• Support PF! Buy your school textbooks, materials and every day products Here!

Termodynamics doubt

  • Thread starter jaumzaum
  • Start date
  • #1
302
13
I was solving the question below :

"A cylinder have a piston whose mass is insignificant and moves without friction. One mole of an ideal gas is confined into the cylinder. If we expand isotermally the gass at temperature 298K, against a constant external pressure of 0.1 atm, the gas pressure goes from 1atm to 0.1 atm. Calculate the work need for the expansion."

I have 2 answers, and I don't know why the second one is wrong:
1) Pressure is constant. W = P(ext)(V2-V1) = P(ext) nRT(1/P2-1/P1) = 0,9.1.8,31.298 = 2.23kJ

2) Gas Pressure is P, so the difference of pressure is P1=(P-0,1).

W = ∫P1.dv.
V = nRT/P
dv = -nRT/P²dP
W = ∫-(P-0.1)nRT/P² dP = 2476(ln(P1/P2) - 0.1(P2-P1)) = 2476(2.3+0.09) = 5,91kJ

Why the second is wrong, and more important, why the first one is right? What is the definition of work?

[]'s
João
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,848
1,645
The Work is the area under the PV diagram.

[edit]
What makes you think (1) is correct and (2) is not?
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
7,583
346
I was solving the question below :

"A cylinder have a piston whose mass is insignificant and moves without friction. One mole of an ideal gas is confined into the cylinder. If we expand isotermally the gass at temperature 298K, against a constant external pressure of 0.1 atm, the gas pressure goes from 1atm to 0.1 atm. Calculate the work need for the expansion."

I have 2 answers, and I don't know why the second one is wrong:
1) Pressure is constant. W = P(ext)(V2-V1) = P(ext) nRT(1/P2-1/P1) = 0,9.1.8,31.298 = 2.23kJ

2) Gas Pressure is P, so the difference of pressure is P1=(P-0,1).

W = ∫P1.dv.
V = nRT/P
dv = -nRT/P²dP
W = ∫-(P-0.1)nRT/P² dP = 2476(ln(P1/P2) - 0.1(P2-P1)) = 2476(2.3+0.09) = 5,91kJ

Why the second is wrong, and more important, why the first one is right? What is the definition of work?

[]'s
João
In 2. you are using the internal pressure of the gas rather than the external pressure. The thermodynamic work here is the work done by the gas on the external environment (ie. against the external pressure). In effect, the excess internal pressure is not used to do mechanical work.

This is a non-reversible expansion because there is a significant difference between the internal gas pressure and the external pressure. This excess pressure causes a very rapid initial outward expansion but does not add to the total work that is done. That work is simply W = Pext(ΔV).

AM
 

Related Threads on Termodynamics doubt

  • Last Post
Replies
4
Views
2K
Replies
28
Views
2K
  • Last Post
Replies
2
Views
922
  • Last Post
Replies
1
Views
768
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
661
  • Last Post
Replies
12
Views
1K
Top