atmosphere, measured perpendicular to the Sun's rays. It can be calculated from the Sun's surface temperature (Stefan-Boltzmann law) and the Earth-Sun distance (inverse square law). The actual solar radiation reaching the Earth's surface is less due to atmospheric absorption, reflection by clouds, and scattering by air molecules (Rayleigh scattering — why the sky appears blue). The Earth's albedo (reflectivity) is about 0.3, meaning 30% of incoming solar radiation is reflected back to space. The remaining 70% is absorbed by the Earth's surface and atmosphere, driving weather patterns, ocean currents, and photosynthesis — the ultimate source of almost all energy on Earth.

Newton's Law of Cooling — Temperature vs Time Time (t) T T0 Cooling curve Troom
Newton's law of cooling: temperature difference decays exponentially with time. The rate of cooling is maximum at the start (when ΔT is largest) and decreases as the body approaches room temperature.
Newton's Cooling — Equal Interval Rule
If a body cools from T1 to T2 in time t, and from T2 to T3 in the same time t, then (T1 − T2) / (T2 − T3) = (T1 + T2 − 2T0) / (T2 + T3 − 2T0). This is a direct consequence of the approximate form and is often tested in NEET. Also, if the time intervals are equal, the temperature differences form a geometric progression in the exact logarithmic form.
UPSC GS Connection: Thermodynamics in Environment & Climate
The greenhouse effect (Stefan-Boltzmann + Wien's law), atmospheric convection (global wind patterns, monsoon formation), and thermal expansion of oceans (sea level rise due to global warming) are key intersections of thermodynamics with environment and geography — important for UPSC GS Paper 1. Understanding black body radiation is essential for analyzing Earth's energy budget and the enhanced greenhouse effect due to anthropogenic CO2 emissions.
Black Body Radiation Spectrum λ Intensity T1 (5000 K) T2 (4000 K)
Black body radiation curves at two temperatures. Higher temperature → higher peak intensity and shorter peak wavelength (Wien's displacement).
Newton's Cooling Exponential Form
The solution of −dT/dt = k(T − T0) is: (T − T0) = (Ti − T0) e−kt. The temperature difference decays exponentially. For equal time intervals, the ratio of successive temperature differences is constant.
Example 10b — Newton's Law of Cooling
A body cools from 80°C to 64°C in 5 minutes when the surrounding temperature is 20°C. How much more time will it take to cool from 64°C to 52°C?
Solution: Using the approximate form: (T1 − T2)/t = K[(T1+T2)/2 − T0]. First interval: (80−64)/5 = K[(80+64)/2 − 20] ⇒ 16/5 = K(72−20) = 52K ⇒ K = 16/(5×52) = 16/260 = 0.0615. Second interval: (64−52)/t = 0.0615[(64+52)/2 − 20] ⇒ 12/t = 0.0615(58−20) = 0.0615×38 ⇒ t = 12/(0.0615×38) = 12/2.337 = 5.13 minutes.
Conduction Series & Parallel Trick
For slabs in series: total thermal resistance R = R1 + R2 + ..., and heat current is same through all. For slabs in parallel: total conductance 1/R = 1/R1 + 1/R2 + ..., and temperature difference is same across all. Use the electrical analogy: V → ΔT, I → Q/t, R → d/(kA).
Example 10c — Conduction in Series
A composite wall has an iron layer (k = 80 W/mK, thickness = 2 cm) and a wood layer (k = 0.12 W/mK, thickness = 3 cm) in series. The outer iron surface is at 100°C and the outer wood surface at 20°C. Find the interface temperature and heat flux. (Area = 1 m²)
Solution: In series, heat current Q/t is same. Riron = d/(kA) = 0.02/(80×1) = 2.5 × 10−4 K/W. Rwood = 0.03/(0.12×1) = 0.25 K/W. Total R = 0.25025 K/W. Q/t = ΔT/R = (100−20)/0.25025 = 319.7 W. Temperature drop across iron: ΔTiron = (Q/t) × Riron = 319.7 × 2.5×10−4 = 0.08°C. Interface temperature = 100 − 0.08 = 99.92°C. Most of the drop occurs across the poor conductor (wood).
Key Formulas — Thermodynamics at a Glance
TopicFormulaVariables
Linear ExpansionΔL = αL0ΔTα = expansion coeff.
