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Change In Internal Energy Of An Ideal Gas

  1. Dec 5, 2015 #1
    • OP warned about not using the homework template
    An ideal gas is compressed from a volume of Vi = 4.50 L to a volume of Vf = 3.00 L while in thermal contact with a heat reservoir at T = 295 K as in the figure below. During the compression process, the piston moves down a distance of d = 0.120 m under the action of an average external force of F = 21.0 kN.

    (a) Find the work done on the gas, I Solved That its 2.5KJ.

    ideal gas is PV = nRt
    work done is W = F(yf - yi)
    so work is W = 21000N*(.120m)
    W = 2520J or 2.5 KJ


    (b) Find the change in internal energy of the gas.

    (c) Find the thermal energy exchanged between the gas and the reservoir.

    i tried to solve b and c but with no pressure or moles i have no formula that can get me the change in internal energy.

    for b the change is internal energy is given by: U = Q + W
    i have the W as 2.5kJ but i am still Missing Q the energy transferred to the Gas by Heat
     
    Last edited: Dec 5, 2015
  2. jcsd
  3. Dec 5, 2015 #2

    DrClaude

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    Don't you have an equation for the internal energy of an ideal gas?
     
  4. Dec 5, 2015 #3
    yes its (3/2)nR(tf - ti)
    but i dont have moles and i cant get them from the ideal gas law because i dont have the pressure
    and i cant get it from the W = PAChangeinY because i dont have A
     
  5. Dec 5, 2015 #4

    DrClaude

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    I had missed that you didn't have the number of moles. Could you make some assumption, like that the system is initially at atmospheric pressure?
     
  6. Dec 5, 2015 #5
    im not sure that is the whole question but for the sake of learning to do this problem i can

    so if i assume that the pressure is 101325PA then
    101325PA * .0045 m3 = n8.31T where t is still unknown
     
  7. Dec 5, 2015 #6

    DrClaude

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  8. Dec 5, 2015 #7
    so because it is contact with the heat reservoir it can be assumed that it is at the temperature always?
     
  9. Dec 5, 2015 #8

    DrClaude

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    Yes. That's in essence the definition of a "reservoir."
     
  10. Dec 5, 2015 #9
    so that means that 101325PA * .0045 m3 = n8.31(273K)
    gives n = .185
    then (3/2)(.185)*8.31*295 = 683.94
    but because the temperature never changes does that still hold ?
    allowing me to solve it all again with vf and then subtract both finding the answer
     
  11. Dec 5, 2015 #10

    DrClaude

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    Sorry, I'm slow tonight. I just realized that you are asked for the change in internal energy. So what happens when you calculate ΔU = (3/2)nR(Tf- Ti)? Hint: you don't need to know n.
     
  12. Dec 5, 2015 #11
    i am not sure because if it is contact with the heat reservoir that would that mean that temperature doesn't change and it would be 0
     
  13. Dec 5, 2015 #12

    DrClaude

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    Exactly, the internal energy of the gas doesn't change.
     
  14. Dec 5, 2015 #13
    Right so if the internal energy doesn't change that means that for part c Q = W or 2.5KJ

    Thank you very much for your help
     
  15. Dec 5, 2015 #14

    DrClaude

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    Your welcome. And sorry again, that should've been quicker.
     
  16. Dec 6, 2015 #15
    Not exactly. It means that the temperature of the gas in the final thermodynamic equilibrium state of the system (295) is the same as in the initial thermodynamic equilibrium state of the system (295). During the transition from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state, the temperature of the gas does not have to be uniform at 295 throughout. In fact, even the average temperature of the gas can differ from 295 during the transition.

    Chet
     
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