Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Canot Cycle, Work done

  1. Jan 18, 2007 #1

    G01

    User Avatar
    Homework Helper
    Gold Member

    Hi. I have a qualitative question about the work done from state 1 to state 2 of a Carnot cycle. This is the Isothermal Expansion Phase. Is work done by the engine on the environment during THIS PHASE ONLY equal to the energy extracted from the hot reservoir to keep the temperature constant. I feel that it should be less. Is my reasoning incorrect?
     
  2. jcsd
  3. Jan 18, 2007 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Yes.

    Since the temperature does not change, the heat flow into the gas is equal to the work done: ([itex]\Delta Q = \Delta U + \Delta W = \Delta U + \int PdV[/itex]; Since [itex]\Delta U = 0, \Delta Q = \int PdV[/itex])

    The work done is the area under the PV graph for the isothermal expansion phase. Since the isotherm is T = PV/nR = constant, it is of the form P = K/V where K is constant = nRT. The area under that graph ([itex]\int PdV = K\int dV/V[/itex]) is nRTln(Vi/Vf).

    AM
     
    Last edited: Jan 18, 2007
  4. Jan 18, 2007 #3

    G01

    User Avatar
    Homework Helper
    Gold Member

    Thanks alot Andrew. Man I love having this website:smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Canot Cycle, Work done
  1. Work done (Replies: 16)

Loading...