thermodynamics formulas pyqs
Complete Class 11 Physics Thermodynamics Formula Sheet, NCERT Examples, NCERT Exercises, PYQs, Numericals and Quick Revision Notes.
This final revision page collects the working formulae from thermal equilibrium, heat, work, internal energy, first law, thermodynamic processes, heat engines, Carnot cycle and entropy. Use it as a last-day sheet and as a practice bank: first identify the correct formula, check units, convert temperature to kelvin where needed, then solve.
How to Use This Final Sheet
Thermodynamics questions usually become easy when you first identify the process, then choose the correct energy balance.
Step 1: Identify Process
Check whether the process is isothermal, adiabatic, isochoric, isobaric, cyclic, or a heat-engine/refrigerator cycle.
Step 2: Apply First Law
Use ΔU = Q - W. Decide signs: heat supplied positive, work done by system positive.
Step 3: Use Graph Meaning
Area under P-V graph gives work. Area inside a closed clockwise loop gives positive net work.
Step 4: Convert Units
Use kelvin temperature in ideal-gas, Carnot and entropy formulae. Convert litres to m3 if using SI pressure.
Step 5: Check Limitation
Heat engines cannot be 100% efficient. Refrigerators need work input to pump heat from cold to hot.
Step 6: Interpret Result
Negative work usually means compression. Positive entropy of universe means a natural irreversible process.
Complete Formula Sheet
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: Q is heat, m is mass, c is specific heat capacity, ΔT is temperature change.
Use: Used when temperature changes but no phase change occurs.
Exam tip: Use kelvin or degree Celsius for ΔT; the numerical change is the same.
Symbols: m is mass and L is latent heat.
Use: Used during change of state at constant temperature.
Exam tip: Do not add mcΔT during pure melting or boiling at constant temperature.
Symbols: C is heat capacity of the whole body.
Use: Heat needed per unit temperature rise of the body.
Exam tip: Do not confuse C with c; C depends on mass.
Symbols: c is specific heat capacity.
Use: Heat needed per kg per kelvin rise.
Exam tip: Check units: J kg-1 K-1.
Symbols: W is work done by gas, P is pressure, dV is small volume change.
Use: General work formula for expansion/compression.
Exam tip: Area under P-V graph gives work.
Symbols: ΔU is change in internal energy, Q is heat supplied, W is work done by system.
Use: First law with physics sign convention.
Exam tip: If work is done on gas, W is negative.
Symbols: P pressure, V volume, n moles, R gas constant, T kelvin temperature.
Use: Ideal gas equation.
Exam tip: Temperature must be in kelvin.
Symbols: U is internal energy of ideal gas.
Use: Internal energy depends only on temperature for ideal gas.
Exam tip: For ideal gas, ΔU = nCVΔT.
Symbols: CP and CV are molar heat capacities.
Use: Mayer's relation for ideal gas.
Exam tip: Use molar heat capacities, not total heat capacities.
Symbols: γ is adiabatic index.
Use: Used in adiabatic relations.
Exam tip: For monoatomic ideal gas γ = 5/3; diatomic near room temperature γ = 7/5.
Symbols: η is efficiency, W work output, QH heat absorbed.
Use: Heat engine efficiency.
Exam tip: Efficiency is always a fraction or percentage, not COP.
Symbols: ΔS entropy change for reversible heat transfer at T.
Use: Basic entropy formula.
Exam tip: Use reversible path for entropy change.
Heat Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: m mass, c specific heat capacity, ΔT final minus initial temperature.
Use: Sensible heat formula.
Exam tip: Use this only when no phase change occurs.
Symbols: Lf is latent heat of fusion.
Use: Heat during melting/freezing.
Exam tip: Melting Q positive for system; freezing Q negative for system.
Symbols: Lv is latent heat of vaporization.
Use: Heat during boiling/condensation.
Exam tip: Vaporization needs larger heat than fusion for water.
Symbols: calorie and joule are heat units.
Use: Unit conversion.
Exam tip: In school problems, 1 cal ≈ 4.2 J is often acceptable.
Symbols: J is mechanical equivalent of heat.
Use: Relates mechanical work and heat.
Exam tip: Historical value: J ≈ 4.186 J cal-1.
Symbols: Heat lost by hotter body equals heat gained by colder body in an isolated mixture.
Use: Calorimetry principle.
