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.
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.
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.
Expansion: work by system → ΔU decreases.
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.
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:
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.
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.
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 = 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.
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:
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
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
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
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
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
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
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
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 = _____
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
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%
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.
Model Answer
- Isothermal expansion (absorb heat H₁ from source at T₁)
- Adiabatic expansion (temperature falls to T₂)
- Isothermal compression (reject heat H₂ to sink at T₂)
- 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
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.
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.
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.
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
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 _____.
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
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
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.
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.
State the zeroth law of thermodynamics and explain how it justifies thermometers.
Solution — step by step with formulas
- If A is in equilibrium with B and B with C, then A with C.
- 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
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).
A gas absorbs 500 J of heat and expands doing 200 J of work. Using NIOS form ΔQ = ΔU + ΔW, find ΔU.
Solution — step by step with formulas
- NIOS notes: ΔQ = ΔU + ΔW with ΔW = work by the system.
- ΔQ = +500 J, ΔW = +200 J (expansion).
- ΔU = ΔQ − ΔW = 500 − 200 = +300 J.
Final answer: ΔU = +300 J
Formulas used in this problem
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).
Why is ΔU = 0 for ideal-gas isothermal expansion, and how are heat and work related?
Solution — step by step with formulas
- Ideal gas: internal energy depends only on temperature.
- Isothermal ⇒ ΔT = 0 ⇒ ΔU = 0.
- ΔQ = ΔU + ΔW ⇒ ΔQ = ΔW.
Final answer: ΔU = 0; ΔQ = ΔW
Formulas used in this problem
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).
In rapid adiabatic compression of an ideal gas, temperature rises. Explain with ΔQ = ΔU + ΔW.
Solution — step by step with formulas
- Adiabatic: ΔQ = 0 ⇒ ΔU = −ΔW.
- Compression: work is done on the gas, so work by the system ΔW is negative.
- 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
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).
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?
Solution — step by step with formulas
- Cycle: ΔU = 0 ⇒ W = H₁ − H₂ = 400 J (as in NIOS notes).
- η = (H₁ − H₂)/H₁ = 0.40 = 40%.
- Carnot: η = 1 − T₂/T₁ (absolute temperatures).
Final answer: W = 400 J; η = 40%; Carnot η = 1 − T₂/T₁
Formulas used in this problem
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).
Why is a heat engine of 100% efficiency impossible according to the second law?
Solution — step by step with formulas
- Kelvin–Planck: cannot convert heat completely to work cyclically without rejecting heat.
- η = 1 would need Q₂ = 0.
Final answer: Violates second law (must reject some heat)
Formulas used in this problem
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).