Second Law Thermo

8 MCQs9-step worked example

Source: NCERT ThermodynamicsPYQ coverage: NEET 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The second law of thermodynamics addresses what the first law cannot: direction. The first law says energy is conserved — it does not say which processes actually happen. The second law fills that gap.

Kelvin–Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work (NCERT Class 11 Physics Chapter 12, page 10). A heat engine must reject some heat to a cold reservoir; 100% efficiency is impossible.

Clausius statement: No process is possible whose sole result is the transfer of heat from a colder body to a hotter body. Spontaneous heat flow goes hot → cold; reversing it requires external work (a refrigerator).

These two statements are equivalent — violating one implies violating the other. NEET occasionally tests this equivalence as a conceptual recall item.

Why it matters for NEET: Questions on the second law typically test whether you can distinguish what the law forbids from what it permits. A common confusion is assuming the second law forbids heat flowing from cold to hot entirely — it does not. It forbids this as the sole result, without any other change. A refrigerator transfers heat from cold to hot, but it consumes work to do so, which is permitted.

Another frequent confusion: aspirants conflate "efficiency < 100%" (a consequence of the second law) with "efficiency depends on the working substance." The Carnot theorem — also a consequence of the second law — states that efficiency depends only on reservoir temperatures, not on the working substance, for reversible engines.

Watch out: when a question asks "which of the following violates the second law," look for the option claiming a sole result. If external work or another compensating change is mentioned, the process may be allowed.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following is the Kelvin–Planck statement of the second law of thermodynamics?

MCQ 2Easy RecallPractice

The Clausius statement of the second law of thermodynamics states that:

MCQ 3Easy RecallPractice

The Kelvin–Planck and Clausius statements of the second law of thermodynamics are:

MCQ 4Concept TrapPractice

A refrigerator transfers heat from a cold compartment to a warmer room. Does this violate the second law of thermodynamics?

MCQ 5Concept TrapPractice

An inventor claims to have built a heat engine that absorbs 1000 J of heat from a single thermal reservoir and converts all of it into 1000 J of work, with no other effect. This claim:

MCQ 6Direct ApplicationPractice

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is the maximum efficiency of this engine?

MCQ 7Direct ApplicationPractice

A Carnot engine has an efficiency of 40%. If the temperature of the hot reservoir is 500 K, what is the temperature of the cold reservoir?

MCQ 8Direct ApplicationPractice

For a Carnot engine operating between 800 K and 400 K, the engine absorbs 2000 J from the hot reservoir per cycle. How much heat is rejected to the cold reservoir per cycle?

Worked Example

1
Given
A Carnot engine operates between a source at T_H = 727 °C and a sink at T_C = 227 °C. The engine absorbs Q_H = 1.00 × 10³ J per cycle from the source.
2
Required
(a) The efficiency of the engine. (b) The work done per cycle. (c) The heat rejected to the sink per cycle.
3
Concept
The second law implies that no engine operating between two reservoirs can exceed the Carnot efficiency, which depends only on the reservoir temperatures (not the working substance).
4
Formula
Carnot efficiency: η = 1 − T_C / T_H (temperatures must be in kelvin). Work: W = η × Q_H. Heat rejected: Q_C = Q_H − W.
5
Substitution
Convert temperatures: T_H = 727 + 273 = 1000 K, T_C = 227 + 273 = 500 K. η = 1 − 500/1000 = 1 − 0.500 = 0.500 (50.0%). W = 0.500 × 1.00 × 10³ = 5.00 × 10² J. Q_C = 1.00 × 10³ − 5.00 × 10² = 5.00 × 10² J.
6
Calculation
The arithmetic is completed above. Note: 273 (the Celsius-to-Kelvin offset) and the temperature values 727, 227 are exact given values in this problem; they do not limit significant figures. The heat Q_H = 1.00 × 10³ J has 3 significant figures, so answers are reported to 3 significant figures.
7
Final answer
(a) η = 50.0% (b) W = 5.00 × 10² J (c) Q_C = 5.00 × 10² J
8
Common trap
Forgetting to convert °C to K before applying the Carnot formula. Using T_H = 727 and T_C = 227 directly gives η = 1 − 227/727 ≈ 68.8%, which is wrong. Always convert to absolute temperature first.
9
Similar NEET-style question
A Carnot engine works between 527 °C and 127 °C. If 600 J of heat is supplied per cycle, find the work output and the heat rejected. ---

Before solving, remember these

Law

Second law of thermodynamics

Kelvin-Planck statement: No process whose sole result is to extract heat from a reservoir and convert it entirely into work. Clausius statement: No process whose sole result is to transfer heat from cold to hot reservoir.

-- NCERT, p. 10

Formulas

4 formulas — click to collapse

Adiabatic relations for ideal gas

P V^{γ} = const; T V^{γ - 1} = const

Relations holding during reversible adiabatic process. gamma = Cp/Cv.

Symbol	Quantity	SI Unit
P	pressure	Pa
V	volume	m^3
T	temperature	K
gamma	adiabatic index	-

Valid when

Q = 0 (no heat exchange)
Quasi-static (reversible)
Ideal gas

First law of thermodynamics

Δ U = Q - W

Change in internal energy = heat ADDED minus work DONE BY the system. Energy conservation including thermal energy.

Symbol	Quantity	SI Unit
Delta_U	change in internal energy	J
Q	heat added to system	J
W	work done BY system	J

Valid when

Closed system (no mass exchange)
Sign convention: Q>0 heat in, W>0 system does work

Work done in isothermal process (ideal gas)

W = n R T ln \frac{V _{f}}{V _{i}}

Work done by ideal gas during isothermal expansion. Q = W (since Delta_U = 0). Reverse for compression.

Symbol	Quantity	SI Unit
n	moles	mol
R	gas constant 8.314	J/mol/K
T	temperature	K
V_i, V_f	initial/final volume	m^3

Valid when

Ideal gas
Quasi-static (reversible) isothermal

Mayer's relation (Cp - Cv = R)

C_{p} - C_{v} = R

For ideal gas: difference of molar specific heats equals gas constant R. Useful for converting between Cp and Cv.

Symbol	Quantity	SI Unit
Cp	molar specific heat at const P	J/mol/K
Cv	molar specific heat at const V	J/mol/K
R	8.314	J/mol/K

Valid when

Ideal gas
Per mole basis

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

2 items — click to collapse

Trap

Adiabatic vs isothermal P-V curve

Category: Graph Interpretation

Student misidentifies which P-V curve is adiabatic (steeper) vs isothermal.

When it triggers

P-V graph showing one or more processes; question asks for process type.

How to avoid

Adiabatic curve is STEEPER than isothermal at the same point (slope ratio = γ). Adiabat: PV^γ; isotherm: PV = const.

Mistake

Identifies adiabatic curve as gentler than isothermal on P-V diagram (or vice versa).

Root cause: graph misread

Correction

Adiabatic curve is STEEPER than isothermal at the same P-V point. Slope ratio at same point: adiabatic / isothermal = γ. Memorise: 'adiabatic angles down sharper'.