Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ideal gas PV diagram heat transfer process

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    There is a Pressure-Volume diagram of an ideal gas. The processes make up a (rough) triangle. At constant pressure, we have point A to point C to the right of it to point D to the right of C. Then above A we have point B. C goes to B and D goes to B.. thus forming something that resembles a triangle. C to B is adiabatic while D to B is isothermal.


    2. Relevant equations
    deltaE = Q + W


    3. The attempt at a solution
    My question is just on one part of a larger problem: The professor in class deduced that the energy change from point A to B (vertical) is equal to the energy change from point A to D (horizontal).

    Now, why is that? I understand that AD is doing more work than AC, which does more work than AB (which does no work since it doesn't have a change in volume). From just that work statement, he concluded that the deltaE(AB) = deltaE (AD).

    I believe this small part will help unravel the confusion in the rest of the problem. Thanks :)
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You seem to have one triangle: ABD the sides of which are straight lines and an adiabatic curve between B and C, C being a point between A and D.

    I will assume you are looking at the change in internal energy. Do they give you the values of the changes in P from A to B and the changes in volume from A to C and A to D? If not, you cannot answer the question.

    The internal energy is a function of temperature. So just work out the temperatures at B and at D using PV=nRT. If they are the same, then the internal energy is the same.

    You know that the internal energy at B is greater than at A or C since the temperatures are lower at A and C (we know it is lower at C since AC is adiabatic, so there is work done by the gas in going from A to C but there is no heat flow into the gas).

    AM
     
  4. Sep 7, 2011 #3
    We are not given any numbers of any kind. It's meant to be a conceptual question and should be able to be answered conceptually, according to the professor.
     
  5. Sep 7, 2011 #4
    AB is a constant volume process where delta U=Q=Cv(T2-T1)

    AD is a constant pressure process where delta U=Q-W; Q=Cv(T2-T1)+R(T2-T1) and W=R(T2-T1)

    R is a constant
     
  6. Sep 8, 2011 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Sorry, I was confused by your description of the graph. It is an L shaped graph with two curves from the base to the top point (B). I missed the part about the graph from D to B being isothermal.

    Since B and D are at the same temperature, there is no difference in internal energy between D and B. So going from A to B or A to D results in the same change in internal energy.

    AM
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook