1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics Enthalpy Problem

  1. Apr 19, 2014 #1
    A frictionless piston cylinder device contains a saturated liquid (refrigerant-134a). The refrigerant is heated until it reaches to some temperature. Determine the work done and heat transferred to the refrigerant during this process.

    I just want to make sure that I understand this topic. I will have to use tables for this.


    (a)The saturated liquid is turned into super heated gas.
    (b) Heat transferred is the change in the internal energy u.
    (c)Work done is the change in ΔP*V where P is pressure and V is volume

    h=u+PV

    Are those 3 assumptions correct?

    Thanks
     
    Last edited: Apr 19, 2014
  2. jcsd
  3. Apr 19, 2014 #2
    Is the pressure held constant during the process or is the volume held constant?

    Chet
     
  4. Apr 19, 2014 #3
    Pressure is constant as the piston is free to move.
     
  5. Apr 19, 2014 #4
    Then assumptions b and c are incorrect.

    Chet
     
  6. Apr 19, 2014 #5
    Why not a? There is a rise in temperature. So, vaporization is process completed. Or is it because my explanation wasn't adequate?
     
  7. Apr 19, 2014 #6
    Assumption a is OK.
     
  8. Apr 19, 2014 #7
    So could you tell me where I am wrong? I believe my assumptions are correct. Energy transfer by work is in terms of force(PV) and by heat is in terms of temperature (internal energy, or is it enthalpy?)
     
    Last edited: Apr 19, 2014
  9. Apr 19, 2014 #8
    It's enthalpy for this problem.

    The first law applied to this problem gives:

    ΔU=Q-PΔV

    What does that tell you about what Q is equal to? What does that tell you about what the work is equal to?

    Based on this information, how would you go about using your tables to answer the problem questions?

    Chet
     
  10. Apr 19, 2014 #9
    I see what you mean. Thanks a lot for making me realize my mistake. But there is one more thing that I am confused: The concept of enthalpy. It is hard to define it that's why I have difficulties of using it.

    I mean, if u is the internal energy, which is a function of Temperature, how can you add that up with PV, which is again a function of Temperature (PV=RT). So u and PV are not independent from each other. Moreover, increase/decrease in temperature will have the same effect on each of them. It comes up to my mind as they are two different reflections of the same thing. Then why adding two same things to each other?
     
  11. Apr 19, 2014 #10
    It isn't anything fundamental. The fundamental function is U. But, the enthalpy has been found to be very convenient function to work with in many types of problems.

    Actually, an increase/decrease in temperature will have a different effect on each of them. For an ideal gas,

    dU = CvdT

    but

    dH = dU + d(PV)= CvdT + RdT = (Cv+R)dT = CpdT

    As you get more experience solving thermo problems, you will get a better appreciation of why it is often much more convenient to work with the enthalpy.

    Chet
     
  12. Apr 19, 2014 #11
    You have been really helpful. Thanks for you time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thermodynamics Enthalpy Problem
  1. A thermodynamics problem (Replies: 18)

  2. Thermodynamics Problem (Replies: 5)

  3. THermodynamics problem (Replies: 13)

Loading...