How to Calculate Work Done from a PV Diagram?

Click For Summary
SUMMARY

The work done by a system from point A to point C on a PV diagram is calculated as the area under the curve. Each square on the diagram represents an area of 1.5 atm·L, equivalent to 152 J. To accurately determine the work, it is essential to divide the area into rectangles and triangles, calculating each area separately using appropriate formulas. The final calculated work done is 5020 J after applying these methods.

PREREQUISITES
  • Understanding of PV diagrams and their components
  • Knowledge of pressure and volume units (atm and L)
  • Familiarity with calculating areas of geometric shapes (rectangles and triangles)
  • Basic principles of thermodynamics related to work done
NEXT STEPS
  • Study the concept of work in thermodynamics and its relation to PV diagrams
  • Learn how to calculate areas under curves using integration techniques
  • Explore the implications of work done in different thermodynamic processes
  • Review examples of PV diagrams in real-world applications, such as engines and refrigeration cycles
USEFUL FOR

Students studying thermodynamics, physics educators, and professionals involved in mechanical engineering or energy systems who need to understand work calculations from PV diagrams.

KJ22
Messages
2
Reaction score
0

Homework Statement



Look at the PV diagram of a system above. The units for P and V are atmospheric pressure atm and liter L. Process A to B and B to C are straight lines on it.
How much is the work done by the system from A to C in J?

https://tycho-s.physics.wisc.edu/cgi/courses/shell/common/showme.pl?courses/phys103/fall08/homework/12/pv/pv.gif

HELP: In general, the work done in an expansion from some initial state to some final state is the area under the curve on a PV diagram. So determine the area under the ABC-curve.

HELP: Each square on the diagram has an area of: 1.5 atm * 1 L = 1.5 atm L = 152 J. Estimate the numbers of squares below the ABC-curve and multiply that number by 152 J. Good luck

The Attempt at a Solution




Wab= pressure x change in volume= 303975 Pa(.004meters cubed)= 1215.9
Wbc= pressure x change in volume= 607950 Pa(.006 meters cubed)= 3647.7

W= Wab +Wbc

but this is not correct. Can anyone help me?
 
Last edited by a moderator:
Physics news on Phys.org
I did what you said, i.e. counted the squares under the curve and multiplied that number by 152, but I do not get the same answer as you. Try dividing the area into two rectangles and two triangles and calculate those four areas separately. Use the formula for the area of a triangle rather than guessing how many squares make up the triangle. Hope that helps.
 
thanks! that worked.

I got the Area of the triangles to be 4 and 3 and the area of the rectangles to be 8 and 18. multiplied by 152 and got an answer of 5020 J.

thanks for your help!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
535
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K