How can a reversible adiabatic process (theoretically) be brought about since the system is insulated? A $\mathrm dP$ change in pressure will lead to a $\mathrm dT$ change in temperature thus disturbing the thermodynamic equilibrium. Because there is a temperature difference between the system and the surrounding which isn't possible in an reversible process.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Hello and welcome to Physics SE! I have tried to improve the clarity of your question. If I misunderstood your intent, please edit further. Thanks! $\endgroup$– jng224Commented Aug 7, 2021 at 9:12
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
7
I tried to explain this in connection with your other answer. But here's another try.
A reversible process is one in which no entropy is generated. Causes of entropy product include, but are not necessarily limited to, the following:
- Heat transfer occurs across a finite temperature difference.
This is what is meant by thermal disequilibrium. There is no thermal disequilibrium between two objects having different temperatures if there is no opportunity for heat to transfer between the two object. Insulating the system from the surroundings in an adiabatic process prevents the opportunity of heat transfer across a finite temperature difference.
- Work transfer involving a finite pressure difference.
For a reversible adiabatic process the difference in pressure is always infinitesimal so that the pressures can be considered the same, i.e., the system and surroundings are in mechanical equilibrium. This, plus the absence of any mechanical friction, makes the adiabatic process reversible.
Hope this helps.
-
-
$\begingroup$ It will help in my jee advanced prep thks once again $\endgroup$– BhaumikCommented Aug 9, 2021 at 6:24
-
$\begingroup$ Glad it helped. So then was my answer acceptable? $\endgroup$– Bob DCommented Aug 11, 2021 at 22:14
-
$\begingroup$ i dont know how to accept an answer as a i new to tyhis platform, $\endgroup$– BhaumikCommented Aug 14, 2021 at 10:13
-
$\begingroup$ Next to each answer you should see an “accept” button $\endgroup$– Bob DCommented Aug 14, 2021 at 10:31
$\begingroup$
$\endgroup$
7
When they say adiabatic, they mean perfectly insulated.
You can have a temperature difference between the system and the environment, but you just pretend that zero heat flows. That’s just a given. You are told to assume it.
Maybe that is not always realistic, especially when there are big temperature differences. But we assume it’s true to do the theoretical problem and learn. And sometimes is ok assumption in real life situations if close to true. But yes there is no perfect insulation. In reality, there would be heat lost through the insulation (and also into the insulation to set-up a temperature gradient, because the insulation is not zero mass, and is not zero “thermal mass”). Idealized case.
-
$\begingroup$ Basically my point is that can there be Temperature difference btw system and surrounding in a reversible process if no then how the temperature of the system will rise .how can the system be in thermodynamic equilibrium if there is temp difference btw system and surrounding $\endgroup$– BhaumikCommented Aug 7, 2021 at 9:52
-
$\begingroup$ I did not understand what are u are trying to say plz explain by few egs and also plz be little bit specific as i have raised many ques $\endgroup$– BhaumikCommented Aug 7, 2021 at 10:20
-
$\begingroup$ Adiabatic means no heat energy flows in or out. We might still gain or lose energy by changing the volume. If the volume of the system expands, then the system does work on the environment, and the system loses some energy that way. If it reduces in volume it gains energy. Make sense so far? $\endgroup$– Al BrownCommented Aug 7, 2021 at 12:14
-
$\begingroup$ Yes ot makes sense so far but ques remains unanswered how temp of any system can rise in an reversible process if surr3temp remains constant ? Will it not hamper thermal equilibrium condition $\endgroup$– BhaumikCommented Aug 7, 2021 at 16:40
-
$\begingroup$ @Bhaumik Oooh. I see. I edited my answer! $\endgroup$– Al BrownCommented Aug 7, 2021 at 20:39