Thermodynamics

Concept Core

Four laws, five processes, two key engines — the thermodynamics framework.

The Four Laws of Thermodynamics

Law	Statement	Implication
Zeroth	Systems in thermal equilibrium with a third have equal temperatures	Defines temperature
First	ΔU = Q − W (energy conservation)	Energy can't be created or destroyed
Second	Heat flows naturally from hot to cold; entropy of universe never decreases	Defines direction of processes
Third	Entropy → 0 as T → 0 K	Absolute zero is unattainable

Five Thermodynamic Processes

Process	Constant	Key Result
Isothermal	T	ΔU=0, Q=W, PV=const
Adiabatic	Q=0	ΔU=−W, PVᵞ=const
Isochoric	V	W=0, ΔU=Q
Isobaric	P	W=PΔV, Q=ΔU+PΔV
Cyclic	ΔU=0	Q_net=W_net

Work Done in Thermodynamic Processes

W = ∫P dV = Area under P-V graph

Isothermal: W = nRT ln(V₂/V₁)

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

Isobaric: W = P(V₂−V₁) = nRΔT

Isochoric: W = 0 (no volume change)

Heat Engines & Efficiency

An engine absorbs Q₁ from hot reservoir, does work W, rejects Q₂ to cold reservoir:

η = W/Q₁ = 1 − Q₂/Q₁ = 1 − T_cold/T_hot

The last equality holds only for a Carnot (ideal) engine. Carnot engine has maximum possible efficiency for given temperatures.

Specific Heats of Gases

Cᵥ = molar heat capacity at constant volume. Cₚ = at constant pressure.

Cₚ − Cᵥ = R (Mayer's relation) γ = Cₚ/Cᵥ

Monoatomic (Ar, He): γ = 5/3. Diatomic (O₂, N₂): γ = 7/5 = 1.4. Polyatomic: γ = 4/3.

Entropy

Entropy S = measure of disorder. Change in entropy:

ΔS = Q_reversible / T

For reversible process: ΔS_universe = 0. For irreversible: ΔS_universe > 0. The universe always tends toward maximum entropy (Second Law).

Formula Vault

All thermodynamics formulas — processes, engines, and specific heats.

First Law

ΔU = Q − W

Q positive in, W positive out

Isothermal Work

W = nRT ln(V₂/V₁)

ΔU = 0; Q = W

Adiabatic Law

PVᵞ = const; TVᵞ⁻¹ = const

Q = 0; ΔU = −W

Adiabatic Work

W = (P₁V₁−P₂V₂)/(γ−1)

= nCᵥ(T₁−T₂)

Isobaric Work

W = PΔV = nRΔT

Constant pressure

Carnot Efficiency

η = 1 − T_c/T_h

Maximum possible efficiency

Mayer's Relation

Cₚ − Cᵥ = R

R = 8.314 J/mol/K

Ratio of Specific Heats

γ = Cₚ/Cᵥ

Mono:5/3; Di:7/5; Poly:4/3

Entropy Change

ΔS = Q_rev/T

ΔS_universe ≥ 0 always

COP Refrigerator

COP = Q_c/W = T_c/(T_h−T_c)

Coefficient of Performance

Worked Examples

5 problems covering process identification, Carnot efficiency, and PV diagrams.

Easy100 J heat given to gas at constant volume — find ΔU▾

100 J of heat is supplied to a gas at constant volume. Find the change in internal energy.

Isochoric (constant V): W = 0

First Law: ΔU = Q − W = 100 − 0 = 100 J

All heat goes into increasing internal energy when volume is constant.

✓ ΔU = 100 J

EasyCarnot engine between 127°C and 27°C — find efficiency▾

A Carnot engine operates between a hot reservoir at 127°C and a cold reservoir at 27°C. Find its efficiency.

Convert: T_h = 400 K, T_c = 300 K (always Kelvin!)

η = 1 − T_c/T_h = 1 − 300/400 = 1 − 0.75 = 0.25 = 25%

✓ Efficiency = 25%

MediumGas expands isothermally — which process does more work: isothermal or adiabatic?▾

Gas expands from V₁ to V₂. Which process does more work: isothermal or adiabatic? (same initial conditions)

In isothermal: temperature stays constant (heat input maintains pressure). P-V curve is a hyperbola.

In adiabatic: no heat input → temperature falls → pressure drops faster. P-V curve falls more steeply.

Area under isothermal curve > area under adiabatic curve for same V₁ to V₂.

More area under P-V curve = more work done → isothermal does more work

✓ Isothermal process does more work than adiabatic for same expansion

EAPCET LevelFind heat absorbed in one cycle given PV diagram area = 300 J▾

A gas undergoes a cyclic process. The area enclosed by the P-V diagram is 300 J. Find the net heat absorbed in one cycle.

For a cyclic process: ΔU = 0 (system returns to initial state)

First Law: ΔU = Q_net − W_net → 0 = Q_net − W_net

Q_net = W_net = area of P-V diagram = 300 J

✓ Net heat absorbed = 300 J

Trap QuestionCan a Carnot engine have 100% efficiency?▾

A Carnot engine absorbs 1000 J from a hot reservoir. What conditions give 100% efficiency? ⚠️ Conceptual trap.

η = 1 − T_c/T_h = 1 only when T_c = 0 K (absolute zero).

The trap: Students say 'yes, just reject no heat' — but the Second Law forbids converting all heat to work.

Even Carnot (ideal, reversible) engine requires T_c = 0 K for 100% efficiency.

The Third Law says T = 0 K is unattainable. So 100% efficiency is theoretically impossible.

✓ Impossible — would require cold reservoir at absolute zero (T_c = 0 K), which the Third Law forbids

Mistake DNA

4 thermodynamics errors that cost marks — sign conventions and process confusion.

🌡️

Using Celsius Instead of Kelvin in Carnot Efficiency

Carnot efficiency formula requires absolute temperature in Kelvin. Using °C gives completely wrong answers.

❌ Wrong

T_h=127°C, T_c=27°C: η = 1−27/127 = 0.787 ✗ (used Celsius directly)

✓ Correct

T_h=400K, T_c=300K: η = 1−300/400 = 0.25 ✓ Always convert to K

The Carnot formula is η = 1 − T_c/T_h where T is in Kelvin (absolute temperature). Using °C gives a ratio that has no physical meaning.

🔄

First Law Sign Convention: Q and W Signs

Different textbooks use different conventions. EAPCET uses ΔU = Q − W (Q in, W out positive).

❌ Wrong

Work done ON gas W=−30J: ΔU = Q − W = 50 − (−30) = 80? ✗ (sign confusion)

✓ Correct

W by gas = −W on gas ΔU = Q − W_by_gas = 50 − (−30) = 80 J ✓ Clarify the sign first

ΔU = Q − W where W = work done BY the gas. If gas is compressed (work done ON it), W is negative, so ΔU increases more than Q alone.

📊

Adiabatic vs Isothermal: Confusing Which Curve Falls Faster

Adiabatic curve (PVᵞ=const) falls more steeply than isothermal (PV=const) on P-V diagram.

❌ Wrong

Adiabatic: same slope as isothermal on P-V graph ✗ (they cross, not parallel)

✓ Correct

Adiabatic falls steeper ✓ (γ > 1; more T drop) Isothermal is gentler ✓ (T stays constant)

Adiabatic: PVᵞ = const (γ>1). Isothermal: PV = const (exponent=1). Higher exponent → steeper fall → less work done in adiabatic expansion.

🔥

Thinking the Second Law Allows Perpetual Motion Machines

No device can convert all heat from a single reservoir into work (Kelvin-Planck statement of 2nd Law).

❌ Wrong

Machine absorbs 1000J from hot reservoir, does 1000J work, rejects 0J heat ✗ (violates 2nd law)

✓ Correct

Must reject some heat Q_c ✓ W = Q_h − Q_c < Q_h ✓ Efficiency < 100% always ✓

The Second Law (Kelvin-Planck statement): it is impossible to construct a heat engine that absorbs heat from a single reservoir and converts it entirely to work in a cyclic process.

Chapter Intelligence

Thermodynamics links Physics and Chemistry — the concepts repeat in both subjects.

EAPCET Weightage (2019–2024)

Carnot efficiency

First Law / ΔU = Q−W

Processes identification

Work from P-V diagram

Specific heats γ

High-Yield PYQ Patterns

Carnot η from temperaturesΔU in isochoric processWork done in isothermal expansionNet work in cyclic processWhich process does more work?COP of refrigerator

Exam Strategy

For any thermodynamics question: first identify the process (isothermal, adiabatic, isochoric, isobaric). This tells you which quantities are zero.
Carnot efficiency: always convert to Kelvin before substituting. η = 1 − T_c/T_h.
Cyclic process: ΔU = 0 always (return to initial state). Therefore Q_net = W_net = area of P-V loop.
Adiabatic vs isothermal on P-V diagram: adiabatic is steeper. The isothermal process does more work for same expansion.
This chapter repeats in Chemistry as Chemical Thermodynamics (ΔH, ΔG, ΔS, Hess's Law). The First and Second Laws are identical — study them together.

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

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