Can adiabatic process be isothermal process?

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Discussion Overview

The discussion revolves around the relationship between adiabatic and isothermal processes in thermodynamics. Participants explore definitions, clarify concepts, and examine the conditions under which these processes occur, focusing on the implications of heat transfer and thermal equilibrium.

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant defines an adiabatic process as one with no heat transfer, suggesting that if no heat transfer occurs, the system must be in thermal equilibrium, implying it could be isothermal.
  • Another participant challenges the notion of two states being in thermal equilibrium, arguing that thermal equilibrium refers to the absence of heat transfer between two different systems, not between states of the same system.
  • A further clarification is made that a system cannot be in thermal equilibrium with itself, as equilibrium requires comparison with another system.
  • One participant introduces the idea that there can be adiabatic processes that are also isothermal, although this claim is not elaborated upon.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of thermal equilibrium in the context of adiabatic and isothermal processes. No consensus is reached on whether adiabatic processes can simultaneously be isothermal.

Contextual Notes

Participants highlight the importance of precise definitions and the distinction between states of a system versus interactions with external systems, which may affect the understanding of thermal equilibrium.

Who May Find This Useful

This discussion may be of interest to students and professionals in thermodynamics, particularly those exploring the nuances of heat transfer processes and thermodynamic definitions.

sphyics
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i'm entirely confused with this.

with analyzing each definition:

adiabatic process : a thermodynamic process in which there is no transfer of heat

if a system is in state (P1, V1, T1) → (P2, V2, T2) if it is adiabatic no heat transfer occurs, if no heat transfer occurs the two states must be in temperature equilibrium i.e the system will be in thermal equilibrium (isothermal process).
 
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sphyics said:
... the two states must be in temperature equilibrium i.e the system will be in thermal equilibrium (isothermal process).

Note that when you mention 'two states', these are two states of the SAME system. That is the system is passing from one state to the other. At no time is the system in both of these states. So I do not think one can talk about two states being in thermal equilibrium.
 
grzz said:
Note that when you mention 'two states', these are two states of the SAME system. That is the system is passing from one state to the other.
yes..not an issue
grzz said:
At no time is the system in both of these states. So I do not think one can talk about two states being in thermal equilibrium.
please be more precise :)
 
Think of your system of being in a thermo can while performing the adiabatic process. Even if it would like to (because it has a different temperature than its environment) it can't exchange heat with the surrounding because it is thermally isolated from it.
Nevertheless there are also adiabatic processes which are at the same time isothermal.
 
Let me explain what I meant when I said, 'At no time is the system in both of these states. So I do not think one can talk about two states being in thermal equilibrium'.

'Thermal equilibrium between two systems' implies that there is no net transfer of heat between these two systems. Now the OP was considering a system passing from one state to another state. Part of a system can transfer heat to another part of the system but can a system transfer heat form its own initial state to its own final state?
 
@sphics
A system can't be in thermal equilibrium with itself. It can only be in equilibrium with some other system - e.g. the Lab. That would constitute an isothermal change and not an adiabatic one.
 
Now i understand :)
Thanks all for ur help and effort appreciate it :)
 

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