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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.

Draw the PV diagram for the process.

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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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hmm, my question does involve isothermal process, it ask me to find the work done on gas.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

so for isothermal process we have to use the second equation? but why?

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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/Vhmm, 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?

The work involved going from V

so for isothermal change W = ∫nRT/V dV

W= nRT∫dV/V

W= nRT ln(V

If in doubt about the mathemeatics you can always draw the curve of P against V and calculate the area between V

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Simon Bridge

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... then you should not be getting different answers unless there is not just an isothermal process.hmm, my question does involve isothermal process,

Perhaps you have misapplied the first equation.

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Chestermiller

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Both your equations give the same result, right? If not, please show your work.

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i'm sorry for the late reply, was busying on my assignment. thanks for your answer, it have sort thing out for me :)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 V_{1}to V_{2}is given by ∫P dV

so for isothermal change W = ∫nRT/V dV

W= nRT∫dV/V

W= nRT ln(V_{1}/V_{2})

If in doubt about the mathemeatics you can always draw the curve of P against V and calculate the area between V_{1}and V_{2}by 'counting squares' !! (another way to integrate)

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i'm sorry for the late reply, was busying on my assignment. i will check on it, thanks for answering me :D... then you should not be getting different answers unless there is not just an isothermal process.

Perhaps you have misapplied the first equation.

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