Thermodynamic Processes
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Direct answer
Every process obeys the first law, Q = delta U + W (heat supplied = change in internal energy + work done by the gas), with delta U = n Cv delta T for an ideal gas depending on temperature alone. Work done equals the area under the curve on the pressure–volume (P–V) diagram, which is why the standard processes are best held in mind as graphs: isothermal (constant T), adiabatic (Q = 0, P V^gamma constant), isochoric (constant V, zero work) and isobaric (constant P).
What you must remember
- Isothermal (ideal gas): delta U = 0, so Q = W = n R T ln(V2/V1); needs a slow process through conducting walls.
- Adiabatic: Q = 0, so W = −delta U = n Cv (T1 − T2) = (P1 V1 − P2 V2)/(gamma − 1); along the path P V^gamma, T V^(gamma − 1) and T^gamma P^(1 − gamma) stay constant.
- An adiabatic curve is steeper than an isotherm — its slope at any point is gamma times the isothermal slope — because compression without heat escape raises the temperature too.
- Isochoric: W = 0, Q = n Cv delta T. Isobaric: W = P delta V = n R delta T, Q = n Cp delta T, with Cp − Cv = R and gamma = Cp/Cv.
- Over a full cycle delta U = 0, so net Q = net W = the enclosed P–V area, positive for a clockwise loop.
- Molar heat capacities: monatomic Cv = (3/2) R, gamma = 5/3; diatomic at ordinary temperatures Cv = (5/2) R, gamma = 7/5.
- Carnot engine between T1 and T2 (kelvin): efficiency = 1 − T2/T1; run backward as a refrigerator, coefficient of performance = T2/(T1 − T2).
Common confusion
The two evergreen errors are sign errors and misjudged speed. Work by the gas is positive in expansion, negative in compression — reversing this wrecks every energy balance. And rapid changes (a burst tyre, sudden compression) are adiabatic, not isothermal, because heat flow needs time; only slow, well-coupled processes hold the temperature.
Exam-focused takeaway
JEE Main asks work, heat and efficiency in the named processes — mostly resolved by the area-under-curve idea. JEE Advanced prefers multi-step cycles drawn as P–V, P–T or V–T graphs: compute work as enclosed area, use temperature to get delta U leg by leg, distinguish adiabatic from isothermal legs by slope, and handle process-specific heat capacities (zero for adiabatic, infinite for isothermal). Sketch and label the P–V diagram before computing anything.
Frequently asked questions
Why is an adiabatic curve steeper than an isotherm?
Adiabatic compression raises the temperature along with the pressure, so pressure climbs faster than in isothermal compression, where T cannot change; the slope ratio is gamma.
Which of Q, W and delta U are path functions?
Q and W depend on the route between states; delta U depends only on the end states, which is why over a cycle net Q = net W.
Why must kelvin be used in the Carnot formula?
Efficiency = 1 − T2/T1 is a ratio of absolute temperatures; the arbitrary zero of Celsius would change the ratio, since only kelvin makes T proportional to molecular kinetic energy.
What is the work done in a cyclic process?
The area enclosed by the loop on the P–V diagram — positive if clockwise, negative if anticlockwise — and it equals the net heat absorbed over the cycle.
Why is Cp larger than Cv?
At constant pressure the gas also expands and does work; that extra R per mole of heat is why Cp − Cv = R.