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Isothermal process

  1. Apr 28, 2017 #1
    "An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change will occur slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange."

    I cannot understand this part. By adjusting does system comes to the temperature of outside thermal reservoir?


    Thank you.
  2. jcsd
  3. Apr 28, 2017 #2


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    You left out one word or two.
    As you stated the temperature of a system is a constant for an isothermal process.
    The system starts out at a particular temperature and continuously accepts or ejects heat to remain at that same temperature, depending upon the difference between the system temperature and that of the reservoir.

    One way for an isothermal process to occur is to immerse the system in a heat bath, wait for thermal equilibrium, start the process, and proceed slowly so that infinitesimal changes in temperature of the system give rise to a heat flow to or from the reservoir. A way to think of it is as a feedback loop, with the signal being the temperature, and the output being the heat flow - a continuously adjusting closed loop.

    What was underlined could be narrowly interpreted to mean that the system is experiencing a heat flow to either increase or decrease its starting temperature to that of the heat reservoir.
  4. Apr 28, 2017 #3
    It means that, at any instant through the process, the temperature of the gas differs only slightly from that of the reservoir (just enough for a tiny amount of heat to be flowing).
  5. Apr 29, 2017 #4
    Yes, I understand from this explanations that it is slow process and the temperature change is infinitesimal or that of slight different one, assumming no change in temperature. But how can this cause to a state change?

    Thank you.
  6. Apr 29, 2017 #5


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    Imagine a system with a metal cylinder and piston, containing gas inside it. If you pull the piston, the gas does expansion work against the atmospheric pressure and therefore loses some of its internal energy, becoming colder. However, if you pull the piston very slowly, much slower than the time scale where heat conduction through the cylinder walls happens, the gas will remain at almost same temperature with the surroundings through the whole expansion process.
  7. Apr 29, 2017 #6
    Is this process called quasi-equilibrium?

    Thank you.
  8. Apr 29, 2017 #7


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    I've heard the terms quasistatic or reversible being used, but I guess quasi-equilibrium is fine, too.
  9. Apr 29, 2017 #8
    Are you talking about a change from liquid to vapor, for example?
  10. Apr 29, 2017 #9
    In wiki, it is given as change of system so I thought it is change of state of the system ie we always fix by finding two independent intensive property.

    Thank you.
  11. Apr 29, 2017 #10
    OK. Then I don't understand your question. Doesn't a state change include changes in which the temperature is constant, but the pressure increases and the volume decreases? So, holding the temperature constant doesn't cause a state change. It just sets the conditions for a certain type of state change involving pressure and volume.
  12. Apr 29, 2017 #11
    For example? I know only a state change in which temperature is constant is a phase transition but in this case because temperature and pressure are dependent to each other this means that in such a state change pressure is constant as well.

    Thank you.
  13. Apr 29, 2017 #12
    The term "change in state" is used in two entirely different contexts in thermodynamics:

    1. A change of phase for a pure substance at constant temperature and pressure

    2. A change in thermodynamic equilibrium state for a single phase substance, from (T1,P1,V1) to (T2,P2,V2)
  14. Apr 29, 2017 #13
    Does not 2 already include 1 ie 1 is a special case for 2, pertaining to phases? Because phase change take place in constant pressure and temperature and v, the specific volume is different for different phases there should be another intensive property for 2 to include 1 and I think that property should be the specific internal energy. Do you agree?

    Thank you.
    Last edited: Apr 29, 2017
  15. Apr 29, 2017 #14
    1 requires two phases, and 2 only requires 1 phase. In my mind they are very different.
  16. Apr 29, 2017 #15
    I cannot remember at this moment. For clarification would you please give an example of a state change in which temperature is constant, but the pressure and the volume changes?

    Thank you.
  17. Apr 29, 2017 #16
    Isothermal expansion of an ideal gas.
  18. Apr 29, 2017 #17
    If this isothermal expansion takes place in constant pressure, can there still be a state change? Of course volume will change but what is the situation according to your definition 2 for a state change?
    Thank you.
  19. Apr 29, 2017 #18
    It depends. Would you consider the following isothermal irreversible expansion process to also be a constant pressure process?

    I have an ideal gas in a cylinder with a frictionless massless piston. The gas is in thermodynamic equilibrium at temperature T1, pressure P1, and volume V1. The cylinder is in contact with a constant temperature reservoir at temperature T1 (so the surface of the cylinder in contact with the gas is always going to be at temperature T1). I am applying a constant force per unit area ##P_{ext}## to the piston equal to P1.

    At time zero, I suddenly drop the external force per unit area applied to the piston to a new lower value ##P_{ext}=P2<P1##, and hold ##P_{ext}## constant at this new value for all subsequent times. This allows the gas to expand irreversibly, and eventually to re-equilibrates at a new volume V2, at temperature T1 and pressure P2.

    So, what do you think? Would you consider this process both isothermal (at T1) and at constant pressure (at P2)?
  20. Apr 30, 2017 #19
    I think this would not be an isothermal expansion process we have been discussing here but be an irreversible expansion because you change the external pressure suddenly which implies that it is not slow enough to be reversible. What should be tempreture of this process? Average of T1 and coldest value of gas undergoing the process?

    Thank you.
  21. Apr 30, 2017 #20
    So, in other words, you are only looking at reversible processes. There is no reversible process involving an ideal gas that is both isothermal and isobaric, unless there is chemical reaction occurring.
  22. Apr 30, 2017 #21
    I have never thought of this but when you said this I remebered the ideal gas equation ##PV=mRT## which clear show us an ideal gas cannot expand both isobaric and isothermal. But your information is still a matter of question for me because you mention reversible processes. Is not this equation correct for irreversible process? And R is already constant and if the system is a closed system then m is costant but for an open system ##m## will not be constant so cannot any process be found as both ##isobaric## and ##isothermal## in which the volume change will be balanced by mass change?

    Thank you.
    Last edited: Apr 30, 2017
  23. Apr 30, 2017 #22
    It is only valid at the initial and final states of an irreversible process. In an irreversible expansion or compression, the temperature and pressure within the system are not even uniform spatially.
    And R is already constant and if the system is a closed system then m is costant but for an open system ##m## will not be constant so cannot any process be found as both ##isobaric## and ##isothermal## in which the volume change will be balanced by mass change?

    Thank you.[/QUOTE]
    It's not really a process, as you describe it. Certainly, if you if you have a container with a barrier in the middle, and the gas on either side of the barrier is at the same temperature and pressure, when you remove the barrier, nothing will have really have happened. Yet x number of moles of gas from gas B will have been added to gas A, and both gases will now combine to occupy the entire volume.
  24. Apr 30, 2017 #23
    We dismissed the temperature of this this process. Temperatures for process are their initial and and final temperatures.

    No, I didn't exactly described but asked about if it can be valid for open system, ie if mass can change. Then I understand that there is no process both isothermal and isobaric. By there is no reversible process involving an ideal gas that is both isothermal and isobaric do you mean it can be possible for irreversible processes?

    Do you mean by "it is only valid", ##PV=mRT## only valid at the initial and final states of an irreversible process because the temperature is not uniform?

    Thank you.
  25. Apr 30, 2017 #24
    It is possible only if you accept the definitions that I gave in a previous post. If not, then no.
    Also, typically, the pressure. In an irreversible process, there are also viscous stresses that contribute to the gas force acting on the piston face (over and above the thermodynamic pressure).
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