1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics, calculating W

  1. May 13, 2012 #1
    n = 9.95 moles of ideal gas are slowly heated from initial pressure Po = 93398.6 Pa and initial volume Vo = .679 m^3 to final pressure Pf = 117508 Pa and final volume Vf = 1.073 m^3. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system:

    If the plot is a straight line then it is an isobaric process so W = nRTb (1 - Va/Vb)

    Using n = PV/RT, Tb=1524.172269K

    W = (9.95)(8.314)(1524.172269)(1 - .679/1.073) = 46.3 kJ

    Can someone please advise why/how this is wrong??
     
  2. jcsd
  3. May 13, 2012 #2
    Really trying here and really struggling...please, please help.
     
  4. May 13, 2012 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It did not say horizontal line. Plot the two points on a PV diagram. They are not at the same pressure so how can the process be isobaric?

    Is there a geometric aspect of the graph that represents the work done? What is it? Why?

    AM
     
  5. May 14, 2012 #4
    The area under the curve but that is not correct either.
     
  6. May 14, 2012 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    What is the mathematical equation for the area under the graph of a straight line that is not parallel to the bottom axis?

    AM
     
  7. May 14, 2012 #6
    Area of the rectangle + Area of the triangle.
     
  8. May 14, 2012 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    So plot the two points, draw the line between them (straight) and work out the area. That is the amount of work done by the system. Caution: be careful to get the sign right. The question asks for the work done ON the system. How is this related to the work done BY the system?

    AM
     
  9. May 14, 2012 #8
    It would appear that Wrev=Won gas-Wby gas. We can calculate Wby gas but since the system is slowly heated we can't assume it is adiabatic. I don't see how we calculate Wrev and without we can't calculate Won gas. What am I missing?
     
  10. May 14, 2012 #9

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    This is not correct. The process is very slow so it will be close to a reversible process so Wrev ≈ Wactual ≈ ∫PdV where W is the work done by the gas in the reversible and actual situation. If internal pressure of the gas is infinitessimally higher than the external pressure on the gas for the duration of the process the reversible and actual work done would be equal.


    Calculate ∫PdV where P is the internal pressure of the gas which is determined by the co-ordinates of the PV diagram at all points between the beginning and end points that are given.

    AM
     
  11. May 14, 2012 #10
    The temperature is controlled in such a way that the pressure versus volume plot is a straight line. You can use the ideal gas law to determine precisely how the temperature is being controlled. Lucky for you, the p vs v plot is a straight line.

    Chet
     
  12. May 14, 2012 #11
    The process you described is not isobaric: isobaric means the pressure is constant, but the pressure - as you said in the problem - changes.

    The work associated with a system like this is equal to the negative area of the P-V graph. Negative work means that work is done by the system, and positive work means work is done on the system.

    Sounds like you're finding the area of a trapezoid. Your trapezoid height will be the difference between the volumes (Vf - Vi), one base will be (Pi - 0) and the other base will be (Pf - 0).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook