# Free energy in the free expansion of an ideal gas

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1. Jan 23, 2015

### MexChemE

Hello PF! Consider the free expansion of an ideal gas. The process occurs at constant temperature, therefore, ΔU = 0, Q = 0, and W = 0. Suppose we are given the initial and final pressures of the gas, and we calculate ΔG = nRT ln(P2/P1). As P2 < P1, ΔG < 0. This is intuitive, as a free expansion is clearly spontaneous. My question is, ΔG is commonly (physically) interpreted as the opposite of the "useful work" done by the system, but in this case the system does zero work. Does this mean ΔG is not an actual physical quantity, but a helpful theoretical measure of the useful work that can be extracted from the system?

Thanks in advance for any input!

2. Jan 23, 2015

### Bystander

All it really tells you is the difference in free energy between the two states. It does not include any specification of path, or work done.

3. Jan 24, 2015

### Staff: Mentor

ΔG is commonly interpreted as the opposite of the maximum possible useful work (for a flow process) between the initial and final states.

Chet

4. Jan 24, 2015

### MexChemE

Got it, so the system doesn't have to do work for it to have a negative change in free energy.