MixedJEE Physics · Original learning card5 original chapter questions

Irreversible, Sudden, and Free Adiabatic Changes

For an irreversible adiabatic process Q = 0 but generally PV^γ is not constant; boundary work is evaluated from the external pressure, and ideal-gas free expansion into vacuum has W = 0, ΔU = 0, and ΔT = 0.

Why this shows up in the exam

Partition removal into vacuum · Sudden piston loading or release · Distinguishing adiabatic from isentropic motion

Learn the idea

Adiabatic does not imply reversible: sudden motion and free expansion require external-pressure energy accounting. An insulated gas can change without following a Poisson curve; in vacuum it does no boundary work and an ideal gas keeps the same temperature.

🧠 Memory hook: No heat is not enough for Poisson; reversibility is the missing key.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • W_by = ∫ P_ext dV — irreversible boundary work determined by external, not equilibrium gas, pressure
  • Q = 0 => ΔU = -W_by — first-law result with work by the system positive
  • free expansion into vacuum: W = 0, ΔU = 0 — ideal-gas result for an insulated rigid outer container

How to approach it

  1. 1Decide whether the path is quasistatic
  2. 2Write work using external pressure
  3. 3Use Q = 0 in the first law before invoking any gas relation

Common slip-ups that cost marks

  • •Applying PV^γ to every insulated process
  • •Using gas pressure instead of external pressure for sudden motion
  • •Assuming all adiabatic changes alter temperature

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?

Take a timed JEE Physics sectional mock