Mathematical interpretation of work done by a gas

AI Thread Summary
The discussion centers on the mathematical interpretation of work done by a gas in thermodynamics, specifically the expression W = ∫PdV. The user initially assumes pressure is constant, leading to the simplified expression W = P(V2-V1) for isobaric processes. However, they later realize that pressure can vary with volume, necessitating the use of the integral W = ∫P(V)dV. This understanding clarifies their confusion regarding the treatment of pressure as a function. The user expresses enthusiasm for thermodynamics and plans to focus their undergraduate research on the subject.
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: W = \int_{V_1}^{V_2} PdV I suppose pressure is being treated like a constant in the above equation. If so, if we integrate we would get this expression: W = P(V_2-V_1), 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: W = P(V_2-V_1), 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:
W = \int_{V_1}^{V_2} P(V)dV
 
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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