Reversible Adiabatic Ideal-Gas Processes
For a quasistatic reversible adiabatic process of a calorically perfect ideal gas, Q = 0 and PV^γ, TV^(γ−1), and T^γP^(1−γ) remain constant, with γ = C_P/C_V.
Why this shows up in the exam
Insulated quasistatic piston motion · Atmospheric compression and expansion estimates · Finding γ from endpoint data
Learn the idea
A reversible adiabatic ideal gas changes temperature as it does work with no heat exchange. With no heat crossing the boundary, expansion work comes from internal energy and cools the gas; compression reverses this.
🧠 Memory hook: Adiabatic says Q zero, not T constant.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- PV^γ = constant — Poisson relation for a reversible adiabatic ideal gas
- TV^(γ-1) = constant — temperature-volume form under the same assumptions
- W_by = (P_iV_i - P_fV_f)/(γ-1) — work by the gas; positive for ordinary adiabatic expansion
How to approach it
- 1Verify both adiabatic and quasistatic assumptions
- 2Choose the Poisson form matching known variables
- 3Use Q = 0 to cross-check the work against ΔU
Common slip-ups that cost marks
- •Using Poisson relations for irreversible free expansion
- •Dropping absolute temperatures
- •Giving compression work the expansion sign
🌟 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.
A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?
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