Isochoric Processes
An isochoric process has dV = 0 and hence quasistatic pressure-volume work W = ∫P dV = 0; for an ideal gas with constant C_V, Q = ΔU = nC_VΔT.
Why this shows up in the exam
Heating gas in a sealed rigid vessel · Vertical legs of a P-V diagram · Cooling at fixed piston position
Learn the idea
At fixed volume, boundary work vanishes and supplied heat changes internal energy. A rigid container cannot move its boundary, so energy entering as heat stays as microscopic internal energy.
🧠 Memory hook: No volume change means no P-dV work.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- W = ∫ P dV = 0 — boundary work for a constant-volume process
- Q_V = ΔU = n C_V ΔT — heat supplied to a fixed amount of ideal gas at constant volume
How to approach it
- 1Confirm that volume is fixed
- 2Set boundary work to zero
- 3Use the first law and ideal-gas temperature relation
Common slip-ups that cost marks
- •Calling pressure constant in a rigid vessel
- •Using C_P for fixed-volume heating
- •Forgetting that pressure can change while work remains zero
🌟 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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