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Entropy Change of a Monatomic Ideal Gas Compressed Adiabatically

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data

    A monatomic ideal gas is compressed adiabatically. What is ΔS of the gas?

    n = 1.00
    vi = 10.00 L
    vf = 4.00 L
    Constant External Pressure = 50.00 atm


    2. Relevant equations

    ΔU = q + w = 0 + (-PextΔV)
    ΔS = ∫[(qrev)/T]dQ


    3. The attempt at a solution

    In order to calculate ΔS you have connect two thermodynamic states with a reversible path. I assumed that the ΔS of the system is 0 because dQ is 0 for adiabatic processes). Then second state focuses on ΔS of the surroundings that has an external pressure associated with it.

    I found Tf = 322.06 K by using the ΔU equation

    I am not sure what to do at this point, because every single attempt I made has led me to a negative ΔS.
     
  2. jcsd
  3. Apr 1, 2013 #2

    Andrew Mason

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    You cannot assume that the ΔS = 0 because the process is not reversible. Since external pressure is constant and internal pressure varies, the compression is not quasi-static. A reversible path between the initial and final states is not adiabatic - there is heat flow.

    I am not sure how you are calculating the final temperature. Are you given the initial temperature or the initial (internal) pressure of the gas?

    AM
     
  4. Apr 1, 2013 #3
    Yeah I started thinking it over and it wouldn't make sense. I thought I had read somewhere that if it is reversible it is 0. I am sorry I thought I had specified that initial temperature. It is at 298 K

    I used

    ΔU = -PextΔV
    ΔU = =-(50.00 atm)(4.00 L - 10.00 L) = 300 J
    300 J = n(3/2)R(Tf - Ti)
    300 J = (1.00 mole)(3/2)(8.314 J/Kmol)(Tf - 298 K)
    Tf = 322.06 K

    But the question is given that it is compressed adiabatically so that's where my problem lies
     
  5. Apr 2, 2013 #4

    Andrew Mason

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    If the process is reversible the ΔS of the system + surroundings will be 0. As I said, this is not a reversible process. To calculate entropy change you have to first find a reversible path between the beginning and end states. So you have to find the pressure, volume and temperature of the gas initially and at the end and then find a reversible path from the initial to final states.

    Whoa! You need MKS units to get an answer in Joules: Pressure is in Pascals and Volume is in m3. 1 atm = 101.325 x 103 Pa. and 1 L = 10-3 m3.

    Your method is ok. You are just out by a few orders of magnitude with the units. (R = 8.314 J/mol-K). Tf is much higher than 322K.

    Apart from the units, I am not sure where you are having a problem. Adiabatic processes do not have to be quasi-static. Compression in an internal combustion engine is a good example of an irreversible adiabatic compression.

    Welcome to PF by the way!

    AM
     
  6. Apr 2, 2013 #5
    I came to that realization not too long ago thankfully. I am glad for the clarification. I thought I was going mad for a second. Thank you so much!
     
  7. Apr 2, 2013 #6
    You still need to get the entropy change. You have all the information you need about the initial and final states. You just need to figure out a reversible path from the initial to the final state and calculate dq/T for that path, or (less preferrably) you need to remember the equation for the change of entropy for an ideal gas in terms of the volume ratio and the temperature ratio. Have you seen an equation like that?

    Chet
     
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