Physics 27 First Law of Thermodynamics (22 of 22) Work Done By A Gas

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  • Опубліковано 20 жов 2024

КОМЕНТАРІ • 70

  • @MichelvanBiezen
    @MichelvanBiezen  10 років тому +41

    Thank you all for comments. It is great to see that these videos are helping.

  • @rovinorodrigues5963
    @rovinorodrigues5963 8 місяців тому +5

    I'm 48. I did not have this clarity when doing my engineering. Great teachers are a rarity, but in you, the internet has found a wonderful teacher!

    • @MichelvanBiezen
      @MichelvanBiezen  8 місяців тому +3

      Thank you so much for your kind words. Glad you are enjoying the videos.

  • @Skittix447
    @Skittix447 5 років тому +7

    Just got through your Thermodynamics playlist and I've got to say thank you. You've gotten me through my first semester of Physics and now my 2nd. Your videos are extremely helpful. THANKS

  • @zekrweymedamien
    @zekrweymedamien 9 років тому +2

    I've watched all the 22 videos & I fully understand the whole subject now. Thank you, sir!!

  • @chriskoperniak784
    @chriskoperniak784 9 років тому +1

    Hands down the best lectures on thermodynamics on the net!! You're amazing.

  • @katherine4386
    @katherine4386 5 років тому

    I'm so happy that you take the time to educate us when our professors fail to do so. Thank you

  • @princessdianemagboo7096
    @princessdianemagboo7096 9 років тому +2

    I've watch all your videos, i just wanna say Thank you so much! now i feel so eager to learn about thermodynamics. Wish i could passed the exam! :) you're such a great professor!!!! thank you!!

  • @blueu127
    @blueu127 7 місяців тому

    thank you soo much for this! i didnt understand the first law of thermodynamics, with the formulas changing depending on the state. i never knew i didnt have to memorize the formulas in order to understand. i just have to understand the concepts from your teaching. thank you professor!

  • @AbigailSabandal
    @AbigailSabandal 4 роки тому +1

    Wow this playlist helped me a lot to understand these processes specially now that we are undergoing distance learning. I could not help but thank you, sir. You are an amazing educator.

    • @MichelvanBiezen
      @MichelvanBiezen  4 роки тому +2

      Thank you for writing. We are glad you find the videos helpful.

  • @mellowwhlzki4906
    @mellowwhlzki4906 9 років тому +1

    you're absolutely fantastic. I love your videos. thanks so much!! you're doing what my 'first class, elite' university lecturer couldn't make me do... understand!

    • @MichelvanBiezen
      @MichelvanBiezen  9 років тому

      mellow whlzki I am glad these videos are helping. Keep you your studies.

  • @xoloxcuintle05
    @xoloxcuintle05 11 років тому +8

    great vids!! I enjoyed the whole series. Thanks

    • @timothywilliam438
      @timothywilliam438 10 років тому +2

      I enjoyed the whole series too !!! Thanks a lot sir !

    • @MichelvanBiezen
      @MichelvanBiezen  10 років тому +1

      Timothy William Great! Thanks for letting me know.

  • @marychristinedelarosa8323
    @marychristinedelarosa8323 9 років тому

    I wish we could have a professor as awesome as you! These videos are very helpful!

  • @sjay4281
    @sjay4281 10 років тому

    Cannot describe in enough words how much these series helped me with thermodynamics. Thank you. Undergrad :)

  • @ddd7731
    @ddd7731 5 років тому +1

    You make learning fun and enjoyable!! Thanks a lot.

  • @franksonbenedict544
    @franksonbenedict544 Рік тому +1

    You deserve an award🎉

  • @fadlihafni3109
    @fadlihafni3109 2 роки тому +1

    thank you sir,you really helps me alot,which me luck on upcoming thermodynamic paper

  • @johnnygriffith3061
    @johnnygriffith3061 9 років тому

    Thanks very much. You have helped me when others couldn't!

  • @herwin4946
    @herwin4946 8 років тому +1

    Thank you. Really helpful ! I believe I've got 90% of understading. there other 10% is for practices.

  • @mustafaalmohsen5741
    @mustafaalmohsen5741 7 років тому +1

    they are very helpful videos. Thank you very much

  • @neworld2.073
    @neworld2.073 7 років тому +1

    A steam turbine in a power station develops 1000 kW power. The amount of heat
    supply into the boiler is 2800 kJ/kg, and the heat rejection from the condenser is
    2100 kJ/kg. The power required to pump the steam to the boiler is 5 kW. Consider
    the system is stationary and neglect the change of internal energies.
    Applying the energy balance relation above, calculate the rate at which steam
    circulates through the network of the power plant.

