Isochoric and isobaric processes
Compare isochoric and isobaric processes using the first law, including work signs and ideal-gas temperature relations.
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ContinueA rigid vessel and a freely moving loaded piston can receive heat yet follow different paths. Isochoric means constant volume; isobaric means constant pressure. The constraint tells you which boundary-work term is available, while the first law tells you where the transferred energy goes.
Build on gas heat capacities. Use Q > 0 into the gas and Wₒₙ > 0 for work into it, so Δ U = Q + Wₒₙ. Work output is Wₒᵤₜ = -Wₒₙ. We consider a fixed amount of ideal gas, negligible changes in bulk kinetic and potential energy, and only pressure–volume work unless a question states otherwise.
Read the boundary-work term from the motion
For a quasistatic mechanically balanced piston path, Wₒᵤₜ = ∫ P dV and Wₒₙ = -∫ P dV. Expansion gives positive work output and negative work input; compression reverses the signs. In a non-equilibrium expansion the gas may lack one well-defined pressure, and boundary work needs the appropriate external resisting load. Do not integrate an invented equilibrium gas curve.
| Constraint | Boundary work | Heat relation for the stated model |
|---|---|---|
| Constant volume | Wₒᵤₜ = 0 | Q = Δ U = n∫ CV(T) dT |
| Quasistatic constant pressure | Wₒᵤₜ = PΔ V = nRΔ T | Q = Δ U + Wₒᵤₜ = n∫ CP(T) dT |
Constant volume removes pressure–volume work, not electrical or paddle work. Constant pressure does not mean constant volume or constant internal energy. The gas may change temperature and expand, with both energy storage and boundary work involved.
Connect the path to the ideal-gas state relation
At fixed volume and amount, Pf/Pᵢ = Tf/Tᵢ. At fixed pressure and amount, Vf/Vᵢ = Tf/Tᵢ. Both use absolute temperatures. The internal-energy change of an ideal gas depends on the temperature interval, so either path uses Δ U = n∫ CV(T) dT even though the constant-pressure heat relation uses CP.
For constant capacities, the integrals become nCVΔ T and nCPΔ T. If capacity changes, keep the integral. At constant pressure the ideal-gas work relation nRΔ T remains valid; it does not depend on whether CV is constant.
Guided practice 1
Close a piston’s energy balance
Problem
A quasistatic constant-pressure path expands a monatomic ideal gas at 120 kPa from 2.0 L to 3.5 L. Use the classical CV = 3R/2, assume only pressure–volume work and find the work done on the gas, its internal-energy change and the heat entering it. You are not given the amount or either temperature: explain why their product with the temperature change is enough.
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Convert the volume change to cubic metres. The ideal-gas state relation supplies nRΔT without requiring n and ΔT separately.Show solution step by step
Calculate boundary work
Method
Use Wₒₙ = -PΔ V.Reason
The piston moves outwards against the fixed pressure.Working
Answer
Δ V = (3.5-2.0) × 10⁻³ = 1.5 × 10⁻³ m³, so Wₒₙ = -(120000)(0.0015) = -180 J.
Close the energy account
Method
Use Δ U = (3/2)nRΔ T = (3/2)PΔ V.Reason
The ideal-gas state change links expansion and temperature under this fixed-pressure constraint.Working
Answer
Δ U = 270 J and Q = Δ U-Wₒₙ = 270-(-180) = 450 J. Check: 450-180 = 270. Of the heat entering, 60% raises internal energy and 40% leaves as expansion work in this monatomic model. That split is not universal for every gas or process.
Two short process examples
Heating in a rigid tank
Worked example 2
Rigid-tank energy gain
Problem
An ideal gas in a rigid tank receives 420 J of heat. With no other energy transfers and only pressure–volume work possible, find its internal-energy change.
Show full solution
Remove the boundary-work term
Method
Set W = -∫ P dV = 0.Reason
The rigid tank cannot change volume, so its boundary does not move.
Working
Answer
Δ E = Q + W = 420 + 0 J.State the energy gain
Working
Answer
Δ E = 420 J All the heat raises the internal energy; constant volume does not imply zero heat transfer.
Work in constant-pressure expansion
Worked example 3
Expansion work
Problem
Show full solution
Convert the volume change
Method
Calculate final minus initial volume and convert litres to cubic metres.Reason
Pressure in Pa needs volume in m³ to give work in J.Working
Answer
Δ V = 1.5 × 10⁻³ m³State the signed work
Working
Answer
W = -PΔ V = -(1.2 × 10⁵)(1.5 × 10⁻³) = -180 J The work-on sign is negative because the gas supplies 180 J to its surroundings.
Compare two constraints from one starting state
Check your understanding 1
Two 1.0 mol samples of classical monatomic ideal gas start in the same state at 300 K. Heat each by 30 K, one in a rigid vessel and one with a quasistatic constant-pressure piston. Use R = 8.314 J/(mol K). Find each heat input and signed work input, and describe the final pressure and volume relative to their starting values. Do they reach the same thermodynamic state?
Show answer
Both have Δ U = (3/2)(8.314)(30) = 374.13 J. The rigid sample has Wₒₙ = 0 and Q = 374.13 J; its volume is unchanged and pressure rises by 330/300 = 1.10. The piston sample has Wₒₙ = -nRΔ T = -249.42 J and Q = 623.55 J; its pressure is unchanged and volume rises by 1.10. They finish at equal temperatures and ideal-gas internal energies, but different pressures and volumes, so they are different states.
Practice set
Check your understanding 2
A gas cools at constant volume with Q = -260 J and no non-boundary work. Find Δ E.
Show hint
Show answer
W = 0, so Δ E = Q = -260 J. The heat loss is entirely an internal-energy decrease.
Check your understanding 3
Gas is compressed quasistatically at 2.0 × 10⁵ Pa from 5.0 L to 3.0 L. Find work done on it.
Show hint
Keep the negative volume change in W = -PΔ V.
Show answer
Δ V = (3.0-5.0) × 10⁻³ = -2.0 × 10⁻³ m³, so W = -(2.0 × 10⁵)(-2.0 × 10⁻³) = +400 J. Compression supplies work to the gas.
Check your understanding 4
Which condition guarantees zero pressure–volume boundary work: constant volume or constant pressure?
Show hint
Boundary work requires a volume displacement.
Show answer
Constant volume: dV = 0, so -∫ P dV = 0. Constant pressure alone still allows volume change and nonzero work.
What to keep
Apply the constraint to boundary motion, then the first law to heat, work and internal energy. For an ideal gas, the temperature interval fixes Δ U; the path fixes boundary work and hence heat. Rigid walls remove only pressure–volume work, and constant pressure alone does not remove work.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.