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When is ΔH=ΔU

  1. Feb 9, 2016 #1
    In thermodynamics when can we say that ΔH = ΔU?
     
  2. jcsd
  3. Feb 9, 2016 #2
    Well, for cases where ##\Delta(PV)=0##
     
  4. Feb 10, 2016 #3
    That is true but when I asked my teacher she also said that in a closed system this is possible. Why is that?
     
  5. Feb 10, 2016 #4
    Do you think that delta PV can't be zero in a closed system? What about a process involving an ideal gas for which there is vo temperature change?
     
  6. Feb 10, 2016 #5
    I know that in a closed system there is no change in volume and we can assume that pressure is remaining constant too but then using the ideal gas equation moles come out to be constant too so in such a case moles formed must be equal to the moles reacted which does not happen in all cases. I just wanted to know if someone could give me an example of this so that i could understand it better. An example of an equation would help.
     
  7. Feb 10, 2016 #6
    You never mentioned that you were exclusively talking about situations involving chemical reaction. Is that what you want to focus on?
     
  8. Feb 10, 2016 #7
    Yes, i am sorry i forgot to mention that in the post.
     
  9. Feb 12, 2016 #8
    If it is a reaction involving ideal gases, then it happens if the number of moles of products is equal to the number of moles of reactants (stoichiometrically).
     
  10. Feb 14, 2016 #9
    There is no flow work in a close system ΔPV=0, since ΔH=ΔU+ΔPV. The gas can not expand, so what happens is, internal energy of the gas can be measured by simply the difference in its temperature→ CvΔT
     
    Last edited: Feb 14, 2016
  11. Feb 14, 2016 #10
    Are you really saying that Δ(PV) = 0 for any process in a closed system?
    Are you really saying that, when a chemical reaction occurs in a closed system, the change in internal energy is CvΔT?
     
  12. Feb 14, 2016 #11
    Yes, Chester. Why? Is there something wrong with it?
    Yes Chester, could there by any more energy unaccounted by it?
     
    Last edited: Feb 14, 2016
  13. Feb 15, 2016 #12
    Hi Ronie,

    Actually neither of your answers is correct. I was just checking to see if you really meant what you seemed to indicate. Your answers come very close to bordering on misinformation, which is a no-no at Physics Forums. Please, in the future, if your expertise is not sufficient to address a member's questions, do not offer answers. All this does is confuse the member further. Understand that misinformation can lead to infraction points and eventual banning.

    Chet
     
  14. Feb 15, 2016 #13
    How and why? I am delighted and loved to see the reasons why its bordering misinformation. Isn't that suppose the goal of the forum, to educate people in the right track. I have no problem getting wronged, when it's proven.
    Here is my reference analysis for "bomb calorimeter" from Ohlone College.
     
    Last edited: Feb 15, 2016
  15. Feb 16, 2016 #14
    My problem is not with your desire to learn in Physics Forums, but with the authoritative manner in which you gave your answer in post #9 (considering it was incorrect), and your lack of qualifying statements for what you were saying.

    With regard to your response in the first statement, Δ(PV) is not generally equal to zero even in an isolated system, let alone in a closed system. If the volume is constant in a closed system, then Δ(PV)= VΔP, and if the pressure is constant in a closed system, then Δ(PV)=PΔV. And then there are all those other cases in which both P and V change in a closed system.

    Now for the second statement you agreed to: when a chemical reaction occurs in a closed system, the change in internal energy is CvΔT. Now, we're not talking about the calorimeter here (which contains the reacting system), we are talking about the actual set of chemical species that constitute the reaction mixture. You are aware that, even if the temperature and pressure of a system are unchanged between the initial and final states of the system, its internal energy and enthalpy can still change, correct?

    I assume you have heard of changes in phase which occur at constant temperature and pressure? Examples are melting and vaporization. You are aware that both the enthalpy and internal energy of the material change during a phase change at constant temperature and pressure? You have heard of the terms heat of vaporization and heat of melting, correct? So, for these, the internal energy change is not equal to CvΔT.

    In the case of chemical reactions, there are energy effects associated with making and breaking of chemical bonds. You are aware that both the enthalpy and the internal energy of the reacting mixture change during a chemical reaction at constant temperature and pressure (or constant temperature and volume)? You have heard of the terms heat of reaction, heat of combustion, and heat of formation, correct? So, for these, the internal energy change is not equal to CvΔT.

    Further questions?
     
  16. Feb 16, 2016 #15
    I am sorry Chet, we seemed to have a disconnect here.
    In post #9, I was responding to OP's query "In thermodynamics when can we say that ΔH = ΔU?". To cut the chase, my answer was direct to the point. CvΔT(sensible heat) refers to an Isochoric process.

    You caught me like a lawyer on that one, for the term "close" system, probably, lost the case. Isothermal & Isobaric expansion also happens in a close system, I agree, but it's far from the point of ΔH = ΔU which is in the first place the, OP's concern.

    Yes, I am well aware of latent heats of medium(pure substance) which occurs at constant pressure & temperature regions--" internal energy of the material change during a phase change at constant temperature and pressure"

    Nope and yes, one more. Then, how would you measure or determine the values of specific heat in the lab for "heat of reaction, heat of combustion, and heat of formation"? I suppose you have a good understanding on this.
     
