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Mathematical interpretation of work done by a gas

  1. Mar 13, 2014 #1
    Hello everybody! First of all, I would like to say that this is my first post in this forum, even though I've occasionally read some posts before. I'm a ChemE major from Mexico!

    I am currently taking Thermodynamics I, and I have trouble figuring out the expression: [tex]W = \int_{V_1}^{V_2} PdV[/tex] I suppose pressure is being treated like a constant in the above equation. If so, if we integrate we would get this expression: [itex]W = P(V_2-V_1)[/itex], right? But then I read in HyperPhysics that we express work as an integral when pressure varies too, so, wouldn't we need an integral of two variables?

    I know we express work as an integral because it is the area under the PV curve, but as you can see I'm currently lost between concepts. I hope my question makes sense. Thanks in advance!
     
  2. jcsd
  3. Mar 13, 2014 #2

    Doc Al

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    Staff: Mentor

    No. In general, the pressure could be a function of volume.

    Right. For an isobaric process.

    You'd need to evaluate:
    [tex]W = \int_{V_1}^{V_2} P(V)dV[/tex]
     
  4. Mar 13, 2014 #3
    P is the function that is being integrated. V is the variable of integration.
     
  5. Mar 13, 2014 #4
    I never considered pressure as a function, that was the problem. It makes sense now. *sigh* Thank you both!
     
  6. Mar 13, 2014 #5
    Hi MexChemE. Welcome to Physics Forums!!! I too am a ChE.

    For more on this subject, see my Blog on my PF personal page. I treat a lot of issues that people who are new to Thermodynamics get confused about. It is only a couple of pages long.

    Chet
     
  7. Mar 13, 2014 #6
    Thanks Chet! I am definitely going to check your blog, thermodynamics is actually my favorite branch of physics. Here in Mexico, we have the option of presenting an undergraduate research thesis in order to obtain our degree, and I'm planning on focusing my research on thermodynamics. I'm just a freshman currently, though.
     
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