NCERT Plus One (Class 11) Physics • Chapter 11

Thermodynamics

Q: State the Zeroth Law of Thermodynamics. Which physical quantity is defined by it?

Statement: If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. It defines Temperature .

Q: What is internal energy? Write the mathematical statement of the First Law of Thermodynamics and explain each symbol.

Internal energy U is the sum of molecular kinetic and potential energies inside a system (excluding bulk motion). First Law: ΔQ = ΔU + ΔW , where ΔQ = heat supplied to system, ΔU = rise in internal energy, and ΔW = work done BY the system.

Q: Derive Mayer's relation Cp − Cv = R for one mole of an ideal gas.

At constant volume ( ΔV = 0 ): ΔQᵥ = Cᵥ ΔT = ΔU . At constant pressure: ΔQₚ = Cₚ ΔT = ΔU + P ΔV . Using PV = RT at constant P : P ΔV = R ΔT , and since ΔU = Cᵥ ΔT depends only on T : Cₚ ΔT = Cᵥ ΔT + R ΔT ⇒ Cₚ − Cᵥ = R .

Q: Derive an expression for work done during an isothermal expansion of μ moles of an ideal gas from volume V₁ to V₂ at temperature T.

W = ∫(V₁→V₂) P dV . Using P = μRT / V with constant T : W = μRT ∫(V₁→V₂) (1/V) dV = μRT ln(V₂ / V₁) = 2.303 μRT log₁₀(V₂ / V₁) .

Q: Show that the adiabatic curve is steeper than the isothermal curve at the same point (P, V).

For isothermal ( PV = k ): slope (dP/dV)_iso = −P/V . For adiabatic ( PVᵞ = k ): slope (dP/dV)_adia = −γ P/V . Since γ = Cₚ/Cᵥ > 1 , |slope_adia| = γ |slope_iso| > |slope_iso| .

Q: A Carnot engine has an efficiency of 40% when its sink is at 27 °C. By how many degrees must the source temperature be raised to make efficiency 50%?

Sink T₂ = 27 + 273 = 300 K . Initially: 0.40 = 1 − 300/T₁ ⇒ 300/T₁ = 0.60 ⇒ T₁ = 500 K . For 50%: 0.50 = 1 − 300/T₁' ⇒ T₁' = 600 K . Required rise = 600 − 500 = 100 K (or 100 °C) .

Q: An electric heater supplies heat to a system at 100 W. If the system performs work at 75 J/s, at what rate is the internal energy increasing?

Per second: ΔQ = +100 J , ΔW = +75 J . By First Law: ΔU = ΔQ − ΔW = 100 − 75 = 25 J/s (25 W) .

Q: State Kelvin-Planck and Clausius statements of the Second Law of Thermodynamics. Name the four processes of a Carnot cycle in order.

Kelvin-Planck: No process is possible whose sole result is absorbing heat from a reservoir and converting it completely into work. Clausius: No process is possible whose sole result is transferring heat from a colder body to a hotter body. Carnot steps: (1) Isothermal expansion, (2) Adiabatic expansion, (3) Isothermal compression, (4) Adiabatic compression.

Thermal equilibrium, Zeroth & First Laws, Cp − Cv = R, isothermal vs adiabatic PV curves, cyclic processes & Carnot engine.

📄 Official NCERT Textbook PDF

Topic 11.2 – 11.3: Thermal Equilibrium & Zeroth Law of Thermodynamics

Imagine placing a piping-hot cup of Sulaimani tea next to a cold steel spoon on your kitchen table in Thrissur. Touch the spoon ten minutes later — it is no longer icy, and the tea is no longer boiling! Both have met at one shared comfort temperature. Even smarter: if your thermometer reads 37 °C for you and 37 °C for your sister, you both share the same temperature without ever touching foreheads!

THERMAL EQUILIBRIUMMacroscopic variables (P, V, T) stop changing with time — no net heat flow!

TWO WALL TYPESAdiabatic wall = blocks heat; Diathermic wall = lets heat pass freely!

ZEROTH LAWIf A ↔ C and B ↔ C are in thermal equilibrium, then A ↔ B are too!

🧠 Memory Trick: Why "Zeroth" Law? Scientists realized Temperature (the shared equality tag T) is so basic it had to sit BEFORE the First and Second Laws — at number Zero!

💡 Simple Kid-Check: Unlike mechanics, where we track every single ball's position and speed, Thermodynamics looks at the whole gas crowd from outside using bulk quantities like Pressure (P), Volume (V), and Temperature (T).

1
Adiabatic vs Diathermic Walls: An insulating flask wall (like a thermos flask) is adiabatic — gases on either side keep independent states. A thin copper sheet is diathermic — heat flows until temperatures match.
2
Fowler's Zeroth Law (1931): Two systems in thermal equilibrium with a third system (like a thermometer C) are in thermal equilibrium with each other.
3
What Zeroth Law Gives Us: It defines Temperature (T) — the one physical quantity that equals out when two bodies reach thermal equilibrium!

Exam Trap: Never say the First Law defines temperature! Zeroth Law defines Temperature, while the First Law introduces Internal Energy (U).

Topic 11.4 – 11.5: Internal Energy, Heat, Work & First Law of Thermodynamics

Pump air rapidly into a bicycle tyre in Kochi and touch the brass nozzle — it gets surprisingly hot even without a stove! Pushing the pump handle does mechanical work on the trapped air, boosting its internal energy. Similarly, heating a pressure cooker transfers heat, raising steam energy until the whistle lifts up and does work!

FIRST LAW EQUATIONΔQ = ΔU + ΔW (Conservation of Energy!)

STATE VS PATHU is a state variable; Q & W are path variables!

NCERT SIGN RULEΔQ > 0 heat IN; ΔW = PΔV > 0 work BY gas (expansion)!

🧠 Memory Trick: Think of your bank wallet! ΔQ = Salary deposited, ΔW = Money spent outside, ΔU = Change inside your pocket (ΔU = ΔQ − ΔW)!

💡 Simple Kid-Check: Inside any gas cylinder, billions of molecules are zipping and spinning around. All their tiny microscopic kinetic and potential energies added together is called Internal Energy (U) — and for an ideal gas, U depends ONLY on temperature T!

First Law: ΔQ = ΔU + ΔW where ΔW = P ΔV (for small quasi-static steps)

🧒 Super-Easy Example: You supply 500 J of heat to a gas cylinder, and the expanding gas pushes a piston out doing 200 J of work. How much did its internal energy rise?
ΔU = ΔQ − ΔW = 500 − 200 = +300 J!

Watch the Chemistry vs Physics sign trap! In NCERT Physics, ΔW is work done BY the system (+PΔV on expansion), so ΔQ = ΔU + ΔW. Never accidentally write ΔQ = ΔU − ΔW with Physics conventions!

Topic 11.6 – 11.7: Specific Heat of Gases (Cp − Cv = R) & State Variables

Try heating 1 mole of air inside two cylinders by 1 K. The first cylinder has its lid welded shut (constant volume), while the second cylinder has a loose sliding lid pushing against the atmosphere (constant pressure). The sliding-lid cylinder demands MORE heat because it has to heat the air AND lift the lid! That exact extra bit of heat per mole per kelvin is R = 8.314 J mol⁻¹ K⁻¹!

MAYER'S RELATIONCₚ − Cᵥ = R (for 1 mole of any ideal gas!)

EXTENSIVE VS INTENSIVEExtensive (scales with mass): V, U, m; Intensive (unchanged): P, T, ρ

EQUATION OF STATEPV = μRT connects equilibrium state variables!

🧠 Memory Trick: Split a room in half with a paper wall! Anything that cuts in half (V, U, m) is Extensive; anything that stays identical (P, T, density) is Intensive!

💡 Simple Kid-Check: Why are there TWO molar specific heats for gases when solids basically use one? Because gases expand hugely when heated at constant pressure, wasting extra energy doing expansion work PΔV!

1
At Constant Volume (ΔV = 0): Work ΔW = 0. So ΔQᵥ = ΔU = μ Cᵥ ΔT. For 1 mole, Cᵥ = ΔU / ΔT.
2
At Constant Pressure (ΔP = 0): Both U and volume V rise! For 1 mole, ΔQₚ = Cₚ ΔT = ΔU + P ΔV.
3
Using Ideal Gas Law (PV = RT): At constant pressure, P ΔV = R ΔT. Since ΔU depends ONLY on ΔT, Cₚ ΔT = Cᵥ ΔT + R ΔT, which instantly yields Cₚ − Cᵥ = R!

Exam Trap: If a question gives specific heats in per gram (lowercase cₚ, cᵥ) instead of per mole (Cₚ, Cᵥ), Mayer's relation becomes cₚ − cᵥ = R / M (or r), NOT R! Also check units: calories vs Joules (R/J).

Topic 11.8.1 – 11.8.2: Isothermal vs Adiabatic Processes (PV Work & Curves)

Pull a bicycle pump handle out super slowly in metal shade — surrounding air keeps the temperature steady (Isothermal). Now slam the pump handle down lightning fast — heat has zero time to escape, trapping all that work inside as intense heat (Adiabatic)! Even though both curves look smooth on a PV graph, the adiabatic plunge is visibly steeper!

ISOTHERMAL (T = CONST)PV = const • ΔU = 0 • W = μRT ln(V₂/V₁)

ADIABATIC (ΔQ = 0)PVᵞ = const • W = (P₁V₁ − P₂V₂)/(γ − 1)

SLOPE RULESlope_adia = γ × Slope_iso (Adiabatic curve is ALWAYS steeper!)

🧠 Memory Trick: Iso-THERMAL = Slow with copper walls (keeps T flat, ΔU = 0); Adia-BATIC = Sudden/Insulated (blocks Q, ΔQ = 0, cools on expansion)!

💡 Simple Kid-Check: On a P–V graph, the area underneath the curve IS the work done by the gas! Since an adiabatic expansion cools the gas down, pressure drops faster than in an isothermal expansion — making its curve steeper and giving slightly LESS work!

Isothermal Slope: dP/dV = −P/V | Adiabatic Slope: dP/dV = −γ P/V
Adiabatic Relations: PVᵞ = const | TVᵞ⁻¹ = const | Tᵞ P¹⁻ᵞ = const

🧒 Why does a bursting tyre feel icy cold? Because air shoots out almost instantaneously (adiabatic expansion, ΔQ = 0). The gas pays for expansion work out of its own wallet: ΔU = −ΔW, so temperature drops sharply!

Classic Graph Question: Starting from the same state (P₁, V₁), which curve lies LOWER during expansion and HIGHER during compression? Adiabatic! Because γ > 1 makes adiabatic curves γ times steeper everywhere!

Topic 11.8.3 – 11.8.5: Isobaric, Isochoric & Cyclic Processes

Boil water in an open pot — pressure stays fixed at 1 atmosphere while steam expands freely (**Isobaric**). Now lock food inside a rigid sealed vessel — volume cannot budge even a millimetre (**Isochoric**), so ZERO work gets done! And if a piston goes around a closed loop and returns to the exact start (**Cyclic**), it runs like a real engine!

ISOBARIC (P = CONST)V ∝ T • W = P(V₂ − V₁) = μR ΔT • ΔQ = μ Cₚ ΔT

ISOCHORIC (V = CONST)P ∝ T • W = 0 (Vertical line on PV!) • ΔQ = ΔU = μ Cᵥ ΔT

CYCLIC LOOPReturns to start ⇒ ΔU = 0 ⇒ Q_net = W_net = Enclosed Area

🧠 Memory Trick: On a P–V loop, ClockWise = positive work (+CW engine); Anti-Clockwise = negative work (refrigerator/work ON gas)!

💡 Simple Kid-Check: Work needs movement! If volume never changes (isochoric), the piston never moves an inch, so work is strictly ZERO. And in a complete round trip (cyclic), you end up at the same temperature as you started, so ΔU = 0!

Process	Held Constant	PV Graph Shape	Work Done (W)	First Law Form
Isobaric	Pressure (P)	Horizontal line	`P(V₂ − V₁)`	`ΔQ = ΔU + PΔV`
Isochoric	Volume (V)	Vertical line	`0`	`ΔQ = ΔU`
Isothermal	Temp (T)	Rectangular hyperbola	`μRT ln(V₂/V₁)`	`ΔQ = W`
Adiabatic	Heat (Q)	Steeper curve	`(P₁V₁−P₂V₂)/(γ−1)`	`ΔU = −W`
Cyclic	Start = End	Closed loop	Enclosed loop area	`Q_net = W_net`

Watch out for Cyclic PV loops with V on the Y-axis and P on the X-axis! On standard P vs V (P on Y, V on X), clockwise is positive work; if axes are swapped, clockwise becomes NEGATIVE work! Always check axis labels first!

Topic 11.9 – 11.11: Second Law, Reversible Processes & Carnot Engine Efficiency

A dropped glass shatters into pieces, yet shattered pieces NEVER leap back together by themselves! Nature prefers a one-way street (**Irreversible processes**). Even the best petrol engine on Earth takes heat from burning fuel, turns part of it into wheel motion, and MUST dump leftover waste heat out the exhaust — Sadi Carnot proved why **100% efficiency is impossible**!

SECOND LAW STATEMENTSKelvin-Planck: No 100% heat engine! Clausius: Heat won't flow cold → hot by itself!

CARNOT EFFICIENCYη = W/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁ (Always use KELVIN!)