Volume ExpansionΔV = γV0ΔTγ ≈ 3α
Specific HeatQ = mcΔTc in J/kg·K
Latent HeatQ = mLLf or Lv
Ideal GasPV = nRTR = 8.314 J/mol·K
RMS Speedvrms = √(3RT/M)M in kg/mol
Internal EnergyU = (f/2)nRTf = degrees of freedom
Molar Specific HeatsCv = (f/2)R, Cp = Cv + Rγ = Cp/Cv
First LawΔU = Q − WSign convention matters!
Isothermal WorkW = nRT ln(V2/V1)ΔU = 0
AdiabaticPVγ = constQ = 0
Carnot Efficiencyη = 1 − Tc/ThT in Kelvin
ConductionQ/t = kA(ΔT)/dk = thermal conductivity
Stefan-BoltzmannP = εσAT4σ = 5.67×10−8
Wien's LawλmT = bb = 2.898×10−3 mK
Top 5 Common NEET Mistakes in Thermodynamics
1. Forgetting to convert °C to K in gas equations and efficiency formulas.
2. Getting the sign of W wrong in ΔU = Q − W (W positive = work BY system).
3. Mixing up Cp and Cv — remember Cp > Cv by R.
4. Using γ = 5/3 for diatomic gases (it is 7/5 = 1.4, not 5/3 = 1.67).
5. Confusing isothermal and adiabatic curves on PV diagrams — adiabatic is steeper.
Example 10g — Mixed Concept: First Law + Ideal Gas
An ideal diatomic gas (1 mole) at 300 K expands isobarically to double its volume. It is then cooled isochorically to its original temperature. Find (a) total work done, (b) total heat exchanged, and (c) total change in internal energy. (Cv = 5R/2)
Solution: (a) Process AB (isobaric): WAB = PΔV = RΔT = R(600−300) = 300R. Process BC (isochoric): WBC = 0. Total W = 300R = 300 × 8.314 = 2494 J. (b) QAB = nCpΔT = 1 × (7R/2) × 300 = 1050R = 8730 J. QBC = nCvΔT = 1 × (5R/2) × (−300) = −750R = −6236 J. Total Q = 8730 − 6236 = 2494 J. (c) Since the system returns to the original temperature, ΔUtotal = 0. Check: ΔU = Q − W = 2494 − 2494 = 0. ✓
Cyclic Process on PV Diagram V P A → B: Isobaric C → A: Isochoric B → C: Cooling A B C
A cyclic process on a PV diagram: A→B (isobaric expansion), B→C (isochoric cooling), C→A (compression back to initial state). For a complete cycle, ΔU = 0 and Qnet = Wnet.
Thermodynamic Cycle Checklist
When solving cyclic process problems: (1) Identify each leg of the cycle. (2) For each leg, identify the type of process (isobaric, isochoric, isothermal, adiabatic). (3) Calculate Q, W, ΔU for each leg using the appropriate formulas. (4) Sum them up — ΔUtotal should be 0 for a cycle. (5) Verify Qnet = Wnet.
Example 10d — Wien's Displacement
The Sun's surface temperature is approximately 5800 K. Find the peak wavelength of its radiation. (b = 2.898 × 10−3 m K)
Solution: Wien's law: λmT = b ⇒ λm = b/T = 2.898 × 10−3 / 5800 = 5.0 × 10−7 m = 500 nm (green light). This is why the Sun's peak radiation is in the visible spectrum. A cooler star (3000 K) would peak in the infrared at λm = 966 nm.
Example 10e — Net Radiation
A spherical black body of radius 5 cm is at 327°C in a room at 27°C. Find the net rate of heat loss by radiation. (σ = 5.67 × 10−8 W m−2 K−4)
Solution: Tbody = 327 + 273 = 600 K, Troom = 27 + 273 = 300 K. For a black body, ε = 1. Surface area A = 4πr² = 4π(0.05)² = 0.0314 m². Pnet = σA(T14 − T24) = 5.67 × 10−8 × 0.0314 × (6004 − 3004) = 5.67 × 10−8 × 0.0314 × (1296 × 108 − 81 × 108) = 5.67 × 10−8 × 0.0314 × 1215 × 108 = 5.67 × 0.0314 × 1215 = 216.3 W.
NEET Previous Year Question Trend
In the last 5 years of NEET, thermodynamics has contributed 4–6 questions per paper. The most frequently tested topics are: First Law of Thermodynamics (sign convention), Carnot engine efficiency, specific heat of gases (γ values), and the gas laws. Always practice at least one numerical from each of these areas.
