Thermodynamics- pV^n = constant.

In summary, the problem is asking for the value of n in the equation pV^n = constant, given p1 = 20 lb/in^2, V1 = 10 ft^3, p2 = 100 lb/in^2, and V2 = 2.9 ft^3. After rearranging and taking the logarithm of both sides, the equation becomes n*ln(V1) = ln(P2/P1) + n*ln(V2/V1). With this equation, n can be solved for by isolating it on one side of the equation.
  • #1
NBAJam100
146
0

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.
 
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  • #2
The logarithm function turns exponential factors into multiplicative factors...
 
  • #3
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 :tongue:
 
  • #4
You're not taking the logarithm correctly.
 
  • #5
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)
 
  • #6
Better!
 
  • #7
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!
 

1. What is the significance of the pV^n = constant equation in thermodynamics?

The pV^n = constant equation, also known as the polytropic process equation, is used to describe the relationship between pressure (p) and volume (V) in a system undergoing a reversible process. It is often used to analyze the behavior of gases and determine the work done by or on a gas during a process.

2. How is the value of n determined in the pV^n = constant equation?

The value of n in the pV^n = constant equation depends on the type of process being studied. For an isothermal process (constant temperature), n = 1. For an adiabatic process (no heat transfer), n = γ, where γ is the ratio of specific heats for the gas. For an isobaric process (constant pressure), n = 0. For an isochoric process (constant volume), n = ∞.

3. Can the pV^n = constant equation be applied to all thermodynamic processes?

No, the pV^n = constant equation can only be applied to reversible processes. In real-world scenarios, most processes are irreversible, meaning they involve some kind of energy loss or dissipation, and therefore cannot be described by this equation.

4. How does the pV^n = constant equation relate to the first law of thermodynamics?

The pV^n = constant equation is a mathematical representation of the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or transformed. The equation shows the relationship between the work done on or by a gas and the change in its internal energy.

5. What are some practical applications of the pV^n = constant equation?

The pV^n = constant equation has many practical applications, including in the design and analysis of engines, turbines, and compressors. It is also used in the study of atmospheric and oceanic processes, as well as in the production and storage of gases in industrial processes. Additionally, the equation is used in the measurement of gas properties and the development of thermodynamic models.

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