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Class 11 Physics Chapter 11 Thermodynamics — Formulas & Key Points

Q: What is the difference between the first and second laws of thermodynamics?

The first law is energy conservation: ΔQ = ΔU + ΔW, stating energy cannot be created or destroyed. The second law introduces directionality and efficiency limits, stating heat cannot spontaneously flow from cold to hot and no engine can be 100 % efficient.

Q: Why is Cₚ always greater than Cᵥ for an ideal gas?

At constant pressure, the gas does expansion work while being heated, so extra energy is needed beyond just raising temperature. Hence Cₚ = Cᵥ + R, making Cₚ larger by exactly R.

Q: How do I remember which process uses which work formula?

Use mnemonics: isobaric (constant P) → W = P ΔV; isothermal (constant T) → W = nRT ln(V₂/V₁); adiabatic (ΔQ=0) → W = −ΔU; isochoric (constant V) → W = 0. Write these on your formula sheet.

Q: What are common sign convention mistakes in the first law?

Students often forget that work done by the system is positive and work done on the system is negative. Always define ΔW based on whether the gas expands (+) or is compressed (−), then apply ΔQ = ΔU + ΔW carefully.

Q: Can a heat engine have 100 % efficiency?

No. The second law (Kelvin-Planck statement) forbids a heat engine from converting all absorbed heat into work. Some heat Q₂ must always be rejected to the cold reservoir, so η = 1 − Q₂/Q₁ < 1.

Q: How is the Carnot engine different from real engines?

A Carnot engine is a theoretical reversible engine with maximum efficiency η = 1 − T₂/T₁. Real engines have friction, turbulence and finite-time processes, so their efficiency is always less than the Carnot value for the same temperatures.

Q: What is the coefficient of performance (COP) for a refrigerator?

COP = Q₂/W, where Q₂ is heat removed from the cold space and W is work input. For an ideal Carnot refrigerator, COP = T₂/(T₁ − T₂). Higher COP means better performance; COP can exceed 1.

Q: Why must I use Kelvin in Carnot efficiency and COP formulas?

Carnot formulas are derived from absolute thermodynamic temperature. Using Celsius gives incorrect ratios (even negative efficiencies). Always convert: T(K) = T(°C) + 273 before substituting into η = 1 − T₂/T₁ or COP = T₂/(T₁−T₂).

Q: What is the relation between P, V and T in an adiabatic process?

For a reversible adiabatic process of an ideal gas: PVᵞ = constant, TVᵞ⁻¹ = constant and T P⁽¹⁻ᵞ⁾/ᵞ = constant. Use the relation that contains your known and unknown variables.

Q: How does CBSETUTOR.ai help with thermodynamics numericals?

Upload a photo of any NCERT or board-level problem and the AI tutor provides a step-by-step solution with formula identification, sign checks and unit conversions. The flat ₹999/month plan covers all chapters for Classes 6–12, with a three-day free trial to test it first.

Thermodynamics is the first chapter in NCERT Class 11 Physics where macroscopic energy transformations take centre stage. Board exams consistently ask 3–5 mark numericals on work done in various processes, efficiency of Carnot engines and application of the first law. This formula sheet organises every equation, sign rule and constant you need, so you can revise the entire chapter in one sitting and tackle any CBSE or competitive exam question with confidence.

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Key takeaways

✓First Law of Thermodynamics (ΔQ = ΔU + ΔW) links heat, internal energy and work; sign convention is critical.
✓Zeroth Law establishes thermal equilibrium and the concept of temperature measurement.
✓Second Law dictates the direction of heat flow and defines efficiency limits for heat engines.
✓Specific heat at constant pressure (Cp) is always greater than at constant volume (Cv); γ = Cp/Cv.
✓Work done in isothermal, adiabatic, isobaric and isochoric processes follows distinct formulae.
✓Heat engine efficiency η = 1 − Q₂/Q₁; Carnot efficiency η = 1 − T₂/T₁ (temperatures in Kelvin).
✓Coefficient of performance (COP) for refrigerators is Q₂/W; higher COP means better cooling per unit work.

