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Learn beyond limits
Thermodynamics in NEET Physics is largely about applying one core relationship — the first law — correctly across different processes. Most mistakes come from mixing up sign conventions or forgetting which quantity is held constant in a given process, not from not knowing the formulas at all.
NEET weightage: High · NCERT Physics Class 11, Part 2, Chapter 12
ΔQ = ΔU + ΔW, where ΔQ is heat supplied to the system, ΔU is the change in internal energy, and ΔW is work done BY the system. It is essentially a statement of energy conservation applied to thermal systems: heat added either raises internal energy or does work, or both.
Temperature stays constant (ΔT = 0), so for an ideal gas ΔU = 0 and all heat supplied goes into work done: ΔQ = ΔW. Requires the process to happen slowly enough, or with a perfectly conducting boundary, for temperature to stay uniform.
No heat is exchanged with the surroundings (ΔQ = 0), so ΔU = −ΔW. Any work done by the gas comes entirely at the expense of its internal energy (and hence its temperature falls during adiabatic expansion).
Isobaric: pressure is constant; work done W = PΔV. Isochoric (constant volume): ΔV = 0, so no work is done (W = 0) and all supplied heat changes internal energy: ΔQ = ΔU.
Cv is the specific heat at constant volume, Cp at constant pressure. For an ideal gas, Cp − Cv = R (Mayer's relation) — Cp is larger because, at constant pressure, some of the supplied heat also does expansion work rather than only raising temperature.
A Carnot engine is an idealised, reversible heat engine operating between two temperatures. Its efficiency η = 1 − T₂/T₁ (temperatures in Kelvin) is the maximum possible efficiency for any engine operating between those two temperatures — no real engine can exceed it.
First law of thermodynamics
ΔQ = ΔU + ΔW
Isothermal process
ΔU = 0, so ΔQ = ΔW
Adiabatic process
ΔQ = 0, so ΔU = −ΔW
Isochoric (constant volume)
W = 0, so ΔQ = ΔU
Isobaric work done
W = PΔV
Mayer's relation
Cp − Cv = R
Carnot engine efficiency
η = 1 − T₂ / T₁
An ideal gas absorbs 500 J of heat and does 200 J of work while expanding. What is the change in its internal energy?
Using ΔQ = ΔU + ΔW: 500 = ΔU + 200, so ΔU = 300 J. The internal energy increases by 300 J — only part of the absorbed heat went into external work; the rest raised the gas's internal energy.
A gas undergoes an adiabatic expansion and does 150 J of work on its surroundings. What happens to its internal energy, and by how much?
Adiabatic means ΔQ = 0, so from the first law, ΔU = −ΔW = −150 J. The internal energy decreases by 150 J, which is why gas temperature drops during adiabatic expansion (e.g. in a rapidly expanding gas cylinder).
A gas is compressed isothermally, and 300 J of work is done on the gas. How much heat is released by the gas?
300 J. For an isothermal process ΔU = 0, so ΔQ = ΔW. Work done ON the gas is −300 J of work done BY the gas, so ΔQ = −300 J, meaning the gas releases 300 J of heat to keep its temperature constant.
In an isochoric process, a gas absorbs 400 J of heat. What is the work done by the gas, and what is the change in internal energy?
Work done W = 0, because volume doesn't change (no PΔV term). By the first law, ΔQ = ΔU + ΔW = ΔU + 0, so ΔU = 400 J — all the absorbed heat goes into increasing internal energy.
A Carnot engine operates between a source at 500 K and a sink at 300 K. What is its efficiency?
η = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.4, i.e. 40%. This is the maximum theoretical efficiency for any engine operating between these two temperatures.
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