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!

How to better understand thermodynamics? With statistical mechanics?

  1. Jun 14, 2007 #1
    Does anyone understand thermodynamics? There are so many terms that I feel that I am doing the maths but not really understanding the physics.

    Is it better to do stuff from a stat physics way (which makes more sense) and derive the thermodynamic relations from there?
  2. jcsd
  3. Jun 15, 2007 #2
    I was a bit stunned as I was reviewing my notes from a 3 week break. I then reviwed 1st year thermodynamics material and the stuff came back to me a bit and now I am back into it again. So there is physical understanding after all.
  4. Jun 15, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes - people, particularly physicists and many engineers, understand thermodynamics. It does help to have a firm grounding in basic physics, particular an understanding of energy/work and momentum, and force.
  5. Jun 15, 2007 #4
    It seems that the concept of potential energy doesn't arise in thermodynamics because of the huge number of atoms in the system. Or is it because thermodynamics deals with equilibrium situations and so no foce is acted on the system (i.e. forces all cancel).

    Or is it because the system is not usually conserved as the 1nd law suggests that total energy of the system can change. So keeping a potential energy in the system would be meaningless. It is only meaningful to account for the kinetic energy of molecules which comes into full fruition when relating it to the temperture in the system.
    Last edited: Jun 15, 2007
  6. Jun 15, 2007 #5
    Huh? That's not the 1st law that I remember.
  7. Jun 15, 2007 #6
    Oh, dear. Please open your thermo book and reread it cover to cover.
  8. Jun 15, 2007 #7
    what about studying thermodynamics from a chemistry perspective?

    that might help you get a more conceptual understanding.
  9. Jun 15, 2007 #8


    User Avatar

    Staff: Mentor

    I'm not sure what the number of atoms has to do with anything, but if you mean gravitational potential energy, it can come into play but doesn't often because you don't generally have large changes in elevation with basic thermodynamic cycles.
    Huh? Total energy of an isolated system must be conserved. That's what the first law says. If you drop a rock off a cliff, you convert potential energy to kinetic, during the fall, the total energy never changes.

    If you mean potential energy in terms of a compressed gas, it's the same. You convert it to kinetic energy (minus the ever-present entropy), but the total stays the same.
  10. Jun 15, 2007 #9
    Change in U=Q-W. So yes, the total energy of the system can change according to how much heat and work that has occured in the larger system (the system is a subset of the larger system). However the total energy in the larger system (i.e universe) is usually constant.
  11. Jun 15, 2007 #10
    The book suggested that when the number of atoms is large, the individual trajectories of atoms are discarded. Normally in smaller system accounting for a countable number of particles with forces, it is convenient to use kinetic and potential energy if the system is isolated.

    The 1st law is an extension of the isolated system and considering a system inside a larger system, i.e. bath or universe. Hence the system as oppossed to the bath is not isolated. That is much more realistic wouldn't you say?

    If the system dosen't interact with the surrounding than the 1st law is reduced to the isolated system's case of total energy = constant.

    So this would be the case of a gas trapped inside a cyclinder without any outside contact. However W can be nonzero as the gas expands and contracts but is 0 if the system is inside a vacuum. Q=0 always in this case.

    In fact it makes a lot of sense because suppose we had an oscillating spring in an air filled room. If we let it oscillate by starting its position from a non equilibrium position, it will eventually slow down because it is doing work on the air molecules hence losing total energy. It is directly losing kinetic energy thereby not springing to as far a distance as before hence decreasing its potential energy as well and the cycle spirals until no energy is left.
    Last edited: Jun 16, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook