- #36
member 731016
Thank you for your reply @Steve4Physics! Sorry I meant part (b) and (d)!Steve4Physics said:I’m coming in rather late but I hope this helps.I see no part d)!
Can I pull together a few points...
The ‘horizontal’ radius of the semicircle is 1.0L.
The ‘vertical’ radius of the semicircle is (about) 1.5 atm.
As @kuruman showed in post #26, we get the official answer (160J) if we assume the pressure scale is incorrectly drawn such that the ‘vertical’ radius is in fact 1.0 atm.
As @Chestermiller showed in Post #19, if we use 1.5 atm as the ‘vertical’ length, then we get 240J.
A general point (if not already apparent):
The area inside a closed loop on a PV diagram represents the work done by (or on) the gas. And this equals the amount of heat entering (or leaving).
Going clockwise round the loop, the area represents the work done by the gas in the cycle and this equals the amount of heat entering;
Going anticlockwise round the loop, the area represents the work done on the gas in the cycle and this equals the amount of heat leaving.
If you wanted to solve this by integration, it would be simply an exercise to find half the area of a circle (or ellipse) by integration. You would start by finding the equation of the curve.
Yeah I will try to solve using integration formulae
Many thanks!