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Internal Energy Thermodynamics

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    CO2 is at P=3atm, T = 295K and V=1.2m3.
    It is isobarically heated to T = 500K.
    Find ΔU and ΔH

    2. Relevant equations
    dU = cpdT

    3. The attempt at a solution
    I am having a hard time in general in this class. I understand that in this problem, ΔP = 0. Does this mean that there must be a ΔV? Also, I am having a hard time understanding how to use cp if in this case, it changes with temperature. I do not think I need to use integration, but I am not sure.
  2. jcsd
  3. Feb 20, 2017 #2


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    Staff: Mentor


    If ##c_p## changes with temperature, then you need to integrate.
  4. Feb 20, 2017 #3

    Andrew Mason

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    No. dU = nCvdT

    CO2 can be treated as an ideal gas, so: PV = nRT. If T increases and P stays the same, what happens to V?
    Cp is the heat capacity at constant pressure. So multiplying Cp by the change in temperature and number of moles gives you .......?. Since H = U + PV, can you determine the ΔH? (Hint: for constant pressure changes, how is it related to ΔQ?). Use ΔU = nCvΔT to determine change in internal energy.

  5. Feb 20, 2017 #4


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    dU = cpdT wasn't that dH :smile: ?
    Yes. Ideal gas law is good enough.
    Either you integrate (stepwise), or you look it up. it indeed changes with T

    That's OK. With thermodynamics the sequence is: completely disoriented, then gradually more confident, and -- by the time you Master it -- complete disorientation again :smile:

    [edit] wow, three responses !
  6. Feb 20, 2017 #5
    Okay so I am familiar with the PV=nRT formula. In our class, we usually use PV=mRT, and use the specific R for the gas in question. Would my first step be to calculate the mass of CO2 in state 1 using this formula?
  7. Feb 20, 2017 #6

    Andrew Mason

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    Yes. Does the mass change?

  8. Feb 20, 2017 #7
    No I wouldn't think that mass would change. I went ahead and calculated the volume of the second state, although I'm not sure if it was needed. Would I then have to use dU = mcvdT ? And since it cv is not constant, would I need to integrate both sides of this equation?
  9. Feb 20, 2017 #8
    I am confused because I do not understand which value to choose for Cp if it changes with volume. Even if I were to integrate I would have:
    ∫dU = m∫Cp dT
    But how can I do that? The Cp is really throwing me off, and I have no idea how to use it in this problem. The same goes for Cv, since that also changes with temperature. Any help?
  10. Feb 20, 2017 #9
    For an ideal gas, Cv and Cp are both functions only of temperature. They are related to the changes in internal energy and enthalpy by:
    If the heat capacities are functions of temperature, then, to be exact, you need to integrate. But, often, if the temperature interval is small, the heat capacities can be approximated as being constant.
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