I am going to lose it I need a hint

  • Thread starter Roger Wilco
  • Start date
In summary, the conversation discusses deriving a formula for work done in an adiabatic expansion of an ideal gas using the fact that pV^{\gamma}=C. There is a hint given to rewrite the formula and redo the integration, followed by a discussion on converting units to standard SI units.
  • #1
Roger Wilco
30
0

Homework Statement


just to get me started! This is killing me...

I need to derive the formula for an work done W of an adiabatic expansion of an ideal gas: [tex]W=\frac{1}{1-\gamma}[p_2V_2-p_1V_1][/tex]

Using the fact that [tex]p_1V_1^{\gamma}=p_2V_2^{\gamma}=constant[/tex]

thanks
RW
 
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  • #2
I know that [tex]W=\int pdV[/tex] also...
 
  • #3
This is going nowhere extremely fast...
 
  • #4
Roger Wilco said:
I know that [tex]W=\int pdV[/tex] also...
Here's a hint.

Rewrite: [tex]pV^{\gamma}=C[/tex]

As [tex]p=C/V^{\gamma}[/tex]
 
  • #5
Okay from now on Will use n in place of gamma for simplicity.

So now if I have [tex] W=c\int\ V^{-n}dV[/tex]

[tex]\Rightarrow W=c\frac{V^{-n+1}}{-n+1}[/tex]...

Now I am stuck again. ...
 
Last edited:
  • #6
Redo that integration more carefully. (Check by taking the derivative of your answer.)
 
  • #7
Doc Al said:
Redo that integration more carefully. (Check by taking the derivative of your answer.)

Gotcha...now I am having trouble putting that in terms of p1 and p2...I now have it in terms of c...and I still have 1-n up in th exponent...do I need a natural log somewhere?

Thank you
RW
 
  • #8
You can replace c with what it equals.
 
  • #9
Doc Al said:
You can replace c with what it equals.

But which do I use...I know [tex]p_1V_1^n=p_2V_2^n[/tex]

oh..wait... how about I distibute c first and then replace it...it's clear now.

Thanks Doc!
You're the best!

RW
 
  • #10
So to actually evaluate this expression when I am given pressure in atms and volume in cubic centimeters do i need to convert these units into SI.

I see that the coefficient of 1/(1-n) is dimensionless. And I had assumed that since p1v1-p2v2 had the SAME dimensions it would be okay to NOT convert...

i think I see my error though...I need Joules when all is said and done. I may have answered my own question.

RW
 
  • #11
Use standard SI units.
 

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