Core Laws of Thermodynamics — Statements & Mathematical Forms

The four laws form the theoretical backbone of this chapter. The Zeroth Law ensures we can define temperature consistently across systems. The First Law is energy conservation tailored for thermal systems. The Second Law introduces entropy and irreversibility, forbidding perpetual motion machines of the second kind. The Third Law (not in NCERT Class 11 syllabus but worth knowing) states that absolute zero is unattainable. For CBSE exams, focus on precise statements and when each law applies. The table below captures the essence and mathematical expressions where applicable.

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. Basis of thermometry.
First Law: ΔQ = ΔU + ΔW. Heat supplied equals change in internal energy plus work done by the system.
Second Law (Kelvin-Planck): No process is possible whose sole result is absorption of heat and complete conversion into work.
Second Law (Clausius): Heat cannot spontaneously flow from a colder body to a hotter body without external work.

Sign Conventions for Heat, Work and Internal Energy

Sign errors cost students full marks in CBSE numericals. NCERT follows the convention that heat supplied to the system is positive, work done by the system is positive and increase in internal energy is positive. Conversely, heat removed is negative, work done on the system is negative and decrease in internal energy is negative. Always write ΔQ, ΔW, ΔU with their signs explicitly before substituting into the first law. Mixing conventions—like taking expansion work as negative—leads to wrong answers in cyclic process or PV diagram questions.

ΔQ > 0 when heat is added to the system; ΔQ < 0 when heat leaves the system.
ΔW > 0 when system does work on surroundings (expansion); ΔW < 0 when surroundings compress the system.
ΔU > 0 when internal energy rises (temperature increase for ideal gas); ΔU < 0 when it falls.
In cyclic processes ΔU = 0, so ΔQ = ΔW over the full cycle.

Work Done in Different Thermodynamic Processes

The formula for work depends on which variable (P, V or T) is held constant. In an isobaric process pressure is constant, so W = P ΔV. In an isothermal process temperature is constant and W = nRT ln(V₂/V₁) for an ideal gas. In an adiabatic process no heat is exchanged, so ΔQ = 0 and W = −ΔU. In an isochoric process volume is constant hence ΔV = 0 and W = 0. Memorising these four formulas covers almost every CBSE numerical. The table below lists them with typical applications.

Isobaric (constant P): W = P (V₂ − V₁) = nR ΔT. Used in piston problems with constant external pressure.
Isothermal (constant T): W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂). Applicable to slow expansions with heat reservoir.
Adiabatic (ΔQ = 0): W = (P₁V₁ − P₂V₂)/(γ−1) = nR(T₁ − T₂)/(γ−1). Fast processes, insulated containers.
Isochoric (constant V): W = 0. All heat changes internal energy: ΔQ = nCᵥ ΔT.

Specific Heat Capacities and Mayer's Relation

Every ideal gas has two principal molar specific heats: Cᵥ (at constant volume) and Cₚ (at constant pressure). Because work is done during isobaric heating, Cₚ is always larger than Cᵥ. The difference is exactly R, the universal gas constant, as given by Mayer's relation: Cₚ − Cᵥ = R. The ratio γ = Cₚ/Cᵥ appears in adiabatic process equations (PVᵞ = constant). For monoatomic gases γ ≈ 1.67, for diatomic γ ≈ 1.4 and for polyatomic γ ≈ 1.3. These values are essential for numerical problems on adiabatic expansion or compression.

Cᵥ = (f/2)R where f is degrees of freedom (3 for monoatomic, 5 for diatomic, 6 for polyatomic).
Cₚ = Cᵥ + R (Mayer's relation).
γ = Cₚ/Cᵥ. For monoatomic ideal gas, Cᵥ = (3/2)R, Cₚ = (5/2)R, γ = 5/3 ≈ 1.67.
Heat supplied at constant V: ΔQ = nCᵥ ΔT; at constant P: ΔQ = nCₚ ΔT.

Adiabatic Process Equations and Relations

When a system is thermally insulated or a process happens so fast that heat exchange is negligible, the process is adiabatic. For an ideal gas undergoing a reversible adiabatic change, three relations hold simultaneously: PVᵞ = constant, TVᵞ⁻¹ = constant and Pᵞ⁻¹T⁻ᵞ = constant. These can be derived from the first law with ΔQ = 0 and the ideal gas law. In numericals you often know two variables and must find the third; pick the relation that eliminates the unknown. Remember that temperature drops during adiabatic expansion (system does work at the expense of internal energy) and rises during adiabatic compression.

PVᵞ = P₁V₁ᵞ = P₂V₂ᵞ. Most commonly used when P and V are given.
TVᵞ⁻¹ = T₁V₁ᵞ⁻¹ = T₂V₂ᵞ⁻¹. Useful when T and V are known.
T P⁽¹⁻ᵞ⁾/ᵞ = constant. Less common, but helps when P and T are direct variables.
Work done: W = (P₁V₁ − P₂V₂)/(γ−1) = nR(T₁ − T₂)/(γ−1). Always use Kelvin for T.

Heat Engines — Efficiency and Key Formulas

A heat engine absorbs heat Q₁ from a hot reservoir, converts part into work W and rejects Q₂ to a cold reservoir. By energy conservation, W = Q₁ − Q₂. Efficiency η = W/Q₁ = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁. Real engines have η < 1 because some heat must always be rejected (second law). The Carnot engine is an ideal reversible engine operating between two temperatures T₁ (hot) and T₂ (cold) with maximum possible efficiency η_Carnot = 1 − T₂/T₁. Temperatures must be in Kelvin. CBSE often asks you to compare real engine efficiency with Carnot efficiency or calculate heat rejected given efficiency and Q₁.

Efficiency η = Work output / Heat input = W/Q₁ = 1 − Q₂/Q₁.
Carnot efficiency η = 1 − T₂/T₁ (T in Kelvin). Maximum efficiency for given T₁, T₂.
For a real engine, η_real < η_Carnot because of irreversibilities (friction, turbulence).
If η = 40 % and Q₁ = 1000 J, then W = 400 J and Q₂ = 600 J.

Refrigerators and Coefficient of Performance (COP)

A refrigerator is a heat engine running in reverse: it absorbs heat Q₂ from a cold reservoir (inside the fridge), work W is supplied externally (by the compressor) and heat Q₁ = Q₂ + W is rejected to the hot reservoir (room). Performance is measured by the coefficient of performance, COP = Q₂/W = Q₂/(Q₁ − Q₂). Higher COP means more cooling per unit work. For a Carnot refrigerator, COP = T₂/(T₁ − T₂). Unlike efficiency (which is always less than 1), COP can be greater than 1. CBSE numericals often give Q₂ and W and ask for COP, or give temperatures and ask for ideal COP. Remember to use absolute temperatures (Kelvin) in the Carnot formula.

COP = Q₂/W = Heat extracted / Work input.
For Carnot refrigerator, COP = T₂/(T₁ − T₂) where T₂ < T₁.
Relation: Q₁ = Q₂ + W, so W = Q₁ − Q₂.
Higher COP is better; real refrigerators have COP 2 to 5, ideal Carnot COP can be much higher if T₁ − T₂ is small.

