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L-11: Thermodynamics

Physics — Class 12 · NIOS Code 312 · Module 3 · Source: 312_Physics_Eng_Lesson11.pdf

Thermodynamics — Heat, Work, and Energy Flow

Thermodynamics is the phenomenological science of thermal energy transfer between bodies at different temperatures. Rubbing hands, inflating a bicycle tyre, or ice melting in warm water all involve conversion between mechanical work and thermal effects.

NIOS objectives: indicator diagrams; thermodynamic equilibrium; Zeroth law; internal energy; first law and limitations; triple point; second law; Carnot cycle and efficiency.

11.1 Concept of Heat and Temperature

11.1.1 Heat

Heat is energy transferred between two systems (or a system and surroundings) because of a temperature difference. Direction of flow is always from higher to lower temperature until thermal equilibrium is reached.

Joule showed equivalence of heat and mechanical work — molecular motion is associated with heat. Units:

  • calorie (cal): heat to raise 1 g water from 14.5°C to 15.5°C
  • 1 kcal = 10³ cal
  • 1 cal = 4.18 J

11.1.2 Temperature

Temperature is the property that determines whether a body is in thermal equilibrium with others. All bodies in thermal equilibrium share the same temperature.

11.1.3 Thermodynamic Terms

  • System: definite quantity of matter separated by a boundary (real or imaginary).
  • Surroundings: everything outside the boundary.
  • Open system: exchanges mass and energy.
  • Closed system: exchanges energy only.
  • Isolated system: exchanges neither.
  • Variables: P, V, T describe the thermodynamic state.
  • Indicator diagram: P–V graph showing how pressure varies with volume during a process.
ΔW = P ΔV
Small expansion step (Eq. 11.1)
Work done by system = area of shaded strip on P–V diagram.
Total work V₁ → V₂ = area under entire path.
Important: work depends on path, not just initial/final states.
Fig 11.1 — Indicator Diagram (Work = Area) V P PΔV W = area under curve
Fig 11.1 — Work done during expansion equals area under the P–V indicator diagram

Work done on the system is taken as negative; work done by the system is positive.

11.2 Thermodynamic Equilibrium and Processes

A liquid at 60°C left in a room reaches room temperature — thermal equilibrium. When unbalanced forces cease → mechanical equilibrium. When chemical reactions stop → chemical equilibrium. All three together → thermodynamic equilibrium.

11.2.1 Types of Processes

  • Reversible: slow, through equilibrium states; can be retraced (melting ice ↔ refreezing; spring loading). Ideal only.
  • Irreversible: cannot retrace same equilibrium path (friction, dissolving sugar, rusting). All natural processes.
  • Isothermal: constant T; heat flows to/from surroundings to maintain temperature.
  • Adiabatic: no heat exchange — ΔQ = 0.
  • Isobaric: constant P.
  • Isochoric: constant V → no expansion work.
  • Cyclic: returns to initial state → ΔU = 0.
Adiabatic: ΔQ = 0 → ΔU = −ΔW
Compression: work on system → ΔU increases.
Expansion: work by system → ΔU decreases.
Isochoric: ΔW = 0 → ΔQ = ΔU
All heat changes internal energy only.
Cyclic: ΔU = 0 → ΔQ = ΔW
Net heat absorbed = net work done in one cycle.

11.2.2 Zeroth Law of Thermodynamics

If 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 provides the basis for temperature as a measurable quantity.

Phase Diagram and Triple Point

Matter exists as solid, liquid, and gas (phases). A phase diagram plots P vs T showing fusion curve (melting), vaporization curve (boiling), and sublimation (hoarfrost) curve. Where extended curves meet is the triple point — all three phases coexist.

  • Latent heat of fusion: heat to convert unit mass solid → liquid at melting point.
  • Latent heat of vaporization: heat to convert unit mass liquid → gas at boiling point.
Fig 11.2 — Phase Diagram & Triple Point T P fusion vaporization sublimation Triple point P solid · liquid · vapour coexist
Fig 11.2 — Fusion, vaporization and sublimation curves meet at the triple point

Triple point of water is the upper fixed point in the Kelvin thermometric scale — uniquely stable T and P.

11.3 Internal Energy of a System

Internal energy U = kinetic energy of molecules + potential energy from intermolecular forces. For a metal: electron KE + atomic PE + lattice vibration energy.

U is a state function — depends only on state variables (P, V, T), not the path. For adiabatic work W done on system:

Uᵢ − Uf = −W
Work done on system increases internal energy.
Work done by system decreases U.

11.4 First Law of Thermodynamics

Conservation of energy for thermodynamic systems: heat supplied to a system equals change in internal energy plus work done by the system.

ΔQ = ΔU + ΔW
First law (Eq. 11.3b)
Also: ΔU = ΔQ − ΔW (when −ΔW is work done on system).
All quantities in SI units (joules).

Sign conventions:

  • ΔW positive when system expands (work by system).
  • ΔW negative when system compressed (work on system).
  • ΔQ positive when heat added to system.
  • ΔU positive when internal energy increases.

ΔQ and ΔW depend on path; ΔU depends only on initial and final states.

11.4.1 Limitations

  • Does not forbid heat flowing from cold to hot — fails to give direction of heat flow.
  • Does not state how much heat can be converted to work — needs second law.

11.5 Second Law of Thermodynamics

Kelvin-Planck: It is impossible for any system to absorb heat from a reservoir at fixed temperature and convert all of it into work.

Clausius: It is impossible for heat to flow from a colder body to a hotter body without external work.

A heat engine needs: (i) hot source, (ii) cold sink, (iii) working substance. It absorbs H₁ from source, does work W, rejects H₂ to sink.

Fig 11.4 — Carnot Cycle on P–V Diagram A B C D W = area ABCD T₁ isothermal T₂ isothermal
Fig 11.4 — Carnot cycle: two isothermals + two adiabatics; net work = enclosed area

11.5.1 Carnot Cycle

         CARNOT CYCLE (reversible ideal engine)
         ====================================
    A ──isothermal expansion──► B   (absorb H₁ from source T₁)
    B ──adiabatic expansion───► C   (T falls to T₂)
    C ──isothermal compression► D   (reject H₂ to sink T₂)
    D ──adiabatic compression─► A   (T rises back to T₁)
                    |
                    v
         Net work W = area ABCD on P–V diagram
         ΔU = 0 for complete cycle  →  W = H₁ − H₂

11.5.2 Efficiency of Carnot Engine

η = (H₁ − H₂) / H₁ = 1 − H₂/H₁
Efficiency = heat converted to work / heat from source.
η = 1 − T₂/T₁
For Carnot engine: H₂/H₁ = T₂/T₁.
η = 100% only if T₂ = 0 K (impossible).
Efficiency independent of working substance.
No real engine exceeds Carnot efficiency between same T₁, T₂.

Example: Source 400 K, absorbs 200 cal, rejects 150 cal → sink T₂ = 300 K, η = 25%.

11.5.3 Limitations of Carnot Engine

Isothermal steps need infinitely slow piston motion for heat transfer; adiabatic steps need infinitely fast motion to prevent heat leak. These ideal conditions cannot be met — real engines have η < Carnot η.

Quick Revision

  • Heat: energy due to ΔT; 1 cal = 4.18 J.
  • Work: ΔW = PΔV; area under P–V curve.
  • Zeroth law: defines temperature via thermal equilibrium.
  • First law: ΔQ = ΔU + ΔW; U is state function.
  • Second law: Kelvin-Planck + Clausius; limits heat engines.
  • Carnot: η = 1 − T₂/T₁; maximum possible efficiency.
  • Triple point: solid, liquid, vapour coexist.
23 cards · click any card to flip
Heat
Energy transferred between systems due to temperature difference. 1 cal = 4.18 J; 1 kcal = 10³ cal.
Temperature
Property determining thermal equilibrium. Bodies in equilibrium share the same temperature.
Open system
Can exchange both mass and energy with surroundings (e.g. water heater).
Closed system
Can exchange energy but not mass (e.g. gas in piston cylinder).
Isolated system
Exchanges neither mass nor energy (ideal thermos flask).
Indicator diagram
P–V graph for a thermodynamic process. Work done = area under the curve.
Work in expansion
ΔW = P ΔV (small step). Total work from V₁ to V₂ = area under P–V path. Work depends on path, not just endpoints.
Thermodynamic equilibrium
Thermal + mechanical + chemical equilibrium; macroscopic properties constant with time.
Reversible process
All intermediate states are equilibrium states; can be retraced exactly. Idealised, never fully achieved in practice.
Irreversible process
Cannot be retraced through same equilibrium states. All natural processes are irreversible.
Isothermal process
Constant temperature. ΔT = 0. Heat can flow to maintain T (conducting walls).
Adiabatic process
No heat exchange: ΔQ = 0, so ΔU = −ΔW. Compression increases U; expansion decreases U.
Isobaric / isochoric
Isobaric: constant P. Isochoric: constant V → ΔW = 0 → ΔQ = ΔU.
Cyclic process
System returns to initial state → ΔU = 0 → ΔQ = ΔW.
Zeroth law
If A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A and B are in thermal equilibrium.
Triple point
P–T point where solid, liquid and vapour coexist (for water: unique T and P). Used as upper fixed point in Kelvin scale.
Internal energy U
Sum of molecular kinetic + potential energy. State function — depends on P, V, T not path.
First law
ΔQ = ΔU + ΔW. Heat supplied = change in internal energy + work done by system.
First law limitations
Does not give direction of heat flow; does not limit extent of heat→work conversion.
Kelvin-Planck statement
Impossible to absorb heat from a single reservoir at fixed T and convert all of it into work.
Clausius statement
Impossible for heat to flow from colder to hotter body without external work.
Carnot cycle
Four steps: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.
Carnot efficiency
η = 1 − H₂/H₁ = 1 − T₂/T₁. Maximum efficiency for engine between two temperatures. η < 1 always.

Q1. Heat is transferred between bodies because of:

Q2. 1 calorie is equal to:

Q3. Work done by a system during small expansion ΔV at pressure P is:

Q4. Zeroth law of thermodynamics defines:

Q5. In an adiabatic process:

Q6. For an isochoric process:

Q7. First law of thermodynamics is essentially:

Q8. ΔU for a thermodynamic process depends on:

Q9. Kelvin-Planck statement relates to:

Q10. Efficiency of a Carnot engine between T₁ (source) and T₂ (sink) is:

1 cal = 4.18 J
ΔW = P ΔV
W = area under P–V curve
ΔQ = ΔU + ΔW
ΔU = ΔQ − ΔW
Uᵢ − Uf = −W
Adiabatic: ΔQ = 0 → ΔU = −ΔW
Isochoric: ΔW = 0 → ΔQ = ΔU
Cyclic: ΔU = 0 → ΔQ = ΔW
η = (H₁ − H₂) / H₁
η = 1 − H₂/H₁
H₂/H₁ = T₂/T₁
η = 1 − T₂/T₁
W = H₁ − H₂

1. Formulas & Definitions

Full Ch 11 study guide — heat, work, first law, processes, and Carnot efficiency.

1 cal = 4.18 J

Definition: Conversion between calorie (heat unit) and joule (SI energy unit).

Derivation

Joule's mechanical equivalent of heat; 1 cal = heat to raise 1 g water by 1°C (14.5→15.5°C).

Variables

cal · J · 1 kcal = 10³ cal

Why it works

Exam questions mix cal and J — always convert to SI for first-law problems.

Historical context

James Prescott Joule proved heat and work are equivalent forms of energy.

Deep understanding

Use 4.18 J/cal unless problem states otherwise. kcal common in nutrition and older texts.

2. Diagrams & Visuals

1 cal = 4.18 J

Color-coded visual · step-by-step breakdown below

  1. Identify heat in calories
  2. Multiply by 4.18 for joules
  3. Or divide joules by 4.18 for cal

3. Solved Examples

Basic

Q: 100 cal in joules?

Solution: 100×4.18

Answer: 418 J

Intermediate

Q: 836 J in cal?

Solution: 836/4.18

Answer: 200 cal

Advanced

Q: 2 kcal to J?

Solution: 2000×4.18

Answer: 8360 J

Exam

Q: SI unit of heat/energy?

Solution: Joule

Answer: cal is non-SI

ΔW = P ΔV

Definition: Small work done by system during expansion at nearly constant pressure.

Derivation

Eq. 11.1 — work = force × displacement; F = PA → W = PΔV for small step.

Variables

ΔW (J) · P (Pa) · ΔV (m³)

Why it works

Foundation of indicator diagrams — each strip on P–V graph is work.

Historical context

Steam engine indicator diagrams (19th century) visualised this relation.

Deep understanding

Work BY system positive on expansion. Total W = area under full P–V path.

2. Diagrams & Visuals

PΔV ΔW = PΔV

Color-coded visual · step-by-step breakdown below

  1. Read P at expansion step
  2. Find volume change ΔV
  3. ΔW = PΔV
  4. Sum strips or integrate for total

3. Solved Examples

Basic

Q: P=2×10⁵ Pa, ΔV=0.001 m³.

Solution: ΔW=200 J

Answer: 200 J by system

Intermediate

Q: Compression ΔV=−0.002 m³, P=10⁵.

Solution: ΔW=−200 J

Answer: Work ON system

Advanced

Q: Isobaric process meaning?

Solution: P constant

Answer: ΔW = P(V₂−V₁)

Exam

Q: Work sign: expansion?

Solution: ΔW positive (by system)

Answer: Sec 11.1

W = area under P–V curve

Definition: Total work in a process equals area under the indicator diagram from V₁ to V₂.

Derivation

Sum of all PΔV strips; limit → W = ∫P dV along the path.

Variables

W (J) · path on P–V diagram

Why it works

Same initial and final states can give different work for different paths.

Historical context

Path dependence distinguishes thermodynamic work from state functions.

Deep understanding

ΔU is path-independent; ΔW and ΔQ are NOT. Cyclic process: enclosed area = net work.

2. Diagrams & Visuals

W = shaded area

Color-coded visual · step-by-step breakdown below

  1. Plot process on P–V axes
  2. Identify V₁ and V₂
  3. Find area under curve (geometry or integration)
  4. Assign sign: expansion positive W

3. Solved Examples

Basic

Q: Rectangle under graph: P=10⁵, ΔV=0.02.

Solution: W=2000 J

Answer: 2000 J

Intermediate

Q: Same ΔU, different paths?

Solution: Different areas → different W

Answer: Path matters

Advanced

Q: Carnot cycle net work?

Solution: Area enclosed ABCD

Answer: Fig 11.4

Exam

Q: Indicator diagram shows?

Solution: P vs V; work = area

Answer: Fig 11.1

ΔQ = ΔU + ΔW

Definition: First law of thermodynamics — energy conservation for a system.

Derivation

Eq. 11.3b: heat supplied = change in internal energy + work done by system.

Variables

ΔQ · ΔU · ΔW (all in J)

Why it works

Central equation linking heat, internal energy, and mechanical work.

Historical context

Mayer, Joule, Helmholtz; formalised as first law (~1850s).

Deep understanding

ΔU is state function; ΔQ and ΔW depend on path. Sign convention critical.

2. Diagrams & Visuals

ΔQ = ΔU + ΔW heat in = ΔU + work out First Law

Color-coded visual · step-by-step breakdown below

  1. Assign signs (+Q into system, +W by system)
  2. Find two quantities
  3. Solve third from ΔQ = ΔU + ΔW
  4. Check units: joules

3. Solved Examples

Basic

Q: ΔQ=500 J, ΔW=200 J.

Solution: ΔU=300 J

Answer: 300 J

Intermediate

Q: ΔU=−100 J, ΔW=150 J.

Solution: ΔQ=50 J

Answer: 50 J heat added

Advanced

Q: Adiabatic: ΔQ=0?

Solution: ΔU=−ΔW

Answer: Special case

Exam

Q: State function in first law?

Solution: ΔU only

Answer: Not Q or W

ΔU = ΔQ − ΔW

Definition: Alternate first-law form — internal energy change equals heat in minus work out.

Derivation

Rearrangement of ΔQ = ΔU + ΔW.

Variables

Same as first law

Why it works

Some texts define W as work ON system — then ΔU = ΔQ + W_on.

Historical context

Equivalent forms; NIOS uses ΔW positive when system expands.

Deep understanding

Be consistent with sign convention stated in the question.

2. Diagrams & Visuals

ΔU = ΔQ − ΔW

Color-coded visual · step-by-step breakdown below

  1. Know NIOS convention (+W = by system)
  2. Rearrange first law
  3. Substitute known values

3. Solved Examples

Basic

Q: Q=1000 J, W=400 J (expansion).

Solution: ΔU=600 J

Answer: 600 J

Intermediate

Q: Compressed: W=−300 J, Q=0.

Solution: ΔU=300 J

Answer: Adiabatic compression heats

Advanced

Q: ΔU path independent means?

Solution: Same endpoints → same ΔU

Answer: Different Q,W paths OK

Exam

Q: First law limitation?

Solution: No direction of heat flow

Answer: Needs second law

Uᵢ − Uf = −W

Definition: Internal energy change from adiabatic work alone.

Derivation

ΔQ=0 → ΔU = −ΔW; work ON system (negative ΔW) increases U.

Variables

Uᵢ, Uf (J) · W = work on/on system per convention

Why it works

Compressing gas adiabatically heats it — no heat escape.

Historical context

Joule's paddle-wheel experiments measured mechanical heating.

Deep understanding

U is state function: sum of molecular KE + intermolecular PE.

2. Diagrams & Visuals

compress U increases

Color-coded visual · step-by-step breakdown below

  1. Confirm adiabatic (ΔQ=0)
  2. Calculate work W
  3. ΔU = −ΔW (NIOS: +W by system)
  4. Uf − Ui = ΔU

3. Solved Examples

Basic

Q: Adiabatic expansion W=+100 J.

Solution: ΔU=−100 J

Answer: U decreases

Intermediate

Q: Work 50 J on system adiabatically.

Solution: ΔW=−50; ΔU=+50

Answer: U rises

Advanced

Q: U depends on?

Solution: State (P,V,T) only

Answer: Not path

Exam

Q: Internal energy includes?

Solution: Molecular KE + PE

Answer: Sec 11.3

Adiabatic: ΔQ = 0 → ΔU = −ΔW

Definition: No heat exchange with surroundings — insulated or rapid process.

Derivation

First law with ΔQ=0.

Variables

ΔQ = 0

Why it works

Compression heats gas; expansion cools — used in diesel ignition, weather.

Historical context

Adiabatic processes model rapid expansions where heat has no time to flow.

Deep understanding

Ideal adiabatic: perfectly insulated. Approximate: fast compression in engines.

2. Diagrams & Visuals

insulated ΔQ = 0

Color-coded visual · step-by-step breakdown below

  1. Verify adiabatic condition
  2. Set ΔQ=0
  3. ΔU = −ΔW
  4. Solve for unknown

3. Solved Examples

Basic

Q: ΔW=+80 J adiabatic expansion.

Solution: ΔU=−80 J

Answer: −80 J

Intermediate

Q: Carnot has two adiabatic legs.

Solution: B→C and D→A

Answer: Fig 11.4

Advanced

Q: Adiabatic vs isothermal expansion?

Solution: Isothermal: T constant, Q flows

Answer: Adiabatic: Q=0, T changes

Exam

Q: ΔQ=0 implies?

Solution: ΔU=−ΔW

Answer: Process type

Isochoric: ΔW = 0 → ΔQ = ΔU

Definition: Constant volume process — no expansion work.

Derivation

ΔV=0 → ΔW=PΔV=0 → first law gives ΔQ=ΔU.

Variables

ΔV = 0

Why it works

Heating gas in rigid container — all heat raises internal energy.

Historical context

Isochoric heating defines heat capacity at constant volume C_V.

Deep understanding

Electric heater in sealed room: ΔW≈0, Q mostly increases U (and T).

2. Diagrams & Visuals

V fixed

Color-coded visual · step-by-step breakdown below

  1. Confirm constant volume
  2. ΔW=0
  3. ΔQ=ΔU
  4. All heat changes U

3. Solved Examples

Basic

Q: 500 J heat to rigid tank gas.

Solution: ΔW=0; ΔU=500 J

Answer: ΔQ=500 J

Intermediate

Q: Why ΔW=0?

Solution: ΔV=0

Answer: No PΔV work

Advanced

Q: Compare isobaric heating?

Solution: Some Q does work expanding

Answer: Isochoric: all to U

Exam

Q: Isochoric process?

Solution: Constant V

Answer: Sec 11.2.1

Cyclic: ΔU = 0 → ΔQ = ΔW

Definition: System returns to initial state after one complete cycle.

Derivation

U is state function — same start/end state → ΔU=0 → ΔQ=ΔW.

Variables

Net ΔU = 0 per cycle

Why it works

Heat engines operate in cycles; net work = heat in − heat out.

Historical context

Carnot cycle is the ideal reversible cyclic process.

Deep understanding

Enclosed area on P–V diagram = net work per cycle.

2. Diagrams & Visuals

closed loop ΔU=0

Color-coded visual · step-by-step breakdown below

  1. Identify cycle returns to start
  2. ΔU=0
  3. Net ΔQ = net ΔW
  4. Area enclosed = W_net

3. Solved Examples

Basic

Q: Engine absorbs 1000 J, rejects 700 J per cycle.

Solution: W=300 J

Answer: η=30%

Intermediate

Q: Carnot cycle ΔU?

Solution: Zero for full cycle

Answer: Returns to state A

Advanced

Q: Why state function matters?

Solution: ΔU=0 on loop

Answer: Q and W net don't cancel individually

Exam

Q: Cyclic process ΔU?

Solution: Zero

Answer: First law simplification

η = (H₁ − H₂) / H₁

Definition: Thermal efficiency of heat engine — fraction of source heat converted to work.

Derivation

η = W/H₁ = (H₁−H₂)/H₁; energy conservation W = H₁−H₂.

Variables

η (dimensionless or %) · H₁ = heat from hot source · H₂ = heat to sink

Why it works

Measures engine performance — always < 100% (second law).

Historical context

Steam age drove efficiency analysis; Carnot set the theoretical limit.

Deep understanding

H₁ = heat absorbed at T₁; H₂ = heat rejected at T₂. Real engines: η < Carnot η.

2. Diagrams & Visuals

T₁ H₁ T₂ H₂ η=(H₁−H₂)/H₁

Color-coded visual · step-by-step breakdown below

  1. Find H₁ from hot reservoir
  2. Find H₂ rejected to sink
  3. W = H₁ − H₂
  4. η = W/H₁ = (H₁−H₂)/H₁

3. Solved Examples

Basic

Q: H₁=500 J, H₂=350 J.

Solution: η=150/500

Answer: 30%

Intermediate

Q: η=25%, H₁=200 cal.

Solution: W=50 cal

Answer: 150 cal rejected

Advanced

Q: Can η=100%?

Solution: Would need H₂=0

Answer: Second law forbids

Exam

Q: Engine needs three parts?

Solution: Source, sink, working substance

Answer: Sec 11.5

η = 1 − H₂/H₁

Definition: Equivalent efficiency formula — fraction not rejected to sink.

Derivation

Algebra from η = (H₁−H₂)/H₁.

Variables

H₂/H₁ = fraction lost as waste heat

Why it works

Shows efficiency rises when less heat rejected for same H₁.

Historical context

Carnot linked H₂/H₁ to T₂/T₁ for reversible engines.

Deep understanding

Improving η means lowering H₂ or raising T₁ (within material limits).

2. Diagrams & Visuals

η = 1 − H₂/H₁

Color-coded visual · step-by-step breakdown below

  1. Calculate ratio H₂/H₁
  2. Subtract from 1
  3. Express as % if needed

3. Solved Examples

Basic

Q: H₂/H₁ = 0.7.

Solution: η=0.3

Answer: 30%

Intermediate

Q: Reject half the heat?

Solution: H₂/H₁=0.5

Answer: η=50%

Advanced

Q: Same as (H₁−H₂)/H₁?

Solution: Yes, identical

Answer: Rearrangement

Exam

Q: Example: 200 cal in, 150 out?

Solution: η=50/200

Answer: 25%

H₂/H₁ = T₂/T₁

Definition: Carnot relation between heat ratios and absolute temperatures.

Derivation

Reversible isothermal heat transfer Q ∝ T for Carnot cycle.

Variables

T₁, T₂ in kelvin (K) — must use absolute scale

Why it works

Links heat engine efficiency directly to reservoir temperatures.

Historical context

Sadi Carnot (1824) — Réflexions sur la puissance motrice du feu.

Deep understanding

Only for Carnot (reversible) engine. T must be in K, never °C in ratio.

2. Diagrams & Visuals

H₂/H₁ = T₂/T₁ T in kelvin only!

Color-coded visual · step-by-step breakdown below

  1. Convert temperatures to kelvin
  2. Compute T₂/T₁
  3. Set equal to H₂/H₁
  4. Solve for unknown heat or T

3. Solved Examples

Basic

Q: T₁=400 K, T₂=300 K.

Solution: H₂/H₁=0.75

Answer: Ratio 3/4

Intermediate

Q: H₁=200 cal, H₂=150 cal.

