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Himanshu posted an Question
April 24, 2020 • 06:39 am 30 points
  • IIT JAM
  • Physics (PH)

Thermodynamics

In first pic they ask for change in entropy for reversible process which not comes 0 but other things on the other hand in the second pic they considering change in entropy equal to 0 for same reversible process why this difference???

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  • Himanshu Pandey

    Ok Abhishek sir i got your point let me clear one thing why in first case entropy is non zero??

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    okay, it is an isothermal reversible process. isothermal means, dU =0, hence from first law of thermodynamics, dQ= PdV + 0 = PdV. there is heat exchange is taking place, how can entropy change of the system be zero for this reversible process(I am talking about this problem only). hopefully this will help you out. for better understanding, I am attaching a pic of maths behind it in comment below. regards.

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    below*

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    Abhishek singh Best Answer

    Okay, I see what is the problem! it is very similar to the doubt, which I have had during my preparation days (if I am not getting it otherwise). See, if you are comfortable with first pic you have uploaded, where entropy change of the system is non zero. Then it would be easy for me to convince you in the second case. Think in this way, we have two blocks in second case ( let say A and B) here also entropy change of block A is non zero (happy?) but the second block (B) have the same non zero change in entropy with a negative sign. And in the question they have defined the two blocks as a system(with no heat exchange with surrounding, only heat change withing block A and B. it is the sum total of entropy change in second case, which comes out to be zero(becoze system as a whole is not interacting with surrounding. Now see both the questions again. and again read the solution and this comment. Think . again think. Do let me know if you are still facing problem. Best regards.

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    Shanu arora best-answer

    Actually the difference lies there in defining the systems .first of all ,for reversible systems change in entropy is zero ( this is a atal satya, nobody can change it ) ,but the thing is for 1st case the system is gas + sorrounding , and in question we have asked to give ans for change in entropy of GAS only , if the question as to calculate change in entropy of sorrounding ,our ans will be( -ve x entropy change of gas) , Also note this important point ,whenever only one isothermal system is given and entropy have to be consider ,it's always good to keep in mind of sorrounding. Whereas in second case system consists of two things which can exchange heat and temp. And there they are considering some sought of ideal environment where the heat exchange with environment is zero or negligible . Free feel for any doubts.

  • Himanshu Pandey

    but if second case comes under Adiabatic process why there is Change in temperature it should be constant

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    Dhairya sharma best-answer

    the difference is that in second there is ofcourse change in temperature occurs but the temperature is changing in the system only it has nothing to do with surrounding. If this situation happens we deal the problem like adiabatically and in adiabatic processes change in entropy is always zero. both are reversible ofcourse but we have to take care about these situation. hope u will understand it.

    eduncle-logo-app

    adiabatic means not interacting with surrounding. two blocks is in one system. they can intract one another not by surrounding. If we say both blocks are adiabatic then there's prblem . but here two blocks are one system.

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