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Change in internal energy at constant pressure?

  1. Oct 8, 2013 #1
    Change in internal energy at constant pressure??

    Hello everybody,
    I am stuck with a concept in thermodynamics. We know that ΔU=CvΔT. But what if we want the change in internal energy at constant pressure?


    ΔU→Change in internal
    Cv=heat capacity at constant volume
     
  2. jcsd
  3. Oct 8, 2013 #2
    The given equation won't help... As one can see it is related to internal energy change at constant volume... The equation for enthalpy change in terms of internal energy change and work done might help.
    i.e. ΔH=ΔU+PΔV
    where P - the value of constant pressure applied
    ΔV - change in volume

    Regards
     
  4. Oct 8, 2013 #3
    But my text book says that ΔU=CvΔT is always true for any process.Why it is like that?How can we use the Cv for processes other than Constant volume??
     
  5. Oct 8, 2013 #4
    Why don't you try to calculate the change in internal energy for an isobaric process?
    This is the best way to convince yourself that the formula "works" for constant pressure too.
     
  6. Oct 8, 2013 #5
    For the case of an ideal gas, the internal energy is independent of pressure, so the same equation applies. If the material is not an ideal gas, one can obtain the effect of pressure on U from the P-V-T behavior of the material. But you need to learn the derivation of this pressure contribution.

    Chet
     
  7. Oct 8, 2013 #6
    You can understand this in three steps.

    1. ΔU = (f/2)nRΔT .This applies to all kinds of processes ,when dealing with ideal gases.

    2. Next consider an isochoric process(constant volume process) .

    Using FLT, ΔQ = ΔU + ΔW

    Now, ΔQ = nCvΔT ,ΔU = (f/2)nRΔT and ΔW = 0

    So,we have nCvΔT = (f/2)nRΔT

    or, Cv = (f/2)R .i.e molar heat capacity at constant volume for a gas is a constant .

    3. Now come back to isobaric process .

    As we have noted in point 1 , ΔU = (f/2)nRΔT , in an isobaric process for an ideal gas .But Cv for the gas (even though the gas is undergoing a constant pressure process) is equal to (f/2)R .

    So we have ΔU = nCvΔT.

    Thus ,we see that this relation ΔU = nCvΔT applies to all kinds of processes involving an ideal gas ,just like ΔU = (f/2)nRΔT holds.

    Hope that helps
     
  8. Oct 14, 2013 #7
    Tanya your name sounds like you are india,are you??
    I am also from india.
     
  9. Mar 13, 2016 #8
    I was confused that Cv is the heat needed to raise the temperature when volume is constant, so, Cv can only be used when volume is constant. But now I ve understood that to find out internal energy Cv is used regardless of any process, as, internal energy is the function of temperature only (depends on temperature only), and is independent of other properties including pressure.
     
  10. Mar 13, 2016 #9
    Internal energy is independent of pressure only for ideal gases and for incompressible solids and liquids. For real gases, internal energy also depends on pressure.
     
  11. Mar 13, 2016 #10
    Thanks for your reply. I need more help from you. we say that internal energy is function of temperature only, but, when I was calculating air properties by an online property calculator, I noticed that internal energy changes with change in pressure, while keeping constant temperature as you said. so, how to clear this concept? I am still confused.
    Can we find the "amount of heat required to raise the piston to certain height, with constant pressure, adiabatic process ( air is the substance, initial and final temperatures and volumes are given)" by using (1) concept of change in enthalpy or using the formula (2) Q=delta U + delta W = mCv*deltaT+P*deltaV or by the formula (3) Q=mCp*deltaT ?? which one is the correct way? I have noticed a little difference between
    2nd and 3rd method
    . 20160314_004636.jpg
     
  12. Mar 13, 2016 #11
    I don't quite understand this question. For a real gas like air, the internal energy is a function not only of temperature, but also of pressure. The dependence on pressure increases as pressure increases.
    .
    You indicated in your problem specification that the it is an "adiabatic process." But this process certainly isn't adiabatic if heat is being supplied to the gas. In any event, for this process, the amount of heat is equal to the change in enthalpy, irrespective of whether the gas is an ideal gas. So method (1) will always be correct, if you have air tables with enthalpy values. Methods (2) and (3) assume you are dealing with an ideal gas for which both U and H are independent of pressure (assuming that, by Cp, you mean the value of the heat capacity in the limit of low pressures). Method 3 would be correct if the value of Cp used were evaluated at the operating pressure P, rather than at pressures approaching ideal gas behavior)
     
    Last edited: Mar 13, 2016
  13. Mar 17, 2016 #12
    Yes you are right. Actually, heat is not supplied though the walls of cylinder, but it is the electric heater which is doing electric work. Well, thank you very much.
     
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