About the equation for work done on gas

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Discussion Overview

The discussion revolves around the equations for work done on a gas, specifically comparing the use of W = p dV and W = nRT ln(Vi/Vf) in the context of isothermal processes. Participants explore when to apply each equation and the reasons for discrepancies in their results.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about the appropriate contexts for using W = p dV and W = nRT ln(Vi/Vf), noting they received different answers when applying both equations to the same problem.
  • One participant suggests that the work in a process can be represented as the area under the P-V diagram and encourages drawing the diagram for clarity.
  • Another participant points out that the second equation is a special case of the first, specifically for isothermal changes in pressure and volume.
  • There is a discussion about the relationship between pressure, volume, and temperature during isothermal processes, with references to Boyle's law and the integral form of work done.
  • Some participants express uncertainty about the application of the equations, suggesting that misapplication could lead to different results.
  • One participant offers a method of visualizing the work done by counting squares on a graph of P against V as an alternative way to integrate.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the reasons for the differing answers obtained from the two equations. There is acknowledgment of the need for clarity regarding the conditions under which each equation is applied, particularly in isothermal processes.

Contextual Notes

Some limitations are noted, including potential misapplication of the equations and the assumption that the process is strictly isothermal, which may not be the case in all scenarios.

curiosity colour
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when do we use W=p dV and W= nRT ln Vi/Vf? i want to know because i got 2 different answer when i use them in the same question
 
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The work in a process is the area under the P-V diagram.
Draw the PV diagram for the process.
 
curiosity colour said:
when do we use W=p dV and W= nRT ln Vi/Vf? i want to know because i got 2 different answer when i use them in the same question

What was the question you had to solve? Your second equation is a special case of the application of your first equation! It relates to isothermal change in P and V
 
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lychette said:
What was the question you had to solve? Your second equation is a special case of the application of your first equation! It relates to isothermal change in P and V
hmm, my question does involve isothermal process, it ask me to find the work done on gas.
so for isothermal process we have to use the second equation? but why?
 
curiosity colour said:
hmm, my question does involve isothermal process, it ask me to find the work done on gas.
so for isothermal process we have to use the second equation? but why?

For an isothermal change PxV =constant x T P x V = nRT and the graph of P against V is the Boyles law graph and P = nRT/V
The work involved going from V1 to V2 is given by ∫P dV
so for isothermal change W = ∫nRT/V dV
W= nRT∫dV/V
W= nRT ln(V1/V2)

If in doubt about the mathemeatics you can always draw the curve of P against V and calculate the area between V1 and V2 by 'counting squares' ! (another way to integrate)
 
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curiosity colour said:
hmm, my question does involve isothermal process,
... then you should not be getting different answers unless there is not just an isothermal process.
Perhaps you have misapplied the first equation.
 
Both your equations give the same result, right? If not, please show your work.
 
lychette said:
For an isothermal change PxV =constant x T P x V = nRT and the graph of P against V is the Boyles law graph and P = nRT/V
The work involved going from V1 to V2 is given by ∫P dV
so for isothermal change W = ∫nRT/V dV
W= nRT∫dV/V
W= nRT ln(V1/V2)

If in doubt about the mathemeatics you can always draw the curve of P against V and calculate the area between V1 and V2 by 'counting squares' ! (another way to integrate)
i'm sorry for the late reply, was busying on my assignment. thanks for your answer, it have sort thing out for me :)
 
Simon Bridge said:
... then you should not be getting different answers unless there is not just an isothermal process.
Perhaps you have misapplied the first equation.
i'm sorry for the late reply, was busying on my assignment. i will check on it, thanks for answering me :D
 

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