Solution: T₂/T₁=0.75

Answer: If T₁=400 K → T₂=300 K

Advanced

Q: Use °C in ratio?

Solution: WRONG

Answer: Convert to K first

Exam

Q: Carnot example T₂?

Solution: 300 K from 400 K, 150/200

Answer: Sec 11.5.2

η = 1 − T₂/T₁

Definition: Maximum (Carnot) efficiency between two temperatures.

Derivation

Substitute H₂/H₁ = T₂/T₁ into η = 1 − H₂/H₁.

Variables

T₁ = source temp (K) · T₂ = sink temp (K)

Why it works

No engine between same T₁, T₂ can exceed this — design benchmark.

Historical context

Carnot efficiency is the theoretical upper limit for all heat engines.

Deep understanding

η=100% only if T₂=0 K (unattainable). Independent of working substance.

2. Diagrams & Visuals

η = 1 − T₂/T₁ Carnot maximum efficiency T₁=400 K, T₂=300 K → η=25%

Color-coded visual · step-by-step breakdown below

  1. T₁, T₂ in kelvin
  2. η = 1 − T₂/T₁
  3. Compare real engine η (always lower)
  4. Express as percentage

3. Solved Examples

Basic

Q: T₁=500 K, T₂=400 K.

Solution: η=1−0.8

Answer: 20%

Intermediate

Q: Ex: T₁=400 K, T₂=300 K.

Solution: η=1−300/400

Answer: 25%

Advanced

Q: Double T₁ only?

Solution: η increases

Answer: Larger temperature gap helps

Exam

Q: Carnot η with °C?

Solution: Convert to K first

Answer: Common trap

W = H₁ − H₂

Definition: Net work per Carnot (or any) cycle equals heat in minus heat out.

Derivation

First law for cyclic process: ΔQ = ΔW → W = H₁ − H₂.

Variables

W (J or cal) · H₁, H₂

Why it works

Energy not rejected as heat appears as useful work.

Historical context

Also equals enclosed area on Carnot P–V diagram (Fig 11.4).

Deep understanding

For Carnot: W = ηH₁ = H₁(T₁−T₂)/T₁. Real engines lose more to irreversibility.

2. Diagrams & Visuals

W = area ABCD = H₁−H₂

Color-coded visual · step-by-step breakdown below

  1. Heat H₁ from source
  2. Heat H₂ to sink
  3. W = H₁ − H₂
  4. Check η = W/H₁

3. Solved Examples

Basic

Q: H₁=400 J, H₂=280 J.

Solution: W=120 J

Answer: 120 J

Intermediate

Q: η=25%, H₁=200 cal.

Solution: W=50 cal

Answer: H₂=150 cal

Advanced

Q: P–V Carnot cycle work?

Solution: Enclosed area

Answer: Equals H₁−H₂

Exam

Q: Cyclic first law?

Solution: ΔQ_net = W_net

Answer: Sec 11.5

5. Special Features & Extras

Complete study guide for Thermodynamics.

Exam Tips & Tricks

  • Sign convention: +ΔW = work BY system (expansion); +ΔQ = heat INTO system.
  • State function: only ΔU — not Q or W individually.
  • Carnot η: use kelvin — never °C in T₂/T₁.
  • η = 1 − T₂/T₁ is maximum efficiency; real engines are lower.
  • Indicator diagram: work = area under P–V curve (path dependent).
  • Convert cal → J (×4.18) before mixing with SI work units.

Common Student Mistakes

  • Using °C in Carnot efficiency formula
  • Confusing ΔW sign (on vs by system)
  • Assuming η can reach 100% (second law)
  • Treating heat Q as state function
  • Forgetting isochoric → ΔW = 0
  • Mixing calories and joules without conversion

Memory Aids & Mnemonics

First law: "Q goes to U and W" — ΔQ = ΔU + ΔW
Processes: Adiabatic Q=0 · Isochoric W=0 · Cyclic ΔU=0
Carnot: "One minus cold over hot" — η = 1 − T₂/T₁ (kelvin!)
Heat engine: H₁ in → W out + H₂ rejected

Which Formula When?

  • Small expansion step? → ΔW = PΔV
  • Full process work? → Area under P–V curve
  • Energy balance? → ΔQ = ΔU + ΔW
  • No heat exchange? → Adiabatic: ΔU = −ΔW
  • Fixed volume heating? → ΔQ = ΔU
  • Engine cycle? → η = (H₁−H₂)/H₁ or 1−T₂/T₁
  • Cal vs J? → ×4.18

QUICK REFERENCE — Ch 11 Thermodynamics

1 cal = 4.18 JΔW = P ΔVW = area under P–V curveΔQ = ΔU + ΔWΔU = ΔQ − ΔWUᵢ − Uf = −WAdiabatic: ΔQ = 0 → ΔU = −ΔWIsochoric: ΔW = 0 → ΔQ = ΔUCyclic: ΔU = 0 → ΔQ = ΔWη = (H₁ − H₂) / H₁η = 1 − H₂/H₁H₂/H₁ = T₂/T₁η = 1 − T₂/T₁W = H₁ − H₂

Units: J (SI) · 1 cal = 4.18 J · η dimensionless or % · T in K for Carnot

Laws: Zeroth (temperature) · First (ΔQ=ΔU+ΔW) · Second (Kelvin-Planck, Clausius)

Tip: Draw the P–V diagram first — it tells you the process type and the work.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L11 — Thermodynamics only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 11 — Thermodynamics (L11)

20 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. Thermodynamics means — (A) study of the relationship between heat and other forms of energy   (B) study of conversion of chemical energy only   (C) study of mechanical energy only   (D) study of conversion of mechanical energy only

1 mark · Section A Q6 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Answer: (A)

Thermodynamics deals with heat and its conversion to/from mechanical, electrical, chemical energy, etc.

Explanation

NIOS L11 defines thermodynamics as the science of thermal energy transfer and energy interconversion — not limited to one energy form.

PYQ2. Out of the following, a law of thermodynamics actually is — (A) Zeroth law   (B) Faraday's law of thermodynamics   (C) Ideal gas law of thermodynamics   (D) Boyle's law of thermodynamics

1 mark · Section A Q7 (i) · 68/ESS/1-312-A

Model Answer

Answer: (A) Zeroth law of thermodynamics

The four laws are: Zeroth, First, Second, Third. Faraday's, Boyle's and ideal gas law are not thermodynamic laws.

Explanation

Zeroth law defines temperature via thermal equilibrium. Boyle's law and ideal gas equation describe gas behaviour; Faraday's law belongs to electromagnetism.

PYQ3. Out of the following, a type of thermodynamic system is — (A) Open system only   (B) Closed system only   (C) Thermally isolated system only   (D) All of the mentioned

1 mark · Section A Q7 (ii) · 68/ESS/1-312-A

Model Answer

Answer: (D) All of the mentioned

Open (mass + energy), closed (energy only), isolated/thermally isolated (neither) — all are thermodynamic system types.

Explanation

L11 §11.1.3 classifies systems by what crosses the boundary. All three listed types are valid examples.

PYQ4. In a Carnot cycle — (A) isothermal expansion is followed by adiabatic expansion   (B) isothermal compression is followed by isothermal expansion   (C) isothermal expansion is followed by adiabatic compression   (D) isothermal expansion is followed by isothermal compression

1 mark · Section A Q17(a) · 312/TUS/104A (Carnot passage)

Model Answer

Answer: (A)

Full sequence: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.

Explanation

After isothermal expansion (A→B, absorb H₁ at T₁), gas undergoes adiabatic expansion (B→C, T falls to T₂). L11 Carnot flowchart §11.5.1.

PYQ5. In a Carnot engine, heat is — (A) absorbed during isothermal expansion and released during isothermal compression   (B) absorbed during isothermal expansion and released during adiabatic compression   (C) absorbed during adiabatic expansion and released during isothermal compression   (D) absorbed during adiabatic compression and released during isothermal expansion

1 mark · Section A Q17(b) · 312/TUS/104A

Model Answer

Answer: (A)

Heat exchange occurs only in isothermal steps: absorb at T₁ (expansion), reject at T₂ (compression). Adiabatic steps have ΔQ = 0.

Explanation

Carnot cycle: two isothermals (heat transfer with reservoirs) + two adiabatics (no heat leak). Net work W = H₁ − H₂.

PYQ6. The law of thermodynamics which forbids conversion of 100% heat into work is the — (A) zeroth law   (B) first law   (C) second law   (D) third law

1 mark · Section A Q17(a) · 312/MAY/204A–C (first-law passage)

Model Answer

Answer: (C) second law

Kelvin-Planck: no engine converts all heat from a reservoir into work — needs a cold sink.

Explanation

First law allows energy conservation but not direction/limit of conversion. Second law (Kelvin-Planck) forbids 100% efficiency (L11 §11.5).

PYQ7. Match concept with law (any two): (a) Temperature → ?   (b) Conservation of energy → ?   (c) 100% efficiency impossible → ?   (d) No spontaneous heat flow cold→hot → ?   Options: (i) Clausius   (ii) Kelvin-Planck   (iii) First law   (iv) Zeroth law

2 marks (1×2) · Section A Q22 · 312/TUS/104A

Model Answer

(a) Temperature ↔ (iv) Zeroth law

(b) Conservation of energy ↔ (iii) First law

(c) 100% efficiency impossible ↔ (ii) Kelvin-Planck

(d) No self-transfer cold→hot ↔ (i) Clausius

Explanation

Zeroth defines temperature; first law is ΔQ = ΔU + ΔW; second law has two equivalent statements — Kelvin-Planck (engines) and Clausius (refrigerators/heat flow).

PYQ8. Fill in the blanks (any two): (a) Efficiency of a heat engine between T₁ and T₂ (T₁ > T₂) is always less than _____   (b) Work done in thermodynamic process AB on an indicator diagram = _____

2 marks (1×2) · Section A Q27 · 312/TUS/104A

Model Answer

(a) 1 − T₂/T₁ (Carnot efficiency) or (T₁ − T₂)/T₁

(b) Area under the curve AB on the P–V indicator diagram

Explanation

No real engine exceeds Carnot η between the same reservoirs. Work = ∫P dV = shaded area under P–V graph (L11 §11.1, §11.5.2).

PYQ9. Passage — First law limitations: Which energy form is most closely associated with heat? — (A) Potential   (B) Magnetic   (C) Sound   (D) Kinetic

1 mark · Section A Q17(b) · 312/MAY/204A–C

Model Answer

Answer: (D) Kinetic energy

Heat is random thermal motion of molecules — microscopic kinetic energy.

Explanation

Temperature measures average molecular KE. First law treats heat as energy in transit; microscopically it raises disordered KE of particles.

PYQ10. Match (any two): (a) Upper fixed point on Kelvin scale → ?   (b) Ideal heat engine efficiency between ice point and steam point → ?   Options: (i) Boiling point of water   (ii) Triple point of water   (iii) 100%   (iv) 26·9%

2 marks (1×2) · Section A Q27 · 312/MAY/204A–C

Model Answer

(a) Upper fixed point on Kelvin scale ↔ (ii) Triple point of water

(b) Carnot efficiency ice (273 K) to steam (373 K) ↔ (iv) 26·9%

η = 1 − 273/373 ≈ 0.268 ≈ 26.9%

Explanation

Kelvin scale upper fixed point is the triple point of water (273.16 K definition context in NIOS). Carnot sets maximum η = 1 − T₂/T₁ for given reservoirs.

PYQ11. Name the four strokes of the Carnot cycle in sequence.

2 marks · Section B Q33 · 312/MAY/204A (also Q30/32 in 204B/C)

Model Answer

  1. Isothermal expansion (absorb heat H₁ from source at T₁)
  2. Adiabatic expansion (temperature falls to T₂)
  3. Isothermal compression (reject heat H₂ to sink at T₂)
  4. Adiabatic compression (temperature rises back to T₁)

Explanation

Standard NIOS sequence A→B→C→D→A on P–V diagram. Two isothermals for heat exchange; two adiabatics so no heat leak (L11 §11.5.1).

PYQ12. In each Carnot cycle, the heat converted into work equals — choose and explain: (A) heat absorbed   (B) heat released   (C) heat absorbed − heat released   (D) heat absorbed + heat released

2 marks · Section A Q17(c) · 312/TUS/104A

Model Answer

Answer: (C) heat absorbed − heat released

For a cyclic process ΔU = 0 ⇒ ΔQ = ΔW. Net work W = H₁ − H₂ = area enclosed on P–V diagram.

Explanation

First law for a cycle: net heat in = net work out. Some heat must be rejected to the sink — cannot all become work (second law).

