# I am going to lose it! I need a hint

1. Dec 7, 2007

### Roger Wilco

1. The problem statement, all variables and given/known data
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

2. Dec 7, 2007

### Roger Wilco

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

3. Dec 7, 2007

### Roger Wilco

This is going nowhere extremely fast...

4. Dec 7, 2007

### Staff: Mentor

Here's a hint.

Rewrite: $$pV^{\gamma}=C$$

As $$p=C/V^{\gamma}$$

5. Dec 7, 2007

### 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. ....

Last edited: Dec 7, 2007
6. Dec 7, 2007

### Staff: Mentor

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

7. Dec 7, 2007

### Roger Wilco

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. Dec 7, 2007

### Staff: Mentor

You can replace c with what it equals.

9. Dec 7, 2007

### Roger Wilco

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

10. Dec 7, 2007

### 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

11. Dec 7, 2007

### Staff: Mentor

Use standard SI units.