Thermodynamics- pV^n = constant.

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving the equation pV^n = constant, where a closed system of gas is analyzed with given pressures and volumes. The original poster is tasked with finding the value of n based on the provided conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to rearrange the equation to isolate n but expresses difficulty in doing so. They explore using logarithmic properties but feel stuck with the cancellation of terms. Other participants suggest the correct application of logarithms to facilitate the isolation of n.

Discussion Status

Participants are actively engaging in clarifying the logarithmic approach to the problem. Some guidance has been offered regarding the correct manipulation of the logarithmic expressions, and the original poster acknowledges the assistance received, indicating a positive direction in the discussion.

Contextual Notes

The original poster mentions feeling rusty after a break, indicating a potential lack of recent practice with the material. There is also a light-hearted tone in the discussion, suggesting a supportive environment among participants.

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Homework Statement


A closed system consisting of 2 lb of a gas undergoes a process in which pV^n=constant. For: p1=20 lb/in^2 , V1=10 ft^3 and p2=100lb/in^2 V2=2.9 ft^3.

Find n.

Homework Equations



pV^n=constant

The Attempt at a Solution



Im just having trouble realizing how to find n. This seems very basic and maybe I am just rusty from having the summer off, this is a review question.

(p1)(V1)^n = (p2)(V2)^n

Now I am having trouble realizing how to re-arrange that to find n... I've tried using the relation p1V1^n=5P1(.29V1)^n but everything keeps canceling out. Can someone push me in the right direction?

Ive also tried solving for n in terms of C and plugging that back in and that just gives me a jumbled mess.
 
Physics news on Phys.org
The logarithm function turns exponential factors into multiplicative factors...
 
Mapes said:
The logarithm function turns exponential factors into multiplicative factors...

Yah I understand that. But Even if I do that won't I be left with:

n*ln(P1V1)=n*ln(P2V2)

So... All I can do is cancel things out. I can't really isolate n... right? unless I want #n=0.

Feel free to call me a moron if I am missing something obvious :-p
 
You're not taking the logarithm correctly.
 
Mapes said:
You're not taking the logarithm correctly.

Oh God I suck...

So does this sound good to you?

ln(P1(V1)^n)=ln(P2(V2)^n)

ln(P1) + n*ln(V1) = ln(P2) + n*ln(V2)
 
Better!
 
Mapes said:
Better!

Awesome, thanks Mapes! How did I not see that :blushing:

I got to say, I appreciate your help. I haven't been on the boards in quite a while but i remember you helping me in the past as well. Thanks again Mapes!
 

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