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Thermodynamics
Physics · Class 11NCERT pp. 226–2392 sets × 25 questions
Key facts and formulas for Thermodynamics, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| First law | ΔQ = ΔU + ΔW (heat in, work done by gas positive) |
| Work at constant pressure | W = PΔV (P in Pa, V in m³; 1 L = 10⁻³ m³) |
| Ideal gas and Mayer | PV = μRT, C_p − C_v = R, γ = C_p/C_v |
| Isothermal (ideal gas) | ΔU = 0, Q = W = μRT ln(V₂/V₁), P₁V₁ = P₂V₂ |
| Adiabatic (ideal gas) | Q = 0, T V^(γ−1) = const, W = μR(T₁ − T₂)/(γ − 1) |
| Degrees of freedom | Monatomic: C_v = 3R/2, γ = 5/3. Rigid diatomic: C_v = 5R/2, γ = 7/5 |
| Cycles on P–V | Net work = enclosed area; clockwise is positive, anticlockwise is negative |
| Carnot engine | η = 1 − T₂/T₁ = W/Q₁ (kelvin), Q₁ = W + Q₂ |
| Refrigerator and heat pump | α = Q₂/W = T₂/(T₁ − T₂); heat pump COP = α + 1 |
| Entropy and second law | ΔS = Q/T at constant T; no engine can have 100 percent efficiency |
Three sample questions
Q1. Count Rumford observed that the heat produced while boring a cannon depended on
- the sharpness of the drill, not on the work done
- the work done by the horses, not on the sharpness of the drill
- the mass of the water around the cannon
- the colour of the metal
Hint and solution
Hint: Rumford kept boring with a blunt drill and still got heat. Ask what was being spent to make that heat.
- Heat produced ∝ work done — Rumford noticed that more work by the horses gave more heat, whatever the state of the drill.
- Blunt drill still gave heat — If heat were a fluid inside the metal, a blunt drill would give little. It did not.
- Heat is a form of energy — So heat comes from mechanical work, not from the sharpness of the drill.
- Answer: work done by the horses — The statement that heat depends on the work done is the second one. Option B.
Answer: B. Heat followed the work done. If heat were a fluid stored in the metal, a sharper drill would release more.
Q2. Which of the following is NOT used in a thermodynamic description of a gas?
- Pressure
- Temperature
- Mass and composition
- The velocity of each molecule
Hint and solution
Hint: Thermodynamics deals with the gas as a whole. Which item belongs to a single molecule?
- Thermodynamics = macroscopic — It describes a system with large scale variables like pressure, volume and temperature.
- Macroscopic: P, V, T, mass, composition — Options A, B and C are all such bulk quantities.
- Microscopic: velocity of one molecule — Tracking each molecule is the job of kinetic theory, not thermodynamics.
- Answer: velocity of each molecule — So the item not used is the molecular velocity. Option D.
Answer: D. Thermodynamics is macroscopic. It uses P, V, T, mass and composition, not individual molecules.
Q3. Statement I: An adiabatic wall must be rigid.
Statement II: An adiabatic wall does not allow heat to flow across it.
Choose the correct option.
- I is false and II is true
- Both are true
- I is true and II is false
- Both are false
Hint and solution
Hint: An adiabatic wall is defined by what it blocks. Does blocking heat also mean it cannot move, like a piston?
- Adiabatic wall: no heat crosses — This is the only requirement, so Statement II is true.
- Piston can be adiabatic and movable — A piston that blocks heat can still move and let work be done.
- Statement I is false — Rigidity is a different property from blocking heat.
- Answer: I false, II true — So the correct choice is the first option. Option A.
Answer: A. An adiabatic wall only blocks heat. It can be movable, like a piston.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.