Why Can Volume Be Used Instead of Specific Volume in PV Diagrams?

  • Thread starter Thread starter dan28029
  • Start date Start date
  • Tags Tags
    Diagram Work
AI Thread Summary
The discussion centers on the use of volume instead of specific volume in the equation for net work in PV diagrams, specifically in the context of an ideal gas. Participants express confusion about the omission of mass (m) in the equation W=R*T*ln(Vb/Va), suggesting it should include m, leading to W=m*R*T*ln(Vb/Va). Clarifications indicate that the incorrect formula is not used in calculations, and the omission may stem from a typo in the reference material. Additionally, the use of a slash through the V's is explained as a means to differentiate between volume and specific volume, which initially caused confusion. Overall, the conversation highlights the importance of clarity in thermodynamic equations and the potential for typographical errors in educational resources.
dan28029
Messages
5
Reaction score
0
1. Problem Statement

The problem asks to find the net work from one complete cycle A-B-C. Problem 13, starting on page 66 of the slide show here:

https://engineering.purdue.edu/~xe/Files/FE%20Review/Thermo/FEThermoReview2013.pdf

Homework Equations


ideal gas law: PV=mRT
Boyle's Law: W=RTln(v2/v1)

3. Attempt at solution

I understand most of the solution, but on page 67, says that W=R*T*ln(Vb/Va) is equivalent to W=p*Va*ln(Vb/Va), assuming ideal gas and constant mass. I don't understand why you can use volume here instead of specific volume.

From the ideal gas law: PV=mRT and RT=PV/m. So I thought it should be W=p*Va/m*ln(Vb/Va). Why can we eliminate m here?

Thank you!
 
Last edited by a moderator:
Physics news on Phys.org
There should be an m in the equation W=R*T*ln(Vb/Va). It should read W=m*R*T*ln(Vb/Va).
 
  • Like
Likes 1 person
What's wrong is "W = RT ln(Vb/Va)". It should be W = mRT ln(Vb/Va).
Notice that the wrong formula is never used in the computations.

P.S. why do they put a slash thru the V's?
 
  • Like
Likes 1 person
One equation for isothermal work is W = m*R*T* ln(Vb/Vb). He has either made a mistake by omitting the number of moles or he has calculated the heat for 1 mole.
 
  • Like
Likes 1 person
Thanks for the responses. Good to know I'm not crazy.

Chestermiller said:
There should be an m in the equation W=R*T*ln(Vb/Va). It should read W=m*R*T*ln(Vb/Va).

In the reference material for the fundamentals of engineering exam, the formula also does not include mass. I guess its a typo. Maybe that's where the author of the presentation got the formula from.

rude man said:
P.S. why do they put a slash thru the V's?

I think it is to differentiate between volume and specific volume. It confused me at first though.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top