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[Thermo] One m^3 of an ideal gas expands in an isothermal

  1. Sep 26, 2015 #1
    1. The problem statement, all variables and given/known data
    One m3 of an ideal gas expands in an isothermal process from 760 to 350 kPa. Determine the specific work done by the gas.

    2. Relevant equations
    ω=W/m
    1W2=mRTln(P1/P2) = P1V1ln(P1/P2)

    P1V1=P2V2

    3. The attempt at a solution
    P1V1ln(P1/P2) = (760)(1)ln(760/350) = 589.29kJ

    V2=(760)(1)/(350) = 2.17m3

    I tried doing PV=mRT using STP and my mass calculated was .31kg but it doesn't seem right. How do I find mass?
     
  2. jcsd
  3. Sep 26, 2015 #2
    Are you sure you're not given the molar mass of the substance? (or allowed to suppose its a certain element, like Nitrogen for example). Also, in case you didn't know in ##PV=mRT## , ##m## referes to moles and not mass... So its more usually written as ##PV=nRT##
     
  4. Sep 26, 2015 #3
    Oh alright I'll remember that. I just have it written down differently. Unfortunately, that's all I'm given. Maybe there was a typo and she meant to just say work instead of specific work?
     
  5. Sep 26, 2015 #4
    Maybe she meant work per mole, which you can find.
     
  6. Sep 26, 2015 #5
    How?
     
  7. Sep 26, 2015 #6
    Well work along an isothermal path is ##W_{1\to2} = -nRT \ln\frac{V_2}{V_1}##. You can see that ##\frac{W{1\to2}}{n}=-RT \ln\frac{V_2}{V_1}## would be the work done per mole.

    You can do manipulations like the ones you did in your original post (such as P1V1=P2V2) to figure out more compact expressions that match the given data.
     
  8. Sep 28, 2015 #7
    She said it was supposed to be 1 m3/kg, so how does this change it? I know that gives us specific volume, not sure what to do with it.
     
  9. Sep 28, 2015 #8
    Well I would say that makes no sense. Specific work has units J/kg.
     
  10. Sep 28, 2015 #9

    SteamKing

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    Mass density, ρ, is the reciprocal of specific volume, v , such that ρ = 1 / v.

    https://en.wikipedia.org/wiki/Ideal_gas_law
     
  11. Sep 28, 2015 #10
    Okay so if I have 1 m3/kg of an ideal gas, the mass density of that gas is 1 kg/m3, what do I do with that and how does that get me to find my work? I've manipulated the equations as much as I know how and now I'm even more lost than I was when I posted this problem.
     
  12. Sep 30, 2015 #11
    Hey guys, this still is unanswered, I could really use some help. Please?

    Given:
    specific volume (v1) = 1 m3/kg - ideal gas
    P1 = 760 kPa
    P2 = 350 kPa

    Find specific work.

    Sorry to nag, I just really want to know how to do this.
     
  13. Sep 30, 2015 #12
    In my opinion, you need someway to know the molar mass of this substance. The ideal gas equations don't care about how much mass you've got, but care about the number of particles; which is why it is expressed as PV=nRT where n is the number of moles.

    I feel you're teacher is hiding something of great importance here. (The molar mass)
     
  14. Sep 30, 2015 #13
    I'm not sure you know any calculus yet but here is the reasoning for what I've just said:

    Work is ##-pdV##

    Pressure is ##P=\frac{nRT}{V}=\frac{\frac{m}{M}RT}{V}##

    So work is ##dW=-\frac{m}{M}RT\frac{dV}{V}##

    And specific work is ##d\omega =-\frac{1}{M}RT\frac{dV}{V}##

    And we conclude you need the molar mass of the substance....

    Are you sure that your problem doesn't say "1 m3 of Nitrogen gas..." or something along those lines?
     
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