Common Mistakes, Units and Notation Pitfalls

Students often substitute temperature in Celsius instead of Kelvin into Carnot formulas, yielding absurd efficiencies above 100 %. Always convert: T(K) = T(°C) + 273. Another pitfall is confusing the sign of work in compression versus expansion. Write down ΔV explicitly; if ΔV > 0 (expansion) then W > 0; if ΔV < 0 (compression) then W < 0 under the NCERT convention. Mixing up Q₁ and Q₂ in engine versus refrigerator problems is common—draw a simple diagram labeling hot and cold reservoirs. Finally, remember that γ is dimensionless, R = 8.31 J mol⁻¹ K⁻¹ and all energies should be in joules (or convert kJ to J) before plugging into formulas.

Always use Kelvin in efficiency and COP formulas: η_Carnot = 1 − T₂/T₁ requires absolute temperature.
Sign of W: expansion → W > 0; compression → W < 0 (NCERT convention).
Do not confuse γ (gamma, Cₚ/Cᵥ) with specific heat itself.
Check units: Pressure in Pa, Volume in m³, n in mol, R in J mol⁻¹ K⁻¹. Convert litres to m³ (1 L = 10⁻³ m³).

Memory Tricks and Mnemonics for Thermodynamics

Remembering which process keeps which variable constant can be aided by simple mnemonics. 'ISO-BAR-IC' sounds like 'bar' → pressure (isobaric = constant P). 'ISO-CHOR-IC' rhymes with 'core' → volume at the core stays fixed (constant V). 'ISO-THERM-AL' has 'therm' → temperature constant. For the first law, the mnemonic 'QUW' reminds you Q = U + W (heat equals internal energy change plus work). To recall Mayer's relation Cₚ − Cᵥ = R, think 'P comes after V, so Cₚ is Cᵥ plus R'. For efficiency, 'one minus cold over hot' (1 − T₂/T₁) is the Carnot formula. These tricks save precious seconds in the exam hall and reduce silly errors.

'QUW': ΔQ = ΔU + ΔW (First Law).
'Isobaric = constant Pressure' (bar → P).
'Isochoric = constant Volume' (chor → V).
'Isothermal = constant Temperature' (therm → T).
'Carnot efficiency = 1 minus cold/hot' → η = 1 − T₂/T₁.
'COP-Cool': COP of refrigerator = Cooling/Work = Q₂/W.

Three Solved Mini-Examples Applying Key Formulas

Working through compact numericals cements formula usage. Example 1 demonstrates the first law in an isobaric process. Example 2 shows adiabatic work calculation using the PVᵞ relation. Example 3 computes Carnot engine efficiency and compares it with a real engine. Each solution highlights the formula invoked, units checked and the logical flow from given data to answer. Practice these types repeatedly; CBSE board papers recycle similar structures every year, just with different numbers or gas types.

One-Glance Last-Minute Revision Box

On the morning of your exam, scan this box to activate all critical formulas and laws. The Zeroth Law defines temperature; First Law ΔQ = ΔU + ΔW governs energy flow; Second Law sets the efficiency ceiling. For processes: isobaric W = P ΔV, isothermal W = nRT ln(V₂/V₁), adiabatic ΔQ = 0 so W = −ΔU, isochoric W = 0. Specific heats: Cₚ − Cᵥ = R, γ = Cₚ/Cᵥ. Heat engine η = 1 − Q₂/Q₁ = 1 − T₂/T₁ (Carnot). Refrigerator COP = Q₂/W = T₂/(T₁−T₂) (Carnot). Always use Kelvin. Sign: heat in +, work by system +, ΔU increase +. These ten lines cover 90 % of CBSE numericals.

**Zeroth Law:** Thermal equilibrium is transitive.
**First Law:** ΔQ = ΔU + ΔW.
**Isobaric:** W = P ΔV; **Isothermal:** W = nRT ln(V₂/V₁); **Adiabatic:** PVᵞ = const, W = nR(T₁−T₂)/(γ−1); **Isochoric:** W = 0.
**Mayer:** Cₚ − Cᵥ = R; γ = Cₚ/Cᵥ.
**Engine η:** 1 − Q₂/Q₁ (real), 1 − T₂/T₁ (Carnot).
**Refrigerator COP:** Q₂/W (real), T₂/(T₁−T₂) (Carnot).
**Sign convention:** Heat in +, Work by system +, ΔU increase +.

