Homework Help: Using area of pv graph to find work

1. Jul 6, 2009

Iwanttolearn7

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

An ideal graph is shown in a p-v diagram which looks like one small rectangle inside one large rectangle, with point one on the top right hand corner of the large rectangle leading negatively down to the right hand corner of the small rectangle.

The points of the small rectangle are:
p=2.0^5 Pa
v=0.5 m^3

The points of the large rectangle are:
p=5.0^5
v=1.0 m^3

2. Relevant equations

My professor said to use the area of the triangle and of the rectangle to calculate the Work.

So which triangle and rectangles do I find the areas of?

Thanks!

3. The attempt at a solution

I have tried different combinations of triangles and rectangle areas for the graph, and cannot come up with the correct answer, which is -1.8^5 J.

Examples:

area small rectangle: 2x10^5*0.5=1x10^5
area large triangle: (5x10^5*1)/2=2.5x10^5
total = 3.5^5 (wrong)

another combination:

area small triangle: (2x10^5*0.5)/2=5x10^4
area large rectangle: 5x10^5*1=5x10^5
total = 5.5x10^5

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

2. Relevant equations

3. The attempt at a solution

2. Jul 6, 2009

cepheid

Staff Emeritus
I can't make heads or tails of your description without the diagram. Also, how do two "points" define a rectangle? Were these meant to be their lengths and widths?

3. Jul 6, 2009

LowlyPion

I can't really see your diagram, but as described isn't the line joining the two corners the P-V profile in moving from one volume to the other?

In which case, by taking the integral under the path of the P-V from the Volume of the small rectangle to the large you have a) a small rectangle from .5 to 1 of height 2. That area is 1/2 * 2 = 1

On top of that you have the additional area of the triangle that sits on top of height (5 - 2)*(1 - .5)*1/2 = .75

4. Jul 6, 2009

Iwanttolearn7

I really wish I had a picture of the graph. It is like no other on google relating to physics. Trust me, I tried. There are only two points on the graph. My description, including the coordinates, will draw out the correct graph, but I realize you may not have the time to draw it out.

I am confused as to how the last poster is getting the area of triangles by subtracting. I have learned that the equation is base*height/2. I wish I could understand where the subtracting is coming from.

I appreciate your attempts, I just can't figure this one out. If you have time, the description of the graph will let you draw out the exact two rectangles on the graph as well as the coordinate points. I posted on another forum too and someone came up with an answer that was correct, but used the wrong coordinates, so I have no idea what is going on.

5. Jul 6, 2009

LowlyPion

Noting that the you are wanting to find the work to go from one P-V coordinate to the other along a straight line, I summed two areas (taking the integral) under the P-V function, namely a triangle stacked on top of a rectangle.

Figure that F = P*A and W = F*Δx then W = P*ΔV

But your problem has P(v) as a function of volume. So figuring Work from F*Δx you need to take the integral over the change in volume. Which is what my calculation was intended to do.

6. Jul 7, 2009

Iwanttolearn7

Thank you very much! I also finally got to class and asked my Professor about it, and he said that I was supposed to use the AVERAGE of p and v to find the volume, which was my main problem. I was trying to use both p and both v points to find the volume when I should have been averaging them. But you are right, because after averaging the coordinates I needed to multiply them together. Wow, that was a difficult one. Thank you for your help!