Recent content by rajendra123

  1. R

    Gibbs free energy -- mathematical expression

    okay. yes it's the maximum non-PV work for closed system.
  2. R

    Gibbs free energy -- mathematical expression

    So, you mean to say, if a chemical reaction occurring in closed container releasing ΔH amount of heat somehow we convert (ΔH-Tsystem*ΔSsystem) to work i.e. without interacting with surrounding. This work is ΔG?
  3. R

    Gibbs free energy -- mathematical expression

    I have found this link which says "The whole Gibbs relationship or function is about entropy change." http://2ndlaw.oxy.edu/gibbs.html
  4. R

    Gibbs free energy -- mathematical expression

    Okay, so to calculate ΔG we have to have such a device which will transfer heat at such a rate that Tsystem will be constant and heat transfer shall take place between infinitesimal temperature difference.
  5. R

    Gibbs free energy -- mathematical expression

    If Tsystem = Tsurrounding then how can heat transfer take place?
  6. R

    Gibbs free energy -- mathematical expression

    YES, both of you are right, though I didn't mean that, it's also a way of showing it.
  7. R

    Gibbs free energy -- mathematical expression

    First of all thank you for your quick reply Friends. Do you mean that Gibbs Free Energy is calculated for ISOLATED systems only? If the way I have mentioned in the question is not right then how else could we get mathematical relation. Yes, but to maintain temperature constant there must be...
  8. R

    Gibbs free energy -- mathematical expression

    I am not able to understand the mathematical expression of "change in Gibbs free energy", For a chemical reaction occurring at constant temperature and constant pressure, (ΔS)total = (ΔS)system + (ΔS)surrounding Considering that reaction is exothermic, ΔH be the heat supplied by system to...
  9. R

    Confusion on entropy change calculations for irreversible process

    Hello dear Chet , I would like to join the discussion, At equilibrium both slabs will reach a temperature (Hot slab will lose heat and cold body will gain heat) Change in entropy of hot body = C ln(T/Th1) Change in entropy of hot body = C ln(T/Tc1) Where, Th1 and Tc1 are initial temperatures...
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