Mathematical interpretation of work done by a gas

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Discussion Overview

The discussion revolves around the mathematical interpretation of work done by a gas in thermodynamics, specifically focusing on the expression W = ∫_{V_1}^{V_2} PdV. Participants explore the implications of treating pressure as a constant versus a variable and the appropriate mathematical treatment of work in different thermodynamic processes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the expression W = ∫_{V_1}^{V_2} PdV, initially assuming pressure is constant.
  • Another participant clarifies that pressure can be a function of volume, indicating that the integral must account for this variability.
  • A further clarification states that for an isobaric process, the work can be simplified to W = P(V_2-V_1).
  • Participants discuss the need for evaluating W = ∫_{V_1}^{V_2} P(V)dV when pressure varies with volume.
  • One participant acknowledges their misunderstanding regarding pressure as a function, indicating a shift in their understanding.

Areas of Agreement / Disagreement

Participants generally agree on the need to treat pressure as a function of volume in certain contexts, but there is no consensus on the implications for different thermodynamic processes or the best approach to integrate the work done by a gas.

Contextual Notes

Participants do not fully resolve the implications of treating pressure as a function versus a constant, nor do they clarify the conditions under which each approach is applicable.

Who May Find This Useful

This discussion may be useful for students studying thermodynamics, particularly those grappling with the mathematical concepts related to work done by gases and the role of pressure in these calculations.

MexChemE
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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!
 
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MexChemE said:
I suppose pressure is being treated like a constant in the above equation.
No. In general, the pressure could be a function of volume.

If so, if we integrate we would get this expression: [itex]W = P(V_2-V_1)[/itex], right?
Right. For an isobaric process.

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?
You'd need to evaluate:
[tex]W = \int_{V_1}^{V_2} P(V)dV[/tex]
 
P is the function that is being integrated. V is the variable of integration.
 
I never considered pressure as a function, that was the problem. It makes sense now. *sigh* Thank you both!
 
MexChemE said:
I never considered pressure as a function, that was the problem. It makes sense now. *sigh* Thank you both!
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
 
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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