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Entropy & Enthalpy - A Weird Deduction from Class

  1. Jun 8, 2010 #1
    In class my professor for thermodynamics made the following deduction:

    We write [tex] dE = \left( \frac{dE}{dT} \right)_V dT + \left( \frac{dE}{dV} \right)_T dV[/tex]

    And from the first law: [tex]dE = dQ - PdV[/tex] *

    Equating and cancelling dE gives:
    (1) [tex]dQ = \left( \frac{dE}{dT} \right)_V dT + \left( \left( \frac{dE}{dV} \right)_T + P \right) dV[/tex]

    Realizing that [tex]\left( \frac{dE}{dT} \right)_V = C_V[/tex] and writing that [tex]dV = \left( \frac{dV}{dP} \right)_T dP + \left( \frac{dV}{dT} \right)_P dT[/tex]

    We can fill this in in (1):
    [tex]dQ = C_V dT + \left( \left( \frac{dE}{dV} \right)_T + P \right) \left( \left( \frac{dV}{dP} \right)_T dP + \left( \frac{dV}{dT} \right)_P dT \right)[/tex]

    Now if we have an isobaric process (P = constant), then [tex]dP = 0[/tex], so then we can write:
    [tex]dQ_P = C_V dT + \left( \left( \frac{dE}{dV} \right)_T + P \right) \left( \frac{dV}{dT} \right)_P dT[/tex]

    Now "dividing both sides by dT" (mathematicians, be aware, I am playing with the ghosts of infinitesimals) and realizing that because dQ is isobaric: [tex]\frac{dQ}{dT} = \left( \frac{dH}{dT} \right)_P = C_P[/tex] we get:
    [tex]C_P = C_V + \left( \left( \frac{dE}{dV} \right)_T + P \right) \left( \frac{dV}{dT} \right)_P[/tex]

    Now, I'm assuming the relationship we got is correct?

    But I'm wondering about the deduction... Is it truly valid?! (look at where I put the red star) Isn't dW = -PdV only true in a reversible change? Now someone might say "well, he actually meant to say dE = TdS - PdV and that is always valid" but then I say no, he cannot have meant that, because he also used that dQ = dH, so unless you want to imply dH = TdS for every process, this cannot be...

    Now my point is not "can I find a completely different deduction that is surely correct" or "can the answer be correct", but really: what he did there, is that allowed? Cause if so, I'm not getting it.

    Thank you,
    mr. vodka
  2. jcsd
  3. Jun 8, 2010 #2


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    The infinitesimal amount of work [itex]w[/itex] (sometimes written [itex]dW[/itex], as your teacher does) always equals [itex]-P\,dV[/itex] (assuming that work done to a system is positive). The derivation looks fine.
  4. Jun 8, 2010 #3
    I agree with Mapes.

    What you might be thinking of is the heat-entropy relation: dQ = T dS.
    This equality holds only for reversible processes. Otherwise it is written as dQ/T < dS.

    Of course, you didn't make use of any of this in your derivation, so no problem! If you're a little distraught though about your very first equation, the assumption is that the system has a state function E = E(V, T). Note this is not a fundamental thermodynamic relation, but is permissible since you are doing an isobaric analysis.
  5. Jun 9, 2010 #4


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    Mr.Vodka, you are completely right. Your prof should have written TdS instead of dQ or dH. The heat isobaric heat capacity - contrary to what it's name might imply - is not dQ/dT in general but T(dS/dT)_P or (dH/dT)_P. So you might see the derivation as a cancellation of two errors.
  6. Jun 9, 2010 #5
    Hm I seem to be getting contradictory replies... Let me try to restate why I don't get the formula I marked with a red star:

    If all infinitesimal work equals dW = -PdV then all work equals W = [tex]\int -PdV[/tex] (cause after all, how useless would our derivation be if we weren't allowed to integrate it...), meaning all work is reversible... Imagine I ran very hard into a piston, then I will have delivered more force than needed to cause its eventual displacement, meaning [tex] W \gneq \int - PdV[/tex]. If your reply is here "but that action can't be divided up into infinitesimal processes", then what you're doing is actually just using infinitesimal change as a synonym for reversible (after all, "reversible" is "an infinitesimal change between equilibrium states"). This would mean ALL the thing you write down in infinitesimals (which is pretty much all we've done in my course of Thermodynamics) imply reversibility...

    Thank you all three for your time.

    EDIT: to show why dW = -PdV implies reversibility: We have dE = Q -PdV
    On the other hand dE = TdS -PdV is always true
    So Q -PdV = TdS -PdV => Q = TdS, which implies reversibility...
    Last edited: Jun 9, 2010
  7. Jun 9, 2010 #6


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    You are also right in observing that w is not equal to -pdV in general. E.g., there are non-equilibrium situations where p may vary in space and even when it is possible to keep the system homogeneous and volume work is the only work done then there will still be a contribution to work from longitudinal viscosity in an irreversible process.
  8. Jun 9, 2010 #7
    Here is a slightly amended version of the derivation.

    I am concerned about the mixing up of partial and total differentials in your derivation, this can lead to unwanted errors since total differentials strictly only apply to state variables, which Q is not.

