0
$\begingroup$

By free expansion, I am referring to gas kept in a piston-cylinder arrangement freely allowed to expand against vacuum.

It is clear to me that free expansion is an irreversible process because if it were not then it could get compressed as well at equilibrium, but we know it doesn't as it would be violating the 2nd law of thermodynamics.

I know that all the parameters work 'W', Change in Internal Energy '∆U', and Heat transfer '∆Q' would all be zero in the case of an ideal gas.

But, what I couldn't understand is the difference between isothermal free expansion and adiabatic free expansion?

Also, I want to ask if Joule expansion is the same thing?

P.S. I have asked a different question related to it here Free Expansion - Ideal Gas vs Real Gas

Improve this question
$\endgroup$
1
  • $\begingroup$ I think you're right with your reasoning and all you wrote is true. The problem is that there isn't much point in expanding gas in an unbounded vacuum. You cannot exchange heat with a vacuum. Vaccum doesn't exert pressure. And in equilibrium by definition state variables don't change in time; free expansion implies no equilibrium untill it stops expanding and otherwise changing. $\endgroup$
    – Some Student
    Commented Jan 15, 2021 at 16:16

2 Answers 2

Reset to default
3
$\begingroup$

But, what I couldn't understand is the difference between isothermal free expansion and adiabatic free expansion?

Isothermal means the temperature of the gas is constant during the expansion process so that the ideal gas law can be applied at each point during the expansion. That requires the isothermal expansion to be reversible.

That is not the case for a free expansion. Although the initial and final equilibrium temperatures are the same, the temperature of the gas is not defined during the free expansion which is an irreversible process. Temperature and pressure gradients exist during the expansion.

Also, I want to ask if Joule expansion is the same thing?

The Joule expansion is the same thing in the case of an ideal gas. But for real gases, the initial and final temperatures for the free expansion are not the same because real gases involve intermolecular forces whereas an ideal gas does not.

Hope this helps

Improve this answer
$\endgroup$
13
  • $\begingroup$ But aren't isothermal and adiabatic free expansion same? $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 13:48
  • $\begingroup$ Why create distinctive terms for same thing? $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 13:49
  • $\begingroup$ A free expansion is not isothermal. Isothermal means the gas temperature is constant during a process, not just that the initial and final equilibrium temperatures are the same. The term "isothermal free expansion" makes no sense, at least to me. See :web.mit.edu/16.unified/www/FALL/thermodynamics/notes/… $\endgroup$
    – Bob D
    Commented Jan 15, 2021 at 14:07
  • $\begingroup$ Thanks for the clarification. $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 14:12
  • $\begingroup$ Just wanted to ask one last thing $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 14:12
0
$\begingroup$

The process you described is adiabatic, but it is isothermal only for a substance whose equation of state is such that pressure is directly proportional to temperature at constant volume (like an ideal gas). Otherwise, the process is not isothermal, because the internal energy depends on both temperature and specific volume.

In Joule-Thomson expansion in steady flow through a porous plug or resistive valve, the process is adiabatic, but it is isothermal only for a substance whose equation of state is such specific volume is directly proportional to temperature at constant pressure (like an ideal gas). Otherwise, the process is not isothermal, because the specific enthalpy depends on both temperature and pressure.

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ Isn't the Joule-Thompson expansion (porous plug) different than the Joule expansion (free expansion into a vacuum) referred to by the OP? $\endgroup$
    – Bob D
    Commented Jan 15, 2021 at 16:43
  • $\begingroup$ @bob d Well, one is at constant u and the other is at constant h. $\endgroup$
    – Chet Miller
    Commented Jan 15, 2021 at 16:49
  • $\begingroup$ I want to clear this confusion. Does internal energy change in case of real gas or not? (As temperature is changing) $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 17:07
  • $\begingroup$ And does heat transfer due to friction b/w piston and cyllinder. $\endgroup$
    – InfiniteCool23
    Commented Jan 15, 2021 at 17:15
  • $\begingroup$ @InfiniteCool23 If the walls of the vessel containing the gas are rigid and insulated, then $W=0$ and $Q=0$ and therefore $\Delta U=0$ regardless of whether the gas is real or ideal. $\endgroup$
    – Bob D
    Commented Jan 15, 2021 at 17:30

Your Answer

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

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