# Free expansion of an ideal gas

I am having trouble understanding the transient phase of an ideal gas expanding into vacuum.

Firstly, the pressure of any gas is defined only when there is an instrument (barometer/ wall/ piston) equally resisting the expansion of the gas. The amount of resistance that the instrument has to put up is what we call pressure.

Consider the scenario where an ideal gas is kept from expanding into another adjacent volume V2 through a door (The second chamber is vacuous). Now, suddenly the door vanishes, the ideal gas expands and re-establishes equilibrium in the new volume V1+V2.

The textbooks calculate the change in entropy for this process using the following equations.

Equation 1: PV = nRT Equation 2: dU = TdS - PdV

The textbooks also state that the temperature, and consequently internal energy, do not change for such an expansion.

My confusion pertains to how one calculates this entropy change. The words 'Pressure' and 'Volume' have no meaning when the gas is still expanding into the volume. So how can one consider equation 1 to hold true during the transient phase, and then use it to calculate the change in entropy?

Edit: I have another question related to equation 2; the textbooks have substituted TdS for dQ in the first law of thermodynamics. However, dQ is equal to TdS only for reversible processes. So how does equation 2 make sense in the context of irreversible processes? (Such as free expansion)

The ideal gas equation only applies to an ideal gas in a thermodynamic equilibrium state, and not to one that is rapidly deforming. The 2nd equation you wrote applies only to two closely neighboring thermodynamic equilibrium states of a system, and describes the relationship between the differential changes in U, S, and V between these adjacent states. Neither equation describes the intermediate changes along an irreversible path between two widely separated thermodynamic equilibrium states. So how do you determine the change in entropy between these two end states that are connected in practice by an irreversible path?

Step 1: This is the most important step. Completely forget about the actual irreversible path between the two end states. Focus only on the initial and final end states.

Step 2: Devise a reversible path between the two end states. The reversible path you devise does not have to bear any resemblance whatsoever to the actual irreversible path, as long as it starts and ends at the same two end states. A reversible path consists of a continuous sequence of thermodynamic equilibrium states. So, along the path you choose, the system can be no more than slightly removed from thermodynamic equilibrium at every point along the path. There are an infinite number of reversible paths than can take you from state A to state B. So choose one that is convenient for performing step 3.

Step 3: Calculate the integral of dQ/T for the reversible path that you devised in step 2.

When they say that the change in entropy is equal to the integral of $dQ_{rev}/T$, what they mean is that you have to calculate the integral for a reversible path.

You can alternatively use your equation 2 to determine the change in entropy since it automatically implies a reversible path.

• Perfect! Your argument works because entropy is a state function. Essentially, what the textbooks have done then is to employ a clever trick to bypass an otherwise analytically difficult process. If I understood this right, we can throw away describing the irreversible process altogether and focus on incremental expansions of the gas by a volume dV. Assuming that equilibrium holds at every step, we can use the equation of state to calculate the entropy change for this process. It's much clear now. Thank you! – StrawMan Mar 17 '16 at 12:50
• Yes, exactly. And it doesn't even have to be adiabatic like the actual irreversible path. You have tremendous flexibility in getting reversibly from state A to state B. – Chet Miller Mar 17 '16 at 13:49

In a free expansion it follows from the first law of thermodynamics that the energy change of the system is $\Delta U = Q + W = 0$. Why is this? First, no work is done as the gas is not pushing on anything. Second, there is no heat flow in or out of the system giving the aforementioned result. At the end entropy gets created out of nothing and there is no law which prohibits that.

You can calculate the entropy change using the Sackur-Tetrode formula which is a more detailed equation of state for the classical ideal gas. As the number of particles stays the same the formula reduces simply to $$\Delta S = Nk\ln(\frac{V_f}{V_i}),$$ with $V_i$ and $V_f$ the initial and final volume respectively.

Regarding your edit. I don't think the textbook replaces $dQ$ (which is actually not a true differential) with $TdS$. As mentioned before is $\Delta U = 0$, so equation 2 reduces to $TdS-PdV = 0$. Now, in order to take into account an entropy increase, we get from the previous equation: $TdS = PdV > 0$.

• I fail how to understand how you answer my question. The derivation that most textbook follow is the one mentioned in the question, and that makes little sense to me simply because macro state variables are poorly defined during transient phases. We speak about the pressure, volume and temperature only after any gas equilibrates with its surroundings. Therefore, how can one assume the equation of state to hold good even during transient phases? Yet, this is precisely what the derivation assumes. Thank you for your answer! – StrawMan Mar 17 '16 at 12:33