[Physics] Is free expansion adiabatic

ideal-gasthermodynamics

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?

Best Answer

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.

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