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Thermodynamics - heat, internal energy, work

  1. Dec 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Heat flows into an ideal gas at a constant volume. The pressure increases from 1.5 atm to 5.5 atm. Next the gas is compressed at constant pressure from 5.0 L to 2.5 L and goes back to its original temperature.

    1.What is the total work done on the gas in the process?
    2.What is the total change in internal energy?
    3.What is the total heat flow of the process?

    2. Relevant equations

    U = Q + W
    W = -P*ΔV

    3. The attempt at a solution

    1. No work is done in the first process because the volume is constant. The work done on the gas in the second process is W = -(5.5 atm) * (2.5 L - 5.5 L) = 1390J

    2. The change in internal energy is 0 because there is no net change in temperature.

    3. 1390J of heat must have left the system because U = Q + W.

    Is this right?
     
  2. jcsd
  3. Dec 23, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Ampere.

    Your process looks right, but I don't agree with your conversion from L-atm to Joules.

    [EDIT: It's not the conversion, it's a typo in your equation W = -(5.5 atm) * (2.5 L - 5.5 L) = 1390J
    where you wrote 5.5 L instead of 5.0 L.

    I agree with your answer now.]
     
  4. Dec 23, 2013 #3
    Yes, I meant (2.5 L - 5.0 L). Thanks for the confirmation!
     
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