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Isochoric and Isobaric (Ch.15 problem #26 Wiley+)

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Chapter 15, Problem 26

    The drawing refers to 5.30 mol of a monatomic ideal gas and shows a process that has four steps, two isobaric (A to B, C to D) and two isochoric (B to C, D to A). (a) What is the work done from A to B? (b) What is the heat added or removed from B to C? (c) What is the change in internal energy from C to D? (d) What is the work done from D to A?




    2. Relevant equations

    W= nRT(Vf/Vi)


    3. The attempt at a solution

    W= 5.3mol * 8.31 * 800K *(Vf/Vi) this is where im stick I don't know how to find the volume. I tried using V=nRT/P but I don't know the pressure?

    Thanks in advance, the homework is due tonight.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 19, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF trinot,
    No you don't know the actual value of the pressure, but you know that it remains constant throughout the process :wink:
     
  4. Apr 19, 2008 #3
    I am still not sure what to do, does this mean I can ignore it and just use V=nRT/P?

    Thanks!
     
  5. Apr 20, 2008 #4

    alphysicist

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    Homework Helper

    Hi trinot,

    I believe you are using the wrong equation for the work done by the gas. For an isothermal process the work done by a gas is

    [tex]
    W = nRT \ln\left({\frac{V_f}{V_i}\right)
    [/tex]

    but these processes are not isothermal.

    For an isochoric process the work done is zero; for an isobaric process the work done by the gas is

    [tex]
    W= P (V_f - V_i)
    [/tex]

    Since you know n and T for all four points on the graph you can find the values of ([itex]PV_f[/itex]) and ([itex]PV_i[/itex]).
     
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