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Gas expands in a cylinder

  • Thread starter Lamoid
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  • #1
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Homework Statement


When gas expands in a cylinder with radius r, the pressure at any given time is a function of the volume: P = P (V). The force exerted by the gas on the piston is the product of the pressure and the area: F = π(r^2)P. Show that the work done by the gas when the volume expands from Volume V1 to Volume V2 is




Homework Equations


Equation1.jpg



The Attempt at a Solution



I was having a problem doing this question so I just sort of trudged ahead to see if I could get something and I ended up using substitution with u = n(r^2)x and du = nr^2 and then adjusted the limits of integration which were x+a and x. I am fairly certain this is wrong though and would appreciate help.
 

Answers and Replies

  • #2
Redbelly98
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What is the usual definition of work?
 
  • #3
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Integral of Force times distance
 
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  • #4
Redbelly98
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Yes.

Can you make substitutions and rearrange terms to get that integral in terms of P and V?
 
  • #5
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eqation2.jpg


Ok so thats the integral. So I replace (pi)(r^2)x with V ?
 
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  • #6
Redbelly98
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Um, wait a moment.

If a force F moves an object over a distance dx, then the work done in that small distance is F*dx.

For extended distances, we would integrate F * dx

Try working with that expression ... you were on the right track otherwise.
 
  • #7
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Hmm, so I am confused. This is my best guess.

f * dx = W

πr^2P * dx = W

PV = W

P*dv = W'

Then you integrate it using the change in volume as the limits of integration?
 
  • #8
Redbelly98
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Try this:

[tex]W = \int F dx[/tex]

and do the same substitution F = P * pi * r^2
 
  • #9
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Okay, when I do that I end up, obviously, W = integral P * pi * r^2 * dx. I know I have to get V's in here, but I am unsure if I substitute pi * r^2 * dx with a V or a dv since I am talking about the change in x.
 
  • #10
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Okay, when I do that I end up, obviously, W = integral P * pi * r^2 * dx. I know I have to get V's in here, but I am unsure if I substitute pi * r^2 * dx with a V or a dv since I am talking about the change in x.
V = [itex]\pi r^2 x[/itex] where we take the origin to be wherever the piston starts. What's dV?
 
  • #11
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So dv is pi*r^2*dx. Now, must I do anything to make the limits of intergration V1 to V2?
 
  • #12
Redbelly98
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This is a change-of-variable as taught in integral calculus.

With the substitution V = pi r^2 x, the limits simply change accordingly so that:
x1 becomes V1, where V1 = pi r^2 x1
and similarly for x2 and V2.

In other words,

[tex]
\int^{x2}_{x1} ... \ dx
[/tex]

becomes


[tex]
\int^{\pi r^2 x2}_{\pi r^2 x1} ... \ dV
[/tex]

or just

[tex]
\int^{V2}_{V1} ... \ dV
[/tex]
 
Last edited:

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