Compute state entropy without inventing an actual reversible process

Declare the system and evidence

Entropy is a state function defined for equilibrium states. Our ideal-gas packet has total Cv=12 J/K and nR=8 J/K, initially 300 K and 1 litre. Only the final temperature and volume are specified. A reversible reference path is a mathematical way to evaluate the state difference; it is not a description of an unknown actual history.

Derive the state or rate account

For a simple compressible ideal gas, the reversible identity T dS=dU+p dV gives dS=12 dT/T+8 dV/V. First heat at fixed volume and then expand isothermally at the final temperature. Integrating gives delta S=12 ln(T2/300)+8 ln(V2/V1). Every logarithm has a dimensionless positive argument. Reversing the two reference legs gives exactly the same sum.

Audit the constraint and model

Heat is not a state function. The first reference path absorbs Q=12(T2−300)+8T2 ln(r), whereas expansion at the initial temperature followed by fixed-volume heating gives Q=8×300 ln(r)+12(T2−300). Their work differs by the same amount, leaving delta U fixed. A negative volume contribution during compression can outweigh a positive temperature contribution; the gas alone need not gain entropy.

Predict a changed boundary

Predict a compression r=0.5 at unchanged temperature. Gas entropy is −8 ln 2, not a second-law violation: a complete entropy balance must include the surroundings. For an actual irreversible path use entropy transfer plus generation; dividing an arbitrary measured total heat by one chosen temperature does not reproduce the reference-path integral.

Worked original case

At T2=450 K and r=2, delta S=12 ln 1.5+8 ln 2≈10.410759 J/K and delta U=1800 J. Reference heats are 1800+3600 ln 2≈4295.329850 J and 1800+2400 ln 2≈3463.553233 J. The entropy change is the same.

Investigation

Retain two reference-path integrations and both heat/work accounts. Predict a compression and an expansion before checking; distinguish the gas state difference from total entropy generation.

Changed-case practice

Try the changed case before feedback

At unchanged 300 K, double the gas volume. Find delta S in J/K.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic cases checked 31 October 2026 against the primary teaching reference. No source exercises or diagrams are reproduced. All thermal operations are mathematical proposals or supplied records, not physical heating, pressure-vessel or exposure instructions. Public feedback and retained prose do not certify independent reasoning or actual practical participation. Preserve first work, critique, revision and a new assigned response in the separate course workspace.