How CBSETUTOR.ai Helps You Master Thermodynamics Formulas

Formula sheets are excellent for revision, but real mastery comes from solving diverse problems and getting instant feedback. CBSETUTOR.ai offers a 24×7 AI tutor that lets you snap a photo of any NCERT thermodynamics numerical—whether it is a tricky adiabatic expansion or a Carnot cycle comparison—and receive a step-by-step solution within seconds. The platform covers every class from 6 to 12 at a flat ₹999 per month, so your younger sibling and you can share one subscription. Parents in metros like Bengaluru and Delhi appreciate the consistent quality and the three-day free trial that lets students test the AI tutor on their own chapter doubts before committing. Instead of waiting for weekend coaching or flipping through multiple guide books, you get on-demand help exactly when you are stuck, reinforcing formula application in real time and building exam confidence day by day.

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Frequently asked questions

What is the difference between the first and second laws of thermodynamics?+

The first law is energy conservation: ΔQ = ΔU + ΔW, stating energy cannot be created or destroyed. The second law introduces directionality and efficiency limits, stating heat cannot spontaneously flow from cold to hot and no engine can be 100 % efficient.

Why is Cₚ always greater than Cᵥ for an ideal gas?+

At constant pressure, the gas does expansion work while being heated, so extra energy is needed beyond just raising temperature. Hence Cₚ = Cᵥ + R, making Cₚ larger by exactly R.

How do I remember which process uses which work formula?+

Use mnemonics: isobaric (constant P) → W = P ΔV; isothermal (constant T) → W = nRT ln(V₂/V₁); adiabatic (ΔQ=0) → W = −ΔU; isochoric (constant V) → W = 0. Write these on your formula sheet.

What are common sign convention mistakes in the first law?+

Students often forget that work done by the system is positive and work done on the system is negative. Always define ΔW based on whether the gas expands (+) or is compressed (−), then apply ΔQ = ΔU + ΔW carefully.

Can a heat engine have 100 % efficiency?+

No. The second law (Kelvin-Planck statement) forbids a heat engine from converting all absorbed heat into work. Some heat Q₂ must always be rejected to the cold reservoir, so η = 1 − Q₂/Q₁ < 1.

How is the Carnot engine different from real engines?+

A Carnot engine is a theoretical reversible engine with maximum efficiency η = 1 − T₂/T₁. Real engines have friction, turbulence and finite-time processes, so their efficiency is always less than the Carnot value for the same temperatures.

What is the coefficient of performance (COP) for a refrigerator?+

COP = Q₂/W, where Q₂ is heat removed from the cold space and W is work input. For an ideal Carnot refrigerator, COP = T₂/(T₁ − T₂). Higher COP means better performance; COP can exceed 1.

Why must I use Kelvin in Carnot efficiency and COP formulas?+

Carnot formulas are derived from absolute thermodynamic temperature. Using Celsius gives incorrect ratios (even negative efficiencies). Always convert: T(K) = T(°C) + 273 before substituting into η = 1 − T₂/T₁ or COP = T₂/(T₁−T₂).

What is the relation between P, V and T in an adiabatic process?+

For a reversible adiabatic process of an ideal gas: PVᵞ = constant, TVᵞ⁻¹ = constant and T P⁽¹⁻ᵞ⁾/ᵞ = constant. Use the relation that contains your known and unknown variables.

How does CBSETUTOR.ai help with thermodynamics numericals?+

Upload a photo of any NCERT or board-level problem and the AI tutor provides a step-by-step solution with formula identification, sign checks and unit conversions. The flat ₹999/month plan covers all chapters for Classes 6–12, with a three-day free trial to test it first.

Related resources

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