Closed path term in Thermodynamics

[kuruman]

Never mind it is because if we convert the units beforehand then it is no longer a simi-circle (imagine a graph of Pressure (Pa) vs volume (m^3) for the values, which means that we have to use the ellipse formula. Is this please correct @kuruman?

Many thanks!

You don't need to convert units to anything as long as it is clear what units are appropriate for each axis. And yes, if you rescale the axes by different factors, the shape will not retain its previous proportions. A circle may become an ellipse but it is also possible for an ellipse to become a circle if the axes are scaled just right.

 

[member 731016]

You don't need to convert units to anything as long as it is clear what units are appropriate for each axis. And yes, if you rescale the axes by different factors, the shape will not retain its previous proportions. A circle may become an ellipse but it is also possible for an ellipse to become a circle if the axes are scaled just right.

Thank you for your reply @kuruman!

That is a very interesting way to think about substituting in values!

 

[member 731016]

Dose anybody please know how to solve this problem using integration instead of graphically calculating the area?

Many thanks!

 

[member 731016]

It also looks the author did it again for part(c) and (d). This time they don't state the simi-circular assumption for (c), so I did an elliptical shape from the figure to get 240J which dose not match the solutions. The correct answer according to the solutions for (c) is 160J.

Many thanks!

 

[Steve4Physics]

I’m coming in rather late but I hope this helps.

It also looks the author did it again for part(c) and (d). This time they don't state the simi-circular assumption for (c), so I did an elliptical shape from the figure to get 240J which dose not match the solutions. The correct answer according to the solutions for (c) is 160J.

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.

 

[member 731016]

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!

 

[member 731016]

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!

 

[DrClaude]

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).

 

[member 731016]

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! :)