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.
1
Isochoric (constant V): W = 0
2
First Law: ΔU = Q − W = 100 − 0 = 100 J
3
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.
1
Convert: T_h = 400 K, T_c = 300 K (always Kelvin!)
2
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)
1
In isothermal: temperature stays constant (heat input maintains pressure). P-V curve is a hyperbola.
2
In adiabatic: no heat input → temperature falls → pressure drops faster. P-V curve falls more steeply.
3
Area under isothermal curve > area under adiabatic curve for same V₁ to V₂.
4
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.
1
For a cyclic process: ΔU = 0 (system returns to initial state)
2
First Law: ΔU = Q_net − W_net → 0 = Q_net − W_net
3
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.
1
η = 1 − T_c/T_h = 1 only when T_c = 0 K (absolute zero).
2
The trap: Students say 'yes, just reject no heat' — but the Second Law forbids converting all heat to work.
3
Even Carnot (ideal, reversible) engine requires T_c = 0 K for 100% efficiency.
4
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)
Processes identification~6 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.