Closed path term in Thermodynamics

In summary: English language. It is not a semicircle.It seems to me that the person who drew the graph (as well as the final editor) should be fired.
  • #36
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.
Thank you for your reply @Steve4Physics! Sorry I meant part (b) and (d)!

Yeah I will try to solve using integration formulae

Many thanks!
 
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  • #37
Do anybody please know (without using the integral definition of work) how the area under the SR curve can be negative when it is above the x-axis. I get a positive area (## 2~L~atm##) [post #22].

I do not understand because when the area is negative it is below the x-axis.

Many thanks!
 
  • #38
ChiralSuperfields said:
Do anybody please know (without using the integral definition of work) how the area under the SR curve can be negative when it is above the x-axis. I get a positive area (## 2~L~atm##) [post #22].

I do not understand because when the area is negative it is below the x-axis.

Many thanks!
In itself, the area is positive. The direction of the cycle will determine if the working substance does work or if work is done on the substance, resulting in a positive or negative sign (and there are different conventions as to which case is positive and which is negative).
 
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  • #39
DrClaude said:
In itself, the area is positive. The direction of the cycle will determine if the working substance does work or if work is done on the substance, resulting in a positive or negative sign (and there are different conventions as to which case is positive and which is negative).
Thank you for your help @DrClaude!

I am thankful to know that this is a just a convention! :)
 
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