PYQ13. What is meant by an indicator diagram? Draw an indicator diagram for an isobaric (constant pressure) expansion process.

2 marks · Section B Q30 · 68/ESS/1-312-A

Model Answer

Indicator diagram: P–V graph showing how pressure varies with volume during a thermodynamic process.

Isobaric: horizontal line at constant P from V₁ to V₂. Work = area of rectangle = P(V₂ − V₁).

Explanation

Work done by system = area under P–V curve. Isobaric path is a horizontal segment; ΔW = PΔV (L11 §11.1, Eq. 11.1).

PYQ14. State three limitations of the first law of thermodynamics.

3 marks · Section A Q17 passage · 312/MAY/204A–C

Model Answer

First law (ΔQ = ΔU + ΔW) asserts heat–energy equivalence but fails to:

  • Indicate the direction of heat flow (hot → cold)
  • Give conditions under which heat can be converted into work
  • State how much heat can be converted into work (needs second law)

Explanation

Energy conservation alone would allow a refrigerator to run without work input. Second law supplies direction and efficiency limits (L11 §11.4.1).

PYQ15. Write TRUE or FALSE (any two): (i) In a refrigerator, the source of heat is the environment and the sink is the inner chamber.   (ii) If the door of a working refrigerator is kept open in a closed room, the room becomes cooler.

2 marks (1×2) · Section A Q23 · 68/ESS/1-312-A

Model Answer

(i) FALSE — heat is removed from the cold interior (cold reservoir) and rejected to the warmer surroundings (hot reservoir); work must be supplied.

(ii) FALSE — the compressor dumps net heat into the room; overall temperature rises.

Explanation

Refrigerator is a heat pump running in reverse of an engine. Clausius statement: heat cannot flow cold→hot without external work (L11 §11.5).

PYQ16. Match: (i) Internal energy of an ideal gas depends on → ?   (ii) A gas performs minimum work when it expands → ?   Options: P. Volume   Q. Temperature   R. Isothermally   S. Isochorically

2 marks (1×2) · Section A Q24 · 68/ESS/1-312-A

Model Answer

(i) Internal energy of ideal gas ↔ Q. Temperature only (U ∝ T)

(ii) Minimum work on expansion ↔ R. Isothermally (for positive expansion work at constant T; isochoric gives W = 0 but no expansion)

Explanation

Ideal gas U depends on T alone (not P or V). Comparing expansion paths: isothermal expansion absorbs heat to do work at constant T; adiabatic does less work for same ΔV.

PYQ17. Fill in: Efficiency of a heat engine does not depend on the nature of the _____.

1 mark · Section A Q22(i) · 68/ESS/1-312-A

Model Answer

working substance (working fluid)

Carnot efficiency η = 1 − T₂/T₁ depends only on reservoir temperatures, not whether the fluid is steam, gas, etc.

Explanation

Carnot passage (TUS Q17): ideal engine efficiency is independent of working substance — a key result from Carnot's analysis (L11 §11.5.2).

PYQ18. Thermodynamics means — (A) study of the relationship between heat and other forms of energy   (B) study of conversion of chemical energy to other forms   (C) study of mechanical energy to other forms   (D) study of conversion of mechanical energy to other forms

1 mark · Section A Q6 · Marking Scheme (68/ESS/1-312-A)

Model Answer

Answer: (A)

Thermodynamics deals with heat and its conversion to/from mechanical, electrical, chemical energy.

Explanation

Board marking scheme Q6: branch of physical science linking heat with other energy forms.

PYQ19. (i) Out of the following, a law of thermodynamics actually is — (A) Zeroth law   (B) Faraday's Law   (C) Ideal Gas Law   (D) Boyle's Law   OR   (ii) A type of thermodynamic system is — (A) Open   (B) Closed   (C) Thermally isolated   (D) All of the mentioned

1 mark · Section A Q7 · Marking Scheme (68/ESS/1-312-A)

Model Answer

(i) (A) Zeroth law of thermodynamics

(ii) (D) All of the mentioned (open, closed, thermally isolated)

Explanation

Marking scheme Q7: four laws are zeroth, first, second, third — not Faraday/Boyle/Ideal Gas Law.

PYQ20. Write TRUE or FALSE (any one): (i) In a refrigerator the source of heat is the environment and sink is the inner chamber.   (ii) If the door of a working refrigerator is kept open in a closed room, the room will become cool.

2 marks (1×2) · Section A Q23 · Marking Scheme (68/ESS/1-312-A)

Model Answer

(i) FALSE — source is inner chamber (cold), sink is environment (hot).

(ii) FALSE — net effect heats the room (compressor work + rejected heat).

Explanation

Refrigerator pumps heat from cold interior to warm surroundings; requires external work (second law).

Problem Solving — L11 Thermodynamics

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6Zeroth law

State the zeroth law of thermodynamics and explain how it justifies thermometers.

Thermal equilibrium is transitive

Solution — step by step with formulas

  1. If A is in equilibrium with B and B with C, then A with C.
  2. A thermometer defines a temperature scale shared by all bodies in equilibrium with it.

Final answer: Thermometers rest on transitive thermal equilibrium.

Formulas used in this problem

Thermal equilibrium is transitive

Textbook formal language

The zeroth law establishes temperature as the label of thermal-equilibrium classes of systems.

Working formula set for this problem: Thermal equilibrium is transitive. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

If two objects each match the same thermometer reading, they match each other.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Thermal equilibrium

Temperature equality is the condition for no net heat flow between systems in contact.

Link to chapter notes (L11 — Thermal equilibrium): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Thermal equilibrium is transitive. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write Thermal equilibrium is transitive before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6First law

A gas absorbs 500 J of heat and expands doing 200 J of work. Using NIOS form ΔQ = ΔU + ΔW, find ΔU.

ΔQ = ΔU + ΔW (NIOS: ΔW = work by system)
ΔW > 0 for expansion (work by system)

Solution — step by step with formulas

  1. NIOS notes: ΔQ = ΔU + ΔW with ΔW = work by the system.
  2. ΔQ = +500 J, ΔW = +200 J (expansion).
  3. ΔU = ΔQ − ΔW = 500 − 200 = +300 J.

Final answer: ΔU = +300 J

Formulas used in this problem

ΔQ = ΔU + ΔW (NIOS: ΔW = work by system)
ΔW > 0 for expansion (work by system)

Textbook formal language

The first law is energy conservation for a thermodynamic system. In NIOS convention, heat added to the system is positive and work done by the system is positive, giving ΔQ = ΔU + ΔW. Internal energy U is a state function; ΔQ and ΔW depend on path.

Working formula set for this problem: ΔQ = ΔU + ΔW; ΔW > 0 for expansion (work by system). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

500 J of heat goes in; 200 J is spent expanding. The leftover 300 J increases the gas’s internal energy.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — First law of thermodynamics (NIOS)

Special cases in notes: adiabatic ΔQ = 0 ⇒ ΔU = −ΔW; isochoric ΔW = 0 ⇒ ΔQ = ΔU; cyclic ΔU = 0 ⇒ ΔQ = ΔW. Work for a small step is ΔW = P ΔV (area under P–V curve).

Link to chapter notes (L11 — First law of thermodynamics (NIOS)): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: ΔQ = ΔU + ΔW; ΔW > 0 for expansion (work by system). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write ΔQ = ΔU + ΔW; ΔW > 0 for expansion (work by system) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Isothermal

Why is ΔU = 0 for ideal-gas isothermal expansion, and how are heat and work related?

Ideal gas: U = U(T) only
ΔT = 0 ⇒ ΔU = 0
ΔQ = ΔW

Solution — step by step with formulas

  1. Ideal gas: internal energy depends only on temperature.
  2. Isothermal ⇒ ΔT = 0 ⇒ ΔU = 0.
  3. ΔQ = ΔU + ΔW ⇒ ΔQ = ΔW.

Final answer: ΔU = 0; ΔQ = ΔW

Formulas used in this problem

Ideal gas: U = U(T) only
ΔT = 0 ⇒ ΔU = 0
ΔQ = ΔW

Textbook formal language

Because U of an ideal gas depends only on T, an isothermal process has ΔU = 0. The first law then requires heat absorbed to equal work done by the gas.

Working formula set for this problem: Ideal gas: U = U(T) only; ΔT = 0 ⇒ ΔU = 0; ΔQ = ΔW. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Fixed temperature means no change in stored internal energy for an ideal gas, so heat in equals work out during expansion.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Isothermal process

Along an ideal-gas isotherm, PV is constant. Do not set ΔU = 0 for a real gas without stating the ideal-gas model.

Link to chapter notes (L11 — Isothermal process): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Ideal gas: U = U(T) only; ΔT = 0 ⇒ ΔU = 0; ΔQ = ΔW. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write Ideal gas: U = U(T) only; ΔT = 0 ⇒ ΔU = 0; ΔQ = ΔW before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Adiabatic

In rapid adiabatic compression of an ideal gas, temperature rises. Explain with ΔQ = ΔU + ΔW.

ΔQ = 0
ΔU = −ΔW

Solution — step by step with formulas

  1. Adiabatic: ΔQ = 0 ⇒ ΔU = −ΔW.
  2. Compression: work is done on the gas, so work by the system ΔW is negative.
  3. Thus ΔU = −(negative) > 0 ⇒ U and T increase for an ideal gas.

Final answer: ΔQ = 0 ⇒ ΔU = −ΔW; compression raises U and T

Formulas used in this problem

ΔQ = 0
ΔU = −ΔW

Textbook formal language

An adiabatic process exchanges no heat. Energy change is only through work. Compression inputs energy, raising internal energy and temperature of an ideal gas.

Working formula set for this problem: ΔQ = 0; ΔU = −ΔW. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

No heat in or out. Squeezing the gas adds energy to the molecules—the gas warms (bicycle-pump effect).

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Adiabatic process

Notes: ΔW = PΔV for quasi-static steps; work is area under the P–V path. Ideal reversible adiabatic: PV^γ = constant.

Link to chapter notes (L11 — Adiabatic process): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: ΔQ = 0; ΔU = −ΔW. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write ΔQ = 0; ΔU = −ΔW before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Engine

A heat engine takes H₁ = 1000 J from a hot reservoir and rejects H₂ = 600 J each cycle. Find W and efficiency. What is the Carnot limit between temperatures T₁ > T₂?

η = (H₁ − H₂)/H₁ = 1 − H₂/H₁
η_Carnot = 1 − T₂/T₁

Solution — step by step with formulas

  1. Cycle: ΔU = 0 ⇒ W = H₁ − H₂ = 400 J (as in NIOS notes).
  2. η = (H₁ − H₂)/H₁ = 0.40 = 40%.
  3. Carnot: η = 1 − T₂/T₁ (absolute temperatures).

Final answer: W = 400 J; η = 40%; Carnot η = 1 − T₂/T₁

Formulas used in this problem

η = (H₁ − H₂)/H₁ = 1 − H₂/H₁
η_Carnot = 1 − T₂/T₁

Textbook formal language

For a cycle the working substance returns to its initial state so ΔU = 0 and net work equals net heat. Efficiency is work per heat input. No engine between two reservoirs exceeds Carnot efficiency.

Working formula set for this problem: η = (H₁ − H₂)/H₁ = 1 − H₂/H₁; η_Carnot = 1 − T₂/T₁. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Useful work 400 J from 1000 J input is 40% efficient. Even ideal engines cannot beat 1 − T_cold/T_hot.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Heat engine efficiency

Second law (Kelvin–Planck) forbids a cyclic engine that converts heat entirely into work with no rejection to a cold sink.

Link to chapter notes (L11 — Heat engine efficiency): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: η = (H₁ − H₂)/H₁ = 1 − H₂/H₁; η_Carnot = 1 − T₂/T₁. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write η = (H₁ − H₂)/H₁ = 1 − H₂/H₁; η_Carnot = 1 − T₂/T₁ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Second law

Why is a heat engine of 100% efficiency impossible according to the second law?

No 100% conversion of heat to work in a cycle

Solution — step by step with formulas

  1. Kelvin–Planck: cannot convert heat completely to work cyclically without rejecting heat.
  2. η = 1 would need Q₂ = 0.

Final answer: Violates second law (must reject some heat)

Formulas used in this problem

No 100% conversion of heat to work in a cycle

Textbook formal language

The second law forbids a perfect converter of heat to work operating in a cycle.

Working formula set for this problem: No 100% conversion of heat to work in a cycle. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

You always dump some heat to a cold sink; nature forbids a perfect heat-to-work engine.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Second law

Entropy of an isolated system does not decrease in real processes.

Link to chapter notes (L11 — Second law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: No 100% conversion of heat to work in a cycle. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write No 100% conversion of heat to work in a cycle before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).