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We know that in free expansion of an ideal gas, no heat enters or leaves the system.

We also know that

$P_\text{initial}V_\text{initial}=P_\text{final}V_\text{final}$

is valid.

If heat exchange is zero, then we can call this process to be adiabatic.

Then why the following is not valid?

$P_\text{initial}{V_\text{initial}}^γ=P_\text{final}{V_\text{final}}^γ$

Also, if I am wrong above, are isothermal free expansion and adiabatic free expansion different?

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  • $\begingroup$ See Wikipedia page. For free expansion of an ideal gas initial and final temperature are same. $\endgroup$ – Tyrion Lannister Apr 19 '16 at 19:37
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Are isothermal free expansion and adiabatic free expansion different?

No. They are the same.

Your mistake is in thinking that $PV^\gamma = \text{constant}$ applies to a free expansion. That expression is for a reversible (i.e., isentropic) adiabatic process. A gas that has undergone a free expansion has more entropy after the expansion is complete than it did before the expansion started. Free expansion is not isentropic, and therefore $PV^\gamma = \text{constant}$ does not apply to free expansion.

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  • $\begingroup$ Can you tell me why Free Expansion is quasi-static despite being irreversible? $\endgroup$ – Tyrion Lannister Apr 20 '16 at 12:22
  • $\begingroup$ @BelalAhmed - Free expansion occurs when (for example) a valve is opened between one chamber containing gas and another at vacuum. This is anything but quasi-static. The gas does not have a well-defined thermodynamic state during the expansion. Are you confusing free expansion for something else? $\endgroup$ – David Hammen Apr 20 '16 at 16:02
  • $\begingroup$ No, I understand free expansion. Oh, I got it. Actually, I confused quasi-static with reversible. Any reversible process is necessarily a quasistatic one. However, some quasistatic processes are irreversible. Am I right? $\endgroup$ – Tyrion Lannister Apr 20 '16 at 16:22
  • $\begingroup$ Right. If the valve David Hammen is talking about is very small, the process is quasi-static (well-defined thermodynamic states in each chamber) but still not reversible. $\endgroup$ – L. Levrel Apr 21 '16 at 9:48
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You can not classify free expansions into any of those categories as free expansion is not a reversible process and hence the intermediate states are not well defined. The equations are not working because they find the area under the p-v graph but here no such graph can be made.

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