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!

Internal Energy and Heat

  1. Mar 8, 2008 #1
    [SOLVED] Internal Energy and Heat

    1. The problem statement, all variables and given/known data
    Ten joules of heat energy are transferred to a sample of an ideal gas at constant temperature. As a result, the internal energy of the gas:
    a) increases by 10 J
    b) increases by less than 10 J
    c) increases by more than 10 J
    d) remains unchanged


    2. Relevant equations
    delta U = Q + W


    3. The attempt at a solution
    I thought the internal energy of the gas remains unchanged since internal energy is proportional to absolute temperature and the temperature remains the same. Can anybody confirm this please?
     
  2. jcsd
  3. Mar 8, 2008 #2

    olgranpappy

    User Avatar
    Homework Helper

    confirmed.
     
  4. Mar 8, 2008 #3
    So the total internal energy remains unchanged? Then what changes?
     
  5. Mar 8, 2008 #4

    olgranpappy

    User Avatar
    Homework Helper

    Are you asking because you don't know, or for pedagogical reasons?

    If the former: heat and work both are non-zero, the system does work. If the latter: the OP didn't ask what changes he just asked for confirmation.
     
  6. Mar 8, 2008 #5
    I ask because I'm confused. So because the gas remains at a constant temperature, its internal energy remains unchanged. By the first law, that means that W is non-zero and so either the gas's pressure decreased and/or its volume increased?
     
  7. Mar 9, 2008 #6

    olgranpappy

    User Avatar
    Homework Helper

    Yeah, that's correct, since the temperature is constant we must have zero change in energy and so the heat Q which flows into the gas must exactly equal the work W done by the gas.
    [tex]
    Q=W_{\rm by}
    [/tex]

    In most circumstances one uses the expression [itex]dW_{\rm by}=pdV[/itex]. Let's assume--as usual--that the previous equation holds. Then, although it's not generally true, in this case (the case of ideal gas of fixed number of particles at constant temperature) both the volume must increase and the pressure must decrease... and it's not too hard to calculate exactly how much the volume must increase and pressure decrease in terms of Q and the temperature and the number of particles and the initial volume. ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Internal Energy and Heat
Loading...