Thermodynamics for NEET UG
Thermodynamics is the bookkeeping of energy: where heat goes, how much work you get out, and why no engine can ever be perfect.
PhysicsClass 1159 MCQs3 lessons14 flashcards3 memory tricks
What you'll learn
- The first law: energy is only moved, never lost
Use ΔQ = ΔU + ΔW with the right signs to find heat, work or change in internal energy. - Processes: isothermal, adiabatic, isobaric, isochoric
Recognise each process, know what stays constant, and calculate the work done. - Heat engines, refrigerators and the second law
Find the efficiency of an engine and the coefficient of performance of a refrigerator, and explain why neither can be perfect.
Key formulas
| First law | ΔQ = ΔU + ΔW |
|---|---|
| Work at constant pressure | W = PΔV |
| Efficiency of any engine | η = W/Q₁ = 1 − Q₂/Q₁ |
| Carnot (ideal) engine | η = 1 − T₂/T₁ (temperatures in kelvin) |
| Coefficient of performance (refrigerator) | β = Q₂/W = T₂/(T₁ − T₂) for a Carnot refrigerator |
Memory tricks
ΔU = Q − W, with Q positive when heat goes in and W positive when the gas does work on the surroundings.
Isothermal: ΔT = 0, so ΔU = 0 for an ideal gas. Isochoric: W = 0. Adiabatic: Q = 0, and PV^γ = constant.
η = 1 − T_cold/T_hot, with temperatures in kelvin. Only T_cold = 0 K would give 100%, which is impossible.
Try 3 NEET-style questions
Q1. In an isochoric process, the first law of thermodynamics reduces to
- Q = ΔU, since no work is done
- Q = W, since ΔU = 0
- ΔU = −W
- Q = 0
Show answer
Answer: (1) Q = ΔU, since no work is done
At constant volume, W = PΔV = 0, so Q = ΔU: all heat supplied changes the internal energy (and temperature).
Q2. For an ideal gas, the ratio γ = Cp/Cv is 1.4. If Cv = 20.8 J/mol K, then Cp is about
- 14.9 J/mol K
- 29.1 J/mol K
- 20.8 J/mol K
- 12.5 J/mol K
Show answer
Answer: (2) 29.1 J/mol K
Cp = γCv = 1.4 × 20.8 ≈ 29.1 J/mol K (also Cp − Cv ≈ R).
Q3. A gas expands at a constant pressure of 10⁵ Pa from 1 L to 3 L. The work done by the gas is
- 200 J
- 300 J
- 2 × 10⁵ J
- 100 J
Show answer
Answer: (1) 200 J
W = PΔV = 10⁵ × 2 × 10⁻³ = 200 J.
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