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!

Heat Capacity and First Thermodynamic Law

  1. Mar 18, 2008 #1
    I have a question about the deduction of First Thermodynamic Law. The book that I have is written by Paul A. Tipler and Gene Mosca and it is called Physics: For Scientists and Engineers.

    The way to deduct it is given here:

    @constant volume,
    Because W=0, E2-E1=Cv(T2-T1)
    After all, Cv=d(E2-E1)/d(T2-T1)

    @constant pressure,
    Given that E2-E1=Qp-P(V2-V1) and Qp=Cp(T2-T1)
    We have Cp(T2-T1)=(E2-E1)+P(V2-V1)
    Now, the author replaces E2-E1 with Cv(T2-T1), I cannot understand this because he incites something under constant volume into a formula under constant pressure.

    My question is what determines the internal heat in a system is defined by CvP, which can even be used in a situation in which a different condition is given.

    Please help me if you know! Thank you in advance!
  2. jcsd
  3. Mar 18, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi FriedrichLuo, welcome to PF. Yours is a very common question, as it is disconcerting to see a "constant-volume" parameter being used in a constant-pressure process. But energy is a state variable (its value is process independent), and for this system the relationship E2-E1 = Cv(T2-T1) holds for all processes.
  4. Mar 18, 2008 #3
    Hey, Mapes. Thank you for your reply! I come down to this after reading your post:

    dU=dQ+dW, dU is a state parameter and dQ and dW are process parameters. Therefore, dQ(or Cv(T2-T1) holds for any situations where process is involved. And dU is a result when a new state is reached. In other words, dQ and dW are what is really happening; dU is only concept to show the resultant of two true processes.

    I hope I get it right. :D
  5. Mar 18, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This sounds like a fine way of thinking about things, but make sure you don't mix your differential and finite values. You can say

    [tex]\delta Q=C_V\,dT[/tex]



    but not mix them.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook