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Homework Help: Find the Mass of the Air, and the Work and Heat Transfer

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2m3. The air undergoes a process to a state where it pressure is 1 bar, during which the pressure-volume relationship is pv = constant. Assuming idea gas behavior for the air, determine the mass of the air, in kg and the work and heat transfer, each in kJ

    2. Relevant equations


    W = ∫p dv

    3. The attempt at a solution

    I found the constant C to be:

    pivi = (200,000 Pa)(2m3)

    C = 400,000 Pam3

    vf = C/pf

    Hence vf = 400,000 Pa m3/100,000 Pa

    vf = 4m3

    W = ∫P dv = ∫24C/v dv

    W = 277258.9 kJ

    Mass of air:

    pv =nrt

    n = (200,000 Pa)(2m3)/(8.314 J/K mol)(300K)

    n = 160.372 Pa m3mol/J

    I'm a little fuzzy on moles and such, but unless I'm much mistaken, I should be able to find the mass of the air by dividing by Avogadro number, right?

    After this, I can calculate the heat transfer with Q = mcΔT in which I can find the ΔT with the relationship:

    pivi/Ti = pfvf/Tf

    Am I making sense?
  2. jcsd
  3. Feb 6, 2013 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Units of Pressure x volume = Joules

    Since PV = nRT and PV = constant what kind of process is this (hint: what happens to T during this process?)?

    Use the mass of one mole of dry air and multiply by the number of moles you have found (which is correct).

    You can look up the molar mass of dry air. Or you can work it out. Dry air consists of mostly nitrogen (78%) and oxygen (21%). There is about 1% argon and traces of other gases. You have to know that the Nitrogen and oxygen gases are diatomic and argon is monatomic. 14x2(.78)+ 16x2(.21) + 40x1(.01) = 29 g/mol

    You don't need to work out the temperature change. It is easily seen what happens to T from PV = nRT = constant.

    What you need to do here is apply the first law:Q = ΔU + W

    Given ΔT what can you say about ΔU? So what is Q?

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