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Thermodynamics cycle work question

  1. Jun 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A mass of gas occupying volume V1 = 2 m3 at the pressure P1 = 4*10^5 Pa performs the cycle represented in the Figure that i have uploaded.What is the work of gas in this cycle, knowing that the pressure P2 = 10^5 Pa
    [​IMG]

    2. Relevant equations
    Work=1/2 * (P1 - P2) * (V1 - V3)

    3. The attempt at a solution
    The work is the area within the triangle.
    1/2 base * height
    = 1/2 * (P1 - P2) * (V1 - V3)


    The question that i have is how do i find V3?
     

    Attached Files:

  2. jcsd
  3. Jun 20, 2017 #2
  4. Jun 20, 2017 #3

    Ssnow

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    The cartesian equation of the line from the step ##3## to the step ##2## is ##P(V)=2\cdot 10^5 \cdot V## so ## V_{3}=0,5 m^3##.
     
  5. Jun 20, 2017 #4
    Can you please explain this to me?

    For example how did you get 2*10^5?
     
    Last edited: Jun 20, 2017
  6. Jun 20, 2017 #5

    Ssnow

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    The number ##2\cdot 10^5## is the angular coefficient of the line from the process ##3## to ##2##. The line pass through the origin so the line has equation of the form ##P=m\cdot V## in order to find ##m## you can use the point ##(V_{1},P_{1})=(2,4\cdot 10^5)##, so ##4\cdot 10^5=m\cdot 2## and ##m=2\cdot 10^5##...
    Ssnow
     
  7. Jun 20, 2017 #6
    What is m? Is this some kinda new thing like volume or pressure or not?
    also now that i know m how do i find V3

    and also i apologize for being obnoxious and wasting your time but..... why is the line passing through the origin means that the equation is of the form P=m*V?
    again sorry for so many questions but i am genuinly curios and really want to learn before school starts
     
  8. Jun 21, 2017 #7
    bump for visibility
     
  9. Jun 23, 2017 #8

    Ssnow

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    In a Cartesian plane, with coordinates ##x## and ##y## you can rapresent a line (a direct proportionality law between ##x## and ##y##) using the equation ##y=mx+q## where ##m## is called the angular coefficient and ##q## is the pont where the line intersect the ##y## axis. In your case instead ##y## and ##x## there are ##P## and ##V## so the equation of your line is ##P=mV+q##. Now ##q## must be ##0## because the line pass in the origin, so from ##0=m\cdot 0 + q## you find ##q=0##, and ##P=mV## . In order to find the line equation you must determine the coefficient ##m##. You will use the other point on the line so as I said in the previous post ##P=2\cdot 10^5\cdot V##. Now you have the general equation of the line and using the point ##(P_{2},V_{3})=(10^5,V_{3})## you can find ##V_{3}##...
    Ssnow
     
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