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Question about u and h and Its Differential Changes for Ideal Gases

  1. Feb 24, 2014 #1
    Hello

    1. I was wondering why internal energy is usually expressed as a function of temperature and specific volume and enthalpy a function of temperature and pressure (ie why is u(T,v) and h(T,p)) and some other set of two properties?

    2. For du = [itex]\frac{∂u}{∂T}[/itex]dT + [itex]\frac{∂u}{∂v}[/itex]dv and dh = [itex]\frac{∂h}{∂T}[/itex]dT + [itex]\frac{∂h}{∂p}[/itex]dp, why are the [itex]\frac{∂u}{∂v}[/itex] and [itex]\frac{∂h}{∂p}[/itex] terms zero for ideal gases only?

    3. Why are the specific heats for ideal gases a function of temperature only? For real gases does this become cv(T,v) and cp(T,p)?

    Thanks very much
     
  2. jcsd
  3. Feb 25, 2014 #2

    DrClaude

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    Staff: Mentor

    Are these homework questions?
     
  4. Feb 25, 2014 #3
    No, I am trying to brush up on my thermodynamics knowledge and found myself wondering about these questions that I hadn't considered before. I am currently not in any type of school.
     
  5. Feb 25, 2014 #4

    DrClaude

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    Staff: Mentor

    The first law can be written as
    $$
    dU = Q + W
    $$
    where ##Q## is the heat entering the system, and ##W## the work performed on the system. Using
    $$
    dS = \frac{Q}{T}
    $$
    and writing compression/expansion work on an ideal gas as ##W = -P dV##, you get
    $$
    dU = T dS - P dV
    $$
    so it is "natural" to take ##U## as a function of ##S## and ##V## (not of ##T##, as you wrote). Enthalpy is defined as
    $$
    H \equiv U + PV
    $$
    from which you get
    $$
    dH = dU + P dV + VdP = TdS + V dP
    $$
    (which is essentially a Legendre transformation), so ##H## is a function of ##S## and ##P## (again, not of ##T## as you wrote).

    For non-ideal gases, the facts that you can't consider them as point particles and that there are interparticle interactions mean that the energy will depend in a more complicated way on such things as volume. Check for instance a van der Waals gas.

    The specific heats for an ideal gas are constant. For a real gas, they are mostly a function of temperature, but I guess they must be slowly varying functions of volume and pressure also.
     
  6. Feb 25, 2014 #5
    So where/how does temperature come in since the definition of cp and cv come from derivative of u and h as a function of T?

    I guess my confusion is that, how does increasing pressure/specific volume without changing temperature not change the internal energy of the an ideal gas?

    I'm a bit confused by this concept, in my book for, c_p for ideal gases is expressed as a polynomial in terms of temperature c_p/R = A+BT+CT^2+DT^3+FT^4 and the coefficients are tabulated for many ideal gases?
     
  7. Feb 25, 2014 #6
    The internal energy of a real gas is determined by the kinetic energy of the molecules plus the potential energy of interaction between the molecules. In the limit of ideal gas behavior (high specific volume, low pressure), the molecules are far enough apart that the potential energy of interaction between the molecules becomes negligible. This leaves only the kinetic energy, which is a function only of temperature. Since h = u + pv, and, for an ideal gas, pv = RT, h is also a function only of temperature for ideal gases.
    Most physicists regard what they define as an ideal gas as a material having constant heat capacity. We engineers regard the heat capacities of ideal gases to be functions of temperature (which is the way that real gases behave in the limit of high specific volume and low pressures). At pressures beyond the ideal gas range, enthalpy and internal energy of real gases are functions not only of temperature but also pressure. Since cv is defined as the partial derivative of u with respect to T at constant v, cv is also a function of pressure (or specific volume). Since cp is defined as the partial derivative of h with respect to T at constant p, cp is also a function of pressure (or specific volume).

    Chet
     
  8. Feb 25, 2014 #7

    DrClaude

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    Staff: Mentor

    My answer was based on an ideal gase defined as
     
  9. Feb 25, 2014 #8
    That's what I thought. We engineers define what we call an ideal gas differently (I invite you to check our texts and literature). There have been other threads on PF in which engineers and physicists have compared notes and finally realized that this difference exists in what they respectively call an ideal gas. Apparently, the OP is learning the engineering version.

    Chet
     
  10. Feb 25, 2014 #9
    So in engineering, cp and cv for ideal gases are actually models of real gases at high temperature and low pressure where p and v have a negligible effect? Is it purely for the sake of convenience/situation that usually we see u(T,v) and h(T,p) and of other state variables? Lastly, what are the differences between the definition of an ideal gases in physics and engineering?

    Thanks very much
     
  11. Feb 25, 2014 #10
    Not necessarily at high temperature, but, yes, at low pressure.

    In my judgement, yes.

    As far as I know, the only difference is related to the temperature dependence of the heat capacity that we discussed (and the related functionalities of u and h on T).

    Chet
     
  12. Feb 26, 2014 #11

    Maylis

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    Gold Member

    The nice thing about using it as a function of T and P is that those quantities are a lot easier to measure than entropy or volume.
     
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