Exam tip: Apply only if heat losses to surroundings are neglected.
Symbols: θf is final equilibrium temperature.
Use: Mixing two bodies without phase change.
Exam tip: Final temperature lies between initial temperatures.
Symbols: n moles, C molar heat capacity.
Use: Molar form of heat formula.
Exam tip: Use CP at constant pressure and CV at constant volume.
Work Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: P external pressure in quasi-static process, dV volume change.
Use: General work by gas.
Exam tip: Expansion gives positive W; compression gives negative W.
Symbols: P constant pressure.
Use: Work in isobaric process.
Exam tip: Use only when pressure is constant.
Symbols: n moles, R gas constant, T constant temperature.
Use: Isothermal work by ideal gas.
Exam tip: Expansion has V2 > V1, so W positive.
Symbols: Common-log version of isothermal work.
Use: Use when log base 10 is provided.
Exam tip: Do not mix ln and log without 2.303 factor.
Symbols: Adiabatic reversible work by gas.
Use: Useful in adiabatic expansion/compression.
Exam tip: Sign depends on initial and final states.
Symbols: Ideal gas adiabatic work by gas.
Use: For Q = 0, W = -ΔU.
Exam tip: Expansion cools gas, so T1 > T2 and W positive.
Symbols: Closed P-V loop area.
Use: Net work in cyclic process.
Exam tip: Clockwise loop: positive work by engine.
Symbols: Volume constant.
Use: Isochoric work.
Exam tip: No boundary work in constant-volume process.
Internal Energy Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: n moles, CV molar heat capacity at constant volume.
Use: Change in internal energy of ideal gas.
Exam tip: Depends only on temperature for ideal gas.
Symbols: f degrees of freedom.
Use: Internal energy of ideal gas.
Exam tip: Monoatomic f=3; diatomic near room temperature f=5.
Symbols: For monoatomic ideal gas.
Use: Internal energy of monoatomic gas.
Exam tip: Use for He, Ne, Ar ideal-gas problems.
Symbols: For diatomic ideal gas without vibration.
Use: Approximate internal energy near room temperature.
Exam tip: Use when problem says diatomic ideal gas.
Symbols: Temperature constant.
Use: Internal energy change in isothermal ideal-gas process.
Exam tip: If ΔT = 0, ΔU = 0.
Symbols: W = 0 because ΔV = 0.
Use: First law at constant volume.
Exam tip: All supplied heat changes internal energy.
Symbols: Q = 0.
Use: First law in adiabatic process.
Exam tip: Work by gas decreases internal energy.
Symbols: Initial and final states are same.
Use: Cyclic process.
Exam tip: Internal energy is a state function.
First Law Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: Heat supplied equals internal energy change plus work done by system.
Use: First law in differential form.
Exam tip: Use signs carefully.
Symbols: Change in internal energy equals heat supplied minus work done by system.
Use: Most common physics convention.
Exam tip: W positive for expansion.
Symbols: ΔU = 0.
Use: First law for isothermal ideal gas.
Exam tip: Heat absorbed becomes work done.
Symbols: W = 0.
Use: Constant-volume heating or cooling.
Exam tip: P-V graph is vertical line.
Symbols: Pressure constant.
Use: Isobaric first law.
Exam tip: Use W = PΔV.
Symbols: No heat exchange.
Use: Thermally insulated or very fast process.
Exam tip: First law becomes ΔU = -W.
Symbols: ΔUcycle = 0.
Use: For a complete cycle.
Exam tip: Net heat absorbed equals net work output.
Symbols: Won is work done on system.
Use: Alternative chemistry convention.
Exam tip: Do not mix W by system and W on system in same solution.
Process Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: For isothermal ideal gas.
Use: Boyle relation at constant temperature.
Exam tip: Only valid for fixed amount of ideal gas.
Symbols: γ = CP/CV.
Use: Reversible adiabatic relation.
Exam tip: Do not use for irreversible free expansion.
Symbols: T kelvin, V volume.
Use: Adiabatic temperature-volume relation.
Exam tip: Useful when pressure is absent.
Symbols: T kelvin, P pressure.
Use: Adiabatic pressure-temperature relation.
Exam tip: Convert Celsius to kelvin.
Symbols: Pressure constant.
Use: Isobaric ideal-gas relation.
Exam tip: Charles law.
Symbols: Volume constant.
