Lecture 39 : Clausius Inequality and Introduction to Entropy

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  • Опубликовано: 8 янв 2025

Комментарии • 23

  • @jagadeeshgurana4490
    @jagadeeshgurana4490 4 года назад +9

    You are one of the reasons why I love thermodynamics, sir.
    Great teaching, love you sir. Thanks a lot

  • @yesidlee
    @yesidlee 5 лет назад +6

    Love this lecture. Amazing way in dealing with subtle things.

  • @hardiksolanki831
    @hardiksolanki831 2 года назад +1

    Great video thank you to nptel and all IIT's professor and all team who contribute to make nptel's video.

  • @sutopamodak6247
    @sutopamodak6247 4 года назад +1

    This is an awesome lecture and very informative 👍
    Thank you sir 🙏

  • @fatumeshariff7447
    @fatumeshariff7447 3 года назад +1

    VERY GOOD VIDEO
    WOW

  • @dhruvsaxena9740
    @dhruvsaxena9740 3 года назад

    16:17 , its true for the refrigerator, but not for heat pump, as there are different formulae for efficiency in both case. Am I correct?????

    • @dhruvsaxena9740
      @dhruvsaxena9740 3 года назад

      as in refrigerator QL is constant, while in heat pump QH is constant in irreversible process.

  • @minakshidas4168
    @minakshidas4168 4 года назад +4

    Sir, why the cyclic integral of dQ in case of reversible heat pump or refrigerator can be equal to zero??Atleast some positive work input will be required there also.

    • @jagadeeshgurana4490
      @jagadeeshgurana4490 4 года назад

      Sir said the same too. Watch from 14:35, he was erasing 'equal to' symbol at 14:40.

    • @navya3911
      @navya3911 4 года назад

      @jagdeesh he erased in example 4 that is irreversible process but there was an equal to symbol in example 2 which is a refrigerator example reversible process

    • @lsrinivasan242
      @lsrinivasan242 4 года назад

      It is not equal to zero

    • @lsrinivasan242
      @lsrinivasan242 4 года назад

      Integral of dq/t is only equal to zero

    • @abhinavsahu6845
      @abhinavsahu6845 3 года назад

      For a reversible process , Q1/T1=Q2/T2, thus Q1/T1-Q2/T2=0 thus Q1/T1+(-Q2/T2)=0, Q1 is heat accepted from a reservoir at temperature T1 and Q2 is heat rejected to reservoir at temperature T2(therefore , is denoted by "-Q2").

  • @vkvajpai6183
    @vkvajpai6183 4 года назад

    Thanks sir for your kind information.

  • @sutopamodak6247
    @sutopamodak6247 3 года назад

    In Clausius inequality there are explained that T is the temperature of the source in several books but you told that T is the temperature of system. Please explain me the reason sir..

  • @thatsound7057
    @thatsound7057 4 года назад

    clear, but where does deltaQ/T comes from ?

  • @vkvajpai6183
    @vkvajpai6183 4 года назад

    Sir can you kindly explain the question
    Prove that both Kelvin Planck and Clausius statements are contained in Clausius inequality statement that the proceses for which integral of delQ/T >=0 are impossible

  • @nimaipatro2871
    @nimaipatro2871 Год назад

    ❤❤❤❤

  • @studentsacademy9954
    @studentsacademy9954 5 лет назад

    Good

  • @angrypepe5991
    @angrypepe5991 2 года назад

    .

  • @granadierfc1953
    @granadierfc1953 4 года назад +4

    Bakwas