    Last edited: Feb 16, 2016
  17. Feb 16, 2016 #16
    Who says that, in an isochoric process, ΔH=ΔU? For an isochoric process ΔH=ΔU+VΔP.

    ΔH is not equal to ΔU unless Δ(PV)=0.
    Yes I do have a good understanding of it. You would not need to determine the values of specific heat in the lab for heat of reaction, heat of combustion, and heat of formation, because these changes are defined at constant temperature. They represent the amount of heat (per mole) that must be added to the system to hold it at constant temperature in going from pure reactants to pure products.

    Ronie: You need to beef up your background in Thermodynamics (particularly chemical thermodynamics) before you can confidently answer thermo questions on Physics Forums. I don't recommend online sources, because they tend to be incomplete. And Physics Forums cannot provide a complete tutorial on Thermodynamics because of our limited structure. Here are some good references, however, that can be very helpful to you:

    Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness
    Fundamentals of Engineering Thermodynamics by Moran et al
     
    Last edited: Feb 16, 2016
  18. Feb 16, 2016 #17
    So, the analysis of below source is erroneous or probably lack some consideration of VdP? How would that differ with ΔH=ΔU? upload_2016-2-17_12-41-30.png
    Then, when is Δ(PV)=0? Is this condition hypothetical?

    deltah3.gif , you are referring to that (T) temperature right? Generally, for medium involving chemical reaction, right? Consequently, for pure substance that does not undergo chemical reaction the change of number of moles Δn is zero. So, the term becomes zero for pure substance, right? Therefore, analysis for pure substance simplifies into ΔH=ΔU. Do you agree on this? But, for medium which undergoes chemical reaction, Δn≠0 therefore, ΔH=ΔE+ΔngRT, is this correct?

    Thus, we can summarize ΔH for isochoric process as follows
    For (Pure Substance - does not undergo chemical reaction) ----> ΔH=ΔU
    For (Substance that undergoes chemical reaction)------------------> ΔH=ΔE+ΔngRT

    Do you agree on this?
     
    Last edited: Feb 16, 2016
  19. Feb 17, 2016 #18
    This derivation is correct. But I don't see any ΔH in there. So, what makes you think that this proved ΔH=ΔU? Here's what I get by continuing this analysis: $$ΔH=ΔU+Δ(PV)=nC_vΔT+nRΔT=n(C_v+R)ΔT=nC_pΔT$$
    Do you have a problem with this.
    No. For an ideal gas, it is when ΔT=0

    No way. The equation you wrote applies to an ideal gas mixture at CONSTANT TEMPERATURE. So, for a pure ideal gas (or a non-reacting ideal gas mixture), even though Δn=0, if the temperature changes, ΔH≠ΔU. For that case, as we showed above, ##ΔU=nC_vΔT## and ##ΔH=nC_pΔT##
    Yes, but only at constant temperature.
    Obviously not.
     
  20. Feb 17, 2016 #19
    So, what's the difference of ΔH, ΔU & Q in this set-up.? How do you define ΔH then?

    upload_2016-2-17_21-25-8.png

    No, I don't have the problem with that derived equation but, how it's used, I have an issue.

    How can a gas mixture has Δn? Mixture does not react, so Δn=0 and of course ΔH= ΔE only .

    I just do not know, I can not somehow agree with you. May be because we are from different field.

    Enthalpy for me is a equation of a state basing on the First Law of Thermodynamics which is "arbitrarily" set to sum up(internal energy+flow work)at a point, so if there is no flow work done, enthalpy ΔH reduces to internal energy ΔU. That's is what the book said, the way we are taught and we use to run boilers and steam engines. Otherwise if yours is true-that for a closed system at constant volume( ΔH=ΔU+VΔP) , we will have to revised a new steam tables, probably.
     

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  21. Feb 17, 2016 #20
    You're saying you have a degree in Mechanical Engineering, and you don't know how ΔH is defined, correct? H = U + PV, and ΔH = ΔU + Δ(PV). In this setup, ΔH=ΔU+VΔP=ΔU+nRΔT


    I see that you have an issue, but I don't know what that issue is.

    Not if the temperature changes.
    No. Your problem is much more fundamental than this. You just haven't learned thermodynamics properly.
    Enthalpy is defined in all thermo textbooks as H ≡ U + PV. For an open flow system operating at steady state like the type you are describing, the change in enthalpy of the flowing stream is not equal to the sum of the change in internal energy plus the flow work (sometimes called shaft work). For an incompressible fluid passing between entrance and exit of a system like that, the rate of change of enthalpy for the flowing stream is equal to ##\dot{m}\Delta h= \dot {m} (\Delta u+v\Delta P)##, where ##\dot{m}## is the mass flow rate, and the lower case letters represent properties per unit mass. Check your textbook and see.
    Oh yeah? What if I told you that, if you check your steam tables, you will find that the specific internal energy values u differ from the specific enthalpy values h by the product of the specific volume times the pressure vP? That is h = u + pV. What do you think about that? So even your own steam tables back up what I have been saying.

    One more thing. I'm not sure you are aware of it, but I am a Chemical Engineer, not a physicist. And, believe it or not, we ChEs deal with boilers and steam engines all the time. In addition, one other thing we deal with that Mechanical Engineers usually don't are chemically reacting systems. This is our "bread and butter." So please don't tell me about how to model the thermodynamics of chemical reactors. I've modeled many of them during my career.

    Chet
     
    Last edited: Feb 17, 2016
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