    • @neworld2.073
      @neworld2.073 7 років тому +1

      Could you help me with this if possible? Thanks

  • @MohammadkazimHalimy
    @MohammadkazimHalimy 7 місяців тому +1

    Thanks for your great explanation

  • @Reichvarg
    @Reichvarg 9 років тому

    Thankyou for doing this series

  • @hissahfahad5567
    @hissahfahad5567 8 років тому

    thanks a lot! your 22 videos is very helpful ...

  • @norhanbadry6090
    @norhanbadry6090 6 років тому

    These videos are very helpful>>>Thanks a looooot!!!.

  • @shepherdzvavanjanja7253
    @shepherdzvavanjanja7253 2 роки тому +1

    Great stuff

  • @edemisz
    @edemisz 9 років тому

    You sir, rock! Thanks a lot for the videos

  • @SergioPerez-tn8hs
    @SergioPerez-tn8hs 2 роки тому +1

    Thank you so much !!!!

  • @lesterlozano4192
    @lesterlozano4192 10 місяців тому +1

    Sir do you have video of polytropic process?

  • @Waleed9
    @Waleed9 10 років тому

    awesome sir!

  • @nipunasnuwan8171
    @nipunasnuwan8171 3 роки тому

    there is a piston with steam cylinder,expands isothermaly and reversibly , in this case does the change of internal energy can zero or not? can u please help

    • @MichelvanBiezen
      @MichelvanBiezen  3 роки тому +1

      Isotherm, means the same temperature. If the temperature doesn't change then there cannot be a change in internal energy, since the internal energy only depends on temperature.

  • @sabbontuu_haarrarge
    @sabbontuu_haarrarge 3 роки тому

    Thank you very much

  • @totallytechalicious
    @totallytechalicious 4 роки тому

    Amazing series! :)

  • @paulmesler9412
    @paulmesler9412 8 років тому +1

    I am a little confused. In the first isobaric example, shouldn't you be using Cp, not Cv for the molar heat capacity in the delta U equation?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      +PR Mesler
      To calculate the molar heat capacity for the delta U, you must always use Cv regardless of the process.

    • @paulmesler9412
      @paulmesler9412 8 років тому +1

      +Michel van Biezen
      Have you done a video explaining why Cv is always used when using delta U? I mean explaining in detail why this is the case because it's not intuitive to me.

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      +PR Mesler
      Watch the rest of the playlist and you'll get the information. I know this is a source of confusion because students confuse the Q in the first law of thermodynamics with the U. Note that U is the internal energy of the gas and it cannot depend on how you got there. Q is the heat gained or lost by the gas. The amount of heat does depend on the process during which the heat is exchanged. The best way to look at it is to take an isovolumetric process. No work is done because the volume of the gas doesn't change. Thus delta U equals Q which is n*Cv *delta T. In an isobaric process delta U is NOT equal to Q which is then n*Cp *delta T.

    • @paulmesler9412
      @paulmesler9412 8 років тому +1

      +Michel van Biezen
      So, are you saying that it doesn't matter if you have an isobaric, isovolumetric, isotherm,etc. process, delta U always uses Cv as the molar heat capacity?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому

      +PR Mesler That is correct.

  • @kevinmarquez117
    @kevinmarquez117 9 років тому

    GOD BLESS YOU!

  • @pratyushdeshmukh7909
    @pratyushdeshmukh7909 10 років тому

    Thanks a lot ......

  • @peterv6i
    @peterv6i 10 років тому

    thank you!

  • @miko74939
    @miko74939 7 років тому

    Do I always need to convert liter to cubic meter with these processes?

  • @sirisha890
    @sirisha890 7 років тому +1

    A vessel has two compartments of volume V1 and V2,containing an ideal gas at pressures P1 and P2 and temperatures T1 and T2.if the wall separating the compartments is removed,what will be resulting equilibrium temperature

    • @MichelvanBiezen
      @MichelvanBiezen  7 років тому +1

      That is indeed an interesting problem. We are assuming here that both gases have the same specific heat? Then the final number of mols becomes nf= n1 + n2 and the final volume becomes Vf = V1 + V2. Thus Pf/Tf = nf * R / Vf, but that leaves you with a ratio. Next you can use the sum of the internal energies. U = n Cv T and thus Uf = n1 Cv T1 + n2 Cv T2 = U1 + U2 From here the final T can be calculated to be Tf = (n1 * T1 + n2 * T2) / (n1 + n2)

    • @sirisha890
      @sirisha890 7 років тому

      Thankyou so much sirI

    • @sirisha890
      @sirisha890 7 років тому

      Thankyou sir

  • @zahraasamer4411
    @zahraasamer4411 9 років тому +1

    thaaaaaank you you are awsome

  • @nikhilstealth8486
    @nikhilstealth8486 8 років тому

    ISOVOLUMETRIC IS SAME AS ISOCHORIC RI8??

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому

      +NIKHIL STEALTH
      Yes, isovolumetric is the same as isochoric

  • @kishen313
    @kishen313 9 років тому

    respect ^^