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How to find added thermal heat in monoatomic gas?

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data
    IMG_1108.png

    IMG_1107.png


    for number 3,4,5 i'm still trying


    2. Relevant equations
    PV/T = PV/T
    q = ΔU + W
    W = P ΔV

    3. The attempt at a solution

    (3) I used PV/T = PV/T to find the ΔT for each process
    for A→B I find PV/TA = P3V/TB ----- TB = 3TA (T increase)
    for B→C I find P3V/TB = 4P3V/TC ----- TC = 4TB (T increase)
    for C→D I find 4P3V/TC = 4PV/TD ----- TD = ⅓TC (T decrease)
    for D→A I find 4PV/TD = PV/TA ----- TA = ¼TD (T decrease)

    so I drew a conclusion that among 4 process , gas received from outside max at process B→C
    are the answer and the reason right?

    (4) at process B→C there's no ΔV. So, W = 0
    q = ΔU + 0
    q = the amount of heat added or removed from system
    So, the question for (4) Is to find q at process B→C ?
    q for monoatomic gas is 3/2nKT
    but what's the n (mol) known?
    can you give me some clue to answer (4)?

    (5) now the heat that gas emits,
    I thought there are 2 W :
    W at A→B = P.(3V - V) = 2PV
    W at C→D = 4P(V-3V) = -8PV (minus ?)
    can you give me some clue to answer (5)?
     
  2. jcsd
  3. Jun 10, 2017 #2

    I like Serena

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    Homework Helper

    Hi Helly,

    Let's bring in a couple more equations.
    For an ideal gas we have PV=nRT.
    And for a monatomic ideal gas we have U=3/2 nRT, where R is the molar gas constant (R=8.31 J/K.mol).

    The question asks about the thermal heat, which is q, and not the change in temperature.
    Can we calculate q for each of the processes?

    The number of moles is given to be 1 mol.
    And we can use that PV=nRT for an ideal gas (I'm assuming that your K is the same as R).

    The net heat is equal to the area inside the graph, which is also equal to the total work done during the process.
    Can we find those?
     
    Last edited: Jun 10, 2017
  4. Jun 10, 2017 #3
    Thanks Serena for all the clues and explanation, I'll try it :)
     
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