CARNOT THEOREMCarnot engine has max efficiency; depends ONLY on T₁ & T₂, NOT gas!

🧠 Memory Trick: Four strokes of Carnot's ideal cycle: Iso-Expand → Adia-Expand → Iso-Compress → Adia-Compress (Alternate Iso and Adia!).

💡 Simple Kid-Check: A heat engine is like a waterfall waterwheel! Water must fall from a high tank (Hot Source T₁) down to a low river (Cold Sink T₂) to spin the wheel (W = Q₁ − Q₂). Without a cold sink below, nothing flows and the wheel stops!

Efficiency: η = 1 − (T₂ / T₁) (where T₁ = source temp in K, T₂ = sink temp in K)

🧒 Worked Carnot Numerical: An ideal engine runs between 127 °C and 27 °C. Find its efficiency.
Convert to Kelvin first! T₁ = 127 + 273 = 400 K, T₂ = 27 + 273 = 300 K.
η = 1 − 300/400 = 1 − 0.75 = 0.25 → 25%!

NEVER plug Celsius temperatures directly into η = 1 − T₂/T₁! Putting 27/127 instead of 300/400 gives a totally wrong answer. Always add 273 (or 273.15) to convert °C → Kelvin first!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): State the Zeroth Law of Thermodynamics. Which physical quantity is defined by it?

Statement: If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.
It defines Temperature.

Q2 (Kerala HSE Say • 2 Marks): What is internal energy? Write the mathematical statement of the First Law of Thermodynamics and explain each symbol.

Internal energy U is the sum of molecular kinetic and potential energies inside a system (excluding bulk motion).
First Law: ΔQ = ΔU + ΔW, where ΔQ = heat supplied to system, ΔU = rise in internal energy, and ΔW = work done BY the system.

Q3 (Kerala HSE March • 3 Marks): Derive Mayer's relation Cp − Cv = R for one mole of an ideal gas.

At constant volume (ΔV = 0): ΔQᵥ = Cᵥ ΔT = ΔU.
At constant pressure: ΔQₚ = Cₚ ΔT = ΔU + P ΔV.
Using PV = RT at constant P: P ΔV = R ΔT, and since ΔU = Cᵥ ΔT depends only on T:
Cₚ ΔT = Cᵥ ΔT + R ΔT ⇒ Cₚ − Cᵥ = R.

Q4 (Kerala HSE Model • 3 Marks): Derive an expression for work done during an isothermal expansion of μ moles of an ideal gas from volume V₁ to V₂ at temperature T.

W = ∫(V₁→V₂) P dV. Using P = μRT / V with constant T:
W = μRT ∫(V₁→V₂) (1/V) dV = μRT ln(V₂ / V₁) = 2.303 μRT log₁₀(V₂ / V₁).

Q5 (Kerala HSE March • 2 Marks): Show that the adiabatic curve is steeper than the isothermal curve at the same point (P, V).

For isothermal (PV = k): slope (dP/dV)_iso = −P/V.
For adiabatic (PVᵞ = k): slope (dP/dV)_adia = −γ P/V.
Since γ = Cₚ/Cᵥ > 1, |slope_adia| = γ |slope_iso| > |slope_iso|.

Q6 (Kerala HSE Say • 2 Marks): A Carnot engine has an efficiency of 40% when its sink is at 27 °C. By how many degrees must the source temperature be raised to make efficiency 50%?

Sink T₂ = 27 + 273 = 300 K.
Initially: 0.40 = 1 − 300/T₁ ⇒ 300/T₁ = 0.60 ⇒ T₁ = 500 K.
For 50%: 0.50 = 1 − 300/T₁' ⇒ T₁' = 600 K.
Required rise = 600 − 500 = 100 K (or 100 °C).

Q7 (NCERT / DHSE Model • 2 Marks): An electric heater supplies heat to a system at 100 W. If the system performs work at 75 J/s, at what rate is the internal energy increasing?

Per second: ΔQ = +100 J, ΔW = +75 J.
By First Law: ΔU = ΔQ − ΔW = 100 − 75 = 25 J/s (25 W).

Q8 (Kerala HSE March • 3 Marks): State Kelvin-Planck and Clausius statements of the Second Law of Thermodynamics. Name the four processes of a Carnot cycle in order.

Kelvin-Planck: No process is possible whose sole result is absorbing heat from a reservoir and converting it completely into work.
Clausius: No process is possible whose sole result is transferring heat from a colder body to a hotter body.
Carnot steps: (1) Isothermal expansion, (2) Adiabatic expansion, (3) Isothermal compression, (4) Adiabatic compression.