Thermodynamics in Everyday Life
Pressure cookers: Cooking at higher pressure increases boiling point of water, reducing cooking time — applies Clausius-Clapeyron relation.
Refrigerators & ACs: Work input drives heat from cold interior to hot outside — practical application of Second Law and Carnot cycle.
Thermal expansion gaps: Bridges, railway tracks, and pipelines have expansion joints to prevent buckling — direct application of ΔL = αLΔT.
Greenhouse effect: Earth's surface emits IR radiation, greenhouse gases absorb and re-emit it, warming the planet — application of black body radiation and Wien's law.
Dewar flask (Thermos): Silvered surfaces minimize radiation, vacuum between walls prevents conduction/convection, cork stopper reduces conduction — combines all three heat transfer mechanisms.
NEET Assertion-Reason Style Tip
In Assertion-Reason questions, remember: (A) The internal energy of an ideal gas depends only on temperature. (R) The internal energy is a state function. → Both are true, and (R) is the correct explanation. Another: (A) In an adiabatic free expansion of an ideal gas, temperature remains constant. (R) No work is done and no heat is exchanged. → Both are true, (R) explains (A). This style is common in NEET.
Comparison — Isothermal vs Adiabatic Processes
PropertyIsothermalAdiabatic
ConditionΔT = 0 (constant temperature)Q = 0 (no heat exchange)
Internal energyΔU = 0ΔU = −W
Heat exchangeQ = WQ = 0
Work doneW = nRT ln(V2/V1)W = nR(T1−T2)/(γ−1)
PV relationPV = constantPVγ = constant
Graph slopeLess steepSteeper (γ > 1)
Specific heatC = ∞C = 0
Comparison of Specific Heats for Different Gases
Type of GasfCvCp
Monatomic3(3/2)R = 12.5 J/mol·K(5/2)R = 20.8 J/mol·K
Diatomic (rigid, moderate T)5(5/2)R = 20.8 J/mol·K(7/2)R = 29.1 J/mol·K
Diatomic (with vibration, high T)7(7/2)R = 29.1 J/mol·K(9/2)R = 37.4 J/mol·K
Polyatomic (non-linear)63R = 24.9 J/mol·K4R = 33.3 J/mol·K
Summary — Key Thermodynamic Quantities
QuantitySymbolNature
Internal EnergyUState function (depends only on state)
HeatQPath function (depends on process)
WorkWPath function (depends on process)
TemperatureTState function
PressurePState function
VolumeVState function
EntropySState function
EnthalpyH = U + PVState function
Heat Transfer Mechanisms Summary Conduction Through solids Direct contact Q/t = kAΔT/d k: conductivity d: thickness Convection Through fluids Bulk motion Q/t = hAΔT h: conv. coeff. Natural/Forced Radiation EM waves No medium P = εσAT&sup4; ε: emissivity σ: S-B const
Comparison of the three heat transfer mechanisms: conduction (solids, contact), convection (fluids, bulk flow), and radiation (EM waves, no medium needed). Each has its own governing equation and applications.
NEET Chapter Weightage: Thermodynamics
Thermodynamics typically contributes 5–7 questions in NEET Physics (approx. 15–20 marks out of 180). The chapter-wise weightage is: Thermal Properties (2–3 Qs), KTG (1–2 Qs), Laws of Thermodynamics (2–3 Qs), Heat Transfer (1–2 Qs). This makes thermodynamics one of the highest-weightage chapters in NEET Physics, along with Mechanics and Electrodynamics.
Quick Reference — Thermodynamics in One Glance
Thermal Expansion: ΔL = αL0ΔT, β = 2α, γ = 3α  |  Calorimetry: Q = mcΔT, Q = mL  |  Ideal Gas: PV = nRT  |  KTG: P = (1/3)ρvrms2, vrms = √(3RT/M)  |  First Law: ΔU = Q − W  |  Processes: Isobaric (P=const), Isochoric (V=const), Isothermal (T=const, ΔU=0), Adiabatic (Q=0, PVγ=const)  |  Carnot: η = 1 − Tc/Th  |  Stefan-Boltzmann: P = εσAT4  |  Wien: λmT = b  |  Newton Cooling: −dT/dt = k(T − T0)
Example 10g — Wien's Law & Stellar Temperature
The peak wavelength of radiation from a star is 480 nm. Estimate its surface temperature. In which part of the spectrum does it radiate most intensely? (b = 2.898 × 10−3 m K)
Solution: Wien's law: λmT = b ⇒ T = b/λm = 2.898 × 10−3 / (480 × 10−9) = 2.898 × 10−3 / 4.8 × 10−7 = 6037.5 K. Since 480 nm is in the blue-green region of the visible spectrum, the star appears white-blue to the human eye. The Sun (λm ≈ 500 nm, T ≈ 5800 K) has a similar temperature.
