I am deeply confused.... For a thermally insulated ideal gas expanding freely, I think that $PV^{\gamma}=cnst$ must hold. Through the equation $PV=nRT$, it must be that $TV^{{\gamma}-1}=\rm constant$. Because the free expansion changes the volume of the gas, the temperature of the gas must change too. However, any book on thermodynamics says that there is no change in the internal energy of the gas in free expansion. Thus, the temperature of the ideal gas must not change. Why does this discrepancy occur? I am just totally stuck.....
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$\begingroup$ The equation $PV^\gamma = const$ holds for a reversible adiabatic process. The free expansion, however, is not reversible. $\endgroup$– By SymmetryCommented Sep 13, 2018 at 13:16
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2 Answers
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$PV^{\gamma}=C$ applies only to the reversible expansion of an ideal gas. Free expansion is not a reversible expansion because, even though the gas can be returned to its original state, its surroundings can not (without changing something else). The ideal gas law describes the behavior of ideal gases only for thermodynamic equilibrium states. In an irreversible expansion, the gas does not pass through a sequence of thermodynamic equilibrium states.
answered Sep 13, 2018 at 13:13
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It depends on the system you are considering. Let's consider two different cases:
1) An ideal gas is inside an insulated piston which can move. You can write:
$$dU=\delta Q -\delta L \rightarrow dU= -\delta L \rightarrow nc_{v}dT=-dL$$ Now, if the process is reversible you have $dL=pdV$. And so: $$nc_{v}dT=-pdV$$ Also remember: $$pV=nRT \rightarrow p=\frac{nRT}{V} \rightarrow nc_{v}dT=-\frac{nRT}{V}dV$$ Then you integer in $dT$ and in $dV$ and obtain $pV^{\gamma}=kost$. Note that we have got this equation ONLY for reversible processes. Of course you can write this equation in some different ways by mixing it with the equation for ideal gases. If you draw this equation into the Clapeyron plane (p-V), you see that this is an hyperbole.
2)Joule free expansion. You have an ideal gas inside a tank; this first tank is linked to a second tank by a valve. In the second thank there is nothing. By opening the valve the gas will go inside the second tank and finally it will be inside the two tank. Let's analyze this process: $$dU=\delta Q -\delta L \rightarrow dU=\delta Q $$ Note that the work is zero and the process is irreversible. If you measure the temperature of the gas at the end of the process, you will note that it hasn't changed, and for this reason the gas hasn't exchange heat. So: $$\delta Q=dU=0$$ I mean only the second one is a free expansion. How you have seen during a free expansion the system is not insulated however its temperature doesn't change, and it doesn't exchange work. While the first case I've analyzed isn't a free expansion: the gas is insulated, it exchanges work and its temperature changes.