Use: Isochoric ideal-gas relation.
Exam tip: Pressure law.
Symbols: T constant.
Use: Work in isothermal ideal gas process.
Exam tip: Expansion positive, compression negative.
Symbols: P constant.
Use: Work at constant pressure.
Exam tip: Area under horizontal P-V line.
Symbols: V constant.
Use: No volume change.
Exam tip: Vertical P-V line has zero area.
Symbols: No heat exchange.
Use: Adiabatic process.
Exam tip: Temperature can still change.
Carnot Engine Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: Heat engine energy balance.
Use: Net work per cycle.
Exam tip: QC is rejected heat magnitude.
Symbols: Efficiency definition.
Use: Useful work output per heat input.
Exam tip: η is less than 1.
Symbols: Equivalent heat-engine efficiency.
Use: Uses absorbed and rejected heat.
Exam tip: If QC decreases, efficiency increases.
Symbols: T in kelvin.
Use: Maximum possible engine efficiency.
Exam tip: Never use Celsius.
Symbols: Carnot reversible relation.
Use: Connects heat and reservoir temperatures.
Exam tip: Use magnitudes of heat.
Symbols: Refrigerator COP.
Use: Cooling obtained per work input.
Exam tip: COP can be greater than 1.
Symbols: Ideal refrigerator temperatures in kelvin.
Use: Maximum COP of refrigerator.
Exam tip: COP rises when temperature gap is small.
Symbols: Heat pump COP.
Use: Heating obtained per work input.
Exam tip: Heat pump COP = refrigerator COP + 1.
Symbols: Ideal heat pump COP.
Use: Maximum heating performance.
Exam tip: Use when desired output is heating.
Entropy Formulae
Every card gives the formula, symbols, meaning and exam tip so that revision remains quick but not shallow.
Symbols: Reversible heat transfer at constant T.
Use: Basic entropy change.
Exam tip: Use reversible path even for irreversible actual process.
Symbols: Equality reversible, greater than zero irreversible.
Use: Second law in entropy form.
Exam tip: System entropy may decrease, universe entropy cannot.
Symbols: Total entropy balance.
Use: Spontaneity test.
Exam tip: For natural irreversible process, total is positive.
Symbols: Ideal gas entropy change using T and V.
Use: General ideal-gas entropy formula.
Exam tip: Use kelvin temperatures.
Symbols: Ideal gas entropy change using T and P.
Use: Alternative entropy formula.
Exam tip: Pressure ratio must be dimensionless.
Symbols: Isothermal ideal-gas entropy change.
Use: Because ΔT = 0.
Exam tip: Free expansion entropy can be calculated by reversible isothermal path.
Symbols: Phase change at constant T.
Use: Melting or boiling entropy change.
Exam tip: Positive for melting/vaporization, negative for freezing/condensation.
Symbols: Heat Q flows from hot to cold.
Use: Entropy generated in natural heat transfer.
Exam tip: Since TH > TC, total is positive.
NCERT Examples: Important Solved Models
These original NCERT-style models cover the calculation patterns repeatedly used in school exams and entrance exams.
A 0.5 kg metal block cools from 80°C to 30°C. If c = 400 J kg-1 K-1, find heat lost.
Show Solution
Given: m = 0.5 kg, c = 400 J kg-1 K-1, ΔT = 50 K.
Formula: Q = mcΔT.
Solution: Q = 0.5 × 400 × 50 = 10000 J. Since it cools, heat lost by block is 10000 J.
Final Answer: 1.0 × 104 J lost.
How much heat is required to melt 20 g of ice at 0°C? Lf = 3.34 × 105 J kg-1.
Show Solution
Given: m = 0.020 kg, Lf = 3.34 × 105 J kg-1.
Formula: Q = mL.
Solution: Q = 0.020 × 3.34 × 105 = 6680 J.
Final Answer: 6.68 × 103 J.
One mole of ideal gas expands isothermally at 300 K from 2 L to 4 L. Find work done.
Show Solution
Given: n = 1, T = 300 K, V2/V1 = 2.
Formula: W = nRT ln(V2/V1).
Solution: W = 1 × 8.314 × 300 × ln2 = 1728 J approximately.
Final Answer: 1.73 × 103 J.
A gas is heated at constant volume and absorbs 600 J. Find work and internal energy change.
Show Solution
Given: Q = 600 J, ΔV = 0.
Formula: W = 0, ΔU = Q - W.
Solution: At constant volume, W = 0. ΔU = 600 - 0 = 600 J.
Final Answer: W = 0, ΔU = 600 J.
An ideal gas absorbs 500 J heat in an isothermal expansion. Find ΔU and W.
Show Solution
Given: Q = 500 J, isothermal ideal gas.
Formula: ΔU = 0, Q = W.
Solution: For an ideal gas, internal energy depends only on temperature. Since T constant, ΔU = 0 and W = Q = 500 J.
Final Answer: ΔU = 0, W = 500 J.
In an adiabatic expansion, a gas does 300 J work. Find heat exchanged and internal energy change.
Show Solution
Given: W = 300 J, adiabatic.
Formula: Q = 0, ΔU = Q - W.
Solution: ΔU = 0 - 300 = -300 J.
Final Answer: Q = 0, ΔU = -300 J.
In a cycle, a gas absorbs net 700 J heat. Find net work done.
Show Solution
Given: Qcycle = 700 J.
Formula: ΔUcycle = 0, Q = W.
Solution: Since the gas returns to the initial state, ΔU = 0. Hence W = 700 J.
Final Answer: 700 J.
A gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work.
Show Solution
Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.
Formula: W = PΔV.
Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.
Final Answer: 1000 J.
A Carnot engine operates between 500 K and 300 K. Find efficiency.
Show Solution
Given: TH = 500 K, TC = 300 K.
Formula: η = 1 - TC/TH.
Solution: η = 1 - 300/500 = 0.40.
Final Answer: 40%.
An ideal refrigerator works between 250 K and 300 K. Find COP.
Show Solution
Given: TC = 250 K, TH = 300 K.
Formula: COPR = TC/(TH - TC).
Solution: COP = 250/(300 - 250) = 5.
Final Answer: 5.
A system absorbs 900 J reversibly at 300 K. Find entropy change.
Show Solution
Given: Qrev = 900 J, T = 300 K.
Formula: ΔS = Qrev/T.
Solution: ΔS = 900/300 = 3 J K-1.
Final Answer: 3 J K-1.
100 J heat flows from 400 K to 300 K. Find total entropy change.
Show Solution
Given: Q = 100 J, TH = 400 K, TC = 300 K.
Formula: ΔS = -Q/TH + Q/TC.
Solution: ΔS = -100/400 + 100/300 = -0.25 + 0.333 = 0.083 J K-1.
Final Answer: 0.083 J K-1.
NCERT Exercise-Type Practice
These exercise-type questions are designed in the same spirit as Class 11 textbook practice: direct formula use, sign convention and conceptual interpretation.
A 2 kg body with c = 500 J kg-1 K-1 is heated by 20 K. Find heat supplied.
Show Solution
Given: m = 2 kg, c = 500 J kg-1 K-1, ΔT = 20 K.
Formula: Q = mcΔT.
Solution: Q = 2 × 500 × 20 = 20000 J.
Final Answer: 20000 J.
A gas expands at constant pressure 105 Pa by 0.02 m3. Find work.
Show Solution
Given: P = 105 Pa, ΔV = 0.02 m3.
Formula: W = PΔV.
Solution: W = 105 × 0.02 = 2000 J.
Final Answer: 2000 J.
A gas receives 1000 J heat and does 400 J work. Find ΔU.
Show Solution
Given: Q = 1000 J, W = 400 J.
Formula: ΔU = Q - W.
Solution: ΔU = 1000 - 400 = 600 J.
Final Answer: 600 J.
A gas is compressed adiabatically and 250 J work is done on it. Find ΔU.
Show Solution
Given: Q = 0, Won = 250 J, so work by gas W = -250 J.
Formula: ΔU = Q - W.
Solution: ΔU = 0 - (-250) = +250 J.
Final Answer: +250 J.
A cyclic engine absorbs 2000 J and rejects 1200 J. Find work and efficiency.
Show Solution
Given: QH = 2000 J, QC = 1200 J.
Formula: W = QH - QC, η = W/QH.
Solution: W = 800 J. η = 800/2000 = 0.40.
Final Answer: W = 800 J, η = 40%.
A Carnot engine works between 600 K and 450 K. Find efficiency.
Show Solution
Given: TH = 600 K, TC = 450 K.
