Free Expansion and Path Dependence

Introduction

A gas expanding into a vacuum meets no mechanical opposition at the boundary, so its pressure-volume work is zero. This differs from a reversible expansion over the same volume change. If the vessel is also insulated and the gas is ideal, both heat and work are zero and its temperature remains unchanged. The example makes path dependence tangible.

Core explanation

Imagine a rigid insulated vessel divided by a removable partition. One side contains gas and the other is evacuated. Remove the partition: the gas spreads into the empty half. The pressure outside the expanding gas at the moving front is effectively zero, so w = −∫P ext dV = 0 for pressure-volume work. Because the outer vessel is insulated, q ≈ 0. The first law gives ΔU = q + w = 0 for the gas, assuming no other work mode and treating the combined vessel as the closed system.

For an ideal gas, internal energy depends only on temperature. Thus ΔU = 0 implies ΔT = 0 if composition and amount remain fixed. The gas can occupy twice the volume after free expansion without a temperature change in this ideal model. A real gas may cool or warm because intermolecular attractions and repulsions make internal energy depend on volume as well as temperature. The ideal-gas result should not be generalized without that material assumption.

Now consider a reversible isothermal expansion of the same ideal gas from the same initial to the same final volume at the same temperature. It performs negative work, w rev = −nRT ln(V f/V i), and absorbs an equal amount of heat, q rev = −w rev. Both paths have the same initial and final ideal-gas equilibrium states, so both have ΔU = 0. Their q and w values differ. This proves by example that heat and work are path-dependent transfers.

Free expansion is irreversible. Once the partition is removed, gas molecules spread into available space; they do not spontaneously gather back on one side. The final equilibrium state has greater volume accessible to the gas and, for an ideal gas at fixed temperature, greater entropy. Its entropy change can be computed through a convenient reversible path even though the actual free path is irreversible, because entropy is a state function. One must not use the actual free-expansion q = 0 in a reversible-heat entropy formula without understanding the path distinction.

The word “free” refers to negligible opposing pressure, not necessarily to an absence of all energy changes in every real gas. The insulated ideal-gas setup is a carefully specified special case. If the vessel is not insulated, q may be nonzero. If an electrical device operates inside, other work may occur. If the gas reacts or changes phase, internal energy and temperature behavior may be different.

The case also clarifies the pressure in the work equation. Using the gas's large initial pressure times ΔV would falsely predict work during expansion into a vacuum. The boundary's opposing external pressure is the relevant force for this irreversible path.

Step-by-step reasoning

1. Choose the gas and vessel as the closed-system model. 2. Identify the evacuated receiving region, so P ext ≈ 0. 3. Set P–V work to zero and heat to zero only if insulated. 4. Use ΔU = q + w; for an ideal gas infer ΔT from ΔU. 5. Compare with a reversible path between the same equilibrium endpoints.

Visual explanation

Draw a rigid insulated box with gas dots on the left and a vacuum on the right, separated by a removable partition. After removal show dots spread throughout. Place q = 0 and w = 0 outside the rigid outer boundary. Beside it draw a piston expansion to the same volume with opposite nonzero q and w arrows.

Real-world analogy

A crowd released into an empty room spreads without pushing a weighted door, while moving a weighted partition gradually would require force and transfer mechanical energy. The analogy captures the difference in external opposition, though molecular energy distribution is more subtle.

Real-world example

Free expansion is a classic thought experiment for testing ideal-gas assumptions. If an experiment observes a temperature change in a carefully insulated expansion, it can reveal nonideal behavior or imperfections in the setup, rather than overturning energy conservation.

Why?

Why does ΔU stay zero even though the gas occupies more volume in the ideal case? No heat or work crosses the chosen outer boundary, and ideal-gas internal energy depends only on temperature, not directly on volume.

Common misconception

“The gas pressure is high, so it must do PΔV work in a vacuum.” Work depends on the opposing external pressure at the boundary. With no opposing load, P–V work is zero for this free path.

Worked example

One mole of ideal gas at 300 K doubles its volume in a rigid insulated vessel by free expansion. The actual path has q = 0, w = 0, so ΔU = 0 and final T = 300 K. A reversible isothermal path between the same volumes would have w = −nRT ln 2 ≈ −1.73 kJ and q ≈ +1.73 kJ. Equal ΔU with different transfers demonstrates path dependence numerically.

Quick check

1. What is P–V work in expansion into a perfect vacuum? Answer: Zero, because the opposing external pressure is zero.

Exam focus

Name the rigid, insulated and ideal-gas assumptions before concluding q = w = ΔU = ΔT = 0. Separate the irreversible actual path from a reversible comparison used to calculate state-function changes such as entropy.

Advanced insight

For an ideal gas, ΔS between equal-temperature states of volumes V i and V f is nR ln(V f/V i), even when the actual free path has q = 0. Entropy change is evaluated from endpoints via a reversible reference path; actual irreversible heat divided by T is not generally equal to ΔS.

Summary

Free expansion into a vacuum has zero P–V work. In an insulated ideal-gas setup, q = 0, ΔU = 0 and temperature remains fixed. A reversible path between the same endpoints has nonzero opposing heat and work, demonstrating their path dependence.

Practice questions

1. An ideal gas freely expands into a vacuum in an insulated rigid vessel. What are q, w and ΔU? Answer: All are zero under the stated assumptions. 2. Does the same volume change guarantee the same work in a piston path? Answer: No. Work depends on the opposing external pressure along the path. 3. Can entropy increase during insulated free expansion even though q = 0? Answer: Yes. Entropy is a state function and can increase in an irreversible process without heat transfer.