Ideal gas law mechanical work problem

[DrOnline]

Homework Statement

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

Solution says: 505 J

Homework Equations

W=\int_i^f p,dv

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

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?

 

[rude man]

where did you get your 709275 number?

Anyway, I make it about 306 J.

 

[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.

 

[rude man]

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.

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!

 

[DrOnline]

Thank you for your help. Gonna ask the professor about it, he made the task after all.

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

 

[rude man]

Thank you for your help. Gonna ask the professor about it, he made the task after all.

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

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.

 

[DrOnline]

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.

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?

 

[rude man]

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?

That assessment is fully correct!

 

[DrOnline]

Thank you very much!