Quantitative Description of Isothermal (Constant Temperature) Process with Ideal Gas on P-V Diagram

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  • Опубліковано 13 гру 2024

КОМЕНТАРІ • 54

  • @xeno4162
    @xeno4162 3 роки тому +17

    Finally found a satisfying explanation.
    Thanks a lot, my friend.
    I will recommend this channel to my friends too.
    Subbed!

    • @terachad3000
      @terachad3000 2 роки тому +1

      @@Elucyda you are very good but I didn't understood why is the internal energy (change) is zero.

  • @ericsi5032
    @ericsi5032 6 місяців тому +1

    Thank you so much for explaining it so clearly, i was very lost until I found your video. Thank you very much!!

  • @dylanbeck3607
    @dylanbeck3607 3 роки тому +8

    You explain scientific concepts in such a brilliant and succinct way! I'm happy to have discovered your channel and I wish I had the pleasure of you teaching my classes :) Keep being awesome^^^

  • @tarcisiomenezes3318
    @tarcisiomenezes3318 Рік тому +1

    Finally found a good explanation. Thanks man!

  • @shrutidodiya7117
    @shrutidodiya7117 3 роки тому +7

    I am indian but your explanation is so better

    • @Schrödinger38326
      @Schrödinger38326 2 роки тому

      ???

    • @lol311
      @lol311 8 годин тому

      What does your nationality have to do with his explanation?

  • @larryzdanis5377
    @larryzdanis5377 3 місяці тому

    Nice.. since I knew that the enthalpy and internal energy of a gas are only a function of its temperature.. I was struggling to understand how the MechPE manual had an expression for reversible work for an isothermal process - my ?? was, where is the energy coming from to do work.. now I realize I was thinking of a closed system, and that was the error.. with an open system, work can be done or absorbed or by addition or subtrataction of heat through a heat sync .. certainly one of the more subtle/tricky concepts in thermodynamics (which is probably rarely utilized in practical applications) and your video cleared that up for me, thanks

  • @djay10101
    @djay10101 8 місяців тому

    Definitely deserves more subscribers

  • @vasconoronha7676
    @vasconoronha7676 Рік тому

    Bro ur an amazing teacher ! Keep it up

  • @deepanshinandal8961
    @deepanshinandal8961 2 роки тому +5

    I am from India
    I am a student of class 11th
    To which classes thermodynamics is most appropriate to teach in your country?

  • @vanvlogs4
    @vanvlogs4 29 днів тому

    I am Indian but this lecture is amazing 👏

  • @איתימטייב
    @איתימטייב 5 місяців тому

    Very clear and informative thanks!

  • @lilykangai1598
    @lilykangai1598 3 роки тому

    You speak so clearly thanks. .well understood

  • @florencejaynepalma8466
    @florencejaynepalma8466 3 роки тому

    Hi! I'm a new fan.. Thank you for your lectures. I like the way how you explain things. Your voice is also very calm.

  • @jeonjungkook1413
    @jeonjungkook1413 2 роки тому

    Nice explanation
    You keep nice going 👏

  • @aanchal6045
    @aanchal6045 2 роки тому

    Thanku for teaching us in such a easy way..

  • @sureshanchan23
    @sureshanchan23 Місяць тому

    Love from India ❤

  • @shivanshguleria1198
    @shivanshguleria1198 3 роки тому

    Nice video bro👍 well understood keep going 🙏

  • @happyfrezar2813
    @happyfrezar2813 8 місяців тому

    Your explanations are amazing bro❤👍🏼 You know what you are saying 😂

  • @putzak
    @putzak 2 роки тому

    Your'e a great teacher!

  • @mandavasasank3862
    @mandavasasank3862 2 роки тому

    nice .great explanation sir ,thank u .

  • @muhainifahemahmasjuktoc7588
    @muhainifahemahmasjuktoc7588 4 роки тому +1

    Great video!

  • @fehmidasalam2036
    @fehmidasalam2036 3 роки тому

    Thank you, I was having a confusion in that! You are amazing!
    Subscribed

  • @CurlyMaiden
    @CurlyMaiden 2 роки тому

    Baby brother of AK lectures.

  • @pokihoki8699
    @pokihoki8699 3 роки тому

    Amazing!! helps me to understand easily😄

  • @PureQuotes
    @PureQuotes 4 роки тому +2

    Thanks

  • @Nick2014B
    @Nick2014B 3 роки тому

    you're an angel thanks bro

  • @danielsantslv4565
    @danielsantslv4565 2 роки тому

    what if I want to do the isothermal curve on a P - v (specific volume) diagram? I don't know in witch table I should seek for the values to do the curve.

  • @mahimachoudhary8138
    @mahimachoudhary8138 3 роки тому +2

    Thank you. Sir

  • @andriyivanov4810
    @andriyivanov4810 Рік тому

    Hi matey, thanks for the explanation what problem completely doll I cannot understand basic sing why if you’re applying heat temperature remain constant this I believe this is basic stink of iso thermic process but I cannot understand could you please explain

  • @heyyo162
    @heyyo162 2 роки тому

    If I compress the piston, my work is converted into thermal energy inside the box. Then the heat sink takes that energy away. how can then the piston go back up, the energy should have been lost ?

  • @наукажизнь-м9б
    @наукажизнь-м9б 2 роки тому

    Good men

  • @graceoffeibeaacquah8996
    @graceoffeibeaacquah8996 Рік тому

    Akoa aben😂. (Very intelligent)

  • @Gruntol5
    @Gruntol5 3 роки тому

    Can you help me? How much heat needs to be rejected during a true isothermal compression of 3,000 cubic feet of air, when compressed from 15 psi to 4,500 psi using a 3-stage compressor?

  • @saumyaelchatwar1967
    @saumyaelchatwar1967 3 роки тому

    Now I understood the concept

  • @cherryrosedaclitan2089
    @cherryrosedaclitan2089 3 роки тому

    Thank you for this.. 😊

  • @arunprayog7986
    @arunprayog7986 2 роки тому

    U r awesome

  • @certezxgirl
    @certezxgirl 2 роки тому

    Ooh thankyou so much 😍👍

  • @陳允昊
    @陳允昊 Місяць тому

    thanks bro

  • @milagros070728
    @milagros070728 3 роки тому

    Thank you so much!!!

  • @smitagravat1063
    @smitagravat1063 3 роки тому

    I know this every thing but i still can't understand how we can apply heat if temperature is constant

  • @iskandar1023
    @iskandar1023 3 роки тому

    As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air acts as an ideal gas, calculate the work done per mol of the air.
    A. -80.35 J
    B. 80.35 J
    C. -811.93 J
    D. 811.93 J
    (please show calculation)
    can use this formula W=nRT ln(p1/p2)​
    Can you help me..everyone doesnt know how to solve this..i must present this question to my lect..please notice me

  • @anonymously7__
    @anonymously7__ 2 роки тому +1

    ❤️

  • @raimazaman1284
    @raimazaman1284 3 місяці тому

    thank uuuuu

  • @Vinicius-gr3it
    @Vinicius-gr3it Рік тому

    Eu tentando estudar para termoquimica e voce apareceu kkk

  • @Abcdej_ha
    @Abcdej_ha Місяць тому

    Thanks bro🥹

  • @zinc6748
    @zinc6748 2 роки тому +1

    Here,∆U=0. Explain further.

  • @harshass7652
    @harshass7652 3 роки тому

    Bro but in google Δu= Q-W so, when internal energy is zero, it becomes Q=W

    • @Elucyda
      @Elucyda  3 роки тому

      Both sign conventions are correct as long as they are used consistently.

  • @SMAN0
    @SMAN0 3 роки тому

    Why not 𝑊=𝑛𝑅𝑇 ln (𝑉f/Vi)? Why have negative?

    • @Elucyda
      @Elucyda  3 роки тому +1

      The negative sign comes from a sign convention for the work done. It is fine to use a different sign convention as long as you are consistent.