Isothermal Ideal-Gas Processes
For a fixed ideal gas undergoing a quasistatic isothermal process at absolute temperature T, PV is constant, ΔU = 0, and W_by = Q = nRT ln(V_f/V_i), positive for expansion under the stated convention.
Why this shows up in the exam
Slow compression in a heat bath · Comparing isotherms on a P-V plot · Calculating isothermal expansion work
Learn the idea
An ideal gas at constant temperature has zero internal-energy change, so heat equals work. Slow thermal contact can replace exactly the energy that an expanding ideal gas sends out as work.
🧠 Memory hook: Same T makes ΔU zero; heat mirrors work.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- PV = constant — isothermal ideal-gas relation for fixed n and constant kelvin temperature
- W_by = Q = nRT ln(V_f/V_i) — reversible isothermal work and heat; expansion is positive
How to approach it
- 1Confirm ideal gas and constant temperature
- 2Use the endpoint volume ratio inside the logarithm
- 3Check expansion gives positive work by the gas
Common slip-ups that cost marks
- •Using PΔV instead of the logarithmic integral
- •Assuming isothermal means no heat transfer
- •Applying ΔU = 0 to a nonideal gas without justification
🌟 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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