3
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ See Wikipedia page. For free expansion of an ideal gas initial and final temperature are same. $\endgroup$
    – Tyrion Lannister
    Apr 19, 2016 at 19:37

2 Answers 2

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Can you tell me why Free Expansion is quasi-static despite being irreversible? $\endgroup$
    – Tyrion Lannister
    Apr 20, 2016 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, 2016 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, 2016 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, 2016 at 9:48
0
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.