# Ideal gas law mechanical work problem

1. Apr 17, 2013

### DrOnline

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

Calculate the gas' mechanical work on its environment for the thermodynamic process path IF.

Solution says: 505 J

2. Relevant equations

W=$$\int_i^f p,dv$$

i = Vi, initial volume
f = Vf, final volume

3. The attempt at a solution

My approach is to find a function for p(v) between 2 and 4 liters.

$$p(v) = 709275 - \frac{3}{2} * 101325 V (Pa)$$

And then integrate it from 2 to 4 liters, which is 0.002m3 to 0.004m3

Problem is, I get:

Which is way off. I should get 505 J..

I'm stumped. What am I doing wrong?

2. Apr 17, 2013

### rude man

where did you get your 709275 number?

Anyway, I make it about 306 J.

3. Apr 17, 2013

### DrOnline

Oh, I should have stated that. I extended the IF line to where it intersects with the Y-axis, which is 7 atm.

7 * 101325 Pa = 709275 Pa.

You got 306 J... it should be 505, unless the document is wrong.

4. Apr 17, 2013

### rude man

The first time I just calculated area = 1/2 bh and got 306.

This time I did the integration and got ~ 518. And I found your mistake - in your 2nd term under the integral you divided by 2 instead of 0.002.

Still don't know why 1/2 bh didn't work!

5. Apr 18, 2013

### DrOnline

The second term is -(3*2) because that is the gradient of the curve. That should be correct.

6. Apr 18, 2013

### rude man

It's not correct. The second term should be -(3/2) * 101,325/0.001 V.
I wouldn't go to your professor 'till we agree on this.

7. Apr 18, 2013

### DrOnline

My God, that seemed to do the trick.

This seems right to me.

I gather since I converted from atm to Pa, and integrated using m3 instead of liters, I also should have taken that into account for the gradient.

Is that assessment correct?

8. Apr 18, 2013

### rude man

That assessment is fully correct!

9. Apr 18, 2013

### DrOnline

Thank you very much!