Formula: η = 1 - TC/TH.
Solution: η = 1 - 450/600 = 0.25.
Final Answer: 25%.
A refrigerator extracts 1800 J using 300 J work. Find COP and heat rejected.
Show Solution
Given: QC = 1800 J, W = 300 J.
Formula: COP = QC/W, QH = QC + W.
Solution: COP = 1800/300 = 6. QH = 1800 + 300 = 2100 J.
Final Answer: COP = 6, QH = 2100 J.
50 g ice melts at 0°C. Lf = 3.34 × 105 J kg-1. Find heat.
Show Solution
Given: m = 0.050 kg, Lf = 3.34 × 105 J kg-1.
Formula: Q = mL.
Solution: Q = 0.050 × 3.34 × 105 = 16700 J.
Final Answer: 16700 J.
200 J heat is absorbed reversibly at 400 K. Find ΔS.
Show Solution
Given: Qrev = 200 J, T = 400 K.
Formula: ΔS = Qrev/T.
Solution: ΔS = 200/400 = 0.5 J K-1.
Final Answer: 0.5 J K-1.
A gas expands freely into vacuum in an insulated container. What are Q and W?
Show Solution
Given: Insulated container and vacuum expansion.
Formula: Q = 0, W = 0.
Solution: No heat enters and no external pressure is opposed, so no work is done. For an ideal gas, ΔU = 0, but entropy increases.
Final Answer: Q = 0, W = 0; entropy increases.
Two moles of ideal gas expand isothermally at 300 K from V to 3V. Find work.
Show Solution
Given: n = 2, T = 300 K, V2/V1 = 3.
Formula: W = nRT ln(V2/V1).
Solution: W = 2 × 8.314 × 300 × ln3 = 5480 J approximately.
Final Answer: 5.48 × 103 J.
A gas at constant volume is supplied 750 J heat. Find W and ΔU.
Show Solution
Given: Q = 750 J, ΔV = 0.
Formula: W = 0, ΔU = Q - W.
Solution: At constant volume, W = 0. Hence ΔU = 750 J.
Final Answer: W = 0, ΔU = 750 J.
At constant volume, a gas has pressure 1.5 × 105 Pa at 300 K. Find pressure at 450 K.
Show Solution
Given: P1 = 1.5 × 105 Pa, T1 = 300 K, T2 = 450 K.
Formula: P/T = constant.
Solution: P2 = P1T2/T1 = 1.5 × 105 × 450/300 = 2.25 × 105 Pa.
Final Answer: 2.25 × 105 Pa.
At constant pressure, a gas volume is 4 L at 300 K. Find volume at 450 K.
Show Solution
Given: V1 = 4 L, T1 = 300 K, T2 = 450 K.
Formula: V/T = constant.
Solution: V2 = 4 × 450/300 = 6 L.
Final Answer: 6 L.
For an ideal gas, CV = 20.8 J mol-1 K-1. Find CP.
Show Solution
Given: CV = 20.8 J mol-1 K-1, R = 8.314 J mol-1 K-1.
Formula: CP - CV = R.
Solution: CP = 20.8 + 8.314 = 29.114 J mol-1 K-1.
Final Answer: 29.1 J mol-1 K-1.
Find ΔU for 1 mole monoatomic ideal gas when temperature rises by 40 K.
Show Solution
Given: n = 1, ΔT = 40 K.
Formula: ΔU = (3/2)nRΔT.
Solution: ΔU = 1.5 × 1 × 8.314 × 40 = 498.8 J.
Final Answer: 499 J approximately.
For reversible adiabatic expansion, P1V1γ = P2V2γ. If volume doubles and γ = 5/3, write pressure ratio P2/P1.
Show Solution
Given: V2 = 2V1, γ = 5/3.
Formula: P2/P1 = (V1/V2)γ.
Solution: P2/P1 = (1/2)5/3.
Final Answer: (1/2)5/3.
300 J heat flows from 600 K body to 300 K body. Find total entropy change.
Show Solution
Given: Q = 300 J, TH = 600 K, TC = 300 K.
Formula: ΔStotal = -Q/TH + Q/TC.
Solution: ΔS = -300/600 + 300/300 = -0.5 + 1 = 0.5 J K-1.
Final Answer: 0.5 J K-1.
An ideal heat pump operates between 280 K and 350 K. Find COP.
