Can Someone Explain the Calculation of Work Done by Gas Expansion?

Click For Summary
SUMMARY

The discussion centers on the calculation of work done during gas expansion, specifically addressing the equations W = -Fx and W' = -P(external)X. The confusion arises from the negative sign in the equation for work done by the system on the surroundings, which is clarified by understanding that the force exerted by the system is in the opposite direction to the movement. The correct interpretation leads to the conclusion that W' = P(external)X, confirming the validity of the equation despite initial misunderstandings.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with work-energy concepts
  • Knowledge of gas laws and behavior
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the First Law of Thermodynamics
  • Learn about the Ideal Gas Law and its applications
  • Explore the concept of pressure-volume work in thermodynamic systems
  • Investigate the role of external forces in gas expansion scenarios
USEFUL FOR

Students of physics, engineers involved in thermodynamics, and anyone seeking to deepen their understanding of work done by gases during expansion processes.

lipwe
Messages
1
Reaction score
0
I have little confusion about understanding a lecture note for calculation of work done by a expansion of gas.

According to text, Work done against a force w= - Fx

So work done on the system by surrounding when piston moving w = - P(external)(power exert by outside force on the piston) . X

but I did not get this one.

Work done by the system on surrounding. According to text

W' = -Fx
I did not get this point (minus in front of F) as what ever force exert by system on should be in the direction of movement and in this case exact opposite of - P(external) Am I right? ) but equation gives the right answer i.e w'= - (- P(external))x= P(external )X . In-contrast my one is incorrect

Can somebody help me out to clear this . Thanks for your time in advanced
Modify message
Report to moderator 68.228.44.236
 
Last edited:
Physics news on Phys.org
A bit of semantics here. If F is the force you are pushing against, then the force you must exert is -F, thus the work you do on the system is -Fx.

On the other hand, if F is the force you exert, then the work you do is simply Fx.
 

Similar threads

Undergrad Is Mechanical Energy Conservation Free of Ambiguity - follow up
  • · Replies 77 ·
3
Replies
77
Views
6K
Graduate Use internal pressure to compute work done by an expanding gas?
  • · Replies 15 ·
Replies
15
Views
3K
Undergrad Confusion about the work done by an ideal gas
  • · Replies 5 ·
Replies
5
Views
2K
High School Understanding the Work-Energy Theorem
  • · Replies 2 ·
Replies
2
Views
16K
Graduate A single external force doing work on a system of particles
  • · Replies 12 ·
Replies
12
Views
2K
Undergrad Sign convention on work done to a closed system containing gas
  • · Replies 5 ·
Replies
5
Views
4K
Undergrad Work done on the system through a piston?
  • · Replies 6 ·
Replies
6
Views
5K
Undergrad Simple doubt in work done by gas
  • · Replies 3 ·
Replies
3
Views
2K
Graduate Assumed value of pressure during quasistatic compression
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Work done by ideal gas processes
  • · Replies 2 ·
Replies
2
Views
2K