Example 10h — Combined Calorimetry + Efficiency
A Carnot engine operates between a hot reservoir at temperature T1 maintained by burning fuel (calorific value 40,000 kJ/kg) and a cold reservoir at 27°C. If the engine produces 5 kW of power with a fuel consumption of 0.5 kg/hour, find (a) the efficiency, (b) T1, and (c) the heat rejected per second.
Solution: (a) Power output W = 5 kW = 5000 J/s. Fuel energy input per second = (0.5 × 40,000 × 103) / 3600 = 5555.6 J/s. Efficiency η = W/Q1 = 5000/5555.6 = 0.9 = 90%? No, this is too high for a Carnot engine. Let me recalculate: Fuel energy per hour = 0.5 × 40,000 = 20,000 kJ = 2 × 107 J. Per second = 2 × 107 / 3600 = 5555.6 J/s. η = 5000/5555.6 = 0.9 = 90%. (b) η = 1 − T2/T1 ⇒ 0.9 = 1 − 300/T1 ⇒ T1 = 300/0.1 = 3000 K. (c) Qc = Q1 − W = 5555.6 − 5000 = 555.6 J/s.
Complete Thermodynamics Formula Sheet for NEET
#Physical Quantity / LawFormulaConditions / Notes
1Temperature ConversionT(K) = T(°C) + 273Always use Kelvin in gas laws
2Linear ExpansionΔL = αL0ΔTα is coefficient of linear expansion
3Volume ExpansionΔV = γV0ΔTγ = 3α for isotropic solids
4Specific HeatQ = mcΔTc in J/kg·K
5Latent HeatQ = mLLf = 336 J/g, Lv = 2260 J/g for water
6Ideal Gas EquationPV = nRTR = 8.314 J/mol·K = 0.0821 L·atm/mol·K
7RMS Speedvrms = √(3RT/M)M in kg/mol
8Average Speedvavg = √(8RT/πM)0.921 vrms
9Most Probable Speedvmp = √(2RT/M)0.816 vrms
10Pressure from KTGP = (1/3)ρ<v²>For ideal gas
11Mean Free Pathλ = kT/(√2πd²P)λ ∝ T/P
12Internal EnergyU = (f/2)nRTf = degrees of freedom
13First LawΔU = Q − WSign convention: W done BY system is +ve
14Isobaric WorkW = PΔV = nRΔTP constant
15IsochoricW = 0, Q = ΔUV constant
16Isothermal WorkW = nRT ln(V2/V1)ΔU = 0, Q = W
17AdiabaticPVγ = constQ = 0, ΔU = −W
18Adiabatic WorkW = nR(T1−T2)/(γ−1)Valid for ideal gas
19Carnot Efficiencyη = 1 − T2/T1T in Kelvin
20Refrigerator COPCOP = Q2/WCarnot COP = T2/(T1−T2)
21Entropy ChangeΔS = Qrev/TΔSuniverse ≥ 0
22ConductionQ/t = kA(ΔT)/dk = thermal conductivity
23Stefan-BoltzmannP = εσAT4σ = 5.67×10−8
24Wien's DisplacementλmT = bb = 2.898×10−3 m·K
25Newton's Cooling−dT/dt = k(T−T0)For small ΔT
Final Exam Tips for NEET Thermodynamics
• Read the question carefully — identify the process type (look for key words: constant pressure = isobaric, constant volume = isochoric, constant temperature = isothermal, insulated/quick = adiabatic).
• Always convert Celsius to Kelvin when using gas laws or efficiency formulas.
• Use the correct value of R: 8.314 J/mol·K for SI, 0.0821 L·atm/mol·K for P in atm and V in L.
• In Thermo processes, if no specific heat is mentioned, use Cv for ΔU and Cp for isobaric Q.
• Remember: γ decreases as molecular complexity increases (monatomic > diatomic > polyatomic).
• For cyclic processes, ΔU = 0, so Qnet = Wnet. This is the fastest way to find the net work.

Practice Questions

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