Show Solution
Given: TC = 280 K, TH = 350 K.
Formula: COPHP = TH/(TH - TC).
Solution: COP = 350/(350 - 280) = 350/70 = 5.
Final Answer: 5.
A clockwise P-V cycle encloses area 1500 J. Find net work and net heat.
Show Solution
Given: Area enclosed = 1500 J, clockwise cycle.
Formula: Wcycle = area, ΔUcycle = 0, Q = W.
Solution: Clockwise cycle gives positive work by gas. W = 1500 J and Q = 1500 J.
Final Answer: W = 1500 J, Q = 1500 J.
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PYQ Bank and Multi-Curriculum Questions
Answers are hidden. Attempt each one before opening the answer.
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Answer: ΔU = Q - W. Q is positive when heat is supplied to the system. W is positive when work is done by the system during expansion.
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Answer: In an isochoric process volume is constant, so ΔV = 0 and W = ∫P dV = 0.
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Answer: It gives work done by gas during expansion or compression.
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Answer: For ideal gas, intermolecular potential energy is neglected, so internal energy is kinetic and depends only on temperature.
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Answer: Heat cannot be completely converted into work in a cyclic engine; heat cannot flow from cold to hot without work input.
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Answer: (C) zero.
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Answer: (A) Q = 0.
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Answer: η = 1 - 300/600 = 50%.
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Answer: Work is area under P-V curve. The process with higher pressure throughout the expansion gives larger work.
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Answer: It measures heat removed per work input; the work transfers heat rather than being directly converted into cooling.
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Answer: W = nRT ln2 = RT ln2.
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Answer: For a cycle ΔU = 0, so Q = W = 500 J.
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Answer: ΔU = Q - W = 800 - 300 = 500 J.
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Answer: PVγ = constant.
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Answer: 0.25 = 1 - TC/400, so TC = 300 K.
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Answer: Work is path dependent and equals area under P-V path; different paths have different areas.
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Answer: Yes. Since ΔU is state function and cycle returns to initial state, ΔU = 0, but net heat equals net work.
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Answer: It is not quasi-static; intermediate states are not equilibrium states with well-defined pressure of the gas against external pressure.
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Answer: For ideal gas, T remains same because Q = W = ΔU = 0; entropy increases due to larger volume.
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Answer: Reversible engine has maximum efficiency. Irreversible engine has entropy production and lower efficiency.
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Answer: ΔS = Q/T; for the same Q, smaller T gives larger entropy change.
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Answer: Entropy is a thermodynamic state function measuring energy dispersal; disorder is only a rough analogy.
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Answer: Carnot formula depends on absolute thermodynamic temperature.
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Answer: Pressure cooker raises pressure, increasing boiling point of water, so food cooks at higher temperature.
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Answer: Faster molecules escape, taking energy away; remaining liquid has lower average kinetic energy.
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Answer: Temperature remains constant while supplied heat changes phase.
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Answer: At constant volume Q = ΔU = nCVΔT. At constant pressure Q = nCPΔT = ΔU + PΔV = nCVΔT + nRΔT, hence CP - CV = R.
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Answer: ΔS = nR ln(4V/V) = nR ln4.
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Answer: Heat depends on path of energy transfer, not only initial and final states.
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Answer: Both true and reason correctly explains assertion.
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Answer: Both true and reason correctly explains assertion.
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Answer: Assertion as worded is incomplete; it does not violate second law because work is supplied. Reason is true.
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Answer: Both true and reason correctly explains assertion.
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Answer: True.
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Answer: True.
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Answer: False. It is zero.
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Answer: Mathematically yes, but 0 K cannot be achieved physically, so practical 100% efficiency is impossible.
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Answer: True, but entropy of an isolated system cannot decrease.
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Answer: Rapid compression is approximately adiabatic; work done on air increases internal energy and temperature.
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Answer: It rejects heat removed from the room plus compressor work to the outside.
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Answer: It remains near equilibrium and avoids dissipative entropy production.
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Answer: Yes, in adiabatic compression or expansion work changes internal energy while Q = 0.
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Answer: Yes, in isochoric heating or cooling where all heat changes internal energy.
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Answer: QC = QHTC/TH = 1200 × 300/600 = 600 J. W = 600 J.
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Answer: W = QC/COP = 2500/5 = 500 J. QH = 3000 J.