    I am sorry that I have a temporary failure of the pc with Mathtype on so I have had to do it longhand (more relaible anyway)

    Your end result is correct. I have also appended a similarly derived formula incorporating enthalpy. Entropy should not enter into this anywhere.

    Attached Files:

  9. Jun 9, 2010 #8
    Hm, interesting, thank you!

    Odd that there is no need for the first law of Thermodynamics.

    But just for clarity: the use of differentials and/or (partial) derivatives don't imply reversibility, right? I've always found it out though that you can use the derivate of E to T for example for an irreversible process, whilst T is certainly not defined for an irreversible process...
  10. Jun 9, 2010 #9
    Variables, P, V, T etc

    are not defined for the process they are defined for the system. And since they are state variables they can only posses one value at any given state. This can only be done for an equilibrium state.

    Thre is a connection between use of the differentials and reversibility/equilibrium.

    The following 3 statements are mathmatically equivalent.

    The function X is a function of the state of the system: X is a state variable.

    The differential of X - dX - is an exact differential.

    The integral of dX about a closed path in the state or indicator diagram is equal to zero.

    One interpretation of The First Law is that although dq and dw are not exact differentials, their sum is.
  11. Jun 9, 2010 #10
    But if they are not defined during the process, then how can we use them? In my course of thermodynamics I've done nothing else than use quantities like dE/dT in processes, while they're not defined...

    For example, if we say "dE = TdS - PdV is true even for irreversible processes" what does that mean, because T and P don't exist
  12. Jun 9, 2010 #11


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    No, T and P don't exist for systems that are not in equilibrium. If you perform a process sufficiently slowly that the system re-equilibrates faster than you are forcing it to change, then T and P remain well defined. A process being irreversible does not imply it is out of equilibrium, so T and P remain defined for irreversible processes that do not take the system out of equilibrium.
  13. Jun 9, 2010 #12
    "A process being irreversible does not imply it is out of equilibrium"

    Can you give me an example of a process in constant equilibrium yet not reversible?
  14. Jun 16, 2010 #13
    I'm surprised by the end result. We always know that Cp-Cv=R where R is the universal gas constant. Thus the R.H.S of the equation should be a constant.
  15. Jun 16, 2010 #14


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    Coolraj, your "universal formula" only holds true for an ideal gas.
  16. Jun 16, 2010 #15
    Joules law states that for gases which obey PV=nRT, the change of internal energy is dependant upon temperature alone.

    A little thought and substitution of those facts into my equation 5 will yield

    Cp-Cv = R

    But as Dr Du said, this only applies to gases with PV=nRT as their equation of state.
  17. Sep 25, 2010 #16
    Why "dE = TdS - PdV is true even for irreversible processes"? Isn't the RHS bigger in irreversible processes?
  18. Sep 25, 2010 #17
    Nope, the principle is that because E, S, V are state variables, the equation dE = TdS - PdV must always hold. For example, if you draw to distinct points on your PV-diagram, then you could draw a reversible line connecting both or you could imagine an irreversible path: in either way the value of dE = TdS - PdV (integrated...) is only determined by where the two points are, because E,S,V are state variables (which, by definition, says they're only dependent on states, i.e. 'points'), and not the way you got there.

    To be clear, the formula dE = TdS - PdV is derived assuming you're doing it reversible, and indeed dW = -PdV and TdS = dQ is ONLY TRUE FOR REVERSIBLE CHANGES, because dW and dQ are not state variables, but once you change both at the same time only state variables remain and now the formula is always valid.
  19. Sep 25, 2010 #18
    I don't quite understand. dE=dQ-PdV is always true so dQ can be obtained from irreversible process. How come dQ=TdS where dQ is obtained from irreversible?
    Also i don;t think dW=-PdV is only TRUE FOR REVERSIBLE CHANGES.
  20. Sep 25, 2010 #19
    It is always true for reversible changes.

    It may be true for irreversible changes, it may be untrue or it may have no meaning, since P may be undefined during irreversible changes.
  21. Sep 25, 2010 #20
    why untrue? This is the infinisimal change.
    Even for strong friction, the P is the "NET" pressure, right?
  22. Sep 25, 2010 #21
    Think about a container of gas with a piston, subject to an external force equal but opposite to the force applied by the gas internal pressure.

    Now imagine the piston to be slowly withdrawn, maintaining the force equality stated above.

    This is the closest we can get to a reversible process. At all times the external force equals the internal force so we can say that

    Pexternal = Pinternal = P in your equation.

    Now imagine the piston to be withdrawn fairly rapidly.

    This time there will still be opposing forces acting across the piston, but there will be a problem. Turbulence will develop in the gas, and some of the energy (work) input will be lost in this. In this situation engineers usually take P in your equation = Pext and say that the turbulence loss is 'negligable'. However fundamental physicists cannot afford the luxury of this approximation.

    The process is irreversible.

    Now imagine a near instantaneous piston withdrawal.

    On the gas side of the piston there will be a void. The gas will temporarily still be in its old volume so it will be unevenly distributed through the new system boundary.
    We cannot now define a pressure to represent Pint and the equation is meaningless.
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