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In free expansion the system is thermally isolated so $\Delta Q = 0$ throughout the process. This implies that process is adiabatic and follows its equation:
$$P_1V_1^\gamma = P_2 V_2^\gamma$$
But since inital and final temperature is same
$$P_1 V_1 = P_2 V_2$$
Both equations can only hold when $V_1 = V_2$. But that is clearly false. How do I resolve this paradox?
The equation
$$PV^{γ}=Constant$$
Or
$$P_{1}V_{1}^{γ}=P_{2}V_{2}^{γ}$$
Is for a reversible adiabatic process. The free expansion is an irreversible process.
$$P_{1}V_{1}=P_{2}V_{2}$$
Only defines the end points at equilibrium for an ideal gas. It is not the same as
$$PV=constant$$
which describes an isothermal process, the path between the end points. The path between the end points is not defined for a free expansion.
Bottom line: A free expansion in an insulated system is adiabatic, but it is not reversible adiabatic.
Hope this helps.
You appear to be confusing the distinctions between what applies at a state versus what applies along a path, as well as between what happens in reversible and irreversible processes.
Free expansion is a process where the external pressure is zero. Adiabatic expansion is a process with no heat flow between the system and its surroundings. Reversible processes keep the system and surroundings in thermal and mechanical equilibrium at all points during the process.
Free expansion is irreversible. It can be done adiabatically. Adiabatic expansion can be done using a reversible or an irreversible process.
A substance at equilibrium has a defined mechanical equation of state. For an ideal gas, it is $pV = nRT$.
During an irreversible process on a closed system containing an ideal gas, we cannot determine the pressure or density for the gas along the path. Along a reversible path, the equation of state of an ideal gas is constrained as
$$pV^n = \mathrm{constant}$$
where $n$ depends on the restrictions isothermal ($n = 1$), isobaric ($n = 0$), isochoric ($n = \infty$), or adiabatic ($n = \gamma$).
Allow an ideal gas to expand with zero external pressure adiabatically.
Free expansion is irreversible. The expression $pV^\gamma = \mathrm{constant}$ does not apply. The mechanical work done on/by the gas is zero in free expansion since $\delta w = - p_{ext} dV = 0$. The heat flow is zero along the adiabatic path. From the first law, $\Delta U = \delta q + \delta w = 0$. For an ideal gas, internal energy change is only a function of temperature
$$\Delta U = \int C_V(T)\ dT $$
Since $\Delta U = 0$, then the gas has the same temperature at the final state as it did at the initial state. This does not mean, the gas has the same, constant temperature at all points along the irreversible path. Indeed, we have no clue what the internal temperature, pressure, and density of the gas are along the irreversible path.
To add to the existing answer:
Ultimately, the relation between initial and final volume/pressure is process-dependent. Adiabatic processes requires zero heat flux, hence if your gas cannot cool into the environment at all, it will change its parameters according to $PV^\gamma=const.$
If the gas can cool instantly into the environment, with the heat flux assumed to be infinite, then the temperature remains the same pre/post compression and your gas parameters evolve according to $PV=const.$
So it is really a question of how the gas can cool away the compressional heat. To put this into a more physical context, in reallife, one will compute the cooling timescale of the gas $t_{\rm cool}$ (e.g. radiative or conductive) and compare it to the timescale for compressing the gas $t_{\rm compression}$. The ratio of those two numbers will decide if one is in either extreme, of zero or infinite cooling, or somewhere in between, in which case one has to put in more work to compute the final state of the gas.
Considering 1 mol of an ideal gas in an insulated adiabatic expansion engine. As the piston moves down, the volume in the cylinder increases and the pressure falls. In an efficient engine this process is near isentropic. The enthalpy h = u + p.v is reduced by the work extracted.
The relation between p and v is described by p.v^k = constant. k=Cp/Cv. w = R.T1.k/(k-1).[(p2/p1)^((k-1)/k) - 1]. In the isentropic expansion the temperature falls, therefore u (which is a function of T only) is also reduced.
I find it impossible to envisage a reversible free expansion process. But how do the molecules of gas experiencing the expansion know or care whether the piston is doing any work or not? This is my dilemma.
My doubts about the statement found everywhere that 'free expansion is isothermal and irreversible' stem from an analogous process where a pressure vessel containing gas at high pressure is blown-down or depressurised. Theory and practice confirm that there is a significant drop in the temperature of the gas remaining.
The portion of gas that has not so far been vented, is effectively being allowed to expand and does so in a process that is (much) closer to isentropic than isothermal. (The core gas temperature is lower than gas at the periphery because heat flows in from the containing vessel walls.)
This process does not appear much different from the free expansion.
I suspect at least part of the answer is that the expanding gas is not ideal. I attended the lecture describing the experimental and theoretical work (Haque, Richardson and Saville in 1992), and asked 'where does the energy extracted from the expanding gas go?' the response at the time was in accelerating the vented gas through a small restriction, and dispersed in turbulence downstream and noise.
In free expansion the intial and final temperature are same but temperature is not constant through out the process, hence it is not an isothermal process. So we can't apply the isothermal process equation.