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Answer: Work done by gas W = -100 J. ΔU = Q - W = 300 - (-100) = 400 J.
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Answer: ΔU = (3/2)nRΔT = 1.5 × 2 × 8.314 × 10 = 249.4 J.
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Answer: ΔS = -100/500 + 100/250 = -0.2 + 0.4 = 0.2 J K-1.
A domestic refrigerator removes heat from the cold chamber and rejects heat to the room. The compressor needs electrical work.
Show Answer
Answer: QC, the heat removed from the cold chamber.
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Answer: Heat is being transferred from cold to hot, against natural direction.
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Answer: QH = QC + W.
A petrol engine absorbs heat from burning fuel, produces shaft work and rejects waste heat to surroundings.
Show Answer
Answer: Some heat must be rejected to the sink.
Show Answer
Answer: Work W.
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Answer: η = W/QH = 1 - QC/QH.
An ideal reversible engine operates between a source at TH and sink at TC.
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Answer: η = 1 - TC/TH.
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Answer: Kelvin temperatures.
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Answer: All processes are reversible and no entropy is produced.
Heat flows naturally from a hot block to a cold block inside an insulated box.
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Answer: It increases.
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Answer: ΔS = -Q/TH + Q/TC.
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Answer: At thermal equilibrium.
A pressure cooker traps steam and increases pressure above atmospheric pressure.
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Answer: Higher pressure requires higher temperature for boiling.
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Answer: Water and steam are hotter than 100°C.
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Answer: Yes, phase change and pressure-temperature relation are involved.
Air is compressed rapidly before fuel injection.
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Answer: Adiabatic compression.
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Answer: Temperature rises.
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Answer: Compressed air becomes hot enough for ignition.
An AC removes heat from the room and rejects heat outdoors.
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Answer: Refrigerator.
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Answer: Electrical work to compressor.
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Answer: It rejects room heat plus compressor work.
Quick Revision Notes
One-Page Revision Sheet
Heat and Calorimetry
- Q = mcΔT for temperature change.
- Q = mL for phase change.
- Heat lost = heat gained in insulated mixing.
- Temperature remains constant during phase change.
Work and Graphs
- W = ∫P dV.
- Area under P-V graph gives work.
- Clockwise cycle gives positive net work.
- Isochoric process has W = 0.
First Law
- ΔU = Q - W.
- Isothermal ideal gas: ΔU = 0.
- Adiabatic: Q = 0.
- Cyclic: ΔU = 0 and Q = W.
Processes
- Isothermal: PV = constant.
- Adiabatic: PVγ = constant.
- Isobaric: P constant.
- Isochoric: V constant.
Heat Engines
- W = QH - QC.
- η = W/QH.
- Carnot: η = 1 - TC/TH.
- 100% efficiency is impossible in practice.
Entropy
- ΔS = Qrev/T.
- For isolated system, ΔS ≥ 0.
- Reversible: ΔSuniverse = 0.
- Irreversible: ΔSuniverse > 0.
Most Important Exam Tips
- Always convert Celsius to kelvin in ideal gas, Carnot and entropy formulae involving absolute temperature.
- Write sign convention at the start if the problem involves work done on gas versus by gas.
- For cyclic processes, immediately write ΔU = 0.
- For ideal gas isothermal process, immediately write ΔU = 0 and Q = W.
- For adiabatic process, immediately write Q = 0.
- COP is not efficiency; it can be greater than one.
- Entropy is a state function; heat and work are path functions.
Most Common Mistakes
- Using litre with pascal without converting to m3.
- Using Celsius in η = 1 - TC/TH.
- Forgetting that W is negative during compression when W is work done by gas.
- Using adiabatic formulas for irreversible free expansion.
- Confusing heat capacity C with specific heat c.
- Opening answer before attempting the question. That one is technically not thermodynamics, but it ruins the score.
Last-Day Revision Order
- Revise formula cards in this order: heat, work, first law, processes, Carnot, entropy.
- Practice five graph questions and five sign-convention questions.
- Revise special cases: isothermal, adiabatic, isochoric, isobaric, cyclic.
- Practice Carnot temperature conversion questions.
- End with assertion-reason and case-study questions.
If Thermodynamics is not clear and you are looking for a Physics Tutor, contact Kumar Sir.
Phone: +91-9958461445 Email: kumarsirphysics@gmail.com