60 solved GATE ME questions on Thermodynamics, drawn from 8 exam years and grouped by year. Every question shows the official answer and a step-by-step solution.
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Revision companion
Begin with Laws & Properties: identify the system, label transfers, and distinguish states from processes. These original ideal-gas exercises build selected foundations, not a complete syllabus. Audit signs and units before calculating.
Our study notes and original examples support the PYQs below; they are not official exam questions or a replacement for the current syllabus.
A closed system exchanges energy but not mass. Here Q is positive into the gas; W is positive for work by the gas. Expansion gives positive boundary work; compression gives negative boundary work.
Check yourself: Which transfers cross the boundary, and which direction is positive?
The ideal-gas equation pV = mRT uses absolute pressure and temperature. Ideal gases are compressible; do not impose constant density. Their internal energy depends only on temperature; heat and work depend on the path.
Check yourself: Have you confused a property difference with an energy transfer?
With negligible kinetic and potential energy changes, ΔU = Q − W. Boundary work integrates resisting pressure over volume change. A quasistatic frictionless piston balances gas and resisting pressures.
Check yourself: Does the stated process justify using a constant pressure in W = pΔV?
For an ideal gas with constant specific heats, ΔU = m cᵥ ΔT and cₚ = cᵥ + R. For this model, constant-pressure heating with only boundary work gives Q = m cₚ ΔT. A rigid boundary instead eliminates displacement work, not necessarily every possible work mode.
Check yourself: Are there any shaft or electrical work transfers left after fixing the volume?
Adiabatic means Q = 0; isothermal means constant temperature. Neither implies the other. Reversible adiabatic ideal-gas relations additionally need reversibility and an appropriate heat-capacity model; conservation alone is insufficient.
Check yourself: Which assumption supplies each process relation you intend to use?
Original teaching example · not a PYQ
Original example: 1 kg of ideal gas has constant R = 0.3 kJ/(kg·K) and cᵥ = 0.7 kJ/(kg·K). A closed, quasistatic, frictionless piston maintains 100 kPa absolute while the uniform gas temperature rises from 300 to 400 K. Only boundary work occurs; kinetic and potential energy changes are negligible. Find ΔU, W, and Q.
ΔU = +70 kJ; W = +30 kJ; Q = +100 kJ.
Separate original check: a stationary, rigid, closed tank contains 0.5 kg of ideal gas initially at a uniform 300 K, with constant cᵥ = 0.8 kJ/(kg·K). It rejects 20 kJ of heat. No shaft, electrical, or other work occurs; potential energy is unchanged. Find W and the final equilibrium temperature.
W = 0 kJ; final temperature = 250 K.
The fixed boundary cannot do displacement work. Q = −20 kJ, so ΔU = −20 kJ and ΔT = −20/(0.5 × 0.8) = −50 K. Cooling does not require a volume change.
Supplemental reading, not an official GATE reading list or an endorsement of these notes.
Free supplemental lecture and transcript. The chemistry convention counts work on the system as positive; translate its sign before comparing equations here.
This page shows 60 recent questions from the released archive, newest first. For older questions and complete papers, browse all GATE ME papers. Questions can carry more than one subject tag; counts are not marks weightage.
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