Entropy change in the free expansion of a gas

Question

Consider the adiabatic free expansion of a gas since there is no external Pressure hence Work done on the system is 0 and since the walls are insulated (hence adiabatic) the heat absorbed is 0. However since this is a irreversible process then entropy change > 0 hence dQ > 0 . However there is no heat absorption. What am I missing ?

Answer 1

What am I missing ?

Entropy can be generated without there being heat transfer, i.e., when $Q=0$. That's the case for a free expansion into a vacuum. The classic example given is an ideal gas located in one side of a rigid insulated vessel with a vacuum in the other side separated by a rigid partition. An opening is created in the partition allowing the gas to expand into the evacuated half of the vessel. $W=0$, $Q=0$, $\Delta T=0$ (for an ideal gas) and therefore $\Delta U=0$. Although no heat transfer has occurred, the process is obviously irreversible (you would not expect the gas to be able to spontaneously return to its original location) and entropy increases.

You can calculate the entropy increase by assuming any convenient reversible process that can bring the system back to its original state (original entropy). The obvious choice is to remove the insulation and insert a movable piston. Then conduct a reversible isothermal compression until the gas is returned to its original volume leaving a vacuum in the other half. All properties are then returned to their original state. The change in entropy for the isothermal compression is then, where $Q$ is the heat transferred to the surroundings by the isothermal compression,

$$\Delta S=-\frac{Q}{T}$$

Since the system is returned to its original state, the overall change in entropy is zero, meaning the original change in entropy due to the irreversible expansion has to be

$$\Delta S=+\frac{Q}{T}$$

Hope this helps.

Answer 2

The equality $dS=dQ/T$ is only valid for reversible processes, so it does not apply in a free expansion.

Answer 3

The reason is explained by others. Mathematically, for free expansion:

$ΔS_{system}=nR\ln\left(\frac{V_{2}}{V_{1}}\right)\ ;\ ΔS_{surroundings}=0$

Hence overall entropy of universe increases.

Answer 4

Substitute the irreversible spontaneous expansion with an appropriate reversible process. This can be done thanks to entropy being a state function.

Since in the spontaneous expansion the temperature remains constant, you can choose a reversible isothermal expansion with the same initial and final state as in the spontaneous expansion. Entropy change ΔS of reversible isothermal expansion is described as follows: ΔS=Q/T.Consider the process to be isothermal.

Internal energy does not change with constant temperature, therefore heat absorbed by the system during reversible isothemal expansion is equal to work done by the gas.

During a spontaneous expansion, the pressure and volume of the gas changes unpredictably. Since pressure p is not constant during reversible isothermal expansion and it is a volume function. Find work done for isothermal process. ΔS=Q/T=W/T

Answer 5

For the Entropy change on a free gas expansion. I like to imagine a kind of "Maxwell demon" holding a movable virtual wall separating a Volume V1 at atmosferic pressure from a volume V2 initialy in Vavuum. The "Demon" is able to move the wall in quasi-static steps, using the potential elastic energy on the movable wall, allowing the gas to double the number of occupied microstates. There is nothing new here, but an option to imagine the physics of the Entropy change (inspired on the Maxwell demon). Obviouly to calculate the Entropy change one should consider a reversible expansion.