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I know that for an isothermal process heat transfer is necessary so process should be slow and walls should be conducting for heat transfer to occur as a process cannot be isothermal and adiabatic at same time. But why slow compression of piston need to be isothermal, can't it be neither isothermal not adiabatic? For example:

In figure,the walls of the container and the piston are weakly conducting. The initial pressure, volume and temperature of the gas are 200 KPa, 800 cm³ and 100 K resp. Find the pressure and the temperature of the gas if it is (a) slowly compressed (b) suddenly compressed to 200 cm³ (y = 1.5)

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Why in part (a) we consider process as isothermal instead of neither isothermal not adiabatic?

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    $\begingroup$ You’re free to try to solve part (a) assuming an intermediate condition between isothermal and adiabatic, if you like. Is that problem solvable with the information you’re given? What do you think the question writer intended for you to do? When “slowly” is mentioned without an actual speed, what do you think is implied? $\endgroup$
    – Chemomechanics
    Commented Mar 29 at 18:18
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    $\begingroup$ If that illustration is from a problem in a textbook, then it probably describes an idealized model of (some part of the cycle of) some heat engine. $\endgroup$
    – Solomon Slow
    Commented Mar 29 at 18:20
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    $\begingroup$ When the word "slow" is used here, it means slow compared to the rate of heat transfer through the walls of the container: that is the implied context. Therefore, it can't be adiabatic, because heat transfer will necessarily happen, exactly because the process is "slow" compared to the rate of heat transfer. If the process is slow "enough", it can be treated as an isothermal process (to a pretty good approximation, depending on the how slow it really is): that is the intent of the problem. It seems like you already get this, so I'm not sure what your question really is. $\endgroup$
    – march
    Commented Mar 29 at 18:31
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    $\begingroup$ If the piston is moved rapidly, do you think that the temperature and pressure of the gas will be uniform during the charge? $\endgroup$
    – Chet Miller
    Commented Mar 29 at 18:32
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    $\begingroup$ That is right. So you have to calculate the work in an alternate way. $\endgroup$
    – Chet Miller
    Commented Mar 29 at 19:06

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But why slow compression of piston need to be isothermal, can't it be neither isothermal not adiabatic?

To be isothermal in the case of an ideal gas, the compression needs to be carried out slow enough that the energy transferring into the system by the compression work exactly equals the energy transferring out in the form of heat so that the temperature of the gas remains constant throughout the process.

The process can be neither isothermal nor adiabatic if the process is carried out not too quickly so that there is some heat transfer during the process.

Your last paragraph comes under the category of homework and exercise which we are asked not to answer.

Hope this helps.

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