ThermodynamicsNEET Physics · Class 11 · NCERT Chapter 11

You can expect 1 to 2 questions from Thermodynamics in NEET 2027. The chapter has high PYQ frequency. The first law, isothermal vs adiabatic processes, Cp and Cv (Mayer relation), Carnot efficiency and refrigerator COP are the most repeated topics.

The first law is energy conservation for a system: Q equals ΔU plus W. Heat Q added to a system goes into changing the internal energy ΔU and into doing work W on the surroundings. NEET uses the IUPAC sign convention: Q is positive when heat is added to the system, W is positive when the system does work on the surroundings (gas expands).

Isothermal: temperature stays constant, so ΔU equals 0 and Q equals W. The PV curve is a hyperbola (PV equals constant). Adiabatic: no heat is exchanged with surroundings, so Q equals 0 and W equals minus ΔU. The PV curve is steeper (PV to the gamma equals constant). For the same compression, an adiabatic process raises temperature; an isothermal process does not.

For one mole of an ideal gas, Cp minus Cv equals R, where R is the universal gas constant 8.314 J per mol per K. The reason: at constant pressure, the gas also does work pushing the piston, so it needs more heat than at constant volume for the same temperature rise. Gamma equals Cp over Cv: 5 over 3 for monoatomic, 7 over 5 for diatomic.

eta equals 1 minus T_cold over T_hot, with both temperatures in kelvin. This is the maximum efficiency any heat engine can achieve operating between two reservoirs. A Carnot engine running between 600 K and 300 K has eta equals 1 minus 300 over 600 equals 0.5 (50 percent). Real engines fall well below the Carnot limit.

Two equivalent statements: (1) Kelvin Planck, no engine can convert heat completely into work; some heat must always be rejected to a cold reservoir. (2) Clausius, heat cannot flow from a colder to a hotter body without external work. The second law sets the direction of natural processes: entropy of an isolated system never decreases.

Coefficient of performance COP equals Q_cold over W, where Q_cold is the heat removed from the cold space and W is the work input. For an ideal (Carnot) refrigerator, COP equals T_cold over (T_hot minus T_cold). A typical home fridge has COP around 2 to 4. A heat pump (opposite use) has COP_HP equals 1 plus COP_refrigerator.