I am going to lose it I need a hint

[Roger Wilco]

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: W=\frac{1}{1-\gamma}[p_2V_2-p_1V_1]

Using the fact that p_1V_1^{\gamma}=p_2V_2^{\gamma}=constant

thanks
RW

 

[Roger Wilco]

I know that W=\int pdV also...

 

[Roger Wilco]

This is going nowhere extremely fast...

 

[Doc Al]

I know that W=\int pdV also...

Here's a hint.

Rewrite: pV^{\gamma}=C

As p=C/V^{\gamma}

 

[Roger Wilco]

Okay from now on Will use n in place of gamma for simplicity.

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

\Rightarrow W=c\frac{V^{-n+1}}{-n+1}...

Now I am stuck again. ...

 

[Doc Al]

Redo that integration more carefully. (Check by taking the derivative of your answer.)

 

[Roger Wilco]

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

 

[Doc Al]

You can replace c with what it equals.

 

[Roger Wilco]

You can replace c with what it equals.

But which do I use...I know p_1V_1^n=p_2V_2^n

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

Thanks Doc!
You're the best!

RW

 

[Roger Wilco]

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

 

[Doc Al]

Use standard SI units.