Integral: Solve (nRT)/(V-nb)dV from Vi to Vf

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Homework Help Overview

The problem involves evaluating the integral of the expression (nRT)/(V-nb) with respect to V, from limits Vi to Vf. The context is rooted in calculus, particularly focusing on integration techniques and variable substitution.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to clarify which variables are constants and which are functions of V. There are suggestions for using substitution to simplify the integral, specifically u = V - nb. Some participants question whether the integral should be treated with respect to V while holding other variables constant.

Discussion Status

The discussion is ongoing, with various participants offering insights on substitution methods and the treatment of constants. There is no explicit consensus, but several productive lines of reasoning have been explored regarding the integration process.

Contextual Notes

Participants note the importance of defining constants in the problem and the implications of the variable substitution on the limits of integration. One participant mentions that the problem is being worked on for a physical chemistry class, indicating a context for the discussion.

Riles246
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1. The problem statement, all variables and given

integral of : (nRT)/(V-nb)dV from Vi to Vf[tex]\int[/tex] [tex]^{Vf}_{Vi}[/tex] (nRT)/(V-nb)dV

I am really stuck on this one!
Do you need to move the denominator up first?
 
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Assuming that everything except V is a constant for this problem, try the substitution

u = V -nb
 
You need to tell us which are constants and which are functions of V in this question. Otherwise, the problem becomes very simple when you factor nRT out.
 
Since the integral says "dV" doesn't that mean it only needs to be integrated with respect to the variable V while holding all other variables constant?
 
Assuming everything else is constant.

factor the nRT out. Let u = V-nb.
du/dv = 1 if nb is a constant.
This means that when u differentiate the denominator u get the numerator and its safe to straightaway ln the denominator.


I think it should be ln|Vf-nb| - ln|Vi-nb|

i might be wrong though.

Btw, is it just a practice question or are u doing this for chem? just curious. ;).
 
Also don't forget if you substitute u = V - nb, you must also write the limits of integration in terms of u.
 
I am doing these problems to touch up on my Calc for a class I am taking (physical chem)
 

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