About the equation for work done on gas

  • #1
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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  • #2
Simon Bridge
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The work in a process is the area under the P-V diagram.
Draw the PV diagram for the process.
 
  • #3
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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
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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  • #4
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?
 
  • #5
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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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  • #6
Simon Bridge
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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.
 
  • #7
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Both your equations give the same result, right? If not, please show your work.
 
  • #8
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 :)
 
  • #9
... 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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