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Can adiabatic process be isothermal process?

  1. Nov 4, 2011 #1
    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).
     
  2. jcsd
  3. Nov 4, 2011 #2
    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.
     
  4. Nov 4, 2011 #3
    yes..not an issue
    please be more precise :)
     
  5. Nov 4, 2011 #4

    DrDu

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    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.
     
  6. Nov 4, 2011 #5
    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?
     
  7. Nov 4, 2011 #6

    sophiecentaur

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    @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.
     
  8. Nov 6, 2011 #7
    Now i understand :)
    Thanks all for ur help and effort appreciate it :)
     
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