1. Not finding help here? Sign up for a free 30min 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 - ideal gas properties

  1. Nov 20, 2011 #1
    Hey guys,

    I'm working on calculating the enthalpy, entropy and internal energy of substances. treating them as an ideal gas. I wonder if I could be pointed in the right direction with some calculations.

    Enthalpy change:
    h2-h1=∫cp dT+ ∫[v-T(∂v/∂T)P]dP

    At ideal gas the pressure integral goes to zero.
    So we have just the integral of Cp with respect to T.

    All correct so far?

    Its with the entropy and internal energy where I get confused.

    ds = (Cp)dT - (∂v/∂T)P dP

    is there a way to simplify this further at ideal gas to make it easily calculated ?

    I guess internal energy can be calculated from h using h = u + Pv.

    Thanks very much

    Look forward to any replies.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 20, 2011 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I am not sure how you get this. If H = U + PV, then dH = dU + PdV + VdP and ΔH = ΔU + ∫PdV + ∫VdP = ΔQ + ∫VdP

    So, if pressure is constant, ∫VdP = 0, so ΔH = ΔQ = ∫CpdT

    Not in general. This is true only if P is constant.
    Since U and PV are state functions, ΔU and Δ(PV) are independent of the path between two states so they are the same whether the path is reversible or irreversible. For a reversible path, ΔH = ΔQ + ∫VdP = ∫TdS + ∫VdP. Consequently, this must be true for all paths.

    So you can use: dH = TdS + VdP

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermodynamics - ideal gas properties
Loading...