Thermodynamics
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Direct answer
Heat added, energy stored, work delivered: the first law, ΔQ = ΔU + ΔW, is bookkeeping across these three, and the process's name tells you which term dies — ΔU = 0 isothermal, ΔQ = 0 adiabatic, ΔW = 0 isochoric. The second law then forbids any engine turning heat fully into work, capping a reversible engine at the Carnot efficiency η = 1 - T2/T1 in kelvins. Between the two laws sit the molar heat relations Cp - Cv = R and γ = Cp/Cv, which NEET-UG quotes in nearly every paper.
What you must remember
- Zeroth law defines temperature through thermal equilibrium; first law ΔQ = ΔU + ΔW with ΔQ positive when heat enters and ΔW positive when the gas does work.
- Isothermal: ΔU = 0, W = nRT ln(V2/V1), needing a perfectly conducting wall and a slow, quasistatic march.
- Adiabatic: ΔQ = 0, PV^γ constant, W = (P1V1 - P2V2)/(γ - 1); an adiabatic expansion always cools the gas.
- Isochoric: W = 0, so ΔQ = ΔU; isobaric: W = P(V2 - V1), the rectangle under the P-V line.
- Mayer's relation Cp - Cv = R per mole; γ = Cp/Cv = 5/3 monatomic, 7/5 diatomic, 4/3 polyatomic — all following from counting degrees of freedom.
- Cyclic process: ΔU = 0 over the loop and net work equals the enclosed area in the P-V diagram, positive if traced clockwise.
- Carnot engine between reservoirs T1 (source) and T2 (sink): η = 1 - T2/T1; refrigerator COP = T2/(T1 - T2); kelvins only.
- Free expansion of an ideal gas into vacuum: W = 0 and Q = 0, hence ΔU = 0 and temperature unchanged.
Comparing the two expansions
Take one mole of a monatomic gas at 300 K expanding from V to 2V, first isothermally, then adiabatically. Isothermal: W = RT ln2 = 8.314 × 300 × 0.693 ≈ 1729 J, all of it supplied as heat, ΔU = 0 throughout. Adiabatic: no heat enters, so the work comes out of internal energy; with γ = 5/3 the temperature falls to T2 = 300 × (1/2)^(2/3) ≈ 189 K and W = R(T1 - T2)/(γ - 1) ≈ 1384 J. Two conclusions worth carrying into the exam: the adiabatic does less work for the same volume doubling, and on a P-V diagram the adiabatic falls more steeply than the isotherm (PV^γ against PV) — steepness alone identifies the process in graph questions. The physical reason for the temperature drop is pure first law: the gas paid for its own expansion because the insulation stopped the till.
The Carnot cap follows the same reasoning at the level of principle. Between 500 K and 300 K, η = 1 - 300/500 = 40%, and no ingenuity of piston or valve can exceed it, because reversing the cycle exactly must return both reservoirs to their starting states — the reversible definition itself.
Where students slip
Sign conventions cause about half of all errors: in the NCERT convention work done by the gas is positive, so compression enters as negative work and care is needed when a cycle mixes legs. Second, Celsius in the Carnot formula: a "300 °C and 30 °C" engine must first become 573 K and 303 K, giving η near 47%, not 90%. Third, the internal energy of an ideal gas is a function of temperature alone — so ΔU = 0 is exactly equivalent to ΔT = 0, whatever else the process does. Fourth, Cp - Cv = R holds per mole: dividing by molar mass converts to specific heats per kilogram, and the units punish silence. Finally, students quote the isothermal work formula for a fast compression: if the process is too quick for heat exchange, it is adiabatic, and the steepness of the P-V trace is the tell the exam uses to catch the difference.
Frequently asked questions
In which thermodynamic process is the work done zero?
Isochoric (constant volume): the piston cannot move, so W = 0 and all heat goes into changing internal energy.
Why does a gas cool during adiabatic expansion?
The gas does work at the expense of its internal energy while no heat can enter through the insulation; ΔU = -W lowers U, and for an ideal gas lower U means lower temperature.
What is the Carnot efficiency between 500 K and 300 K?
η = 1 - T2/T1 = 1 - 300/500 = 40%. Any real engine between these reservoirs performs worse, since real cycles are irreversible.
State Mayer's relation.
For an ideal gas, Cp - Cv = R, where both molar heats are per mole and per kelvin; the difference arises from the extra work of expansion at constant pressure.
What does the area inside a closed loop on a P-V diagram represent?
The net work done by the gas per cycle — positive if the loop runs clockwise, negative if anticlockwise; ΔU over a full cycle is zero.